Exams, Shamash Secondary School

Datum: Approx. January 1, 1951 to December 31, 1960

Aufbewahrung: National Archives and Records Administration (NARA)

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### csp_02c24f5c694a53bbaede6d6dc7718370

B01 · header / latin SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965.

B02 · form / latin Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m.

B03 · paragraph / latin ⟦line⟧ Attempt all questions :

B04 · paragraph / latin 1. (i) If m = (2x + y) / (x + 2y) , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks).

B05 · paragraph / latin 2. (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks).

B06 · paragraph / latin (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks).

B07 · paragraph / latin 3. (i) Compute by logarithms, arranging your work neatly : ⁷√((cos² 18° 47')(sin³ 48° 21')) / ((10.09)³ (0.0002049)) (6 marks).

B08 · paragraph / latin (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks).

B09 · paragraph / latin (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks).

B10 · paragraph / latin (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks).

B11 · paragraph / latin 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ⟦line⟧ (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks).

B12 · paragraph / latin (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks).

B13 · paragraph / latin 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axis and 0.2 inch for your unit on the y-axis. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks).

B14 · paragraph / latin (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks).

B15 · paragraph / latin (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks).

B16 · paragraph / latin (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks).

B17 · footer / latin ⟦line⟧

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965. Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m. ⟦line⟧ Attempt all questions : 1. (i) If m = (2x + y) / (x + 2y) , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks). 2. (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks). 3. (i) Compute by logarithms, arranging your work neatly : ⁷√((cos² 18° 47')(sin³ 48° 21')) / ((10.09)³ (0.0002049)) (6 marks). (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks). 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ⟦line⟧ (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks). 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axis and 0.2 inch for your unit on the y-axis. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks). ⟦line⟧

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### csp_02eb15b7dbde55f5a40a8aa2eb79ff5b

[Marginalia] Kamel

-p.2- Algebra 4th Year. Scientific. 16/9/1965

4. (i) Find the sum of all the numbers between 60 and 600 which are exactly divisible by 13. (10 marks) (ii) An elastic ball is dropped on to a horizontal smooth plate and allowed to go on bouncing in the same vertical line. At each bounce it rises to one-fourth of the height from which it has fallen. If it is originally released from a height of 256 ft., find the total distance the ball has moved altogether, up and down, by the time it strikes the plate for: (1) the 5th time (2) the nth time. (10 marks)

5. (i) Draw the graph of ¼(3x²-5x-4) for values of x from -2 to +3, using a scale of 1 inch to 1 unit on each axis. (6 marks) (ii) Use your graph to find the least value of 3x²-5x-4. (6 marks) (iii) By drawing the appropriate straight line on your graph, solve the equation 3x²-5x-6=0 (8 marks)

------

⟦illegible⟧

**Traduction anglaise —**

Kamel -p.2- Algebra 4th Year. Scientific. 16/9/1965 4. (i) Find the sum of all the numbers between 60 and 600 which are exactly divisible by 13. (10 marks) (ii) An elastic ball is dropped on to a horizontal smooth plate and allowed to go on bouncing in the same vertical line. At each bounce it rises to one-fourth of the height from which it has fallen. If it is originally released from a height of 256 ft., find the total distance the ball has moved altogether, up and down, by the time it strikes the plate for: (1) the 5th time (2) the nth time. (10 marks) 5. (i) Draw the graph of ¼(3x²-5x-4) for values of x from -2 to +3, using a scale of 1 inch to 1 unit on each axis. (6 marks) (ii) Use your graph to find the least value of 3x²-5x-4. (6 marks) (iii) By drawing the appropriate straight line on your graph, solve the equation 3x²-5x-6=0 (8 marks) ⟦line⟧ ⟦illegible⟧

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### csp_02ee07c31dfe57278099eb182e7c134e

SHAMASH SECONDARY SCHOOL Monthly Examination, November 1968 الرقم:: الاسم:: Subject:: General Mathematics Date:: 18/11/1968. Class :: 4th Year Secondary Time:: 8:30 - 10:00 a.m.

1. Give the English Equivalent of the followinh, filling the blanks in this sheet and hand it over with your examination book.

١- ارقام ٢- مراتب ٣- الطرح ٤- العوامل ٥- اس القوة ٦- مضاعف ٧- اعداد زوجية متتالية ٨- اعداد فردية متتالية ٩- الجزء الصحيح من العدد ١٠- اعداد اولية ١١- المقام المشترك الاصغر ١٢- كسر لفظي ١٣- مقلوب العدد ١٤- الكسور العشرية المنتهية ١٥- <del>الكسور</del> العشرية ⟦الدورية⟧ ١٦- الخطأ المئوي ١٧- النسبة والتناسب ١٨- الوسط المتناسب بين عددين ١٩- ربح المساهم ( ربح حامل الاسهم ) ٢٠- البديهية ٢١- الموضوعة ٢٢- زاوية حادة ٢٣- زاوية منفرجة ٢٤- زاوية منعكسة ٢٥- قطعة دائرة ٢٦- قطاع دائرة ٢٧- المعاليم ٢٨- المجاهيل ٢٩- زاويتان متتامتان ٣٠- زاويتان متكاملتان ٣١- مضلع متساوي الاضلاع ٣٢- مثلث متساوي الساقين ٣٣- المعين ٣٤- المحل الهندسي ٣٥- المستقيم القاطع للدائرة ٣٦- ازالة وادخال الاقواس ٣٧- نقل حدود المعادلة من جهة الى الجهة الاخرى ٣٨- متطابقة ٣٩- متباينة

- يتبع -

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Monthly Examination, November 1968 No.:: Name:: Subject:: General Mathematics Date:: 18/11/1968. Class :: 4th Year Secondary Time:: 8:30 - 10:00 a.m. 1. Give the English Equivalent of the followinh, filling the blanks in this sheet and hand it over with your examination book. 1- Digits 2- Places 3- Subtraction 4- Factors 5- Exponent of power 6- Multiple 7- Consecutive even numbers 8- Consecutive odd numbers 9- The integer part of the number 10- Prime numbers 11- Lowest common denominator 12- Vulgar fraction 13- Reciprocal of the number 14- Terminating decimals 15- <del>Decimals</del> ⟦Recurring⟧ 16- Percentage error 17- Ratio and proportion 18- Mean proportional between two numbers 19- Shareholder's profit (dividend) 20- Axiom 21- Postulate 22- Acute angle 23- Obtuse angle 24- Reflex angle 25- Segment of a circle 26- Sector of a circle 27- Knowns 28- Unknowns 29- Two complementary angles 30- Two supplementary angles 31- Equilateral polygon 32- Isosceles triangle 33- Rhombus 34- Locus 35- Secant line of the circle 36- Removing and inserting brackets 37- Transposing terms of the equation from one side to the other 38- Identity 39- Inequality - Continued -

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### csp_02f5541e5e565744ba943caf45a2908f

B01 · other / latin y = 4x - 3 y = x² - 4x - 12

B02 · table / latin x | y -3 | 9 -2 | 0 -1 | -7 0 | -12 1 | -15 2 | -16 3 | -15 4 | -12 5 | -7 6 | 0 7 | 9

B03 · paragraph / latin 1. The ⟦illegible⟧ 2. ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ (2, -16) ⟦illegible⟧ ⟦illegible⟧ x = -2, x = 6 ⟦illegible⟧ ⟦illegible⟧ x² - 8x - 9 < 0 ⟦illegible⟧

B04 · marginalia / latin (2, -16)

**Traduction anglaise —**

y = 4x - 3 y = x² - 4x - 12 x | y -3 | 9 -2 | 0 -1 | -7 0 | -12 1 | -15 2 | -16 3 | -15 4 | -12 5 | -7 6 | 0 7 | 9 1. The ⟦illegible⟧ 2. ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ (2, -16) ⟦illegible⟧ ⟦illegible⟧ x = -2, x = 6 ⟦illegible⟧ ⟦illegible⟧ x² - 8x - 9 < 0 ⟦illegible⟧ (2, -16)

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### csp_03a8f5332966558dbec3e7b3f30e15df

مدرسة فرنك عيني متوسطة وابتدائية بغداد رقم التلفون ٩١٦٩٣ FRANK INY SCHOOL INTERMEDIATE & PRIMARY Baghdad Telephone No. 91693

No.: الرقم: Date: Final Exam. in Algebra Cont. التاريخ:

5. (i) y = 3 cos 60°, ∴ AC = 4 + y = 4 + 3 cos 60° z = 3 sin 60°, ⟦thus⟧ tan α = z / AC = 3 sin 60° / (4 + 3 cos 60°) ∴ tan α = (3√3 / 2) / (4 + 3/2) = 3√3 / (8 + 3) = 3√3 / 11 = (3 × 1.732) / 11 = 5.196 / 11 = 0.47236 tan α = (3 × 0.8660) / (4 + 1.500) = 2.5980 / 5.5 = 0.47236 ∴ α = 25° 17' ∴ θ = 90 - α = 64° 43' Bearing of B from A ⟦Ans.⟧ x / z = sin α ∴ x = z / sin α = 3 sin 60° / sin α = (3 × 0.8660) / 0.4271 = 2.5980 / 0.4271 = 6.0828 Ans.

(ii) Let each side of the equilateral triangle = 2x, then its height AD = x√3 y = x√3 sin 60° = x√3 . √3 / 2 = 3x / 2 sin θ = y / 2x = (3x / 2) / 2x = 3x / 4x = 3 / 4 = 0.7500 <del>θ = 36° 54'</del> Ans. θ = 48° 36' or 48° 35' Ans.

⟦illegible mathematical sketches and faded text in blue ink⟧

**Traduction anglaise —**

Frank Iny School Intermediate and Primary Baghdad Telephone Number 91693 FRANK INY SCHOOL INTERMEDIATE & PRIMARY Baghdad Telephone No. 91693 No.: | الرقم: Date: Final Exam. in Algebra Cont. | التاريخ: 5. (i) y = 3 cos 60°, ∴ AC = 4 + y = 4 + 3 cos 60° z = 3 sin 60°, ⟦thus⟧ tan α = z / AC = 3 sin 60° / (4 + 3 cos 60°) ∴ tan α = (3√3 / 2) / (4 + 3/2) = 3√3 / (8 + 3) = 3√3 / 11 = (3 × 1.732) / 11 = 5.196 / 11 = 0.47236 tan α = (3 × 0.8660) / (4 + 1.500) = 2.5980 / 5.5 = 0.47236 ∴ α = 25° 17' ∴ θ = 90 - α = 64° 43' Bearing of B from A ⟦Ans.⟧ x / z = sin α ∴ x = z / sin α = 3 sin 60° / sin α = (3 × 0.8660) / 0.4271 = 2.5980 / 0.4271 = 6.0828 Ans. (ii) Let each side of the equilateral triangle = 2x, then its height AD = x√3 y = x√3 sin 60° = x√3 . √3 / 2 = 3x / 2 sin θ = y / 2x = (3x / 2) / 2x = 3x / 4x = 3 / 4 = 0.7500 <del>θ = 36° 54'</del> Ans. θ = 48° 36' or 48° 35' Ans. ⟦illegible mathematical sketches and faded text in blue ink⟧

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### csp_04285ee5d26f5dc69a8a2feb5250b5ec

B01 · header / latin Solution to Final Exam. Questions 1 to 4 4th year in Algebra February 6th 1967

B02 · paragraph / latin 1. (i) 3x² - (4a + 2b)x + a² + 2ab = 3x² - (4a + 2b)x + a(a + 2b) = [3x - (a + 2b)][x - a] = (3x - a - 2b)(x - a) Ans.

B03 · paragraph / latin (ii) 8x³ - 27y³ + z³ + 18xyz = (2x)³ + (-3y)³ + z³ - 3(2x)(-3y)(z) = (2x - 3y + z)(4x² + 9y² + z² + 6xy - 2xz + 3yz) Ans.

B04 · paragraph / latin (iii) (a² + b² + c²)(a + 1) + (2ab - 2ac)(a + 1) - 2abc - 2bc ⟦line⟧ a + 1 = (a² + b² + c²)(a + 1) + (2ab - 2ac)(a + 1) - 2bc(a + 1) ⟦line⟧ a + 1 = a² + b² + c² + 2ab - 2ac - 2bc = <del>⟦illegible⟧</del> = a² + b² + (-c)² + 2ab + 2a(-c) + 2b(-c) = (a + b - c)² Ans.

B05 · paragraph / latin 2. (i) { x + y = 2a or x = a + b Now x⁴ + x²y² + y⁴ = x⁴ + 2x²y² + y⁴ - x²y² { x - y = 2b y = a - b = (x² + y²)² - (xy)² 2x = 2(a + b) = (x² + 2xy + y² - 2xy)² - (xy)² 2y = 2(a - b) ∴ xy = (a + b)(a - b) = [(x + y)² - 2xy]² - (xy)² or xy = a² - b² = [(x + y)² - 2xy + xy][(x + y)² - 2xy - xy] = [(x + y)² - xy][(x + y)² - 3xy] = [(2a)² - (a² - b²)][(2a)² - 3(a² - b²)] = (4a² - a² + b²)(4a² - 3a² + 3b²) = (3a² + b²)(a² + 3b²) Ans.

**Traduction anglaise —**

Solution to Final Exam. Questions 1 to 4 4th year in Algebra February 6th 1967 1. (i) 3x² - (4a + 2b)x + a² + 2ab = 3x² - (4a + 2b)x + a(a + 2b) = [3x - (a + 2b)][x - a] = (3x - a - 2b)(x - a) Ans. (ii) 8x³ - 27y³ + z³ + 18xyz = (2x)³ + (-3y)³ + z³ - 3(2x)(-3y)(z) = (2x - 3y + z)(4x² + 9y² + z² + 6xy - 2xz + 3yz) Ans. (iii) (a² + b² + c²)(a + 1) + (2ab - 2ac)(a + 1) - 2abc - 2bc ⟦line⟧ a + 1 = (a² + b² + c²)(a + 1) + (2ab - 2ac)(a + 1) - 2bc(a + 1) ⟦line⟧ a + 1 = a² + b² + c² + 2ab - 2ac - 2bc = <del>⟦illegible⟧</del> = a² + b² + (-c)² + 2ab + 2a(-c) + 2b(-c) = (a + b - c)² Ans. 2. (i) { x + y = 2a or x = a + b Now x⁴ + x²y² + y⁴ = x⁴ + 2x²y² + y⁴ - x²y² { x - y = 2b y = a - b = (x² + y²)² - (xy)² 2x = 2(a + b) = (x² + 2xy + y² - 2xy)² - (xy)² 2y = 2(a - b) ∴ xy = (a + b)(a - b) = [(x + y)² - 2xy]² - (xy)² or xy = a² - b² = [(x + y)² - 2xy + xy][(x + y)² - 2xy - xy] = [(x + y)² - xy][(x + y)² - 3xy] = [(2a)² - (a² - b²)][(2a)² - 3(a² - b²)] = (4a² - a² + b²)(4a² - 3a² + 3b²) = (3a² + b²)(a² + 3b²) Ans.

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### csp_04ae214eebde5ee58df72a4462bd38d2

B01 · header / latin ⟦illegible⟧ 31/12/1948

B02 · paragraph / latin 5. ⟦illegible⟧ ⟦illegible⟧ takes (x/2 + 1) ⟦illegible⟧ Remainder (x/2 - 1) ⟦illegible⟧ and ⟦illegible⟧ takes ⟦illegible⟧ + 1 ⟦illegible⟧ Remainder = (x/2 - 1) - (x/4 + 1/2 + 1) = x/4 - 5/2 ⟦illegible⟧ = x/4 - 5/2 - (x/8 - 5/4 + 1) = x/8 - 9/4 ⟦illegible⟧ who ⟦illegible⟧ + 6 = x - 6 + 42 = x + 42 ⟦illegible⟧ x/2 + x/4 + x/8 + x/16 + ⟦illegible⟧ = x + 42 x = 54 ⟦illegible⟧

B03 · paragraph / latin ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

⟦illegible⟧ 31/12/1948 5. ⟦illegible⟧ ⟦illegible⟧ takes (x/2 + 1) ⟦illegible⟧ Remainder (x/2 - 1) ⟦illegible⟧ and ⟦illegible⟧ takes ⟦illegible⟧ + 1 ⟦illegible⟧ Remainder = (x/2 - 1) - (x/4 + 1/2 + 1) = x/4 - 5/2 ⟦illegible⟧ = x/4 - 5/2 - (x/8 - 5/4 + 1) = x/8 - 9/4 ⟦illegible⟧ who ⟦illegible⟧ + 6 = x - 6 + 42 = x + 42 ⟦illegible⟧ x/2 + x/4 + x/8 + x/16 + ⟦illegible⟧ = x + 42 x = 54 ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_05577c43706b5fadb0d7e0d8144c370b

⟦illegible⟧ Exam. paper in Algebra Page 1 4th year , 18/5/1967

x t = at + b ∴ t(x-a) = b ∴ t = b / (x-a) y = b + a ( b / (x-a) ) = ( b(x-a) + ab ) / (x-a) = ( bx - ab + ab ) / (x-a) = bx / (x-a) y = bx / (x-a) Ans.

(ii) 2^(x+y) = 32 ∴ 2^(x+y) = 2^5 ∴ x+y = 5 3^(x-y) = 9 ∴ 3^(x-y) = 3^2 ∴ x-y = 2 x+y = 5 x-y = 2 2x = 7 ∴ x = 3.5 Ans (7 marks) y = 1.5 Ans

(iii) m/n = 5/4 , p/q = 3/4 , (3m+5p)/(n+q) = ? m = 5n/4 , p = 3q/4 ∴ (3m+5p)/(n+q) = ( 15n/4 + 15q/4 ) / (n+q) = ( 15/4 (n+q) ) / (n+q) = 15/4 = 3 3/4 Ans. (7 marks)

2. (a) When n tables are made for table costs: £ (300 + 8n) / n or 1 table costs: £ (300/n + 8) (b) when (50 + n) tables are made, 1 table costs: £ [300 + 8(50+n)] / (50+n) or 1 table costs: £ ( 300 / (50+n) + 8 )

∴ 300 / (50+n) + 8 + 1 = 300/n + 8 or 300/n - 300 / (50+n) = 1 or 300(50+n) - 300n = n(50+n) or 15000 + 300n - 300n = n² + 50n or n² + 50n - 15000 = 0 ∴ (n+150)(n-100) = 0 ∴ n = -150 (to be discarded) n = 100 Ans. ∴ the cost of one table in case (a) is £ (300/n + 8) = £ (3+8) = £ 11 " " " " " (b) is £ ( 300 / (50+n) + 8 ) = £ (2+8) = £ 10 Ans. (20 marks)

**Traduction anglaise —**

⟦illegible⟧ Exam. paper in Algebra Page 1 4th year , 18/5/1967 x t = at + b ∴ t(x-a) = b ∴ t = b / (x-a) y = b + a ( b / (x-a) ) = ( b(x-a) + ab ) / (x-a) = ( bx - ab + ab ) / (x-a) = bx / (x-a) y = bx / (x-a) Ans. (ii) 2^(x+y) = 32 ∴ 2^(x+y) = 2^5 ∴ x+y = 5 3^(x-y) = 9 ∴ 3^(x-y) = 3^2 ∴ x-y = 2 x+y = 5 x-y = 2 2x = 7 ∴ x = 3.5 Ans (7 marks) y = 1.5 Ans (iii) m/n = 5/4 , p/q = 3/4 , (3m+5p)/(n+q) = ? m = 5n/4 , p = 3q/4 ∴ (3m+5p)/(n+q) = ( 15n/4 + 15q/4 ) / (n+q) = ( 15/4 (n+q) ) / (n+q) = 15/4 = 3 3/4 Ans. (7 marks) 2. (a) When n tables are made for table costs: £ (300 + 8n) / n or 1 table costs: £ (300/n + 8) (b) when (50 + n) tables are made, 1 table costs: £ [300 + 8(50+n)] / (50+n) or 1 table costs: £ ( 300 / (50+n) + 8 ) ∴ 300 / (50+n) + 8 + 1 = 300/n + 8 or 300/n - 300 / (50+n) = 1 or 300(50+n) - 300n = n(50+n) or 15000 + 300n - 300n = n² + 50n or n² + 50n - 15000 = 0 ∴ (n+150)(n-100) = 0 ∴ n = -150 (to be discarded) n = 100 Ans. ∴ the cost of one table in case (a) is £ (300/n + 8) = £ (3+8) = £ 11 " " " " " (b) is £ ( 300 / (50+n) + 8 ) = £ (2+8) = £ 10 Ans. (20 marks)

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### csp_05a4f32294ca578db9e2bd452db6a2d1

Solutions to Conditional Exam in Algebra continued Sept., 1965

2 (ii) cont. ∴ log₁₀ 402 = 2 log₁₀ 2 + 2 + log₁₀ 1.005 = 2 x 0.301030 + 2 + 0.002166 = 0.602060 + 2 + 0.002166 = 2.604226 Ans. (5 marks) ∴ log₁₀ 0.804 = 3 log₁₀ 2 + log₁₀ 1.005 - 2 = 3 x 0.301030 + 0.002166 - 2 = 0.903090 + 0.002166 - 2 = 0.905256 - 2 = -1.094744 Ans. (5 marks)

3. (i) Let the time for train A to overtake train B at pt. C be t seconds ∴ dist AC = 48 (t/3600) mils ∴ dist BC = 32 (t/3600) mils or 16 t / 3600 = 55 / 176 or t = 55 / 176 x 3600 / 16 = 5 / 16 x 450 / 2 = 1125 / 16 = 70.3 Sec = 70 sec to the nearest second (7 marks) Ans.

(ii) x (t/3600) - y (t/3600) = D / 1760 ∴ t (x-y) / 3600 = D / 1760 ∴ t = 360 D / 176 (x-y) or t = 45 D / 22 (x-y) Ans. (7 marks)

(iii) From the above formula x - y = 45 D / 22 t ∴ y = x - 45 D / 22 t Ans. (6 marks)

⟦(5 marks)⟧

**Traduction anglaise —**

Solutions to Conditional Exam in Algebra continued Sept., 1965 2 (ii) cont. ∴ log₁₀ 402 = 2 log₁₀ 2 + 2 + log₁₀ 1.005 = 2 x 0.301030 + 2 + 0.002166 = 0.602060 + 2 + 0.002166 = 2.604226 Ans. (5 marks) ∴ log₁₀ 0.804 = 3 log₁₀ 2 + log₁₀ 1.005 - 2 = 3 x 0.301030 + 0.002166 - 2 = 0.903090 + 0.002166 - 2 = 0.905256 - 2 = -1.094744 Ans. (5 marks) 3. (i) Let the time for train A to overtake train B at pt. C be t seconds ∴ dist AC = 48 (t/3600) mils ∴ dist BC = 32 (t/3600) mils or 16 t / 3600 = 55 / 176 or t = 55 / 176 x 3600 / 16 = 5 / 16 x 450 / 2 = 1125 / 16 = 70.3 Sec = 70 sec to the nearest second (7 marks) Ans. (ii) x (t/3600) - y (t/3600) = D / 1760 ∴ t (x-y) / 3600 = D / 1760 ∴ t = 360 D / 176 (x-y) or t = 45 D / 22 (x-y) Ans. (7 marks) (iii) From the above formula x - y = 45 D / 22 t ∴ y = x - 45 D / 22 t Ans. (6 marks) ⟦(5 marks)⟧

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### csp_05f8edd8213b54658ec9a750c5a5e8a7

⟦illegible⟧ (x - y) (x + y) (x² + xy + y²) (x² - xy + y²) ⟦illegible⟧ (x - y) (x² + xy + y²) (x + y) (x² - xy + y²) ⟦illegible⟧ (x³ - y³) (x³ + y³) ⟦illegible⟧ (x⁶ - y⁶) ⟦illegible⟧ the divisibility ⟦illegible⟧ ⟦illegible⟧ x⁴ + Ax³ + Bx² + x - 6 ⟦illegible⟧ x² + x - 2 ⟦illegible⟧ (x⁴ + x³ - 2x²) ⟦illegible⟧ (A - 1) x³ + (B + 2) x² + x - 6 ⟦illegible⟧ (A - 1) x³ + (A - 1) x² - 2(A - 1) x ⟦illegible⟧ (B - A + 3) x² + (2A - 1) x - 6 ⟦illegible⟧ (B - A + 3) x² + (B - A + 3) x - 2(B - A + 3) ⟦illegible⟧ (3A - B - 4) x + 2B - 2A ⟦illegible⟧ ⟦illegible⟧ 3A - B - 4 = 0 ⟦illegible⟧ 2B - 2A - 6 = 0 ⟦illegible⟧ 3A - B = 4 ⟦illegible⟧ B - A = 3 ⟦illegible⟧ 2A = 7 ⟦illegible⟧ A = 3.5 ⟦illegible⟧ B = 6.5 ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦...⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ x - ⟦illegible⟧ ⟦illegible⟧ x^2 + ⟦illegible⟧ x ⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧ ⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧ ⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧

**Traduction anglaise —**

⟦illegible⟧ x - ⟦illegible⟧ ⟦illegible⟧ x^2 + ⟦illegible⟧ x ⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧ ⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧ ⟦illegible⟧ x^2 + ⟦illegible⟧ x + ⟦illegible⟧

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### csp_060ee2ce01cb5151bb98fe65657387de

Solutions to Conditional Exam in Algebra 4th Year, Sept., 1964. ①

1. (a) Substitute x = 2 in the expression x² + px + q + you get 4 + 2p + q = 0 ... (1) substitute x = 4 " " " " " " " " 16 + 4p + q = 0 ... (2) subtracting equation (1) from (2), we get 12 + 2p = 0 or p = -6 ∴ from (1) 4 - 12 + q = 0 ∴ q = 8 ∴ q = 8 ∴ when x² - 6x + 8 = 48, we get x² - 6x - 40 = 0 or (x - 10)(x + 4) = 0 x = 10 or x = -4 Ans.

(b) x⁴ + ax³ + bx² + cx + 12 will be equal to zero when x = 1, -1, or -3 since (x - 1) and (x + 1) and (x + 3) are factors. Substituting x = 1, we get 1 + a + b + c + 12 = 0 ... (1). Substituting x = -1, we get: 1 - a + b - c + 12 = 0 ... (2). " " x = -3, we get: 81 - 27a + 9b - 3c + 12 = 0 ... (3). Adding equations (1) + (2), we get: 2 + 2b + 24 = 0 or 2b = -26 or b = -13 dividing eq (3) by 3, we get 27 - 9a + 3b - c + 4 = 0 or 27 - 9a + 3(-13) - c + 4 = 0 or 9a + c + 8 = 0 ... (4). But from (1) we get ⟦a + c⟧ = 0 ... (5). Subtracting (5) from (4) we get 8a + 8 = 0 or a = -1. substituting in (1) we get 1 - 1 - 13 + c + 12 = 0 or c = 1 ∴ a = -1, b = -13, c = 1 Ans Hence the expression is x⁴ - x³ - 13x² + x + 12 But (x - 1)(x + 1)(x + 3) = (x² - 1)(x + 3) = x³ + 3x² - x - 3 ∴ the fourth factor = (x⁴ - x³ - 13x² + x + 12) ÷ (x³ + 3x² - x - 3) = x - 4 Ans.

x⁴ - x³ - 13x² + x + 12 | x³ + 3x² - x - 3 x⁴ + 3x³ - x² - 3x | x - 4 ----------------- -4x³ - 12x² + 4x + 12 -4x³ - 12x² + 4x + 12 ---------------------

2. (a) Let the time be x minutes after 11, then the hour hand will be pointing x/12 divisions after 11. Now arc BC = 5 - x/12 ∴ x + arc BC = 45 or x + 5 - x/12 = 45 ∴ 11x/12 = 40 or 11x = 480 or x = 480/11 or x = 43 7/11 min. = 43.63 min = 43 min 38 2/11 seconds <del>⟦illegible⟧</del> ∴ the time is 43 min 38 2/11 sec. after eleven Ans.

[Marginalia] OR x = 11x5 + x/12 - 15 [Marginalia] x = 55 + x/12 - 15 [Marginalia] x = 40 + x/12

**Traduction anglaise —**

Solutions to Conditional Exam in Algebra 4th Year, Sept., 1964. ① 1. (a) Substitute x = 2 in the expression x² + px + q + you get 4 + 2p + q = 0 ... (1) substitute x = 4 " " " " " " " " 16 + 4p + q = 0 ... (2) subtracting equation (1) from (2), we get 12 + 2p = 0 or p = -6 ∴ from (1) 4 - 12 + q = 0 ∴ q = 8 ∴ q = 8 ∴ when x² - 6x + 8 = 48, we get x² - 6x - 40 = 0 or (x - 10)(x + 4) = 0 x = 10 or x = -4 Ans. (b) x⁴ + ax³ + bx² + cx + 12 will be equal to zero when x = 1, -1, or -3 since (x - 1) and (x + 1) and (x + 3) are factors. Substituting x = 1, we get 1 + a + b + c + 12 = 0 ... (1). Substituting x = -1, we get: 1 - a + b - c + 12 = 0 ... (2). " " x = -3, we get: 81 - 27a + 9b - 3c + 12 = 0 ... (3). Adding equations (1) + (2), we get: 2 + 2b + 24 = 0 or 2b = -26 or b = -13 dividing eq (3) by 3, we get 27 - 9a + 3b - c + 4 = 0 or 27 - 9a + 3(-13) - c + 4 = 0 or 9a + c + 8 = 0 ... (4). But from (1) we get ⟦a + c⟧ = 0 ... (5). Subtracting (5) from (4) we get 8a + 8 = 0 or a = -1. substituting in (1) we get 1 - 1 - 13 + c + 12 = 0 or c = 1 ∴ a = -1, b = -13, c = 1 Ans Hence the expression is x⁴ - x³ - 13x² + x + 12 But (x - 1)(x + 1)(x + 3) = (x² - 1)(x + 3) = x³ + 3x² - x - 3 ∴ the fourth factor = (x⁴ - x³ - 13x² + x + 12) ÷ (x³ + 3x² - x - 3) = x - 4 Ans. x⁴ - x³ - 13x² + x + 12 | x³ + 3x² - x - 3 x⁴ + 3x³ - x² - 3x | x - 4 ⟦line⟧ -4x³ - 12x² + 4x + 12 -4x³ - 12x² + 4x + 12 ⟦line⟧ 2. (a) Let the time be x minutes after 11, then the hour hand will be pointing x/12 divisions after 11. Now arc BC = 5 - x/12 ∴ x + arc BC = 45 or x + 5 - x/12 = 45 ∴ 11x/12 = 40 or 11x = 480 or x = 480/11 or x = 43 7/11 min. = 43.63 min = 43 min 38 2/11 seconds <del>⟦illegible⟧</del> ∴ the time is 43 min 38 2/11 sec. after eleven Ans. OR x = 11x5 + x/12 - 15 x = 55 + x/12 - 15 x = 40 + x/12

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### csp_0634567b2a5f5380a286c11b22cb888b

1 (ii) Simplify:

{ (y^½ + y^-½) / (y² - y + 1) - (y^½ - y^-½) / (y² + y + 1) } ÷ { (y^½ + 2y^-½) / (y³ - 1) - (y^½ - 2y^-½) / (y³ + 1) } { (y + 1) / (y^½ (y² - y + 1)) - (<del>⟦illegible⟧</del> (y - 1)) / (y^½ (y² + y + 1)) } ÷ { (y + 2) / (y^½ (y³ - 1)) - (y - 2) / (y^½ (y³ + 1)) } ((y + 1)(y² + y + 1) - (y - 1)(y² - y + 1)) / (y^½ (y² - y + 1)(y² + y + 1)) ÷ ((y + 2)(y³ + 1) - (y - 2)(y³ - 1)) / (y^½ (y³ - 1)(y³ + 1)) = (y³ + y² + y + y² + y + 1 - (y³ - y² + y - y² + y - 1)) / (y^½ (y² - y + 1)(y² + y + 1)) × (y^½ (y³ - 1)(y³ + 1)) / (y⁴ + y + 2y³ + 2 - (y⁴ - y - 2y³ + 2)) = (4 y² + 2) / (y^½ (y² - y + 1)(y² + y + 1)) × (y^½ (y³ - 1)(y³ + 1)) / (4 y³ + 2y) = = (<del>2</del> (2y² + 1)) / (<del>y^½</del> (y² - y + 1)(<del>y² + y + 1</del>)) × (<del>y^½</del> (<del>y - 1</del>)(<del>y² + y + 1</del>) (y + 1)(y² - y + 1)) / (<del>2</del> y (2y² + 1)) = ((y - 1)(y + 1)) / y = (y² - 1) / y = y - 1/y Ans.

**Traduction anglaise —**

2 1 (ii) Simplify: { (y^½ + y^-½) / (y² - y + 1) - (y^½ - y^-½) / (y² + y + 1) } ÷ { (y^½ + 2y^-½) / (y³ - 1) - (y^½ - 2y^-½) / (y³ + 1) } { (y + 1) / (y^½ (y² - y + 1)) - (<del>⟦illegible⟧</del> (y - 1)) / (y^½ (y² + y + 1)) } ÷ { (y + 2) / (y^½ (y³ - 1)) - (y - 2) / (y^½ (y³ + 1)) } ((y + 1)(y² + y + 1) - (y - 1)(y² - y + 1)) / (y^½ (y² - y + 1)(y² + y + 1)) ÷ ((y + 2)(y³ + 1) - (y - 2)(y³ - 1)) / (y^½ (y³ - 1)(y³ + 1)) = (y³ + y² + y + y² + y + 1 - (y³ - y² + y - y² + y - 1)) / (y^½ (y² - y + 1)(y² + y + 1)) × (y^½ (y³ - 1)(y³ + 1)) / (y⁴ + y + 2y³ + 2 - (y⁴ - y - 2y³ + 2)) = (4 y² + 2) / (y^½ (y² - y + 1)(y² + y + 1)) × (y^½ (y³ - 1)(y³ + 1)) / (4 y³ + 2y) = = (<del>2</del> (2y² + 1)) / (<del>y^½</del> (y² - y + 1)(<del>y² + y + 1</del>)) × (<del>y^½</del> (<del>y - 1</del>)(<del>y² + y + 1</del>) (y + 1)(y² - y + 1)) / (<del>2</del> y (2y² + 1)) = ((y - 1)(y + 1)) / y = (y² - 1) / y = y - 1/y Ans.

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### csp_0688bad16a7b5f20abf38529ad13cdb6

B01 · header / mixed ⟦illegible⟧ SCHOOL ⟦illegible⟧ PRIMARY Baghdad ⟦illegible⟧

B02 · paragraph / mixed ⟦illegible⟧ ⟦illegible⟧

B03 · other / latin ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ + ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ tan ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧

B04 · other / latin ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

⟦illegible⟧ SCHOOL ⟦illegible⟧ PRIMARY Baghdad ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ + ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ tan ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_06b90f2578bf565b8ab9674d06e479d9

⟦Final Examination June 1951⟧ Algebra Date: 12/6/1951 Class: 4th Year Secondary Time: 8:00 - 11:00

Attempt all questions

1. How much ⟦illegible⟧ a gross ⟦illegible⟧ for a sovereign lowers the price ⟦illegible⟧ (10 marks)

2. (i) Solve the equation, √(x - a) + √(b + x) = √a + √b (10 marks) (ii) Simplify or express with positive indices: (10 marks) ⟦{ a^(p-q) / √(a^(p^2-q^2)) }^(1/(p-q)) x a^(q/(p+q))⟧

3. (i) Compute by logarithms the following: ⁷√{ (0.001021)² x (4.003)³ / (16.03)⁵ x (2.001)⁴ } (10 marks) (ii) Derive a formula for finding the logarithm of a number to the base 'y' having given tables of logarithms to the base 'x'. Hence use your common logarithmic tables to compute the value of log 70.56. (10 marks) 17

4. (i) A man drives a distance of 14 17/20 miles in 18 minutes. He drives the longest stretch in the first minute; and in every subsequent minute the distance he drives is 1/20 miles shorter than the distance driven in the previous minute. What are the distances driven in the first + last minutes. <del>Consider these distances as an arithmetical progression ⟦illegible⟧</del>. Use Arithmetic Progressions to solve your question. (10 marks) (ii) The sum of n terms of a geometric progression is 4(5ⁿ-1). Find the fifth term and the number of terms that must be taken for their sum to be 312496. (10 marks)

**Traduction anglaise —**

⟦Final Examination June 1951⟧ Algebra Date: 12/6/1951 Class: 4th Year Secondary Time: 8:00 - 11:00 Attempt all questions 1. How much ⟦illegible⟧ a gross ⟦illegible⟧ for a sovereign lowers the price ⟦illegible⟧ (10 marks) 2. (i) Solve the equation, √(x - a) + √(b + x) = √a + √b (10 marks) (ii) Simplify or express with positive indices: (10 marks) ⟦{ a^(p-q) / √(a^(p^2-q^2)) }^(1/(p-q)) x a^(q/(p+q))⟧ 3. (i) Compute by logarithms the following: ⁷√{ (0.001021)² x (4.003)³ / (16.03)⁵ x (2.001)⁴ } (10 marks) (ii) Derive a formula for finding the logarithm of a number to the base 'y' having given tables of logarithms to the base 'x'. Hence use your common logarithmic tables to compute the value of log 70.56. (10 marks) 17 4. (i) A man drives a distance of 14 17/20 miles in 18 minutes. He drives the longest stretch in the first minute; and in every subsequent minute the distance he drives is 1/20 miles shorter than the distance driven in the previous minute. What are the distances driven in the first + last minutes. <del>Consider these distances as an arithmetical progression ⟦illegible⟧</del>. Use Arithmetic Progressions to solve your question. (10 marks) (ii) The sum of n terms of a geometric progression is 4(5ⁿ-1). Find the fifth term and the number of terms that must be taken for their sum to be 312496. (10 marks)

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### csp_08bcc3c68a6650e8a136cacba003ecaa

⟦Shamash Secondary School⟧ Conditional Examination, September 1962

Subject: Mathematics Date: 14/9/1962 Class: 4th Year Secondary Time: 8:00 - 10:30

Attempt all questions:

1. (i) Solve the ⟦following simultaneous⟧ equations for x and y: ax + by = c ----- ① x/a + y/b = 1 ----- ② (10 marks). (ii) Solve the equation 4x² - 12x + 3 = 0, giving the answer correct to two decimal places. (10 marks).

2. (i) Use logarithms to calculate ⟦by the shortest possible way⟧ the value of √(x²+x), when x = 4.836. (10 marks). (ii) When (1 - 2x + x²) is multiplied by (1 - kx + x²) the coefficient of x² is zero. Find the value of k. (10 marks).

3. (i) Find the 50th term and the sum of the first 100 terms of the arithmetical progression whose first term is 20 and whose 3rd term is 21. (10 marks). (ii) Find the first & sixth terms of a geometrical progression whose common ratio is 1/2 when the first four terms add up to 18 3/4. (10 marks).

4. The cost of turfing a piece of lawn for a tennis court at 10 shillings per square yard is £ 12 less than 12 times the cost of fencing it all round at 5 shillings per yard. If the lawn had been 12 ft. longer it would have been twice as long as it is wide. Find the dimensions of the lawn in yards. (20 marks).

5. Draw the graph of x² - 2x between x = -2 and x = 4. From your graph solve approximately the equation x² - 2x = ⟦1.6⟧. With the help of a further graph, solve approximately the equation x² - 2x = x + 1. ⟦(20 marks)⟧

**Traduction anglaise —**

⟦Shamash Secondary School⟧ Conditional Examination, September 1962 Subject: Mathematics Date: 14/9/1962 Class: 4th Year Secondary Time: 8:00 - 10:30 Attempt all questions: 1. (i) Solve the ⟦following simultaneous⟧ equations for x and y: ax + by = c ⟦line⟧ ① x/a + y/b = 1 ⟦line⟧ ② (10 marks). (ii) Solve the equation 4x² - 12x + 3 = 0, giving the answer correct to two decimal places. (10 marks). 2. (i) Use logarithms to calculate ⟦by the shortest possible way⟧ the value of √(x²+x), when x = 4.836. (10 marks). (ii) When (1 - 2x + x²) is multiplied by (1 - kx + x²) the coefficient of x² is zero. Find the value of k. (10 marks). 3. (i) Find the 50th term and the sum of the first 100 terms of the arithmetical progression whose first term is 20 and whose 3rd term is 21. (10 marks). (ii) Find the first & sixth terms of a geometrical progression whose common ratio is 1/2 when the first four terms add up to 18 3/4. (10 marks). 4. The cost of turfing a piece of lawn for a tennis court at 10 shillings per square yard is £ 12 less than 12 times the cost of fencing it all round at 5 shillings per yard. If the lawn had been 12 ft. longer it would have been twice as long as it is wide. Find the dimensions of the lawn in yards. (20 marks). 5. Draw the graph of x² - 2x between x = -2 and x = 4. From your graph solve approximately the equation x² - 2x = ⟦1.6⟧. With the help of a further graph, solve approximately the equation x² - 2x = x + 1. ⟦(20 marks)⟧

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### csp_09e818a26f5b5abdb4ed6b417b43735c

B01 · header / arabic ⟦illegible⟧ الامتحان ⟦illegible⟧

B02 · other / latin ⟦diagram with numerical values: 20, 55, 60, 75, 32.5, θ⟧

B03 · paragraph / latin ⟦illegible⟧ = (20)² + (32.5)² - 2 x 20 x 32.5 cos ⟦illegible⟧ ⟦illegible⟧ = 400 + 1056.25 - 1300 x ⟦illegible⟧ = 1424 - 640 = 784 V = 28 ⟦illegible⟧ 20 / sin θ = 28 / sin 60 sin θ = 20 x 0.866 / 28 = 0.6186 θ = 38° 12' or 38° 13' 75 - 38° 12' = 36° 48' E

**Traduction anglaise —**

⟦illegible⟧ the exam ⟦illegible⟧ ⟦diagram with numerical values: 20, 55, 60, 75, 32.5, θ⟧ ⟦illegible⟧ = (20)² + (32.5)² - 2 x 20 x 32.5 cos ⟦illegible⟧ ⟦illegible⟧ = 400 + 1056.25 - 1300 x ⟦illegible⟧ = 1424 - 640 = 784 V = 28 ⟦illegible⟧ 20 / sin θ = 28 / sin 60 sin θ = 20 x 0.866 / 28 = 0.6186 θ = 38° 12' or 38° 13' 75 - 38° 12' = 36° 48' E

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### csp_0ae2c60b196357a2bb622e36604d5d98

(ii) The side of the first square = 2k " " " second square = k√2 " " " third square = k√2/2 * √2 = k & so on ∴ the sum of the sides of all the squares equals = 4 (2k + k√2 + k + ... to infinity) 8 ∴ Sn = 4 [ 2k / (1 - 1/√2) ] ⟦illegible⟧ the brackets is an infinite geometrical n→∞ progression ⟦illegible⟧ in which: a = 2k, r = k√2/2k = 1/√2 and n = ∞ where S = a / (1-r) ∴ Sn = 4 ( 2k√2 / (√2 - 1) ) n→∞ ∴ Sn = 8k√2(√2+1) / ((√2-1)(√2+1)) = 8k√2(√2+1) Ans. n→∞ = 8 x 18√2(√2+1) = 144√2(√2+1) = 288 + 144√2 inch Ans. ⟦line⟧ 6. (i) ⟦illegible⟧ Ans. 6. (i) ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ (ii) √⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

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### csp_0af4c8f18b5f527e82819b511fa985fe

-p.2- Conditional Exam. in Algebra for the 4th Year, Sept. 1964. (cont'd.) ------

V. (a) Sketch the curve of the function (x⁴-8x²) for values of x from -3 to 3 choosing ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (8 marks)

(b) From this curve find the values of x at which the function (x⁴-8x²) has minimum or maximum values. Find also these minimum and maximum values. (6 marks)

(c) Plot on the same diagram the graph of y= 3x-10 and from these two graphs find the solution of the equation x⁴-8x²+10 = 3x. (6 marks).

--------

[Marginalia] ⟦(8 marks)⟧ [Marginalia] ⟦(6 marks)⟧ [Marginalia] ⟦(6 marks)⟧ [Marginalia] ⟦(6 marks)⟧ [Marginalia] ⟦(6 marks)⟧ [Marginalia] ⟦(6 marks)⟧

**Traduction anglaise —**

-p.2- Conditional Exam. in Algebra for the 4th Year, Sept. 1964. (cont'd.) ⟦line⟧ V. (a) Sketch the curve of the function (x⁴-8x²) for values of x from -3 to 3 choosing ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (8 marks) (b) From this curve find the values of x at which the function (x⁴-8x²) has minimum or maximum values. Find also these minimum and maximum values. (6 marks) (c) Plot on the same diagram the graph of y= 3x-10 and from these two graphs find the solution of the equation x⁴-8x²+10 = 3x. (6 marks). ⟦line⟧ ⟦(8 marks)⟧ ⟦(6 marks)⟧ ⟦(6 marks)⟧ ⟦(6 marks)⟧ ⟦(6 marks)⟧ ⟦(6 marks)⟧

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### csp_0b3bef66cfa05dceb4840bc6230040c7

5 (i) Let y = 3 + 2x - x²

| x | y | | -2 | -5 | | -1 | 0 | | 0 | 3 | | 1 | 4 | | 2 | 3 | | 3 | 0 | | 4 | -5 |

⟦Graph showing parabola and straight line⟧ y = x/2 + 2 (1, 4) A(2, 3) B(-1/2, 1 3/4)

(ii) The roots of the equation x² - 3 = 2x are the same as the roots of the equation 3 + 2x - x² = 0 ∴ the roots are the values of x when y = 0 or the roots are x = -1 } Ans. 2 they are the abscissas of the pts. of intersection with the and x = 3 } x-axis.

(iii) the function 3 + 2x - x² is always positive for all points on the curve above the x-axis i.e. 3 + 2x - x² > 0 when -1 < x < 3 Ans. 3

(iv) the function 3 + 2x - x² is greatest when x = 1 + y_max = 4 Ans. 4

(v) we plot the st. line y = x/2 + 2 of which two points are: (-4, 0) + (0, 2) we join these points + we get the st. line AB which intersects the curve at A(2, 3) + B(-1/2, 1 3/4) ∴ for all values of x between -1/2 and 2, the curve of 3 + 2x - x² lies above the st. line x/2 + 2 ∴ 3 + 2x - x² > x/2 + 2 when -1/2 < x < 2. (Ans. 5)

(vi) The roots of the equation 3 + 2x - x² = x/2 + 2 are the values of x which will make the ordinate of the curve 3 + 2x - x² equal to the ordinate of the st. line x/2 + 2. But the ordinates of the curve + the line are equal at the pts. of intersection A + B. ∴ the roots are the abscissas of A + B or x = -1/2 and x = 2 Ans. 6

**Traduction anglaise —**

5 (i) Let y = 3 + 2x - x² x | y -2 | -5 -1 | 0 0 | 3 1 | 4 2 | 3 3 | 0 4 | -5 ⟦Graph showing parabola and straight line⟧ y = x/2 + 2 (1, 4) A(2, 3) B(-1/2, 1 3/4) (ii) The roots of the equation x² - 3 = 2x are the same as the roots of the equation 3 + 2x - x² = 0 ∴ the roots are the values of x when y = 0 or the roots are x = -1 } Ans. 2 they are the abscissas of the pts. of intersection with the and x = 3 } x-axis. (iii) the function 3 + 2x - x² is always positive for all points on the curve above the x-axis i.e. 3 + 2x - x² > 0 when -1 < x < 3 Ans. 3 (iv) the function 3 + 2x - x² is greatest when x = 1 + y_max = 4 Ans. 4 (v) we plot the st. line y = x/2 + 2 of which two points are: (-4, 0) + (0, 2) we join these points + we get the st. line AB which intersects the curve at A(2, 3) + B(-1/2, 1 3/4) ∴ for all values of x between -1/2 and 2, the curve of 3 + 2x - x² lies above the st. line x/2 + 2 ∴ 3 + 2x - x² > x/2 + 2 when -1/2 < x < 2. (Ans. 5) (vi) The roots of the equation 3 + 2x - x² = x/2 + 2 are the values of x which will make the ordinate of the curve 3 + 2x - x² equal to the ordinate of the st. line x/2 + 2. But the ordinates of the curve + the line are equal at the pts. of intersection A + B. ∴ the roots are the abscissas of A + B or x = -1/2 and x = 2 Ans. 6

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### csp_0b55254e1d7c5f2ead0452ffdb9f4c88

Shamash Secondary School Conditional Examination, Sept. 1964 ⟦line⟧

Subject: Algebra Class: 4th Year Secondary, Scientific Section. Date: 2/9/1964 Time: 8:00 - 11:00 a.m.

⟦line⟧ All questions are to be attempted.

I. (a) The expression x^2+px+q reduces to zero when x equals 2 or 4. Find the values of x for which this expression equals 48. (10 marks) (b) The expression x^4+ax^3+bx^2+cx+12 is factorable into (x-1), (x+1) and (x+3). Find the values of a, b, c and the other factor. (10 marks)

II. (a) At what time between eleven and twelve o'clock will the two hands of a watch be at right angle for the second time ? (10 marks) (b) A car covers the distance between Mosul and Baghdad in five hours, travelling on the route which lies on the right bank of the river Tigris. Another car travelling at an average speed which is less by 20 kilometres than the first, covers the distance between the two cities which lies on the left bank of the Tigris and which is 50 kilometres longer, in 7 1/2 hours. Find the length of each course. (10 marks)

III. (a) If log_2 (4x-4) = 2, find the value of log_4 x. (10 marks) (b) Compute by logarithms, arranging your work neatly: 7√((0.1062)^2 x (0.0071)^3 / (1.005) x (3.007)^5) (10 marks)

IV. (a) If a body falls from rest, (neglecting the friction of the air ) it will fall 16 ft during the first second, 48 ft during the second second, 80 ft during the third second, 112 ft during the fourth second and so on. Find the number of seconds it will take a stone to reach the bottom of a well 1936 ft. deep, if it is dropped from the top of the well. (10 marks) (b) In a Geometric progression, 1023 times the sum of the first five terms is equal to 31 times the sum of the first ten terms. Find the common ratio. (10 marks)

( cont'd.p.2)..

**Traduction anglaise —**

Shamash Secondary School Conditional Examination, Sept. 1964 ⟦line⟧ Subject: Algebra Class: 4th Year Secondary, Scientific Section. Date: 2/9/1964 Time: 8:00 - 11:00 a.m. ⟦line⟧ All questions are to be attempted. I. (a) The expression x^2+px+q reduces to zero when x equals 2 or 4. Find the values of x for which this expression equals 48. (10 marks) (b) The expression x^4+ax^3+bx^2+cx+12 is factorable into (x-1), (x+1) and (x+3). Find the values of a, b, c and the other factor. (10 marks) II. (a) At what time between eleven and twelve o'clock will the two hands of a watch be at right angle for the second time ? (10 marks) (b) A car covers the distance between Mosul and Baghdad in five hours, travelling on the route which lies on the right bank of the river Tigris. Another car travelling at an average speed which is less by 20 kilometres than the first, covers the distance between the two cities which lies on the left bank of the Tigris and which is 50 kilometres longer, in 7 1/2 hours. Find the length of each course. (10 marks) III. (a) If log_2 (4x-4) = 2, find the value of log_4 x. (10 marks) (b) Compute by logarithms, arranging your work neatly: 7√((0.1062)^2 x (0.0071)^3 / (1.005) x (3.007)^5) (10 marks) IV. (a) If a body falls from rest, (neglecting the friction of the air ) it will fall 16 ft during the first second, 48 ft during the second second, 80 ft during the third second, 112 ft during the fourth second and so on. Find the number of seconds it will take a stone to reach the bottom of a well 1936 ft. deep, if it is dropped from the top of the well. (10 marks) (b) In a Geometric progression, 1023 times the sum of the first five terms is equal to 31 times the sum of the first ten terms. Find the common ratio. (10 marks) ( cont'd.p.2)..

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### csp_0ca25c9d0ca2564c910590263aa5e161

[Marginalia] Naim Shahrabani

Shamash Secondary School Final Examination, May, 1966.

--- Answer all f i v e questions:

1. (a) Factor the following: 4 4 2 2 (i) my + 16mx - 12mx y (4 marks) 2 2 2 2 2 2 2 (ii) a b x - a b - 2abx + 2ab + x - 1 (4 marks) 6 9 3 (iii) 27x y + 64y (4 marks) (b) Find the value of p and q which will make the expression 2x³ + px² + qx + 1 divisible by (x-1) and (x+1), and find the third factor. (8 marks)

2. (a) Use the method of completing the square to show that the sum of the 2 roots of the equation ax +bx+c=o is equal to (- b/a) and that their product is equal to ( c/a ). (10 marks) (b) Find the value of x from the following equation: 2x x 3.10 - 13.10 + 4 = o (10 marks)

[Marginalia] 2 sections [Marginalia] only

3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: 2 log (2x -7x) = 2 (10 marks) 3 (ii) Without using tables evaluate: (log 9)(log 32) (10 marks) 2 9 (iii) Compute the value of y by logarithms, arranging your work neatly: _____________________________ 7 / (tan 19°45')² x (cos 77°16')³ y= / ------------------------------- (10 marks) (3.004)⁵ x (50.06)³

(cont'd.p.2)..

**Traduction anglaise —**

Naim Shahrabani Shamash Secondary School Final Examination, May, 1966. ⟦line⟧ Answer all f i v e questions: 1. (a) Factor the following: 4 4 2 2 (i) my + 16mx - 12mx y (4 marks) 2 2 2 2 2 2 2 (ii) a b x - a b - 2abx + 2ab + x - 1 (4 marks) 6 9 3 (iii) 27x y + 64y (4 marks) (b) Find the value of p and q which will make the expression 2x³ + px² + qx + 1 divisible by (x-1) and (x+1), and find the third factor. (8 marks) 2. (a) Use the method of completing the square to show that the sum of the 2 roots of the equation ax +bx+c=o is equal to (- b/a) and that their product is equal to ( c/a ). (10 marks) (b) Find the value of x from the following equation: 2x x 3.10 - 13.10 + 4 = o (10 marks) 2 sections only 3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: 2 log (2x -7x) = 2 (10 marks) 3 (ii) Without using tables evaluate: (log 9)(log 32) (10 marks) 2 9 (iii) Compute the value of y by logarithms, arranging your work neatly: ⟦line⟧ 7 / (tan 19°45')² x (cos 77°16')³ y= / ⟦line⟧ (10 marks) (3.004)⁵ x (50.06)³ (cont'd.p.2)..

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### csp_0cb38154e0945cc88ac3d79f14c5b4d6

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965.

Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m.

Attempt all questions :

1. (i) If m = 2x + y / x + 2y , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks).

2. (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks).

3. (i) Compute by logarithms, arranging your work neatly : ⁷√ (cos² 18° 47') (sin³ 48° 21') / (10.09)³ (0.0002049) (6 marks).

(ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks).

4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ...... (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks).

5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axix and 0.2 inch for your unit on the y-axix. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks).

[Marginalia] ⟦illegible⟧

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965. Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m. Attempt all questions : 1. (i) If m = 2x + y / x + 2y , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks). 2. (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks). 3. (i) Compute by logarithms, arranging your work neatly : ⁷√ (cos² 18° 47') (sin³ 48° 21') / (10.09)³ (0.0002049) (6 marks). (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks). 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ...... (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks). 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axix and 0.2 inch for your unit on the y-axix. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks). ⟦illegible⟧

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### csp_0ee1287497125211979cb2820e004322

Shamash Secondary School Final Examinations, June 1961.

Subject: Algebra Date: 12/6/1961 Class: 4th Year Secondary Time: 8:00-11:00 a.m.

Attempt all questions :

1. How much are pencils a gross when 120 more for a sovereign lowers the price 2d. a score ? ( 20 marks).

2. (i) Solve the equation √a - x + √b + x = √a + √b . (10 marks). (ii) Simplify and express with positive indices : ⟦{ (a^(p-q) / q√(a^(q^2-pq))) * a^(x(p-q)) }^n⟧ (10 marks).

3. (i) Compute by logarithms the following : 7√{ (0.001021)^2 . (4.003) / (16.02)^5 . (3.001)^4 } (10 marks). (ii) Derive a formula for finding the logarithms of a number to the base 'y' having given tables of logarithms to the base 'x'. Hence use your common logarithmic tables to compute the value of log70.56. (10 marks). 17

4. (i) A man drives a distance of 14 17/20 miles in 18 minutes. He drives the longest stretch in the first minute; and in every subsequent minute the distance he drives is 1/20 miles shorter than the distance driven in the previous minute. What are the distances driven in the first and last minutes ? Use arithmetic progressions to solve the question. (10 marks). (ii) The sum of n terms of a geometric progression is 4(5^n -1). Find the fifth term and the number of terms that must be taken for their sum to be 312496. (10 marks).

5. In the equation y = 16x(4 - x), y represents the height in feet risen by a stone thrown vertically upward from a point A on the roof of a building 20 ft. above the level of the ground; and x represents the time in seconds taken by the stone to reach the height y from that point. Draw a graph between x = 0 and x = 5 showing the relationship between y and x. (Take 1 in. = 1 second and 20 ft. respectively). From the graph find : (a) the maximum height above the level of the ground reached by the stone, (b) how long the stone remains at least 68 ft. above the ground, (c) how many seconds elapse from the time the stone was thrown to the time the stone strikes the ground. (20 marks).

**Traduction anglaise —**

Shamash Secondary School Final Examinations, June 1961. Subject: Algebra Date: 12/6/1961 Class: 4th Year Secondary Time: 8:00-11:00 a.m. Attempt all questions : 1. How much are pencils a gross when 120 more for a sovereign lowers the price 2d. a score ? ( 20 marks). 2. (i) Solve the equation √a - x + √b + x = √a + √b . (10 marks). (ii) Simplify and express with positive indices : ⟦{ (a^(p-q) / q√(a^(q^2-pq))) * a^(x(p-q)) }^n⟧ (10 marks). 3. (i) Compute by logarithms the following : 7√{ (0.001021)^2 . (4.003) / (16.02)^5 . (3.001)^4 } (10 marks). (ii) Derive a formula for finding the logarithms of a number to the base 'y' having given tables of logarithms to the base 'x'. Hence use your common logarithmic tables to compute the value of log70.56. (10 marks). 17 4. (i) A man drives a distance of 14 17/20 miles in 18 minutes. He drives the longest stretch in the first minute; and in every subsequent minute the distance he drives is 1/20 miles shorter than the distance driven in the previous minute. What are the distances driven in the first and last minutes ? Use arithmetic progressions to solve the question. (10 marks). (ii) The sum of n terms of a geometric progression is 4(5^n -1). Find the fifth term and the number of terms that must be taken for their sum to be 312496. (10 marks). 5. In the equation y = 16x(4 - x), y represents the height in feet risen by a stone thrown vertically upward from a point A on the roof of a building 20 ft. above the level of the ground; and x represents the time in seconds taken by the stone to reach the height y from that point. Draw a graph between x = 0 and x = 5 showing the relationship between y and x. (Take 1 in. = 1 second and 20 ft. respectively). From the graph find : (a) the maximum height above the level of the ground reached by the stone, (b) how long the stone remains at least 68 ft. above the ground, (c) how many seconds elapse from the time the stone was thrown to the time the stone strikes the ground. (20 marks).

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### csp_101162ba6d1957a593980fe1bad6a0c7

4) 52,320 X 0.6375 + 127,460 X 0.9625 = £ 33,354 + £ 122,680 5s = £ 156,034.25 Average rate = £ 156,034.25 --------------------------------- £ 52,320 + £ 127,460 = £ 156,034.25 --------------------------------- £ 179,780 = £ 0.8679 = £ 17s 4 1/4 d to nearest farthing

**Traduction anglaise —**

4) 52,320 X 0.6375 + 127,460 X 0.9625 = £ 33,354 + £ 122,680 5s = £ 156,034.25 Average rate = £ 156,034.25 ⟦line⟧ £ 52,320 + £ 127,460 = £ 156,034.25 ⟦line⟧ £ 179,780 = £ 0.8679 = £ 1 17s 4 1/4 d to nearest farthing

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### csp_13d5fd2160d25d35a6cac029f9781d72

3. (i) Let x = ⁷√ (cos² 18° 47' × sin³ 48° 21') / (10.09)³ × 0.000 2049

[Marginalia] ⑥

or cos 18° 47' = 0.9465 ∴ log cos 18° 47' = 1.9761 sin 48° 21' = 0.7472 ∴ log sin 48° 21' = 1.8734

2 log cos 18° 47' = 1.9522 3 log sin 48° 21' = 1.6202 log Num. = 1.5724 log Den. = 1.3226 7 log x = 0.2498 log x = 0.03568 = ∴ log x = 0.0357 Correct to 4 dec. pl. ∴ x = 1.086 Ans.

(ii) 2 log a - 5 log b = 3 log c , a = ? log a² - log b⁵ = log c³ or log a²/b⁵ = log c³ ∴ a²/b⁵ = c³ ∴ a² = b⁵.c³ ∴ a = b^(5/2).c^(3/2) Ans.

[Marginalia] ④

(iii) logₐ 4.41 = 2 , a = ? ∴ 4.41 = a² ∴ a = √4.41 = 2.1 Ans.

[Marginalia] ④

(iv) 2^(3-x) = 3^(2x+1) ∴ (3-x) log 2 = (2x+1) log 3 ∴ 3 log 2 - x log 2 = 2x log 3 + log 3 ∴ x(2 log 3 + log 2) = 3 log 2 - log 3 ∴ x = (3 log 2 - log 3) / (2 log 3 + log 2) = (log 8 - log 3) / (log 9 + log 2) = (0.9031 - 0.4771) / (0.9542 + 0.3010) = 0.4260 / 1.2552 = 0.3393... = 0.339 Correct to 3 dec. pl. Ans.

[Marginalia] ⑥

**Traduction anglaise —**

3. (i) Let x = ⁷√ (cos² 18° 47' × sin³ 48° 21') / (10.09)³ × 0.000 2049 ⑥ log cos 18° 47' = 1.9762 | 2 log cos 18° 47' = 1.9524 | 3 log 10.09 = 3.0111 log sin 48° 21' = 1.8734 | 3 log sin 48° 21' = 1.6202 | log 0.0002049 = 4.3115 log 10.09 = 1.0037 | log Num. = 1.5726 | log Den. = 1.3226 log 0.0002049 = 4.3115 | log Den. = 1.3226 | | 7 log x = 0.2500 | | log x = 0.03571 = 0.0357 Correct to 4 dec. pl. | x = 1.086 Ans. or cos 18° 47' = 0.9465 ∴ log cos 18° 47' = 1.9761 sin 48° 21' = 0.7472 ∴ log sin 48° 21' = 1.8734 2 log cos 18° 47' = 1.9522 3 log sin 48° 21' = 1.6202 log Num. = 1.5724 log Den. = 1.3226 7 log x = 0.2498 log x = 0.03568 = ∴ log x = 0.0357 Correct to 4 dec. pl. ∴ x = 1.086 Ans. (ii) 2 log a - 5 log b = 3 log c , a = ? log a² - log b⁵ = log c³ or log a²/b⁵ = log c³ ∴ a²/b⁵ = c³ ∴ a² = b⁵.c³ ∴ a = b^(5/2).c^(3/2) Ans. ④ (iii) logₐ 4.41 = 2 , a = ? ∴ 4.41 = a² ∴ a = √4.41 = 2.1 Ans. ④ (iv) 2^(3-x) = 3^(2x+1) ∴ (3-x) log 2 = (2x+1) log 3 ∴ 3 log 2 - x log 2 = 2x log 3 + log 3 ∴ x(2 log 3 + log 2) = 3 log 2 - log 3 ∴ x = (3 log 2 - log 3) / (2 log 3 + log 2) = (log 8 - log 3) / (log 9 + log 2) = (0.9031 - 0.4771) / (0.9542 + 0.3010) = 0.4260 / 1.2552 = 0.3393... = 0.339 Correct to 3 dec. pl. Ans. ⑥

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### csp_1447b8b1fa3c5a6091ea447505af92a5

SHAMASH SECONDARY SCHOOL ⟦...⟧itional Examination, September, 1969.

Subject: Algebra Class: 4th Year, Secondary Date: 5/9/1969. Time: 8:00-11:00 a.m.

Attempt all questions:

1. (i) Find the values of 'a' and 'b' if 3x³ - ax + b is exactly divisible by (x+1)(x-2). If 'a' and 'b' have these values, factor the expre- ssion completely. (10 marks). (ii) If x³ + 3x + 5 = x(x+1)(x+2) + Ax(x+1) + Bx + C for all values of x, find the values of the constants A, B, and C. (10 marks).

2. (i) The 15th term of an arithmetic progression is 25 and the sum of the first 10 terms is 60. Find the first term of the progression, the common difference and the sum of the first 16 terms. (10 marks). (ii) p+ 3, p + 8, and p + 18, are the 3rd, 4th, and 5th terms of a geometric progression. Find the value of p. Find also the common ratio and the 9th term of the progression. (10 marks).

3. A can walk a mile in 2 minutes less time than B would take. In a walki⟦...⟧ race, B has a start of ¼ mile, and A overtakes B in 10 minutes. Assuming that both men walk at a uniform rate, find their rates of walking in miles per hour. (20 marks).

4. Find the value of x from the following equation : 10²ˣ - 11(10ˣ) + 10 = 0.. (10 marks). (ii) Use logarithms to compute the value of the following expression : ⁷√((0.002013)²(Sin 15° 12')³ / (4.004)³(Cos 42° 13')²) (10 marks).

5. (i) Plot the curve of the equation y = 2x² - x - 3 at half-unit inter- vals between x = -1.5 and x = 2, choosing one inch a⟦...⟧ one unit on each of the two axes. (8 marks). (ii) From your graph, find the roots of the equation x + 3 ⟦...⟧ 2x². ⟦...⟧ marks). (iii) Find the values of x for which the function 2x² - x ⟦...⟧ is always positive. ⟦...⟧ 4 marks). (iv) By drawing another stra⟦...⟧t line on your diagram, fin⟦...⟧ the roots of the equation 2x² - x ⟦...⟧ 1 = 0. ⟦...⟧ 4 marks).

--------------------------------------------------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL ⟦Add⟧itional Examination, September, 1969. Subject: Algebra Class: 4th Year, Secondary Date: 5/9/1969. Time: 8:00-11:00 a.m. Attempt all questions: 1. (i) Find the values of 'a' and 'b' if 3x³ - ax + b is exactly divisible by (x+1)(x-2). If 'a' and 'b' have these values, factor the expre- ssion completely. (10 marks). (ii) If x³ + 3x + 5 = x(x+1)(x+2) + Ax(x+1) + Bx + C for all values of x, find the values of the constants A, B, and C. (10 marks). 2. (i) The 15th term of an arithmetic progression is 25 and the sum of the first 10 terms is 60. Find the first term of the progression, the common difference and the sum of the first 16 terms. (10 marks). (ii) p+ 3, p + 8, and p + 18, are the 3rd, 4th, and 5th terms of a geometric progression. Find the value of p. Find also the common ratio and the 9th term of the progression. (10 marks). 3. A can walk a mile in 2 minutes less time than B would take. In a walki⟦ng⟧ race, B has a start of ¼ mile, and A overtakes B in 10 minutes. Assuming that both men walk at a uniform rate, find their rates of walking in miles per hour. (20 marks). 4. Find the value of x from the following equation : 10²ˣ - 11(10ˣ) + 10 = 0.. (10 marks). (ii) Use logarithms to compute the value of the following expression : ⁷√((0.002013)²(Sin 15° 12')³ / (4.004)³(Cos 42° 13')²) (10 marks). 5. (i) Plot the curve of the equation y = 2x² - x - 3 at half-unit inter- vals between x = -1.5 and x = 2, choosing one inch a⟦s⟧ one unit on each of the two axes. (8 marks). (ii) From your graph, find the roots of the equation x + 3 ⟦=⟧ 2x². ⟦4⟧ marks). (iii) Find the values of x for which the function 2x² - x ⟦- 3⟧ is always positive. ⟦(⟧ 4 marks). (iv) By drawing another stra⟦igh⟧t line on your diagram, fin⟦d⟧ the roots of the equation 2x² - x ⟦-⟧ 1 = 0. ⟦(⟧ 4 marks). ⟦line⟧

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### csp_1556414a8cf4524c8196894adf13c4de

1st Quarter Exam. Subject: Arithmetic & Trigonometry. Date: 1/12/1958 Class: 4th Secondary. Time: 90 minutes. Answer all questions.

1 (a) Decimalise: - £ 9 15s 10¾d £ 5 2s 2¾d £10 10s 8¾d. (b) Express in shillings and pence to the nearest ¼d £0.840, £0.730, £0.910 (c) Express as a compound quantity correct to the nearest unit of the lowest given denomination. 0.6186 of 7½ tons. (tons, cwt., qr.).

2. A man walks from his house to a town 6 miles away at 4 miles per hour and cycles back again at 12 miles per hour. Find his average speed for the double journey.

3. A wirless pole stands at the corner A of a rectangular court ABCD. Its elevation from the corner B is 50°. Find its eleva- tion from the opposite corner C; given the length of AB = ¾ that of AD.

4. A ship steaming S60°E at 10 miles an hour is 10 miles N of a lighthouse at 12.00 noon. At what time will the ship be due E. of the light house ?

5. The area of a rectangular field is 2 acres and its breadth is 88 yds - Find the perimeter of the field and the length of the diagonal correct to the nearest yards

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**Traduction anglaise —**

1st Quarter Exam. Subject: Arithmetic & Trigonometry. Date: 1/12/1958 Class: 4th Secondary. Time: 90 minutes. Answer all questions. 1 (a) Decimalise: - £ 9 15s 10¾d £ 5 2s 2¾d £10 10s 8¾d. (b) Express in shillings and pence to the nearest ¼d £0.840, £0.730, £0.910 (c) Express as a compound quantity correct to the nearest unit of the lowest given denomination. 0.6186 of 7½ tons. (tons, cwt., qr.). 2. A man walks from his house to a town 6 miles away at 4 miles per hour and cycles back again at 12 miles per hour. Find his average speed for the double journey. 3. A wirless pole stands at the corner A of a rectangular court ABCD. Its elevation from the corner B is 50°. Find its eleva- tion from the opposite corner C; given the length of AB = ¾ that of AD. 4. A ship steaming S60°E at 10 miles an hour is 10 miles N of a lighthouse at 12.00 noon. At what time will the ship be due E. of the light house ? 5. The area of a rectangular field is 2 acres and its breadth is 88 yds - Find the perimeter of the field and the length of the diagonal correct to the nearest yards ⟦line⟧

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### csp_165bfa4ca0f55358b86f799b4adc7505

Shamash Secondary School Mid-Year Examination, Feb.1967

Subject: Algebra Date: 6/2/1967 Class: 4th Year,Secondary Time: 8:30 - 11:30 a.m.

----- Attempt all questions:

1. Revolve into factors: (i) 3x²-(4a+2b)x+a²+2ab (6 marks) (ii) 8x³-27y³+z³+18xyz (6 " ) (iii) Divide (a²+b²+c²)(a+1)+(2ab-2ac)(a+1)-2abc-2bc by (a+1) and express the quotient as a perfect square. (8 marks)

2. (i) If x+y = 2a and x-y=2b, find in the shortest possible way, the value of x⁴+x²y²+y⁴. (10 marks) (ii) Find the value of p which will make the expression 2x³+px²-5x+2 divisible by (x+2) and find the other two factors. (10 marks)

3. (i) A man can row upstream at 'a' miles an hour and downstream at 'b' miles an hour. He rows up to a certain point and then returns to his starting point, and finds that his average speed is 's' miles an hour for the double journey. Express each of the letters in terms of the other two. Find the value of 'b' if a=2 and s=3. (10 marks) (ii) Solve the two simultaneous equations: x²+4y²+80 = 15x+30y ........(1) xy = 6 . ........(2) (10 marks)

4. Two men started at the same time to meet each other from points which were 26 miles apart. If one took 4½ minutes longer than the other to walk a mile, and they met 2 hours after starting, find the speed of each in miles per hour. (20 marks)

(cont'd.p.2)...

**Traduction anglaise —**

Shamash Secondary School Mid-Year Examination, Feb.1967 Subject: Algebra Date: 6/2/1967 Class: 4th Year,Secondary Time: 8:30 - 11:30 a.m. ⟦line⟧ Attempt all questions: 1. Revolve into factors: (i) 3x²-(4a+2b)x+a²+2ab (6 marks) (ii) 8x³-27y³+z³+18xyz (6 " ) (iii) Divide (a²+b²+c²)(a+1)+(2ab-2ac)(a+1)-2abc-2bc by (a+1) and express the quotient as a perfect square. (8 marks) 2. (i) If x+y = 2a and x-y=2b, find in the shortest possible way, the value of x⁴+x²y²+y⁴. (10 marks) (ii) Find the value of p which will make the expression 2x³+px²-5x+2 divisible by (x+2) and find the other two factors. (10 marks) 3. (i) A man can row upstream at 'a' miles an hour and downstream at 'b' miles an hour. He rows up to a certain point and then returns to his starting point, and finds that his average speed is 's' miles an hour for the double journey. Express each of the letters in terms of the other two. Find the value of 'b' if a=2 and s=3. (10 marks) (ii) Solve the two simultaneous equations: x²+4y²+80 = 15x+30y ........(1) xy = 6 . ........(2) (10 marks) 4. Two men started at the same time to meet each other from points which were 26 miles apart. If one took 4½ minutes longer than the other to walk a mile, and they met 2 hours after starting, find the speed of each in miles per hour. (20 marks) (cont'd.p.2)...

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### csp_16c745ca5c3d541abdb095d5310f03be

⟦illegible⟧ Algebra ⟦illegible⟧

I. (i) Draw the graph of y = (x-1)(x-3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axis and 0.2 inches for your unit on the y-axis. To get a good drawing of the curve, choose <del>⟦illegible⟧</del> successive values of x at intervals of halves, beginning with -½. (5 marks)

(ii) From this graph find an approximate maximum value and the exact minimum value for y and the correspon-ding values of x which make y a maximum or a minimum. (5 marks)

(iii) By plotting another graph on the same diagram find the roots of the equation (x-1)(x-3)² = 5x - 9. (5 marks)

(iv) From these ⟦two⟧ graphs find the values of x for which the function (x-1)(x-3)² is always greater than (5x-9). (5 marks)

II. ⟦illegible⟧

**Traduction anglaise —**

⟦illegible⟧ Algebra ⟦illegible⟧ I. (i) Draw the graph of y = (x-1)(x-3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axis and 0.2 inches for your unit on the y-axis. To get a good drawing of the curve, choose <del>⟦illegible⟧</del> successive values of x at intervals of halves, beginning with -½. (5 marks) (ii) From this graph find an approximate maximum value and the exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks) (iii) By plotting another graph on the same diagram find the roots of the equation (x-1)(x-3)² = 5x - 9. (5 marks) (iv) From these ⟦two⟧ graphs find the values of x for which the function (x-1)(x-3)² is always greater than (5x-9). (5 marks) II. ⟦illegible⟧

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### csp_17090c0c64575018936f546038e9e725

B01 · header / latin SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965. ⟦line⟧

B02 · form / latin Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m.

B03 · paragraph / latin ⟦line⟧ Attempt all questions : 1. (i) If m = (2x + y) / (x + 2y) , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks). 2. (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks). 3. (i) Compute by logarithms, arranging your work neatly : ⁷√((cos² 18° 47') (sin³ 48° 21')) / ((10.09)³ (0.0002049)) (6 marks). (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks). 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ⟦line⟧ (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks). 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axis and 0.2 inch for your unit on the y-axis. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks). ⟦line⟧

B04 · marginalia / latin ⟦illegible⟧ at the back of this sheet.

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965. ⟦line⟧ Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m. ⟦line⟧ Attempt all questions : 1. (i) If m = (2x + y) / (x + 2y) , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks). 2. (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks). 3. (i) Compute by logarithms, arranging your work neatly : ⁷√((cos² 18° 47') (sin³ 48° 21')) / ((10.09)³ (0.0002049)) (6 marks). (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks). 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ⟦line⟧ (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks). 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axis and 0.2 inch for your unit on the y-axis. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks). ⟦line⟧ ⟦illegible⟧ at the back of this sheet.

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### csp_1848bc771b0253818cc1ea1660798a15

الرقم :: الاسم ::

4th Year - Algebra -p.2- Monthly Quiz.

II. (a) If k = 20ab / (4a+5b) find (i) "a" in terms of "b" and "k" (ii) "b" in terms of "a" and "k". Find also the value of √((k - 4a) / (k - 5b)) in terms of "a" and "b".

(b) If a = 0, b=1, c=-2, d=3, find the value of (3abc - 2bcd) ∛(a³bc - c³bd+3)

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**Traduction anglaise —**

Number :: Name :: 4th Year - Algebra -p.2- Monthly Quiz. II. (a) If k = 20ab / (4a+5b) find (i) "a" in terms of "b" and "k" (ii) "b" in terms of "a" and "k". Find also the value of √((k - 4a) / (k - 5b)) in terms of "a" and "b". (b) If a = 0, b=1, c=-2, d=3, find the value of (3abc - 2bcd) ∛(a³bc - c³bd+3) (c) Find the product of 3/2 x² - ax - 2/3 a² and 3/4 x² - 1/2 ax + 1/3 a² ⟦line⟧

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### csp_1958a42ae73b5693a1bd3117ab7014d9

2 (a) Speed in ft/sec 1000 / (6 1/4 x 60) ÷ (22/7 x 1/3^2) = (1000 x 4 x 7 x 9) / (25 x 60 x 22) = 84/11 = 7 7/11 ft/sec

(b) Sediment 1000 x 4 / 25 x 60 x 48 x 1/2 x 1/16 x 1/2240 = 45/7 = 6 3/7 tons

⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

2 (a) Speed in ft/sec 1000 / (6 1/4 x 60) ÷ (22/7 x 1/3^2) = (1000 x 4 x 7 x 9) / (25 x 60 x 22) = 84/11 = 7 7/11 ft/sec (b) Sediment 1000 x 4 / 25 x 60 x 48 x 1/2 x 1/16 x 1/2240 = 45/7 = 6 3/7 tons ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_195904ffa00b5069813abfcebdd8bc1f

B01 · header / latin Monthly Exam. April 6/4/54 Subj: Trigonometry Class: Fourth year

B02 · paragraph / latin 1. From the base of a tower, 80 ft. high, the angle of elevation of a distant point is 10°, while when viewed from the top of the tower its elevation is 8°. Find the distance of the point from the base of the tower and its height above the base.

B03 · other / latin ⟦diagram showing a tower and angles of elevation 8° and 10°⟧ x/y = tan 82° ∴ x = y tan 82° x/(y+80) = tan 80° ∴ x = y tan 80° + 80 tan 80° ∴ y tan 82° = y tan 80° + 80 tan 80° ∴ y = (80 tan 80°) / (tan 82° - tan 80°) = (80 x 5.6713) / (7.1154 - 5.6713) ∴ y = 453.7040 / 1.4441 = 314.18 ft = 314 correct to the nearest foot ∴ ⟦illegible⟧ = 314 + 80 = 394 ft Ans. I. 394/z = sin 10° ∴ z = 394 / sin 10° = 394 csc 10° = 394 x 5.7588 = 2269.8472 = 2270 ft. correct to the nearest foot. Ans. II

**Traduction anglaise —**

Monthly Exam. April 6/4/54 Subj: Trigonometry Class: Fourth year 1. From the base of a tower, 80 ft. high, the angle of elevation of a distant point is 10°, while when viewed from the top of the tower its elevation is 8°. Find the distance of the point from the base of the tower and its height above the base. ⟦diagram showing a tower and angles of elevation 8° and 10°⟧ x/y = tan 82° ∴ x = y tan 82° x/(y+80) = tan 80° ∴ x = y tan 80° + 80 tan 80° ∴ y tan 82° = y tan 80° + 80 tan 80° ∴ y = (80 tan 80°) / (tan 82° - tan 80°) = (80 x 5.6713) / (7.1154 - 5.6713) ∴ y = 453.7040 / 1.4441 = 314.18 ft = 314 correct to the nearest foot ∴ ⟦illegible⟧ = 314 + 80 = 394 ft Ans. I. 394/z = sin 10° ∴ z = 394 / sin 10° = 394 csc 10° = 394 x 5.7588 = 2269.8472 = 2270 ft. correct to the nearest foot. Ans. II

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### csp_197a9075256b59c38f5edd6bc414ce64

⟦page 2⟧

⟦IV⟧ (i) b) Sum of the Progression S_n = n/2 (a+l) = (z+y-2x)/2(y-x) [x+z] or S_n = (x+z)(y+z-2x)/2(y-x) Ans.

(ii) Let the four terms be a, ar, ar², ar³ . Then : (a) (ar²) = 36 or a² r² = 36 --- (1) } dividing (2) by (1), and (ar)(ar³) = 324 or a² r⁴ = 324 --- (2) } we get r² = 324/36 = 9 ∴ r² = 9 ∴ r = 3 (the negative root is to be excluded) from (1) a² r² = 36 ∴ 9 a² = 36 ∴ a² = 4 ∴ a = 2 . Therefore the four positive numbers are : 2 , 6 , 18 , 54 Ans.

V (i) S = 1 + 2 + 3 + ... + n ∴ S = n/2 (1+n) = n(n+1)/2 T = 1 + 2 + 3 + ... + (n-1) ∴ T = (n-1)/2 [1+(n-1)] = n(n-1)/2 but S² - T² = (S+T)(S-T) = [n(n+1)/2 + n(n-1)/2] [n(n+1)/2 - n(n-1)/2] ∴ S² - T² = (n²+n+n²-n)/2 x (n²+n-n²+n)/2 = n² * n = n³ Ans.

(ii) 3 , 6 , 10 1/2 Let x be the number to be added to each of the three numbers to make a G. P. out of them . Then the three terms are (3+x) , (6+x) , (21/2 + x) . Hence r = (21/2 + x)/(6+x) = (6+x)/(3+x) multiplying across, we get : (6+x)² = (3+x)(21/2 + x) or 36 + 12x + x² = (3+x)(21+2x)/2 or 36 + 12x + x² = (63 + 27x + 2x²)/2 or 72 + 24x + 2x² = 63 + 27x + 2x² or 3x = 9 ∴ x = 3 Ans. ∴ the numbers are 6 , 9 , 27/2 ...

<del>S_6 = 6/2 {2x6 + (6-1) 3/2} = 3 (12 + 15/2) = 3x39/2 = 117/2 = 58 1/2 Ans.</del> S_6 = a(rⁿ - 1)/(r-1) = 6[(3/2)⁶ - 1]/(3/2 - 1) = 6[729/64 - 1]/(1/2) = 12 [665/64] = 1995/16 = ⟦124⟧ 11/16

**Traduction anglaise —**

⟦page 2⟧ ⟦IV⟧ (i) b) Sum of the Progression S_n = n/2 (a+l) = (z+y-2x)/2(y-x) [x+z] or S_n = (x+z)(y+z-2x)/2(y-x) Ans. (ii) Let the four terms be a, ar, ar², ar³ . Then : (a) (ar²) = 36 or a² r² = 36 ⟦line⟧ (1) } dividing (2) by (1), and (ar)(ar³) = 324 or a² r⁴ = 324 ⟦line⟧ (2) } we get r² = 324/36 = 9 ∴ r² = 9 ∴ r = 3 (the negative root is to be excluded) from (1) a² r² = 36 ∴ 9 a² = 36 ∴ a² = 4 ∴ a = 2 . Therefore the four positive numbers are : 2 , 6 , 18 , 54 Ans. V (i) S = 1 + 2 + 3 + ... + n ∴ S = n/2 (1+n) = n(n+1)/2 T = 1 + 2 + 3 + ... + (n-1) ∴ T = (n-1)/2 [1+(n-1)] = n(n-1)/2 but S² - T² = (S+T)(S-T) = [n(n+1)/2 + n(n-1)/2] [n(n+1)/2 - n(n-1)/2] ∴ S² - T² = (n²+n+n²-n)/2 x (n²+n-n²+n)/2 = n² * n = n³ Ans. (ii) 3 , 6 , 10 1/2 Let x be the number to be added to each of the three numbers to make a G. P. out of them . Then the three terms are (3+x) , (6+x) , (21/2 + x) . Hence r = (21/2 + x)/(6+x) = (6+x)/(3+x) multiplying across, we get : (6+x)² = (3+x)(21/2 + x) or 36 + 12x + x² = (3+x)(21+2x)/2 or 36 + 12x + x² = (63 + 27x + 2x²)/2 or 72 + 24x + 2x² = 63 + 27x + 2x² or 3x = 9 ∴ x = 3 Ans. ∴ the numbers are 6 , 9 , 27/2 ... <del>S_6 = 6/2 {2x6 + (6-1) 3/2} = 3 (12 + 15/2) = 3x39/2 = 117/2 = 58 1/2 Ans.</del> S_6 = a(rⁿ - 1)/(r-1) = 6[(3/2)⁶ - 1]/(3/2 - 1) = 6[729/64 - 1]/(1/2) = 12 [665/64] = 1995/16 = ⟦124⟧ 11/16

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### csp_1b97310c33355a309ac1aecdcc4ac612

Solution Algebra 4th Year 12/6/1961

Suppose it is required to find log_y N in terms of logarithms to the base x. let log_y N = z ∴ N = y^z ∴ log_x N = z log_x y ∴ z = log_x N / log_x y ∴ log_y N = log_x N / log_x y ∴ log_17 70.56 = log_10 70.56 / log_10 17 ∴ log_17 70.56 = 1.8486 / 1.2304 = 18486 / 12304 = 1.50243... = 1.5024 correct to 4 decimal places Ans.

4. (i) 14 17/20 = the sum of an arithmetic progression whose No. of terms is 18 + whose common difference is (-1/20). Applying the formula s = ⟦illegible⟧ {2a + (n-1)d} we get : 14 17/20 = 18/2 {2a + 17(-1/20)} or 14 17/20 = 9 (2a - 17/20) ∴ 18a = 14 17/20 + 9x17/20 or 18a = 14 17/20 + 7 13/20 or 18a = 22 1/2 or 18a = 45/2 ∴ a = 45/36 or a = 5/4 ∴ l = a + (n-1)d ∴ l_18 = 5/4 + 17(-1/20) or l_18 = 25/20 - 17/20 or l_18 = 8/20 = 2/5 ∴ distance driven in the 1st minute = a = 5/4 miles and ,, ,, ,, last ,, = l = 2/5 miles Ans.

(ii) s = 4(5^n - 1) . Suppose n=1, ∴ s_1 = the 1st term = 4(5^1 - 1) = 16 Again suppose n=2 ∴ s_2 = 4(5^2 - 1) = 4 x 24 = 96 = 1st + 2nd terms ∴ 2nd term = 96 - 16 = 80 ∴ r = 80/16 = 5 ∴ l_5 = ar^4 = 16(5)^4 = 16 x 625 ∴ l_5 = 10000 Ans. 1 Now if s = 312496, then 312496 = 4(5^n - 1) or 5^n = 312496/4 + 1 or 5^n = 78124 + 1 or 5^n = 78125 or 5^n = 5^7 ∴ n = 7 Ans. 2.

| x | y | | 0 | 0 | | 1 | 48 | | 2 | 64 | | 3 | 48 | | 4 | 0 | | 5 | -80 |

5. y = 16x(4-x) maximum height = 64 + 20 = 84 ft. max height above point A = 64 ft ,, ground = 64 + 20 = 84 ft when y = 48, ⟦illegible⟧ x^2 - 4x + 3 = 0 or x = 1 or 3 ∴ 3 - 1 = 2 seconds when the stone strikes 30 ft above ground level when y = 10 the ground, y = -20 seconds level of ground ∴ 4x^2 - 16x - 5 = 0 x = 16 ± ⟦illegible⟧ / 8 = 2 ± 1/2 √24 = 4.5... sec Graph labels: 70, 60, 30, 0, -30, -60, -90 (y-axis); 1, 2, 3, 4, 5 (x-axis)

**Traduction anglaise —**

Algebra Solution 4th Year 12/6/1961 Suppose it is required to find log_y N in terms of logarithms to the base x. let log_y N = z ∴ N = y^z ∴ log_x N = z log_x y ∴ z = log_x N / log_x y ∴ log_y N = log_x N / log_x y ∴ log_17 70.56 = log_10 70.56 / log_10 17 ∴ log_17 70.56 = 1.8486 / 1.2304 = 18486 / 12304 = 1.50243... = 1.5024 correct to 4 decimal places Ans. 4. (i) 14 17/20 = the sum of an arithmetic progression whose No. of terms is 18 + whose common difference is (-1/20). Applying the formula s = ⟦illegible⟧ {2a + (n-1)d} we get : 14 17/20 = 18/2 {2a + 17(-1/20)} or 14 17/20 = 9 (2a - 17/20) ∴ 18a = 14 17/20 + 9x17/20 or 18a = 14 17/20 + 7 13/20 or 18a = 22 1/2 or 18a = 45/2 ∴ a = 45/36 or a = 5/4 ∴ l = a + (n-1)d ∴ l_18 = 5/4 + 17(-1/20) or l_18 = 25/20 - 17/20 or l_18 = 8/20 = 2/5 ∴ distance driven in the 1st minute = a = 5/4 miles and ,, ,, ,, last ,, = l = 2/5 miles Ans. (ii) s = 4(5^n - 1) . Suppose n=1, ∴ s_1 = the 1st term = 4(5^1 - 1) = 16 Again suppose n=2 ∴ s_2 = 4(5^2 - 1) = 4 x 24 = 96 = 1st + 2nd terms ∴ 2nd term = 96 - 16 = 80 ∴ r = 80/16 = 5 ∴ l_5 = ar^4 = 16(5)^4 = 16 x 625 ∴ l_5 = 10000 Ans. 1 Now if s = 312496, then 312496 = 4(5^n - 1) or 5^n = 312496/4 + 1 or 5^n = 78124 + 1 or 5^n = 78125 or 5^n = 5^7 ∴ n = 7 Ans. 2. x | y 0 | 0 1 | 48 2 | 64 3 | 48 4 | 0 5 | -80 5. y = 16x(4-x) maximum height = 64 + 20 = 84 ft. max height above point A = 64 ft ,, ground = 64 + 20 = 84 ft when y = 48, ⟦illegible⟧ x^2 - 4x + 3 = 0 or x = 1 or 3 ∴ 3 - 1 = 2 seconds when the stone strikes 30 ft above ground level when y = 10 the ground, y = -20 seconds level of ground ∴ 4x^2 - 16x - 5 = 0 x = 16 ± ⟦illegible⟧ / 8 = 2 ± 1/2 √24 = 4.5... sec Graph labels: 70, 60, 30, 0, -30, -60, -90 (y-axis); 1, 2, 3, 4, 5 (x-axis)

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### csp_1c1f31bf60315b8196af448e4ac81df3

Solution to Final Exam. Questions in Algebra to 4th year Comt. page 3. February 6th 1967.

← 26 miles → A * * B → ←

Let the speed of A be x m.p.h. " " " B " y m.p.h. ∴ 2x + 2y = 26 or x + y = 13 or y = 13 - x ---- ① also A walks one mile in 1/x hrs or in 60/x minutes B " " " " 1/y hrs. " " 60/y minutes ∴ 60/x - 60/y = 4 1/2 or 20/x - 20/y = 3/2 or 40y - 40x = 3xy or 40(y - x) = 3xy ---- ② Substitute from ① in ② : 40(13 - 2x) = 3x(13 - x) or 520 - 80x = 39x - 3x² or 3x² - 119x + 520 = 0 or (3x - 104)(x - 5) = 0 or x = 5 and x = 104/3 = 34 2/3 ∴ y = 13 - x or y = 13 - 5 = 8 or y = 13 - 34 2/3 = -21 2/3 inadmissible ∴ x = 5 m.p.h. } Ans. y = 8 m.p.h. }

**Traduction anglaise —**

Solution to Final Exam. Questions in Algebra to 4th year Comt. page 3. February 6th 1967. ← 26 miles → A * * B → ← Let the speed of A be x m.p.h. " " " B " y m.p.h. ∴ 2x + 2y = 26 or x + y = 13 or y = 13 - x ⟦line⟧ ① also A walks one mile in 1/x hrs or in 60/x minutes B " " " " 1/y hrs. " " 60/y minutes ∴ 60/x - 60/y = 4 1/2 or 20/x - 20/y = 3/2 or 40y - 40x = 3xy or 40(y - x) = 3xy ⟦line⟧ ② Substitute from ① in ② : 40(13 - 2x) = 3x(13 - x) or 520 - 80x = 39x - 3x² or 3x² - 119x + 520 = 0 or (3x - 104)(x - 5) = 0 or x = 5 and x = 104/3 = 34 2/3 ∴ y = 13 - x or y = 13 - 5 = 8 or y = 13 - 34 2/3 = -21 2/3 inadmissible ∴ x = 5 m.p.h. } Ans. y = 8 m.p.h. }

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### csp_1ce25868adc15f4e8eebde6d5bf80d1f

Shamash Secondary School Conditional Examination, Sept. 1964 Subject: Algebra Date: 2/9/1964 Class: 4th Year Secondary, Time: 8:00 - 11:00 a.m. Scientific Section. -----

All questions are to be attempted.

I. (a) The expression x² +px+q reduces to zero when x equals 2 or 4. Find the values of x for which this expression equals 48. (10 marks) (b) The expression x⁴ +ax³ +bx² +cx+12 is factorable into (x-1), (x+1) and (x+3). Find the values of a, b, c and the other factor. (10 marks)

II. (a) At what time between eleven and twelve o'clock will the two hands of a watch be at right angle for the second time ? (10 marks) (b) A car covers the distance between Mosul and Baghdad in five hours, travelling on the route which lies on the right bank of the river Tigris. Another car travelling at an average speed which is less by 20 kilometres than the first, covers the distance between the two cities which lies on the left bank of the Tigris and which is 50 kilometres longer, in 7½ hours. Find the length of each course. (10 marks)

III. (a) If log₂ (4x-4) = 2, find the value of log₄ x. (10 marks) (b) Compute by logarithms, arranging your work neatly: ____________________ 7 / (0.1062)² x (0.0071)³ \/ ----------------------- (1.005) x (3.007)⁵ (10 marks)

( cont'd.p.2)..

**Traduction anglaise —**

Shamash Secondary School Conditional Examination, Sept. 1964 Subject: Algebra Date: 2/9/1964 Class: 4th Year Secondary, Time: 8:00 - 11:00 a.m. Scientific Section. ⟦line⟧ All questions are to be attempted. I. (a) The expression x² +px+q reduces to zero when x equals 2 or 4. Find the values of x for which this expression equals 48. (10 marks) (b) The expression x⁴ +ax³ +bx² +cx+12 is factorable into (x-1), (x+1) and (x+3). Find the values of a, b, c and the other factor. (10 marks) II. (a) At what time between eleven and twelve o'clock will the two hands of a watch be at right angle for the second time ? (10 marks) (b) A car covers the distance between Mosul and Baghdad in five hours, travelling on the route which lies on the right bank of the river Tigris. Another car travelling at an average speed which is less by 20 kilometres than the first, covers the distance between the two cities which lies on the left bank of the Tigris and which is 50 kilometres longer, in 7½ hours. Find the length of each course. (10 marks) III. (a) If log₂ (4x-4) = 2, find the value of log₄ x. (10 marks) (b) Compute by logarithms, arranging your work neatly: ⟦line⟧ 7 / (0.1062)² x (0.0071)³ \/ ⟦line⟧ (1.005) x (3.007)⁵ (10 marks) IV. (a) If a body falls from rest, (neglecting the friction of the air ) it will fall 16 ft during the first second, 48 ft during the second second, 80 ft during the third second, 112 ft during the fourth second and so on. Find the number of seconds it will take a stone to reach the bottom of a well 1936 ft. deep, if it is dropped from the top of the well. (10 marks) (b) In a Geometric progression, 1023 times the sum of the first five terms is equal to 31 times the sum of the first ten terms. Find the common ratio. (10 marks) ( cont'd.p.2)..

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### csp_1f47906c1b885aae97b13dcbb82ef4f8

Shamash Secondary School 3rd Quarter Examination

Subject: Algebra Class: 4th Secondary Date: 17/5/1961. Time: 8:30-10:00 a.m.

-------------------- Attempt all Questions:

1. A boy can swim at v m.p.h. in still water. When he swims downstream the current increases his speed by u m.p.h., and when he swims upstream the current decreases his speed by u m.p.h. The difference in his time to swim one mile downstream and one mile upstream is t hours. Find a formula for t in terms of v and u, and find v if u = 2, t = 4/5.

2. A dealer bought a horse, expecting to sell it again at a price that would have given him 10 per cent. profit on his purchase; but he had to sell it for £50 less than he expected, and he then found that he had lost 15 per cent. on what it cost him. What did he pay for the horse ?

3. At what time between 12 o'clock and 1 o'clock are the two hands of a watch at right angles for the second time?

4. By lowering the price of eggs and selling them one penny per egg cheaper, a man finds that he can sell 5 more than he used to do for 5s. At what price per egg did he sell them at first?

5. A cistern can be filled by two pipes running together in 24 minutes. The larger pipe would fill the cistern in 20 minutes less than the smaller one. Find the time taken by each.

--------------------

**Traduction anglaise —**

Shamash Secondary School 3rd Quarter Examination Subject: Algebra Class: 4th Secondary Date: 17/5/1961. Time: 8:30-10:00 a.m. ⟦line⟧ Attempt all Questions: 1. A boy can swim at v m.p.h. in still water. When he swims downstream the current increases his speed by u m.p.h., and when he swims upstream the current decreases his speed by u m.p.h. The difference in his time to swim one mile downstream and one mile upstream is t hours. Find a formula for t in terms of v and u, and find v if u = 2, t = 4/5. 2. A dealer bought a horse, expecting to sell it again at a price that would have given him 10 per cent. profit on his purchase; but he had to sell it for £50 less than he expected, and he then found that he had lost 15 per cent. on what it cost him. What did he pay for the horse ? 3. At what time between 12 o'clock and 1 o'clock are the two hands of a watch at right angles for the second time? 4. By lowering the price of eggs and selling them one penny per egg cheaper, a man finds that he can sell 5 more than he used to do for 5s. At what price per egg did he sell them at first? 5. A cistern can be filled by two pipes running together in 24 minutes. The larger pipe would fill the cistern in 20 minutes less than the smaller one. Find the time taken by each. ⟦line⟧

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### csp_2017b7d3f67f5cc2a7ace67ef9ee5639

[Marginalia] Nadia Jacob.

Shamash Secondary School Final Examination, May, 1966.

Subject: Algebra Date: 29/5/1966 Class: 4th Year, Scientific. Time: 8:00 - 11:00 a.m.

--- Answer all f i v e questions:

1. (a) Factor the following: (i) my⁴ + 16mx⁴ - 12mx²y² (4 marks) (ii) a²b²x² - a²b² - 2abx² + 2ab + x² - 1 (4 marks) (iii) 27x⁶y⁹ + 64y³ (4 marks) (b) Find the value of p and q which will make the expression 2x³ + px² + qx + 1 divisible by (x-1) and (x+1), and find the third factor. (8 marks)

2. (a) Use the method of completing the square to show that the sum of the roots of the equation ax²+bx+c=o is equal to (- b/a) and that their product is equal to ( c/a ). (10 marks) (b) Find the value of x from the following equation: 3.10²ˣ - 13.10ˣ + 4 = o (10 marks)

3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: log₃ (2x²-7x) = 2 (10 marks) (ii) Without using tables evaluate: (log₂ 9)(log₉ 32) (10 marks) (iii) Compute the value of y by logarithms, arranging your work neatly: y = ⁷√[ (tan 19°45')² x (cos 77°16')³ / (3.004)⁵ x (50.06)³ ] (10 marks)

(cont'd.p.2)..

**Traduction anglaise —**

Nadia Jacob. Shamash Secondary School Final Examination, May, 1966. Subject: Algebra Date: 29/5/1966 Class: 4th Year, Scientific. Time: 8:00 - 11:00 a.m. ⟦line⟧ Answer all f i v e questions: 1. (a) Factor the following: (i) my⁴ + 16mx⁴ - 12mx²y² (4 marks) (ii) a²b²x² - a²b² - 2abx² + 2ab + x² - 1 (4 marks) (iii) 27x⁶y⁹ + 64y³ (4 marks) (b) Find the value of p and q which will make the expression 2x³ + px² + qx + 1 divisible by (x-1) and (x+1), and find the third factor. (8 marks) 2. (a) Use the method of completing the square to show that the sum of the roots of the equation ax²+bx+c=o is equal to (- b/a) and that their product is equal to ( c/a ). (10 marks) (b) Find the value of x from the following equation: 3.10²ˣ - 13.10ˣ + 4 = o (10 marks) 3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: log₃ (2x²-7x) = 2 (10 marks) (ii) Without using tables evaluate: (log₂ 9)(log₉ 32) (10 marks) (iii) Compute the value of y by logarithms, arranging your work neatly: y = ⁷√[ (tan 19°45')² x (cos 77°16')³ / (3.004)⁵ x (50.06)³ ] (10 marks) (cont'd.p.2)..

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### csp_2087fbdfd3de55ea985253a801ca49a2

[Marginalia] Name: ⟦illegible⟧ ⟦illegible⟧ Nour Basri 45

-p.2- Algebra. 4th Year Scientific. 29/5/1966. -----

4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks).

(ii) An invalid on a certain day was able to take a single step of 18 inches. If he was each day to walk twice as far as on the preceding day, how long would it be before he can take a walk of 512 yards ? (10 marks)

5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks)

(ii) Using the same axes and scales, draw another graph to find the roots of the equation x³ - 13/4x - 3/2 = 0. ( 7 marks)

(iii) From your diagram, find all values of x which make the expression [x³ - (13/4x + 3/2)] , positive. ( 7 marks). --------

**Traduction anglaise —**

Name: ⟦illegible⟧ ⟦illegible⟧ Nour Basri 45 -p.2- Algebra. 4th Year Scientific. 29/5/1966. ⟦line⟧ 4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks). (ii) An invalid on a certain day was able to take a single step of 18 inches. If he was each day to walk twice as far as on the preceding day, how long would it be before he can take a walk of 512 yards ? (10 marks) 5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks) (ii) Using the same axes and scales, draw another graph to find the roots of the equation x³ - 13/4x - 3/2 = 0. ( 7 marks) (iii) From your diagram, find all values of x which make the expression [x³ - (13/4x + 3/2)] , positive. ( 7 marks). ⟦line⟧

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### csp_2182421e9d0c5cf28ee5de5eb3f03d0d

SHAMASH SECONDARY SCHOOL 4th Quarter Examination, May, 1965.

Subject: Algebra Date: 2/5/1965 Class: 4th Secondary year Time: 8:00-9:30 a.m.

<del>A</del>ttempt all questions.

1. (a) Prove that: (a-a⁻¹)(a⁴/³ + a⁻²/³) = a² - a⁻² / a⁻¹/³ (13 marks) (b) Evaluate: x³/² + xy / xy - y³ - √x / √x-y (1⟦3⟧ marks)

2. Solve the equation: 6 √x - 7 / √x - 1 - 5 = 7 √x - 26 / 7 √x - 21 (25 marks)

3. Find x from the equation: 3²x = 5x+1 (25 marks)

4. Compute by logarithms the value of x, arranging your work neatly: ⁷√ (1.001)² (0.0004061)²/³ / Sin³ 24° 21' Cos² 41° 57' (25 marks)

--------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL 4th Quarter Examination, May, 1965. Subject: Algebra Date: 2/5/1965 Class: 4th Secondary year Time: 8:00-9:30 a.m. <del>A</del>ttempt all questions. 1. (a) Prove that: (a-a⁻¹)(a⁴/³ + a⁻²/³) = a² - a⁻² / a⁻¹/³ (13 marks) (b) Evaluate: x³/² + xy / xy - y³ - √x / √x-y (1⟦3⟧ marks) 2. Solve the equation: 6 √x - 7 / √x - 1 - 5 = 7 √x - 26 / 7 √x - 21 (25 marks) 3. Find x from the equation: 3²x = 5x+1 (25 marks) 4. Compute by logarithms the value of x, arranging your work neatly: ⁷√ (1.001)² (0.0004061)²/³ / Sin³ 24° 21' Cos² 41° 57' (25 marks) ⟦line⟧

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### csp_2390dd186211587c820cbd3b5ddea6c7

B01 · paragraph / latin ⟦illegible⟧ [ 8 + x ] ⟦illegible⟧ = ( ⟦illegible⟧ + 8 ) ⟦illegible⟧ = A ⟦illegible⟧ (ii) Let the four nos. be ⟦illegible⟧. Then ⟦illegible⟧ = 6 ⟦illegible⟧ (1) and (⟦illegible⟧) (⟦illegible⟧) = ⟦illegible⟧ ⟦illegible⟧ (2) ⟦illegible⟧ From (1) ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ Thus ⟦illegible⟧ ⟦line⟧ VII (i) ⟦illegible⟧ = (x+1) ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = [ (1-x) + 1 ] ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ [ ⟦illegible⟧ ] [ ⟦illegible⟧ ] = (T - t) (T + t) = T² - t² ⟦illegible⟧ = T² - t² ⟦line⟧ (ii) ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ Let x be the number to be added to each of the three numbers ⟦illegible⟧. Then the three nos. (x + ⟦illegible⟧), (x + 6), (x + ⟦illegible⟧) are in ⟦illegible⟧. Therefore (x + 6)² = (x + ⟦illegible⟧) (x + ⟦illegible⟧) Simplifying ⟦illegible⟧ x² + 12x + 36 = x² + ⟦illegible⟧ x + ⟦illegible⟧ 12x + 36 = ⟦illegible⟧ x + ⟦illegible⟧ 3x = 9 or x = 3 The numbers are ⟦illegible⟧, 9, ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

⟦illegible⟧ [ 8 + x ] ⟦illegible⟧ = ( ⟦illegible⟧ + 8 ) ⟦illegible⟧ = A ⟦illegible⟧ (ii) Let the four nos. be ⟦illegible⟧. Then ⟦illegible⟧ = 6 ⟦illegible⟧ (1) and (⟦illegible⟧) (⟦illegible⟧) = ⟦illegible⟧ ⟦illegible⟧ (2) ⟦illegible⟧ From (1) ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ Thus ⟦illegible⟧ ⟦line⟧ VII (i) ⟦illegible⟧ = (x+1) ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ = [ (1-x) + 1 ] ⟦illegible⟧ = ⟦illegible⟧ ⟦illegible⟧ [ ⟦illegible⟧ ] [ ⟦illegible⟧ ] = (T - t) (T + t) = T² - t² ⟦illegible⟧ = T² - t² ⟦line⟧ (ii) ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ Let x be the number to be added to each of the three numbers ⟦illegible⟧. Then the three nos. (x + ⟦illegible⟧), (x + 6), (x + ⟦illegible⟧) are in ⟦illegible⟧. Therefore (x + 6)² = (x + ⟦illegible⟧) (x + ⟦illegible⟧) Simplifying ⟦illegible⟧ x² + 12x + 36 = x² + ⟦illegible⟧ x + ⟦illegible⟧ 12x + 36 = ⟦illegible⟧ x + ⟦illegible⟧ 3x = 9 or x = 3 The numbers are ⟦illegible⟧, 9, ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

---

### csp_23a1bb511034500482dc36d0e4ea6684

B01 · header / mixed — ٧ — ١٤٤٩ : ⟦illegible⟧ ١٧ : ⟦illegible⟧

B02 · paragraph / arabic ٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

B03 · paragraph / arabic ٥ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

B04 · paragraph / arabic ٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧

B05 · paragraph / arabic ٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

B06 · paragraph / arabic ٨ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ . ٣٥ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

B07 · paragraph / arabic ٣ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧

B08 · paragraph / arabic ٥ — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧

B09 · header / latin (25 marks) (II) Fill in the blanks in the following equations :-

B10 · form / latin 1. one furlong = ( ) chains = ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds. = ( ) ft. 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre = ( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. = ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones

B11 · footer / latin (25 marks)

**Traduction anglaise —**

— 7 — 1449 : ⟦illegible⟧ 17 : ⟦illegible⟧ 3 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ 3 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ 3 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ 3 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ 3 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ 5 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ 3 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ 3 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ 8 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ . 35 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ 3 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ 5 — ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦line⟧ (25 marks) (II) Fill in the blanks in the following equations :- 1. one furlong = ( ) chains = ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds. = ( ) ft. 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre = ( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. = ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones (25 marks)

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### csp_23c6ce870dab5ae3b921e20baa57ce0d

⟦illegible⟧ Year Secondary, Shamash School

(a) 2x³ - 9x² + 7x + 6 By trial + error when x = 2, then 2x³ - 9x² + 7x + 6 = 2x2³ - 9x2² + 7x2 + 6 = 16 - 36 + 14 + 6 = 0 ∴ (x - 2) is a factor ∴ 2x³ - 9x² + 7x + 6 = 2x²(x - 2) - 5x(x - 2) - 3(x - 2) = (x - 2)(2x² - 5x - 3) ∴ 2x³ - 9x² + 7x + 6 = (x - 2)(2x + 1)(x - 3) Ans.

(b) 4(xy + mn)² - (x² + y² - m² - n²)² = [2(xy + mn) + (x² + y² - m² - n²)][2(xy + mn) - (x² + y² - m² - n²)] = (2xy + 2mn + x² + y² - m² - n²)(2xy + 2mn - x² - y² + m² + n²) = [(x + y)² - (m - n)²][(m + n)² - (x - y)²] = (x + y + m - n)(x + y - m + n)(m + n + x - y)(m + n - x + y) Ans.

2 (a) √2 = 1.414 , √3 = 1.732 , √6 = 2.449 , (3 - √2)(7 + 4√3) / 2√3 - 3 = (3 - √2)(7 + 4√3)(2√3 + 3) / (2√3 - 3)(2√3 + 3) = (21 + 12√3 - 7√2 - 4√6)(2√3 + 3) / 12 - 9 = 42√3 + 72 + ⟦illegible⟧ - 14√6 - 24√2 + 63 + 36√3 - 21√2 - 12√6 / 3 = 135 + 78√3 - 45√2 - 26√6 / 3 = 45 + 26√3 - 15√2 - 26/3 √6 = 45 + 26 x 1.732 - 15 x 1.414 - 26 x 2.449 / 3 = 45 + 45.032 - 21.210 - 21.1466 = 90.032 - 42.3566 = 47.6754 = 47.68 correct to 2 dec. places Ans.

(b) 2√x - 1 / 2√x + 4/3 = √x - 2 / √x - 4/3 or 2√x - 1 / 6√x + 4 = √x - 2 / 3√x - 4 Multiplying across, we have, (6√x + 4)(√x - 2) = (2√x - 1)(3√x - 4) or 6x - 8√x - 8 = 6x - 11√x + 4 ∴ 3√x = 12 ∴ √x = 4 ∴ x = 16 Ans.

3 (a) (√3 √2)ˣ = 36 ∴ (3¹/² x 2¹/²)ˣ = 36 ∴ (3 x 2)ˣ/² = 6² or 6ˣ/² = 6² ∴ x/2 = 2 ∴ x = 4 Ans.

(b) Let x = ⁷√ (0.0002003)²/³ x (0.04031)⁵/³ / 1.004 x 9.006

**Traduction anglaise —**

⟦illegible⟧ Year Secondary, Shamash School (a) 2x³ - 9x² + 7x + 6 By trial + error when x = 2, then 2x³ - 9x² + 7x + 6 = 2x2³ - 9x2² + 7x2 + 6 = 16 - 36 + 14 + 6 = 0 ∴ (x - 2) is a factor ∴ 2x³ - 9x² + 7x + 6 = 2x²(x - 2) - 5x(x - 2) - 3(x - 2) = (x - 2)(2x² - 5x - 3) ∴ 2x³ - 9x² + 7x + 6 = (x - 2)(2x + 1)(x - 3) Ans. (b) 4(xy + mn)² - (x² + y² - m² - n²)² = [2(xy + mn) + (x² + y² - m² - n²)][2(xy + mn) - (x² + y² - m² - n²)] = (2xy + 2mn + x² + y² - m² - n²)(2xy + 2mn - x² - y² + m² + n²) = [(x + y)² - (m - n)²][(m + n)² - (x - y)²] = (x + y + m - n)(x + y - m + n)(m + n + x - y)(m + n - x + y) Ans. 2 (a) √2 = 1.414 , √3 = 1.732 , √6 = 2.449 , (3 - √2)(7 + 4√3) / 2√3 - 3 = (3 - √2)(7 + 4√3)(2√3 + 3) / (2√3 - 3)(2√3 + 3) = (21 + 12√3 - 7√2 - 4√6)(2√3 + 3) / 12 - 9 = 42√3 + 72 + ⟦illegible⟧ - 14√6 - 24√2 + 63 + 36√3 - 21√2 - 12√6 / 3 = 135 + 78√3 - 45√2 - 26√6 / 3 = 45 + 26√3 - 15√2 - 26/3 √6 = 45 + 26 x 1.732 - 15 x 1.414 - 26 x 2.449 / 3 = 45 + 45.032 - 21.210 - 21.1466 = 90.032 - 42.3566 = 47.6754 = 47.68 correct to 2 dec. places Ans. (b) 2√x - 1 / 2√x + 4/3 = √x - 2 / √x - 4/3 or 2√x - 1 / 6√x + 4 = √x - 2 / 3√x - 4 Multiplying across, we have, (6√x + 4)(√x - 2) = (2√x - 1)(3√x - 4) or 6x - 8√x - 8 = 6x - 11√x + 4 ∴ 3√x = 12 ∴ √x = 4 ∴ x = 16 Ans. 3 (a) (√3 √2)ˣ = 36 ∴ (3¹/² x 2¹/²)ˣ = 36 ∴ (3 x 2)ˣ/² = 6² or 6ˣ/² = 6² ∴ x/2 = 2 ∴ x = 4 Ans. (b) Let x = ⁷√ (0.0002003)²/³ x (0.04031)⁵/³ / 1.004 x 9.006 log 0.0002003 = 4.3016 | 2/3 log 0.0002003 = 3.53440 | log 1.004 = 0.0017 log 0.04031 = 2.6054 | 5/3 log 0.04031 = 1.16324 | log 9.006 = 0.9545 log 1.004 = 0.0017 | log Num. = 4.69764 | log Den. = 0.9562 log 9.006 = 0.9545 | log Den. = 0.95620 2 log 0.0002003 = 8.6032 | 7 log x = 3.74144 3 log 0.04031 = 5.8162 | log x = 1.391634 = 1.3916 | x = 0.2463 Ans.

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### csp_241a452ee8f6524eb41eaa16e867c6ab

[Marginalia] متروك

Between 4 p.m Friday and noon the following Wednesday = there ⟦are⟧ 4 5/6 days The fast clock gains 3 x 4 5/6 = 14 1/2 minutes The slow " loses 2 1/2 x 4 5/6 = 12 1/12 " Difference between them = 14 1/2 + 12 1/12 (1) = 26 7/12 minutes Difference per day = 5 1/2 minutes The two clocks will show the same time again when the difference between them has become 12 hours or 12 x 60 = 720 min 720 / 5 1/2 = 720 x 2 / 11 = 130 10/11 days.

11 ) 1440

The fast clock gains 3 minutes per day It will gain 3 x 130 10/11 minutes in 130 10/11 days = 4320 / 11 minutes = 392 8/11 minutes = 6 hours and 32 8/11 minutes Now the correct clock would <del>show</del> read after 130 10/11 days of correct time 1 hr 49 1/11 min p.m The fast clock will therefore read 1 hr 49 1/11 min. + 6 hours 32 8/11 min = 8 hours 21 9/11 minutes p.m The slow clock will read the same.

**Traduction anglaise —**

Left / Abandoned Between 4 p.m Friday and noon the following Wednesday = there ⟦are⟧ 4 5/6 days The fast clock gains 3 x 4 5/6 = 14 1/2 minutes The slow " loses 2 1/2 x 4 5/6 = 12 1/12 " Difference between them = 14 1/2 + 12 1/12 (1) = 26 7/12 minutes Difference per day = 5 1/2 minutes The two clocks will show the same time again when the difference between them has become 12 hours or 12 x 60 = 720 min 720 / 5 1/2 = 720 x 2 / 11 = 130 10/11 days. 11 ) 1440 The fast clock gains 3 minutes per day It will gain 3 x 130 10/11 minutes in 130 10/11 days = 4320 / 11 minutes = 392 8/11 minutes = 6 hours and 32 8/11 minutes Now the correct clock would <del>show</del> read after 130 10/11 days of correct time 1 hr 49 1/11 min p.m The fast clock will therefore read 1 hr 49 1/11 min. + 6 hours 32 8/11 min = 8 hours 21 9/11 minutes p.m The slow clock will read the same.

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### csp_257638bd478a5d468cdec42a6680bee8

(cont'd).. -2- Algebra 4th Year. Scientific 14/5/1969. ---

5. (i) Plot the curve of the function 3+2x-x² for values of x from x=-2 to x=4, choosing one half of an inch for each unit on the axis of x and on the axis of y. (4 marks) (ii) From your graph, find the roots of the equation x²-3=2x. (3 marks) (iii) Find the values of x for which the function 3+2x-x² is always positive. ( 3 marks) (iv) Find from your diagram the value of x at which the function 3+2x-x² is greatest and state the maximum value. (3 marks) (v) By plotting another curve on the same diagram, find the values of x for which 3+2x-x² > x/2 + 2. (4 marks) (vi) From your last diagram, find the roots of the equation 3+2x-x² = x/2 + 2. (3 marks)

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**Traduction anglaise —**

(cont'd).. -2- Algebra 4th Year. Scientific 14/5/1969. ⟦line⟧ 5. (i) Plot the curve of the function 3+2x-x² for values of x from x=-2 to x=4, choosing one half of an inch for each unit on the axis of x and on the axis of y. (4 marks) (ii) From your graph, find the roots of the equation x²-3=2x. (3 marks) (iii) Find the values of x for which the function 3+2x-x² is always positive. ( 3 marks) (iv) Find from your diagram the value of x at which the function 3+2x-x² is greatest and state the maximum value. (3 marks) (v) By plotting another curve on the same diagram, find the values of x for which 3+2x-x² > x/2 + 2. (4 marks) (vi) From your last diagram, find the roots of the equation 3+2x-x² = x/2 + 2. (3 marks) ⟦line⟧

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### csp_25850cef71965e3293f0a510685b15c3

Shamash Secondary School Mid-Year Examination, Feb. 1967 Subject: Algebra Date: 6/2/1967 Class: 4th Year, Secondary Time: 8:30 - 11:30 a.m. -----

Attempt all questions:

2. (i) If x+y = 2a and x-y=2b, find in the shortest possible way, the value of x⁴+x²y²+y⁴ . (10 marks) (ii) Find the value of p which will make the expression 2x³+px²-5x+2 divisible by (x+2) and find the other two factors. (10 marks)

(cont'd.p.2)...

**Traduction anglaise —**

Shamash Secondary School Mid-Year Examination, Feb. 1967 Subject: Algebra | Date: 6/2/1967 Class: 4th Year, Secondary | Time: 8:30 - 11:30 a.m. ⟦line⟧ Attempt all questions: 1. Revolve into factors: (i) 3x²-(4a+2b)x+a²+2ab | (6 marks) (ii) 8x³-27y³+z³+18xyz | (6 " ) (iii) Divide (a²+b²+c²)(a+1)+(2ab-2ac)(a+1)-2abc-2bc by (a+1) and express the quotient as a perfect square. | (8 marks) 2. (i) If x+y = 2a and x-y=2b, find in the shortest possible way, the value of x⁴+x²y²+y⁴ . | (10 marks) (ii) Find the value of p which will make the expression 2x³+px²-5x+2 divisible by (x+2) and find the other two factors. (10 marks) 3. (i) A man can row upstream at 'a' miles an hour and downstream at 'b' miles an hour. He rows up to a certain point and then returns to his starting point, and finds that his average speed is 's' miles an hour for the double journey. Express each of the letters in terms of the other two. Find the value of 'b' if a=2 and s=3. (10 marks) (ii) Solve the two simultaneous equations: x²+4y²+80 = 15x+30y ........(1) xy = 6 . .........(2) | (10 marks) 4. Two men started at the same time to meet each other from points which were 26 miles apart. If one took 4½ minutes longer than the other to walk a mile, and they met 2 hours after starting, find the speed of each in miles per hour. (20 marks) (cont'd.p.2)...

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### csp_26178d2188cf546e9b9c696fb0f61914

(cont'd.).. -p.2- Arith. & Trig. 17/5/67 4th Year Secondary.

⟦line⟧

5. ABC is a triangle with AB = AC = 100 ft. and the angle BAC is 70°. At A is a vertical pole AO = 80 ft. high. Calculate: (i) the length of BC, (ii) the angle of elevation of the top of the pole from B, (iii) the angle of elevation of the top of the pole from the mid-point of BC.

6. A, B and C are three points on a coastline which runs from north to south; B is south of A and C is 1000 yards south of B. A boat is moving in a straight line towards C and when it is at a point P which is 2000 yd. from A on a bearing of (N.60.E.) its bearing from B is ( N.38E.). Calculate: (a) the distance from P to the nearest point X on the coastline. (b) the distance AB. (c) the bearing of P from C.

⟦line⟧

[Marginalia] ⟦illegible⟧ [Marginalia] ⟦illegible⟧ 19.7767 ft. [Marginalia] ⟦illegible⟧ [Marginalia] ⟦illegible⟧ [Marginalia] ⟦illegible⟧ [Marginalia] ⟦illegible⟧ [Marginalia] ⟦illegible⟧

**Traduction anglaise —**

(cont'd.).. -p.2- Arith. & Trig. 17/5/67 4th Year Secondary. ⟦line⟧ 5. ABC is a triangle with AB = AC = 100 ft. and the angle BAC is 70°. At A is a vertical pole AO = 80 ft. high. Calculate: (i) the length of BC, (ii) the angle of elevation of the top of the pole from B, (iii) the angle of elevation of the top of the pole from the mid-point of BC. 6. A, B and C are three points on a coastline which runs from north to south; B is south of A and C is 1000 yards south of B. A boat is moving in a straight line towards C and when it is at a point P which is 2000 yd. from A on a bearing of (N.60.E.) its bearing from B is ( N.38E.). Calculate: (a) the distance from P to the nearest point X on the coastline. (b) the distance AB. (c) the bearing of P from C. ⟦line⟧ ⟦illegible⟧ ⟦illegible⟧ 19.7767 ft. ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_269d3042738a52ccbe0d925cfa2d432b

B01 · header / latin Make-up Examination Fourth Year Secondary

B02 · marginalia / latin Trigonometry 1/2/55

B03 · paragraph / latin 1. Without using tables, find the height and area of an equilateral triangle whose sides are each 7.2 inches long.

B04 · paragraph / latin 2. (i) Find ϕ , if sec 17° + cot 41° = cot ϕ (ii) Evaluate in the shortest possible way: csc 49° / csc 41° , sin 60° sec 60° , csc 41° 31' / sec 48° 29' , csc 75° cos 15°

B05 · paragraph / latin 3. A ladder, 36 ft. long, makes an angle of 50° with the <del>⟦illegible⟧</del> ground and <del>⟦illegible⟧</del> leans against a vertical wall. If the top of the ladder slips down 2 ft., how far will the foot of the ladder move?

B06 · other / unknown ⟦line⟧

**Traduction anglaise —**

Make-up Examination Fourth Year Secondary Trigonometry 1/2/55 1. Without using tables, find the height and area of an equilateral triangle whose sides are each 7.2 inches long. 2. (i) Find ϕ , if sec 17° + cot 41° = cot ϕ (ii) Evaluate in the shortest possible way: csc 49° / csc 41° , sin 60° sec 60° , csc 41° 31' / sec 48° 29' , csc 75° cos 15° A ladder, 36 ft. long, makes an angle of 50° with the <del>⟦illegible⟧</del> ground and <del>⟦illegible⟧</del> leans against a vertical wall. If the top of the ladder slips down 2 ft., how far will the foot of the ladder move? ⟦line⟧

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### csp_26b9b4a37a915f55b3054e5b6b299aad

⟦Solutions to the Final Exam 1953-1954 Questions⟧ Algebra Class : 4th Secondary

1. (i) { a^(p-q) / ⟦illegible⟧ }^(p+q) . { a^(q-r) / ⟦illegible⟧ }^(q+r) . { a^(r-p) / ⟦illegible⟧ }^(r+p) = { a^(p^2-q^2) / a^(p^2-q^2) } . { a^(q^2-r^2) / a^(q^2-r^2) } . { a^(r^2-p^2) / a^(r^2-p^2) } = a^(p^2-q^2+q^2-r^2+r^2-p^2) = a^0 = 1 Ans.

(ii) (6√x - 7) / (√x - 1) - 5 = (√x - 26) / (7√x - 21) ∴ (6√x - 7)(7√x - 21) - 5(√x - 1)(7√x - 21) = (√x - 26)(√x - 1) ∴ 42x - 175√x + 147 - (35x - 140√x + 105) = 7x - 33√x + 26 2√x = 16 ∴ √x = 8 ∴ x = 64 Ans.

2. (i) ∛1.002 ⁷√(0.005001)² / ⁵√(0.03)³ x (4.003)⁵ = x log 1.002 = 0.0003 log 0.005001 = 3.6991 log 0.03 = 2.4771 log 4.003 = 0.6024 1/3 log 1.002 = 0.0001 2/7 log 0.005001 = 1.3426 log Num. = 1.3427 log Den. = 1.8584 log x = 1.4843 ∴ x = 0.3051 Ans.

(ii) 2^x = 8^(y+1) ; 9^y = 3^(x-9) ∴ 2^x = (2^3)^(y+1) ; (3^2)^y = 3^(x-9) ∴ 2^x = 2^(3y+3) and 3^(2y) = 3^(x-9) ∴ x = 3y+3 and 2y = x-9 or x - 3y = 3 } y = 6 } Ans. x - 2y = 9 } x = 21 }

3. (i) l_3 = 15 | l_100 = ? | 15 = a + 2d } ∴ 75 = 15d | ∴ a = 15 - 10 = 5 l_18 = 90 | S_100 = ? | 90 = a + 17d } ∴ d = 5 | ∴ a = 5 , d = 5 l_100 = a + 99d = 5 + 99x5 = 500 Ans. 1 S_100 = n/2(a+l) = 100/2(5+500) = 505 x 50 = 25250 Ans. 2

(ii) (1 + x + x^2 + ... + x^(n-1))(1 - x + x^2 - x^3 + ... + x^(n-1)) = 1 + x^2 + x^4 + ... + x^(2n-2) S_1 = 1(x^n - 1) / (x - 1) , S_2 = 1[1 - (-x)^n] / (1 + x) ; S_3 = 1[(x^2)^n - 1] / (x^2 - 1) = (x^(2n) - 1) / (x^2 - 1) S_1 x S_2 = (x^n - 1) / (x - 1) x (x^n + 1) / (x + 1) = (x^(2n) - 1) / (x^2 - 1) = S_3 Q.E.D.

4. x^2 - 3x + 2 = y ; y = x^2 - 3x + 1 ; y_2 = 1 } x = 2 } Ans. 2 x = 1 } If we subtract 1 from y_1, we get y_2 ∴ the solution of x^2 - 3x + 1 amounts to solving y_1 and y_2 simultaneously where we get: x = 2, 1

⟦Graph showing parabola y = x^2 - 3x + 2 and line y = 1 intersecting at x=1 and x=2⟧

**Traduction anglaise —**

⟦Solutions to the Final Exam 1953-1954 Questions⟧ Algebra Class : 4th Secondary 1. (i) { a^(p-q) / ⟦illegible⟧ }^(p+q) . { a^(q-r) / ⟦illegible⟧ }^(q+r) . { a^(r-p) / ⟦illegible⟧ }^(r+p) = { a^(p^2-q^2) / a^(p^2-q^2) } . { a^(q^2-r^2) / a^(q^2-r^2) } . { a^(r^2-p^2) / a^(r^2-p^2) } = a^(p^2-q^2+q^2-r^2+r^2-p^2) = a^0 = 1 Ans. (ii) (6√x - 7) / (√x - 1) - 5 = (√x - 26) / (7√x - 21) ∴ (6√x - 7)(7√x - 21) - 5(√x - 1)(7√x - 21) = (√x - 26)(√x - 1) ∴ 42x - 175√x + 147 - (35x - 140√x + 105) = 7x - 33√x + 26 2√x = 16 ∴ √x = 8 ∴ x = 64 Ans. 2. (i) ∛1.002 ⁷√(0.005001)² / ⁵√(0.03)³ x (4.003)⁵ = x log 1.002 = 0.0003 log 0.005001 = 3.6991 log 0.03 = 2.4771 log 4.003 = 0.6024 1/3 log 1.002 = 0.0001 2/7 log 0.005001 = 1.3426 log Num. = 1.3427 log Den. = 1.8584 log x = 1.4843 ∴ x = 0.3051 Ans. (ii) 2^x = 8^(y+1) ; 9^y = 3^(x-9) ∴ 2^x = (2^3)^(y+1) ; (3^2)^y = 3^(x-9) ∴ 2^x = 2^(3y+3) and 3^(2y) = 3^(x-9) ∴ x = 3y+3 and 2y = x-9 or x - 3y = 3 } y = 6 } Ans. x - 2y = 9 } x = 21 } 3. (i) l_3 = 15 | l_100 = ? | 15 = a + 2d } ∴ 75 = 15d | ∴ a = 15 - 10 = 5 l_18 = 90 | S_100 = ? | 90 = a + 17d } ∴ d = 5 | ∴ a = 5 , d = 5 l_100 = a + 99d = 5 + 99x5 = 500 Ans. 1 S_100 = n/2(a+l) = 100/2(5+500) = 505 x 50 = 25250 Ans. 2 (ii) (1 + x + x^2 + ... + x^(n-1))(1 - x + x^2 - x^3 + ... + x^(n-1)) = 1 + x^2 + x^4 + ... + x^(2n-2) S_1 = 1(x^n - 1) / (x - 1) , S_2 = 1[1 - (-x)^n] / (1 + x) ; S_3 = 1[(x^2)^n - 1] / (x^2 - 1) = (x^(2n) - 1) / (x^2 - 1) S_1 x S_2 = (x^n - 1) / (x - 1) x (x^n + 1) / (x + 1) = (x^(2n) - 1) / (x^2 - 1) = S_3 Q.E.D. 4. x^2 - 3x + 2 = y ; y = x^2 - 3x + 1 ; y_2 = 1 } x = 2 } Ans. 2 x = 1 } If we subtract 1 from y_1, we get y_2 ∴ the solution of x^2 - 3x + 1 amounts to solving y_1 and y_2 simultaneously where we get: x = 2, 1 ⟦Graph showing parabola y = x^2 - 3x + 2 and line y = 1 intersecting at x=1 and x=2⟧

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### csp_29625438d4a4504697e10a66e1d88f66

Solution to Algebra paper cont. Final Exam 18/5/1967 page 3

⟦4(i)⟧ is 4, 12, 20, 28, ... In this progression a=4 , d=8 ∴ S = n/2 {2x4 + (n-1)x8} or S = n/2 {8 + 8n - 8} or S = n/2 (8n) ∴ S = 4n² or S = (2n)² Ans. ∴ Whatever the value of n is, the sum S is always a perfect square Q.E.D. (10 marks)

⟦7(ii)⟧ 12, x, y, 4 from the statement of the question: 12, x, y are in A.P. ∴ x-12 = y-x ..... ① also x, y, 4 are in G.P. ∴ y/x = 4/y ..... ② from eq. ① : y = 2x-12 ..... ③ ∴ y² = (2x-12)² = 4(x-6)² from eq. ② : y² = 4x ..... ④ or y² = 4(x²-12x+36) ∴ 4(x²-12x+36) = 4x ∴ x²-13x+36 = 0 ∴ (x-4)(x-9) = 0 ∴ x=4 or x=9 when x=4 , y = 2(x-6) = 2(4-6) = -4 when x=9 , y = 2(x-6) = 2(9-6) = 6 ∴ x=4 } Ans. 1 x=9 } Ans. 2 y=-4 y=6 (10 marks) ∴ the terms are either 12, 4, -4, 4 or 12, 9, 6, 4

**Traduction anglaise —**

Solution to Algebra paper cont. Final Exam 18/5/1967 page 3 ⟦4(i)⟧ is 4, 12, 20, 28, ... In this progression a=4 , d=8 ∴ S = n/2 {2x4 + (n-1)x8} or S = n/2 {8 + 8n - 8} or S = n/2 (8n) ∴ S = 4n² or S = (2n)² Ans. ∴ Whatever the value of n is, the sum S is always a perfect square Q.E.D. (10 marks) ⟦7(ii)⟧ 12, x, y, 4 from the statement of the question: 12, x, y are in A.P. ∴ x-12 = y-x ..... ① also x, y, 4 are in G.P. ∴ y/x = 4/y ..... ② from eq. ① : y = 2x-12 ..... ③ ∴ y² = (2x-12)² = 4(x-6)² from eq. ② : y² = 4x ..... ④ or y² = 4(x²-12x+36) ∴ 4(x²-12x+36) = 4x ∴ x²-13x+36 = 0 ∴ (x-4)(x-9) = 0 ∴ x=4 or x=9 when x=4 , y = 2(x-6) = 2(4-6) = -4 when x=9 , y = 2(x-6) = 2(9-6) = 6 ∴ x=4 } Ans. 1 x=9 } Ans. 2 y=-4 y=6 (10 marks) ∴ the terms are either 12, 4, -4, 4 or 12, 9, 6, 4

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### csp_2a182378a8b756bbaae1dc737eab9b5c

Solution to Conditional Exam Cert. Sept. 1962 4th Year Algebra Page 2

3. (i) 1st term a = 20 ----- ① 3rd " a + 2d = 21 ----- ② Subtract ① from ②, ∴ 2d = 1 ∴ d = ½ now a = 20 d = ½ l₅₀ = ? S₁₀₀ = ? l₅₀ = a + (n-1)d = 20 + 49 x ½ = 20 + 24½ = 44½ Ans. 1 S₁₀₀ = n/2 {2a + (n-1)d} = 100/2 {2 x 20 + (100-1) x ½} = 50 (40 + 49½) = 50 x 179/2 = 4475 Ans. 2

(ii) r = ½ a = ? S₄ = 18 ¾ l₆ = ? Sₙ = a(1 - rⁿ) / (1 - r) ∴ S₄ = a(1 - (½)⁴) / (1 - ½) ∴ 18 ¾ = a(1 - 1/16) / ½ or 75/4 = 2a (15/16) ∴ a = (75 x 16) / (4 x 2 x 15) or a = 10 Ans. 1 l₆ = ar⁵ = 10 (½)⁵ = 10/32 Ans. 2

4. Let the length = x yds. " " width = y yds. ∴ Cost of turfing = 10 x y shillings = £(xy/2) also cost of fencing = 5 x 2(x+y) shillings = £(x+y/2) ∴ xy/2 + 12 = 12 x (x+y)/2 -------- ① Now 12ft = 12/3 yds = 4 yds ∴ x + 4 = 2y -------- ② from ①, xy + 24 = 12x + 12y from ②, x = 2y - 4 ∴ (2y - 4)y + 24 = 12(2y - 4) + 12y ∴ 2y² - 4y + 24 = 24y - 48 + 12y or 2y² - 40y + 72 = 0 or y² - 20y + 36 = 0 ∴ (y - 2)(y - 18) = 0 ∴ y = 18 or y = 2 (inadmissible) ∴ x = 2y - 4 = 2 x 18 - 4 = 32 yds. length x = 32 yds width y = 18 yds. } Ans.

x yds y yds

**Traduction anglaise —**

Solution to Conditional Exam Cert. Sept. 1962 4th Year Algebra Page 2 3. (i) 1st term a = 20 ⟦line⟧ ① 3rd " a + 2d = 21 ⟦line⟧ ② Subtract ① from ②, ∴ 2d = 1 ∴ d = ½ now a = 20 d = ½ l₅₀ = ? S₁₀₀ = ? l₅₀ = a + (n-1)d = 20 + 49 x ½ = 20 + 24½ = 44½ Ans. 1 S₁₀₀ = n/2 {2a + (n-1)d} = 100/2 {2 x 20 + (100-1) x ½} = 50 (40 + 49½) = 50 x 179/2 = 4475 Ans. 2 (ii) r = ½ a = ? S₄ = 18 ¾ l₆ = ? Sₙ = a(1 - rⁿ) / (1 - r) ∴ S₄ = a(1 - (½)⁴) / (1 - ½) ∴ 18 ¾ = a(1 - 1/16) / ½ or 75/4 = 2a (15/16) ∴ a = (75 x 16) / (4 x 2 x 15) or a = 10 Ans. 1 l₆ = ar⁵ = 10 (½)⁵ = 10/32 Ans. 2 4. Let the length = x yds. " " width = y yds. ∴ Cost of turfing = 10 x y shillings = £(xy/2) also cost of fencing = 5 x 2(x+y) shillings = £(x+y/2) ∴ xy/2 + 12 = 12 x (x+y)/2 ⟦line⟧ ① Now 12ft = 12/3 yds = 4 yds ∴ x + 4 = 2y ⟦line⟧ ② from ①, xy + 24 = 12x + 12y from ②, x = 2y - 4 ∴ (2y - 4)y + 24 = 12(2y - 4) + 12y ∴ 2y² - 4y + 24 = 24y - 48 + 12y or 2y² - 40y + 72 = 0 or y² - 20y + 36 = 0 ∴ (y - 2)(y - 18) = 0 ∴ y = 18 or y = 2 (inadmissible) ∴ x = 2y - 4 = 2 x 18 - 4 = 32 yds. length x = 32 yds width y = 18 yds. } Ans. x yds y yds

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### csp_2a241468df8050f9a1b9268a07deeed6

B01 · table / latin x | y -3 | -5 -2 | 0 -1 | 3 0 | 4 1 | 3 2 | 0 3 | -5

B02 · other / latin y₁ = 4 - x² y₂ = 5/4 x + 1 C(-1.22, 2.5) D(1.22, 2.5) 1.23 -2.47 y = -1.6 x = ± 2.37

B03 · paragraph / latin (i) From the table above, the graph of y₁ = 4 - x² is plotted as shown also the st. line y₂ = 5/4 x + 1 (ii) The two graphs intersect at two (a) points whose abscissae are: x = -2.47 and x = 1.23 Ans. ∴ 4 - x² > 5/4 x + 1 between -2.47 and 1.23 since between these two values of x the curve of 4 - x² lies above the st. line 5/4 x + 1

B04 · paragraph / latin (ii) (b) We draw the line y = 2.5 This line cuts the graph at two points ⟦illegible⟧ C(-1.22, 2.5) and D(1.22, 2.5) ∴ 4 - x² = 2.5 at x = -1.22 and x = 1.22 Ans. (ii) (c) From y = 4 - x², we get x² = 4 - y or x = ± √4 - y . Now let 4 - y = 5.6 ∴ y = -1.6 ∴ when y = -1.6, x = ± √4 - (-1.6) or x = ± √5.6 and from the graph we find that when y = -1.6, x = ± 2.37 Ans.

**Traduction anglaise —**

x | y -3 | -5 -2 | 0 -1 | 3 0 | 4 1 | 3 2 | 0 3 | -5 y₁ = 4 - x² y₂ = 5/4 x + 1 C(-1.22, 2.5) D(1.22, 2.5) 1.23 -2.47 y = -1.6 x = ± 2.37 (i) From the table above, the graph of y₁ = 4 - x² is plotted as shown also the st. line y₂ = 5/4 x + 1 (ii) The two graphs intersect at two (a) points whose abscissae are: x = -2.47 and x = 1.23 Ans. ∴ 4 - x² > 5/4 x + 1 between -2.47 and 1.23 since between these two values of x the curve of 4 - x² lies above the st. line 5/4 x + 1 (ii) (b) We draw the line y = 2.5 This line cuts the graph at two points ⟦illegible⟧ C(-1.22, 2.5) and D(1.22, 2.5) ∴ 4 - x² = 2.5 at x = -1.22 and x = 1.22 Ans. (ii) (c) From y = 4 - x², we get x² = 4 - y or x = ± √4 - y . Now let 4 - y = 5.6 ∴ y = -1.6 ∴ when y = -1.6, x = ± √4 - (-1.6) or x = ± √5.6 and from the graph we find that when y = -1.6, x = ± 2.37 Ans.

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### csp_2ad30f6dd6df5b9b977db125a2473ee2

B01 · marginalia / arabic ليندا مصري (٤)

B02 · header / latin SHAMASH SECONDARY SCHOOL Final Examination, May 1966 ⟦line⟧

B03 · form / latin Subject: Arithmetic & Trigonometry Date: 18.5.1966 Class: 4th Year Secondary Time: 8:00-10:30 a.m.

B04 · paragraph / latin ⟦line⟧ Attempt five questions only including question (4). 1. A person, having bought a certain amount of 2¾% stock at 95, afterwards sold it, and with the proceeds bought 3½% stock. He obtained £900 less stock than before, but his income was unchanged. How much money did he originally invest? (20 marks)

B05 · paragraph / latin 2. A house holder owns his house which has a rateable value of £44 on which the annual rates are charged at 21s 10d in the £1. He also has to pay an annual property tax at the rate of 9s in the £ on an assessment of £44. Calculate, correct to the nearest penny, the average cost per week of the total of these charges, taking a year as 52 weeks. He subsequently sells his house for £3300, which sum he invests at the rate of 2½% per annum free of tax, and moves into a flat which he rents at £126 per annum. He has however to rent a garage for his car at 7s 6d per week. Find how much per annum he saves by the change. (20 marks)

B06 · paragraph / latin 3. (a) A watch was 5 minutes fast at 9 a.m. on Monday, and 10 minutes slow at 12 noon on the following Wednesday. Find when it was exactly right, assuming that it lost time uniformly. Note:(9 a.m. and 12 noon are correct time). (10 marks)

B07 · paragraph / latin (b) Two clocks sound the first stroke of 12 o'clock at the same instant; one clock allows an interval of 20 secs. between each stroke and the next, and the other allows 25 secs. How many strokes of the slower clock remain after the quicker one has finished striking, and what time will elapse between the 12th stroke of the quicker one and the following stroke of the slower one? (10 marks)

B08 · paragraph / latin 4. (a) A solid consists of a hemisphere, radius 8 cm., joined to a cone of the same base-radius and height 6 cm., so that the plane surfaces coincide. Find (i) the volume, (ii) the total area of the surface of the solid. (Give answer to 3 significant figures). (10 marks)

B09 · paragraph / latin (b) A sphere of radius 3 in. is filed down into the greatest possible cube; find the volume of the material removed. (Give answer to 4 significant figures). (10 marks)

B10 · paragraph / latin 5. Find the difference between the perimeters of a regular pentagon and a regular hexagon, each of which has an area of 24 square inches. (20 marks)

B11 · marginalia / latin ⟦illegible⟧

B12 · paragraph / latin 6. In response to an S O S call from a ship at A, another ship at B, 175 miles due east of A, starts toward A at a speed of 12 miles per hour. At the same time a third ship at C, which is 186 miles from B in a direction bearing 315° west of north, also starts for A at a speed of 16 miles per hour. Which ship will reach A first, and how long will it take ? (20 marks) ⟦line⟧

**Traduction anglaise —**

Linda Masri (4) SHAMASH SECONDARY SCHOOL Final Examination, May 1966 ⟦line⟧ Subject: Arithmetic & Trigonometry Date: 18.5.1966 Class: 4th Year Secondary Time: 8:00-10:30 a.m. ⟦line⟧ Attempt five questions only including question (4). 1. A person, having bought a certain amount of 2¾% stock at 95, afterwards sold it, and with the proceeds bought 3½% stock. He obtained £900 less stock than before, but his income was unchanged. How much money did he originally invest? (20 marks) 2. A house holder owns his house which has a rateable value of £44 on which the annual rates are charged at 21s 10d in the £1. He also has to pay an annual property tax at the rate of 9s in the £ on an assessment of £44. Calculate, correct to the nearest penny, the average cost per week of the total of these charges, taking a year as 52 weeks. He subsequently sells his house for £3300, which sum he invests at the rate of 2½% per annum free of tax, and moves into a flat which he rents at £126 per annum. He has however to rent a garage for his car at 7s 6d per week. Find how much per annum he saves by the change. (20 marks) 3. (a) A watch was 5 minutes fast at 9 a.m. on Monday, and 10 minutes slow at 12 noon on the following Wednesday. Find when it was exactly right, assuming that it lost time uniformly. Note:(9 a.m. and 12 noon are correct time). (10 marks) (b) Two clocks sound the first stroke of 12 o'clock at the same instant; one clock allows an interval of 20 secs. between each stroke and the next, and the other allows 25 secs. How many strokes of the slower clock remain after the quicker one has finished striking, and what time will elapse between the 12th stroke of the quicker one and the following stroke of the slower one? (10 marks) 4. (a) A solid consists of a hemisphere, radius 8 cm., joined to a cone of the same base-radius and height 6 cm., so that the plane surfaces coincide. Find (i) the volume, (ii) the total area of the surface of the solid. (Give answer to 3 significant figures). (10 marks) (b) A sphere of radius 3 in. is filed down into the greatest possible cube; find the volume of the material removed. (Give answer to 4 significant figures). (10 marks) 5. Find the difference between the perimeters of a regular pentagon and a regular hexagon, each of which has an area of 24 square inches. (20 marks) ⟦illegible⟧ 6. In response to an S O S call from a ship at A, another ship at B, 175 miles due east of A, starts toward A at a speed of 12 miles per hour. At the same time a third ship at C, which is 186 miles from B in a direction bearing 315° west of north, also starts for A at a speed of 16 miles per hour. Which ship will reach A first, and how long will it take ? (20 marks) ⟦line⟧

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### csp_2ae18e3e3f9f570c972048cf592141bf

Trigonometry Exam. 13/12/54 Fourth Year

I. Give the English equivalent of the following

II. (a) Find the values of each of the following from the tables sin 51° 40' cos 33° 22' tan 88° 39'

(b) Find the angles from the following equations:

III. The distance between two landmarks on opposite banks of a river is 110 yards, and the line joining them makes an angle of 14° 12' with the banks. Find the breadth of the river.

**Traduction anglaise —**

Trigonometry Exam. 13/12/54 Fourth Year I. Give the English equivalent of the following 5. Generation of the angle | 1. Inaccessible distances 6. Homogeneous algebraic expression | 2. Counter-clockwise rotation 7. We move terms from one side to the other side of the equation and combine like terms | 3. Solving the oblique-angled triangle 8. Base of the power, exponent of the power, the nth power | 4. The circular measurement system for angles II. (a) Find the values of each of the following from the tables sin 51° 40' cos 33° 22' tan 88° 39' (b) Find the angles from the following equations: sin θ = 0.0041 | sin α = 1/2 cos β = 0.1234 | cos γ = 1/2 tan φ = 4.6789 | tan ε = 1 III. The distance between two landmarks on opposite banks of a river is 110 yards, and the line joining them makes an angle of 14° 12' with the banks. Find the breadth of the river.

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### csp_2b90dfd9770d5f82aefd8dce56262e4e

An alternative method x² + xy + 2y² = 8 ...... ① 2x² - 2xy - 3y² = 1 ...... ② multiply Eq (2) by 8 : Subtract { 16x² - 16xy - 24y² = 8 ...... ③ { x² + xy + 2y² = 8 ...... ① 15x² - 17xy - 26y² = 0 ...... ④ ∴ (x - 2y) (15x + 13y) = 0 ∴ y = 1/2 x or y = - 15/13 x When y = x/2 from Eq. ① : x² + x(x/2) + 2(x/2)² = 8 or x² + x²/2 + x²/2 = 8 or 2x² = 8 ∴ x² = 4 ∴ x = ± 2 ∴ y = x/2 = ± 2/2 = ± 1 ∴ x = 2 } Ans. 1 x = -2 } Ans. 2 y = 1 } y = -1 } When y = - 15/13 x from Eq. ① : x² + x(- 15/13 x) + 2(- 15/13 x)² = 8 or x² - 15/13 x² + 450/169 x² = 8 ∴ 169x² - 15x13x² + 450x² = 8 x 169 ∴ 169x² - 195x² + 450x² = 1352 ∴ 424x² = 1352 ∴ x² = 1352/424 ∴ x² = 169/53 ∴ x = ± 13/√53 When x = 13/√53 ∴ y = - 15/13 x or y = - 15/13 (13/√53) = - 15/√53 and when x = - 13/√53 ∴ y = - 15/13 x or y = - 15/13 (- 13/√53) = 15/√53 ∴ x = 13/√53 } Ans. 3 x = - 13/√53 } Ans. 4 y = - 15/√53 } y = 15/√53 }

**Traduction anglaise —**

3 An alternative method x² + xy + 2y² = 8 ⟦line⟧ ① 2x² - 2xy - 3y² = 1 ⟦line⟧ ② multiply Eq (2) by 8 : Subtract { 16x² - 16xy - 24y² = 8 ⟦line⟧ ③ { x² + xy + 2y² = 8 ⟦line⟧ ① 15x² - 17xy - 26y² = 0 ⟦line⟧ ④ ∴ (x - 2y) (15x + 13y) = 0 ∴ y = 1/2 x or y = - 15/13 x When y = x/2 from Eq. ① : x² + x(x/2) + 2(x/2)² = 8 or x² + x²/2 + x²/2 = 8 or 2x² = 8 ∴ x² = 4 ∴ x = ± 2 ∴ y = x/2 = ± 2/2 = ± 1 ∴ x = 2 } Ans. 1 x = -2 } Ans. 2 y = 1 } y = -1 } When y = - 15/13 x from Eq. ① : x² + x(- 15/13 x) + 2(- 15/13 x)² = 8 or x² - 15/13 x² + 450/169 x² = 8 ∴ 169x² - 15x13x² + 450x² = 8 x 169 ∴ 169x² - 195x² + 450x² = 1352 ∴ 424x² = 1352 ∴ x² = 1352/424 ∴ x² = 169/53 ∴ x = ± 13/√53 When x = 13/√53 ∴ y = - 15/13 x or y = - 15/13 (13/√53) = - 15/√53 and when x = - 13/√53 ∴ y = - 15/13 x or y = - 15/13 (- 13/√53) = 15/√53 ∴ x = 13/√53 } Ans. 3 x = - 13/√53 } Ans. 4 y = - 15/√53 } y = 15/√53 }

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### csp_2beea8814f445b528dfcf664efdd2441

B01 · header / latin ⟦illegible⟧ solution to the Conditional Exam in Arithmetic & Trig. 4th year, 14/9/1965

B02 · paragraph / latin Income ( £ 253 15 s / 101 ½ ) 3 ½ = £ 8.75 or £ 8 15 s

B03 · paragraph / latin Amount realised ( £ 253 15 s ) 99 / 101.5 = £ 247 10 s

B04 · paragraph / latin 2 <del>£</del> ( 3 / 2 )^3 = ( 3 / 4 )^3 + 1^3 + x^3 x = radius x^3 = 27 / 8 - 27 / 64 - 1 = 216 / 64 - 27 / 64 - 64 / 64 = 125 / 64 x = 5 / 4 radius Diam; = 2 x = 10 / 4 = 2 ½"

B05 · other / latin ⟦line⟧

**Traduction anglaise —**

⟦illegible⟧ solution to the Conditional Exam in Arithmetic & Trig. 4th year, 14/9/1965 Income ( £ 253 15 s / 101 ½ ) 3 ½ = £ 8.75 or £ 8 15 s Amount realised ( £ 253 15 s ) 99 / 101.5 = £ 247 10 s 2 <del>£</del> ( 3 / 2 )^3 = ( 3 / 4 )^3 + 1^3 + x^3 x = radius x^3 = 27 / 8 - 27 / 64 - 1 = 216 / 64 - 27 / 64 - 64 / 64 = 125 / 64 x = 5 / 4 radius Diam; = 2 x = 10 / 4 = 2 ½" ⟦line⟧

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### csp_2c03c993a7f453c6bf51024c8511d32a

2. 'A' can walk a mile in 2 minutes less time than B would take. In a walking race, 'B' has a start of ¼ mile and A overtakes B in 10 minutes. Assuming both men walk at a uniform rate, find their rates of walking in miles, per hour. (20)

⟦pencil scribbles⟧

**Traduction anglaise —**

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### csp_2cbe19855b4756098ebce6df68250095

⟦Conditional exam July 1957⟧ 4th year secondary : 2

5. (i) Taking 1 inch = 1 unit on the x-axis and 1 inch = 2 units on the y-axis draw the graphs of y = 4 - x² and 4y = 5x + 4 for values of x from -3 to +3. (8 marks) (ii) From your graph find: (a) the range of values of x for which 4 - x² is greater than 5/4 x + 1, (4 marks) (b) the values of x for which 4 - x² = 2.5, (4 marks) (c) the square root of 3.6. (4 marks)

____________________________________

**Traduction anglaise —**

⟦Conditional exam July 1957⟧ 4th year secondary : 2 5. (i) Taking 1 inch = 1 unit on the x-axis and 1 inch = 2 units on the y-axis draw the graphs of y = 4 - x² and 4y = 5x + 4 for values of x from -3 to +3. (8 marks) (ii) From your graph find: (a) the range of values of x for which 4 - x² is greater than 5/4 x + 1, (4 marks) (b) the values of x for which 4 - x² = 2.5, (4 marks) (c) the square root of 3.6. (4 marks) ⟦line⟧

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### csp_2cda1dd236285e26b73ee579381cb508

Shamash Secondary School ------------------------

Subject: Arithmetic & Trigonometry Date: 14/9/1965 Class: 4th Year Secondary Time: 8.00-10.30

------- Attempt all questions.

1. Find the income produced by investing £253 15s. in 3½% stock at 101½ and the amount realised by subsequently selling out at 99.

2. A spherical ball of lead 3 in. in diameter is melted and recast into three spherical balls. The diameters of two of these are 1½ in. ⟦and 2 in.⟧ respectively. What is the diameter of the other?

3. A ladder, 24ft. long, makes an angle of 52° with the ground and leans against a vertical wall. If the top of the ladder slips down 2 ft. how far will the foot of the ladder move?

4. An aeroplane is flying horizontally due E. When it is due N. of an observer its elevation is 42°. Find its elevation when it is N. 55° E. of the observer.

5. A householder has two alternative methods of paying for the electric light and power that he uses during one quarter of a year. Either he pays 7d. per unit for light and 2¾d. per unit for power, or he pays 1½d. per unit for light and for power and also a quarterly charge of £2 5s. 6d. Determine which method is the cheaper, and by how much, for a quarter during which he uses 68 units for light and 196 units for power. A householder paying by the first method used 117 lighting units during a quarter, and his electricity bill for the quarter was £5 3s. 1d. Find the number of units used for power.

------

[Marginalia] ⟦illegible⟧ [Marginalia] 15) 128 [Marginalia] 120 [Marginalia] 8 [Marginalia] 540 [Marginalia] 3 189 [Marginalia] 20 [Marginalia] 134 [Marginalia] ⟦illegible⟧

**Traduction anglaise —**

Shamash Secondary School ⟦line⟧ Subject: Arithmetic & Trigonometry Date: 14/9/1965 Class: 4th Year Secondary Time: 8.00-10.30 ⟦line⟧ Attempt all questions. 1. Find the income produced by investing £253 15s. in 3½% stock at 101½ and the amount realised by subsequently selling out at 99. 2. A spherical ball of lead 3 in. in diameter is melted and recast into three spherical balls. The diameters of two of these are 1½ in. ⟦and 2 in.⟧ respectively. What is the diameter of the other? 3. A ladder, 24ft. long, makes an angle of 52° with the ground and leans against a vertical wall. If the top of the ladder slips down 2 ft. how far will the foot of the ladder move? 4. An aeroplane is flying horizontally due E. When it is due N. of an observer its elevation is 42°. Find its elevation when it is N. 55° E. of the observer. 5. A householder has two alternative methods of paying for the electric light and power that he uses during one quarter of a year. Either he pays 7d. per unit for light and 2¾d. per unit for power, or he pays 1½d. per unit for light and for power and also a quarterly charge of £2 5s. 6d. Determine which method is the cheaper, and by how much, for a quarter during which he uses 68 units for light and 196 units for power. A householder paying by the first method used 117 lighting units during a quarter, and his electricity bill for the quarter was £5 3s. 1d. Find the number of units used for power. ⟦line⟧ ⟦illegible⟧ 15) 128 120 8 540 3 189 20 134 ⟦illegible⟧

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### csp_2d33b95b24eb5fe69059fb3c70cff47f

Solution to Conditional examination in Algebra , Sept., 1967, 4th year ①

1. (i) 2x² - 3x + 4 | 6x³ - 13x² + 18x + k | 3x - 2 | 6x³ - 9x² + 12x ----------------- - 4x² + 6x + k | - 4x² + 6x - 8 ----------------- 8 + k = 0 ∴ k = -8 Ans.

(ii) nth term = 2n+1 / 2n+3 ∴ 1st term = 2+1 / 2+3 = 3/5 , 2nd term = 2x2+1 / 2x2+3 = 5/7 3rd term = 2x3+1 / 2x3+3 = 7/9 (n+1)th term - nth term = 2(n+1)+1 / 2(n+1)+3 - 2n+1 / 2n+3 = 2n+3 / 2n+5 - 2n+1 / 2n+3 = = (2n+3)² - (2n+1)(2n+5) / (2n+5)(2n+3) = 4n² + 12n + 9 - (4n² + 12n + 5) / (2n+5)(2n+3) = 4 / (2n+5)(2n+3) Ans.

2 (i) 3x² - 4x + 5 = a(x - b)² + c ∴ 3x² - 4x + 5 = ax² - 2abx + ab² + c ∴ a = 3 , ∴ -2ab = -4 or 3b = 2 or b = 2/3 ∴ ab² + c = 5 or 3 x 4/9 + c = 5 or c = 5 - 4/3 or c = 11/3 = 3 2/3 ∴ a = 3 , b = 2/3 , c = 11/3 = 3 2/3 Ans. Hence 3x² - 4x + 5 = 3(x - 2/3)² + 11/3 and since (x - 2/3)² is always positive its least value will be zero when x = 2/3 ∴ therefore the least value of 3x² - 4x + 5 is 11/3 when x = 2/3

(ii) ⟦Let the⟧ ⟦illegible⟧ ⟦where x minutes past 3⟧ ∴ x = 15 + ⟦illegible⟧ + x/12 or x = x/12 + 15 ∴ 12x = x + 180 ⟦illegible⟧ 11x = 180 ∴ x = 180/11 = 16 4/11 ∴ 11x = 360 ∴ x = 360/11 = 32 8/11 minutes ∴ x = 32 8/11 minutes past 3 ⟦illegible⟧

⟦Diagram of a clock face showing hands between 3 and 4⟧

**Traduction anglaise —**

Solution to Conditional examination in Algebra , Sept., 1967, 4th year ① 1. (i) 2x² - 3x + 4 | 6x³ - 13x² + 18x + k | 3x - 2 | 6x³ - 9x² + 12x ⟦line⟧ - 4x² + 6x + k | - 4x² + 6x - 8 ⟦line⟧ 8 + k = 0 ∴ k = -8 Ans. (ii) nth term = 2n+1 / 2n+3 ∴ 1st term = 2+1 / 2+3 = 3/5 , 2nd term = 2x2+1 / 2x2+3 = 5/7 3rd term = 2x3+1 / 2x3+3 = 7/9 (n+1)th term - nth term = 2(n+1)+1 / 2(n+1)+3 - 2n+1 / 2n+3 = 2n+3 / 2n+5 - 2n+1 / 2n+3 = = (2n+3)² - (2n+1)(2n+5) / (2n+5)(2n+3) = 4n² + 12n + 9 - (4n² + 12n + 5) / (2n+5)(2n+3) = 4 / (2n+5)(2n+3) Ans. 2 (i) 3x² - 4x + 5 = a(x - b)² + c ∴ 3x² - 4x + 5 = ax² - 2abx + ab² + c ∴ a = 3 , ∴ -2ab = -4 or 3b = 2 or b = 2/3 ∴ ab² + c = 5 or 3 x 4/9 + c = 5 or c = 5 - 4/3 or c = 11/3 = 3 2/3 ∴ a = 3 , b = 2/3 , c = 11/3 = 3 2/3 Ans. Hence 3x² - 4x + 5 = 3(x - 2/3)² + 11/3 and since (x - 2/3)² is always positive its least value will be zero when x = 2/3 ∴ therefore the least value of 3x² - 4x + 5 is 11/3 when x = 2/3 (ii) ⟦Let the⟧ ⟦illegible⟧ ⟦where x minutes past 3⟧ ∴ x = 15 + ⟦illegible⟧ + x/12 or x = x/12 + 15 ∴ 12x = x + 180 ⟦illegible⟧ 11x = 180 ∴ x = 180/11 = 16 4/11 ∴ 11x = 360 ∴ x = 360/11 = 32 8/11 minutes ∴ x = 32 8/11 minutes past 3 ⟦illegible⟧ ⟦Diagram of a clock face showing hands between 3 and 4⟧

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### csp_2d76ea1b8d1f563ab81624f114cc41b1

Page 3

Solution to Conditional Exam cont. Sept. 1962. 4th Year Algebra

5. f(x) = x² - 2x

| x | f(x) | | -1 | 3 | | 0 | 0 | | 1 | -1 | | 2 | 0 | | 3 | 3 | | 4 | 8 |

y = x + 1 (3.3, 4.3) (2.4, 1) y = 1 (0.4, 1) (-0.3, 0.7) ① ②

∴ the solution of the equation x² - 2x = 1 gives x₁ = -0.4 and x₂ = 2.4 } Ans.

also solution of the equation x² - 2x = x + 1 gives:

x₁ = (3 + √13) / 2 = 3.303 x₂ = (3 - √13) / 2 = -0.303 } Ans.

**Traduction anglaise —**

Page 3 Solution to Conditional Exam cont. Sept. 1962. 4th Year Algebra 5. f(x) = x² - 2x x | f(x) -1 | 3 0 | 0 1 | -1 2 | 0 3 | 3 4 | 8 y = x + 1 (3.3, 4.3) (2.4, 1) y = 1 (0.4, 1) (-0.3, 0.7) ① ② ∴ the solution of the equation x² - 2x = 1 gives x₁ = -0.4 and x₂ = 2.4 } Ans. also solution of the equation x² - 2x = x + 1 gives: x₁ = (3 + √13) / 2 = 3.303 x₂ = (3 - √13) / 2 = -0.303 } Ans.

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### csp_2e0ea28f847e585c92ec81e2add054b9

Solution of Algebra ⟦illegible⟧ 14/5/1969 page 2

2 (i) solve the equation: (log x)² = log x⁷ - 10 ∴ (log x)² - 7 log x + 10 = 0 ∴ (log x - 2) (log x - 5) = 0 ∴ log x - 2 = 0 or log x = 2 ∴ x = 10² = 100 Ans. 1 or log x - 5 = 0 or log x = 5 ∴ x = 10⁵ = 100000 Ans. 2

(ii) 3x² + xy - 2y² + 7 = 0 --- ① x² - xy + y² - 7 = 0 --- ② adding, we get: 4x² - y² = 0 ∴ (2x + y) (2x - y) = 0 ∴ either y = 2x --- ③ from ② + ③ : x² - x(2x) + (2x)² - 7 = 0 or y = -2x --- ④ or x² - 2x² + 4x² = 7 or 3x² = 7 ∴ x = ± √7/3 ∴ y = 2x ∴ y = ± 2√7/3 From ② + ④ : x² - x(-2x) + (-2x)² - 7 = 0 or x² + 2x² + 4x² = 7 or 7x² = 7 ∴ x² = 1 ∴ x = ± 1 ∴ y = -2x ∴ y = -2 (± 1) ∴ y = ∓ 2

∴ x = √7/3 } Ans. 1 x = -√7/3 } Ans. 2 y = 2√7/3 y = -2√7/3 x = 1 } Ans. 3 x = -1 } Ans. 4 y = -2 y = 2

3 (i) (5) (4³ˣ⁻¹) (√8)¹⁻ˣ = (√2)ˣ (√50) 5 x 2²⁽³ˣ⁻¹⁾ x 2³⁄₂⁽¹⁻ˣ⁾ = 2ˣ⁄₂ x 5 x 2¹⁄₂ ∴ 2⁶ˣ⁻² x 2³⁄₂⁻³ˣ⁄₂ = 2ˣ⁺¹⁄₂ ∴ 2⁶ˣ⁻²⁺³⁄₂⁻³ˣ⁄₂ = 2ˣ⁺¹⁄₂ ∴ 6x - 2 + 3/2 - 3x/2 = x+1/2 ∴ 9x/2 - 1/2 = x/2 + 1/2 ∴ 4x = 1 ∴ x = 1/4 Ans.

**Traduction anglaise —**

Solution of Algebra ⟦illegible⟧ 14/5/1969 page 2 2 (i) solve the equation: (log x)² = log x⁷ - 10 ∴ (log x)² - 7 log x + 10 = 0 ∴ (log x - 2) (log x - 5) = 0 ∴ log x - 2 = 0 or log x = 2 ∴ x = 10² = 100 Ans. 1 or log x - 5 = 0 or log x = 5 ∴ x = 10⁵ = 100000 Ans. 2 (ii) 3x² + xy - 2y² + 7 = 0 ⟦line⟧ ① x² - xy + y² - 7 = 0 ⟦line⟧ ② adding, we get: 4x² - y² = 0 ∴ (2x + y) (2x - y) = 0 ∴ either y = 2x ⟦line⟧ ③ from ② + ③ : x² - x(2x) + (2x)² - 7 = 0 or y = -2x ⟦line⟧ ④ or x² - 2x² + 4x² = 7 or 3x² = 7 ∴ x = ± √7/3 ∴ y = 2x ∴ y = ± 2√7/3 From ② + ④ : x² - x(-2x) + (-2x)² - 7 = 0 or x² + 2x² + 4x² = 7 or 7x² = 7 ∴ x² = 1 ∴ x = ± 1 ∴ y = -2x ∴ y = -2 (± 1) ∴ y = ∓ 2 ∴ x = √7/3 } Ans. 1 x = -√7/3 } Ans. 2 y = 2√7/3 y = -2√7/3 x = 1 } Ans. 3 x = -1 } Ans. 4 y = -2 y = 2 3 (i) (5) (4³ˣ⁻¹) (√8)¹⁻ˣ = (√2)ˣ (√50) 5 x 2²⁽³ˣ⁻¹⁾ x 2³⁄₂⁽¹⁻ˣ⁾ = 2ˣ⁄₂ x 5 x 2¹⁄₂ ∴ 2⁶ˣ⁻² x 2³⁄₂⁻³ˣ⁄₂ = 2ˣ⁺¹⁄₂ ∴ 2⁶ˣ⁻²⁺³⁄₂⁻³ˣ⁄₂ = 2ˣ⁺¹⁄₂ ∴ 6x - 2 + 3/2 - 3x/2 = x+1/2 ∴ 9x/2 - 1/2 = x/2 + 1/2 ∴ 4x = 1 ∴ x = 1/4 Ans.

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### csp_2ed92f73371956d0ac1975282c0b7796

⟦...⟧ the time at this instant is = (1 + 4/3 + 2/7) hrs after 7:00 a.m. = (1 + 4/3 + 2/7) = (21 + 28 + 6) / 21 = 55 / 21 = 2 13/21 hrs after 7:00 a.m. the time at this instant is 9:37 1/7 a.m. = 9:37 a.m. to the nearest minute (10 marks)

(ii) (log x)² - 4 log x + 4 = 0 ∴ (log x - 2)² = 0 ∴ log x = 2 ∴ x = 100 Ans. (10 marks)

4 (i) Between 92 and 4815 the first number which is divisible by 13 is (92/13 = 7 + 1/13) 8 x 13 = 104 (since 92/13 = 7 1/13). The last number divisible by 13 is 4810 (since 4815/13 = 370 5/13). Hence the numbers required are : 104, 117, 130, ... , 4810 l = a + (n-1)d ∴ 4810 = 104 + (n-1) x 13 ∴ 13n = 4810 - 104 + 13 or 13n = 4719 ∴ n = 4719 / 13 = 363 = no. of terms ∴ S = n/2 (a + l) = 363/2 (104 + 4810) = 363/2 x 4914 = 363 x 2457 = 891891 Ans. (10 marks)

(ii) first x sixth = 99 x fourth ∴ a . ar⁵ = 99 ar³ ∴ ar² = 99 (1) a + ar³ = -286 or a(1 + r³) = -286 (2) Dividing : ar² / a(1 + r³) = 99 / -286 ∴ r² / 1 + r³ = 9 / -26 ∴ 9 + 9r³ = -26r² or 9r³ + 26r² + 9 = 0 by trial + error r = -3 satisfies the equation ∴ (r + 3) is a factor of L.H.S. ∴ (r + 3)(9r² - r + 3) = 0 the second factor is no ⟦...⟧ r = -3 from eq (1) ar² = 99 ∴ 9a = 99 ∴ a = 11 Ans.

P.T.O

**Traduction anglaise —**

2 ⟦...⟧ the time at this instant is = (1 + 4/3 + 2/7) hrs after 7:00 a.m. = (1 + 4/3 + 2/7) = (21 + 28 + 6) / 21 = 55 / 21 = 2 13/21 hrs after 7:00 a.m. the time at this instant is 9:37 1/7 a.m. = 9:37 a.m. to the nearest minute (10 marks) 3 (i) log sin 17° 14' = 1.4717 | 2 log sin 17° 14' = 2.9434 | 3 log 1.003 = 0.0036 log cos 9° 11' = 1.9944 | 3 log cos 9° 11' = 1.9832 | log 15.04 = 1.1772 log 1.003 = 0.0012 | log Num = 2.9266 | log Den = 1.1808 log 15.04 = 1.1772 | log Den = 1.1808 log x = 1.7458 log x = 1.129625 = 1.1296 Correct to 4 dec. p. x = 0.1348 Ans. (10 marks) (ii) (log x)² - 4 log x + 4 = 0 ∴ (log x - 2)² = 0 ∴ log x = 2 ∴ x = 100 Ans. (10 marks) 4 (i) Between 92 and 4815 the first number which is divisible by 13 is (92/13 = 7 + 1/13) 8 x 13 = 104 (since 92/13 = 7 1/13). The last number divisible by 13 is 4810 (since 4815/13 = 370 5/13). Hence the numbers required are : 104, 117, 130, ... , 4810 l = a + (n-1)d ∴ 4810 = 104 + (n-1) x 13 ∴ 13n = 4810 - 104 + 13 or 13n = 4719 ∴ n = 4719 / 13 = 363 = no. of terms ∴ S = n/2 (a + l) = 363/2 (104 + 4810) = 363/2 x 4914 = 363 x 2457 = 891891 Ans. (10 marks) (ii) first x sixth = 99 x fourth ∴ a . ar⁵ = 99 ar³ ∴ ar² = 99 (1) a + ar³ = -286 or a(1 + r³) = -286 (2) Dividing : ar² / a(1 + r³) = 99 / -286 ∴ r² / 1 + r³ = 9 / -26 ∴ 9 + 9r³ = -26r² or 9r³ + 26r² + 9 = 0 by trial + error r = -3 satisfies the equation ∴ (r + 3) is a factor of L.H.S. ∴ (r + 3)(9r² - r + 3) = 0 the second factor is no ⟦...⟧ r = -3 from eq (1) ar² = 99 ∴ 9a = 99 ∴ a = 11 Ans. P.T.O

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### csp_2f0ca9f2b46e521bba143af13cf18231

Solutions to 2nd Quarter Exam in Algebra Cont. 27/12/1966 page 1.

3 (a). _________________5_________________ = _______________5_______________ = _______________5_______________ 6 - _________5_________ 6 - _________5_________ 6 - ____5(6-x)____ 6 - ____5____ 36-6x-5 31-6x 6-x 6-x ∴ _______________5_______________ = ____5 (31-6x)____ = ____155 - 30x____ 186 - 36x - 30 + 5x 156 - 31x 156 - 31x ___________________ 31 - 6x ∴ ____155 - 30x____ = x & 155 - 30x = 156x - 31x² 156 - 31x ∴ 31x² - 186x + 155 = 0 ∴ x² - 6x + 5 = 0 ∴ (x-1)(x-5) = 0 ∴ x = 1 & x = 5 Ans.

[Marginalia] (8 marks)

(b) 3x³ + x² + 4 = 8x ∴ 3x³ + x² - 8x + 4 = 0 when x = 1, then the LHS = 3 + 1 - 8 + 4 = 0 ∴ (x-1) is a factor 3x³ + x² - 8x + 4 = <del>⟦illegible⟧</del> 3x²(x-1) + 4x(x-1) - 4(x-1) 3x³ + x² - 8x + 4 = (x-1)(3x² + 4x - 4) = (x-1)(3x-2)(x+2) = 0 ∴ x = 1 , x = -2 , x = 2/3 Ans.

[Marginalia] (10 marks)

(c) x²y² + 192 = 28xy ----- ① x + y = 8 ----- ② x²y² - 28xy + 192 = 0 ∴ (xy - 16)(xy - 12) = 0 ∴ xy = 16 or xy = 12. squaring eq. ②, x² + 2xy + y² = 64 --- ③ -4xy = -48 or x² + 2xy + y² = 64 __________________ -4xy = -64 ∴ x² - 2xy + y² = 16 or x² - 2xy + y² = 0 (x-y)² = 16 (x-y)² = 0 x - y = ±4 x - y = 0 Now x + y = 8 x + y = 8 x + y = 8 x - y = 4 x - y = -4 x - y = 0 _________ __________ _________ ∴ 2x = 12 2x = 4 2x = 8 x = 6 } Ans.1 x = 2 } Ans.2 x = 4 } Ans.3 y = 2 } y = 6 } y = 4 }

[Marginalia] (10 marks)

**Traduction anglaise —**

Solutions to 2nd Quarter Exam in Algebra Cont. 27/12/1966 page 1. 3 (a). ⟦line⟧5⟦line⟧ = ⟦line⟧5⟦line⟧ = ⟦line⟧5⟦line⟧ 6 - ⟦line⟧5⟦line⟧ 6 - ⟦line⟧5⟦line⟧ 6 - ⟦line⟧5(6-x)⟦line⟧ 6 - ⟦line⟧5⟦line⟧ 36-6x-5 31-6x 6-x 6-x ∴ ⟦line⟧5⟦line⟧ = ⟦line⟧5 (31-6x)⟦line⟧ = ⟦line⟧155 - 30x⟦line⟧ 186 - 36x - 30 + 5x 156 - 31x 156 - 31x ⟦line⟧ 31 - 6x ∴ ⟦line⟧155 - 30x⟦line⟧ = x & 155 - 30x = 156x - 31x² 156 - 31x ∴ 31x² - 186x + 155 = 0 ∴ x² - 6x + 5 = 0 ∴ (x-1)(x-5) = 0 ∴ x = 1 & x = 5 Ans. (8 marks) (b) 3x³ + x² + 4 = 8x ∴ 3x³ + x² - 8x + 4 = 0 when x = 1, then the LHS = 3 + 1 - 8 + 4 = 0 ∴ (x-1) is a factor 3x³ + x² - 8x + 4 = <del>⟦illegible⟧</del> 3x²(x-1) + 4x(x-1) - 4(x-1) 3x³ + x² - 8x + 4 = (x-1)(3x² + 4x - 4) = (x-1)(3x-2)(x+2) = 0 ∴ x = 1 , x = -2 , x = 2/3 Ans. (10 marks) (c) x²y² + 192 = 28xy ----- ① x + y = 8 ----- ② x²y² - 28xy + 192 = 0 ∴ (xy - 16)(xy - 12) = 0 ∴ xy = 16 or xy = 12. squaring eq. ②, x² + 2xy + y² = 64 --- ③ -4xy = -48 or x² + 2xy + y² = 64 ⟦line⟧ -4xy = -64 ∴ x² - 2xy + y² = 16 or x² - 2xy + y² = 0 (x-y)² = 16 (x-y)² = 0 x - y = ±4 x - y = 0 Now x + y = 8 x + y = 8 x + y = 8 x - y = 4 x - y = -4 x - y = 0 ⟦line⟧ ⟦line⟧ ⟦line⟧ ∴ 2x = 12 2x = 4 2x = 8 x = 6 } Ans.1 x = 2 } Ans.2 x = 4 } Ans.3 y = 2 } y = 6 } y = 4 } (10 marks)

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### csp_2fcca1fa7dac53fca2afa4e4628e4082

Shamash Secondary School 1st. Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class : 4th year Secondary. Date: 8/12/1957 ---------- All questions are to be attempted.

(1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m, by 25.00 m.

(2) (a) Decimalise to 3 places the following:

| £ 2 | 12s | 2¾d | | £ 8 | 10s | 8½d | | £ 9 | 5s | 10¾d |

(b) Convert into shillings and pence to nearest ¼d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint.

(3) A watch which gains 5 sec.in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ?

(4) The average age of m boys is b years and of n girls is c years. Find the average age of all together.

(5) At 9 a.m. a ship which is sailing in a direction E.37°S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20°W., find the distance of the fort from the ship at the first observation.

(6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42°7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

**Traduction anglaise —**

Shamash Secondary School 1st. Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class : 4th year Secondary. Date: 8/12/1957 ⟦line⟧ All questions are to be attempted. (1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m, by 25.00 m. (2) (a) Decimalise to 3 places the following: £ 2 | 12s | 2¾d £ 8 | 10s | 8½d £ 9 | 5s | 10¾d (b) Convert into shillings and pence to nearest ¼d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint. (3) A watch which gains 5 sec.in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ? (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together. (5) At 9 a.m. a ship which is sailing in a direction E.37°S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20°W., find the distance of the fort from the ship at the first observation. (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42°7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

---

### csp_32bb201f2058550194e61f04a48200e7

الرقم : Shamash Secondary School Monthly Quiz. الاسم :

Subject:: Algebra Date:: 12/11/1967 Class:: 4th Year Secondary Time:: 1:15- 11:45 a.m.

I. Give the English equivalent to the following and fill in the blanks in this sheet, handing it back with your answer book:

١- في المقدار ٣ س ا معامل س هو 1. ٢- ننقل الحدود من طرف الى الطرف الثاني للمعادلة ونجمع الحدود المتشابهة 2. ٣- نتخلص من الكسور 3. ٤- نوحد مقامات الكسور بابسط مقام مشترك 4. ٥- ان قيمة المقدار ٤٥٠٣٨٤ لاقرب اربعة ارقام معنوية هي 5. ٦- ان حدي الكسر هما بسطه ومقامه 6. ٧- نقيس طول مستقيم فنجد انه يساوي ٥,١١ سم بينما طوله المضبوط هو ٦,٠ سم . وفي هذه الحالة نقول ان الخطأ المطلق هو والخطأ النسبي هو والخطأ المئوي هو 7. ٨- ان المعادلة : ٢ س٢ - ٥ ص٢ + ٤ ع٢ = ٧ هي معادلة من الدرجة الثانية ذات ثلاثة مجاهيل . 8. ٩- في كل عملية قسمة يوجد مقسوم ومقسوم عليه وناتج قسمة ، وفي بعض الحالات باق للقسمة 9. ١٠- مقلوب العدد - زاويتان متتامتان - زاويتان متكاملتان - محيط المضلع 10.

( يتبع ص ٢ ) . .

**Traduction anglaise —**

Number: Shamash Secondary School Monthly Quiz. Name: Subject: Algebra Date: 12/11/1967 Class: 4th Year Secondary Time: 1:15- 11:45 a.m. I. Give the English equivalent to the following and fill in the blanks in this sheet, handing it back with your answer book: 1- In the expression 3x, the coefficient of x is 1. 2- We move the terms from one side to the other side of the equation and combine like terms 2. 3- We eliminate fractions 3. 4- We unify the denominators of the fractions with the simplest common denominator 4. 5- The value of the expression 450384 to the nearest four significant figures is 5. 6- The terms of a fraction are its numerator and its denominator 6. 7- We measure the length of a straight line and find it equals 5.11 cm while its exact length is 6.0 cm. In this case, we say that the absolute error is and the relative error is and the percentage error is 7. 8- The equation: 2x² - 5y² + 4z² = 7 is a second-degree equation with three unknowns. 8. 9- In every division process there is a dividend, a divisor, and a quotient, and in some cases a remainder 9. 10- Reciprocal of the number - complementary angles - supplementary angles - perimeter of the polygon 10. (Continued p. 2) . .

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### csp_33433f05c6bd58fda2a494114c21fbe2

Shamash Secondary School Conditional Exam. Sept.1965

Subject: Arithmetic & Trigonometry Date: 14/9/1965 Class: 4th Year Secondary. Time: 8.00-10.30

--- Attempt all questions.

① Find the income produced by investing £253 15s. in 3½% stock at 101½ and the amount realised by subsequently selling out at 99.

② A spherical ball of lead 3 in. in diameter is melted and recast into three spherical balls. The diameters of two of these are 1½ in. and 2 in. respectively. What is the diameter of the other ?

③ A ladder, 24ft. long, makes an angle of 52° with the ground and leans against a vertical wall. If the top of the ladder slips down 2 ft. how far will the foot of the ladder move ?

4 An aeroplane is flying horizontally due E. When it is due N. of an observer its elevation is 42°. Find its elevation when it is N. 55° E. of the observer.

⑤ A householder has two alternative methods of paying for the electric light and power that he uses during one quarter of a year. Either he pays 7d. per unit for light and 2¾d. per unit for power, or he pays 1½d. per unit for light and for power and also a quarterly charge of £2 5s. 6d. Determine which method is the cheaper, and by how much, for a quarter during which he uses 68 units for light and 196 units for power. A householder paying by the first method used 117 lighting units during a quarter, and his electricity bill for the quarter was £5 3s. 1d. Find the number of units used for power.

[Marginalia] ----- [Marginalia] 240 154 [Marginalia] 12

⟦geometric diagram of a pyramid/triangulation⟧

**Traduction anglaise —**

Shamash Secondary School Conditional Exam. Sept.1965 Subject: Arithmetic & Trigonometry Date: 14/9/1965 Class: 4th Year Secondary. Time: 8.00-10.30 ⟦line⟧ Attempt all questions. ① Find the income produced by investing £253 15s. in 3½% stock at 101½ and the amount realised by subsequently selling out at 99. ② A spherical ball of lead 3 in. in diameter is melted and recast into three spherical balls. The diameters of two of these are 1½ in. and 2 in. respectively. What is the diameter of the other ? ③ A ladder, 24ft. long, makes an angle of 52° with the ground and leans against a vertical wall. If the top of the ladder slips down 2 ft. how far will the foot of the ladder move ? 4 An aeroplane is flying horizontally due E. When it is due N. of an observer its elevation is 42°. Find its elevation when it is N. 55° E. of the observer. ⑤ A householder has two alternative methods of paying for the electric light and power that he uses during one quarter of a year. Either he pays 7d. per unit for light and 2¾d. per unit for power, or he pays 1½d. per unit for light and for power and also a quarterly charge of £2 5s. 6d. Determine which method is the cheaper, and by how much, for a quarter during which he uses 68 units for light and 196 units for power. A householder paying by the first method used 117 lighting units during a quarter, and his electricity bill for the quarter was £5 3s. 1d. Find the number of units used for power. ⟦line⟧ 240 154 12 ⟦geometric diagram of a pyramid/triangulation⟧

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### csp_334f1c4bf45b5ca28b6d2aa7b851a31d

B01 · paragraph / latin ⟦illegible⟧ ⟦illegible⟧ the roots of the ⟦illegible⟧ equation 2x⟦illegible⟧ y = 0 then x = -1 or x = ⟦illegible⟧ the function y ⟦illegible⟧ values x < -1 ⟦illegible⟧ the root of the equation 2x² - x - ⟦illegible⟧ the equation 2x² - x - 3 = -2 Draw the ⟦illegible⟧ ⟦illegible⟧ x = -1 and x = 1 ⟦illegible⟧ ⟦illegible⟧ the points of intersection of the graph of ⟦illegible⟧

B02 · other / latin y 3 2 1 -2 -1 -1/2 1/4 1/2 1 3/2 2 x -1 -2 1/2 - 3

**Traduction anglaise —**

⟦illegible⟧ ⟦illegible⟧ the roots of the ⟦illegible⟧ equation 2x⟦illegible⟧ y = 0 then x = -1 or x = ⟦illegible⟧ the function y ⟦illegible⟧ values x < -1 ⟦illegible⟧ the root of the equation 2x² - x - ⟦illegible⟧ the equation 2x² - x - 3 = -2 Draw the ⟦illegible⟧ ⟦illegible⟧ x = -1 and x = 1 ⟦illegible⟧ ⟦illegible⟧ the points of intersection of the graph of ⟦illegible⟧ y 3 2 1 -2 -1 -1/2 1/4 1/2 1 3/2 2 x -1 -2 1/2 - 3

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### csp_339a84173dbb5af5b72e1f93d7643210

Solutions to Algebra Exam. Final Exam Fourth year, 1965

1. (i) (2x+y)/(x+2y) = m ∴ 2x+y = mx+2my ∴ y(2m-1) = x(2-m) ∴ y = x(2-m)/(2m-1) Ans. If y = mx then mx = x(2-m)/(2m-1) ∴ m(2m-1) = 2-m or 2m² = 2 ∴ m² = 1 ∴ m = ±1 Ans. (ii) c³ - 27b³ + a³ + 9abc = a³ + (-3b)³ + c³ - 3 a (-3b) c = (a-3b+c) (a² + 9b² + c² + 3ab - ac + 3bc) Ans. (iii) 5x² - 14x + 9 into two factors ∴ 5x² - 14x + 9 = (5x - 9)(x - 1) Ans. 5x² - 14x + 9 = (5x - 9)(x - 1) = 5(x - 9/5)(x - 1) = 5(x - 1.8)(x - 1) When 1 < x < 1.8 the factor (x - 1.8) = (-) and the factor (x - 1) = (+) ∴ the product = (-) x (+) = (-) Q.E.D

2. Let A's rate of walking = x miles/hr } ∴ 1/y * 60 - 1/x * 60 = 2 " B's " " " = y miles/hr } since 60/y = No. of minutes for B to cover 1 mile and 60/x = " " " " A " " " ∴ 60/y - 60/x = 2 ⟦line⟧ ① also from the figure : distance AC = 10/60 x = x/6 miles " " " BC = 10/60 y = y/6 " But AC - BC = 1/4 ∴ x/6 - y/6 = 1/4 2x - 2y = 3 or y = (2x-3)/2 ⟦line⟧ ② from eq. ① 30x - 30y = xy ⟦line⟧ ③ and from ② we obtain 30x - 30 ((2x-3)/2) = x ((2x-3)/2) or 60x - 60x + 90 = 2x² - 3x or 2x² - 3x - 90 = 0 or (2x - 15)(x + 6) = 0 ∴ x = -6 inadmissible or x = 15/2 = 7.5 miles/hr. and from eq. ② y = (2x-3)/2 = (2x7.5-3)/2 = 6 ∴ y = 6 miles/hr. Ans.

**Traduction anglaise —**

Solutions to Algebra Exam. Final Exam Fourth year, 1965 1. (i) (2x+y)/(x+2y) = m ∴ 2x+y = mx+2my ∴ y(2m-1) = x(2-m) ∴ y = x(2-m)/(2m-1) Ans. If y = mx then mx = x(2-m)/(2m-1) ∴ m(2m-1) = 2-m or 2m² = 2 ∴ m² = 1 ∴ m = ±1 Ans. (ii) c³ - 27b³ + a³ + 9abc = a³ + (-3b)³ + c³ - 3 a (-3b) c = (a-3b+c) (a² + 9b² + c² + 3ab - ac + 3bc) Ans. (iii) 5x² - 14x + 9 into two factors ∴ 5x² - 14x + 9 = (5x - 9)(x - 1) Ans. 5x² - 14x + 9 = (5x - 9)(x - 1) = 5(x - 9/5)(x - 1) = 5(x - 1.8)(x - 1) When 1 < x < 1.8 the factor (x - 1.8) = (-) and the factor (x - 1) = (+) ∴ the product = (-) x (+) = (-) Q.E.D 2. Let A's rate of walking = x miles/hr } ∴ 1/y * 60 - 1/x * 60 = 2 " B's " " " = y miles/hr } since 60/y = No. of minutes for B to cover 1 mile and 60/x = " " " " A " " " ∴ 60/y - 60/x = 2 ⟦line⟧ ① also from the figure : distance AC = 10/60 x = x/6 miles " " " BC = 10/60 y = y/6 " But AC - BC = 1/4 ∴ x/6 - y/6 = 1/4 2x - 2y = 3 or y = (2x-3)/2 ⟦line⟧ ② from eq. ① 30x - 30y = xy ⟦line⟧ ③ and from ② we obtain 30x - 30 ((2x-3)/2) = x ((2x-3)/2) or 60x - 60x + 90 = 2x² - 3x or 2x² - 3x - 90 = 0 or (2x - 15)(x + 6) = 0 ∴ x = -6 inadmissible or x = 15/2 = 7.5 miles/hr. and from eq. ② y = (2x-3)/2 = (2x7.5-3)/2 = 6 ∴ y = 6 miles/hr. Ans.

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### csp_35c6daf33e7a5863bfa5268c668c87fc

- p.2 - Algebra. 4th Year. 18/5/67 ---

5. (i) Draw the graph of y = x³ from x = -3 to x = +3, using 1 inch for 1 unit on the x-axis and 1 inch for 10 units on the y-axis. (6 marks).

(ii) By drawing two straight-line graphs on the same diagram find the roots of each of the following equations correct to one decimal place. x³ = 3x+2 .......(1) x³-5x-2=0 .......(2) (7 marks)

(iii) Find from the resulting diagram³ the range of values of x for which (3x+2) is greater than x³. (7 marks).

-----

**Traduction anglaise —**

- p.2 - Algebra. 4th Year. 18/5/67 ⟦line⟧ 5. (i) Draw the graph of y = x³ from x = -3 to x = +3, using 1 inch for 1 unit on the x-axis and 1 inch for 10 units on the y-axis. (6 marks). (ii) By drawing two straight-line graphs on the same diagram find the roots of each of the following equations correct to one decimal place. x³ = 3x+2 ⟦line⟧ (1) x³-5x-2=0 ⟦line⟧ (2) (7 marks) (iii) Find from the resulting diagram³ the range of values of x for which (3x+2) is greater than x³. (7 marks). ⟦line⟧

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### csp_362769eb974e5e84913d7b6bdd5f74de

- 2 - Mid-Year Exam. Cont., in Algebra ; 4th Year, Secondary, 17/2/1969. ⟦line⟧ two trains just clear away from each other in the two cases : (i) When the two trains are travelling in opposite directions. (ii) when the two trains are travelling in the same direction. (20 marks). ⟦line⟧ 3. (b) In the following equation, A, B, and C are constants and the equation is true for all values of x. Find the values of A, B and C. A(x² - 2x) + B(x + 4) + C = 3x² + x + 25 . (10 marks) (ii) Solve the following equations simultaneously : x² + xy + 2y² = 8 ........... (1) 2x² - 2xy - 3y² = 1 ........... (2) (10 marks) 4. (i) If x + 1/x = a and y + 1/y = b , find the value of the expression (x² + 1/x² + 2)(y² + 1/y² - 2) in terms of "a" and "b". Hence or otherwise find the value of (x + 1/x + y + 1/y) if a = 1 and b = 2 . (10 marks) (ii) Resolve the expression a² + b² - c² - d² + 2ab + 2cd into two factors one of which is (a + b - c + d). (10 marks) 5. A cask P is filled with 100 gallons of water, and a cask Q with 50 gallons of brandy; x gallons are drawn from each cask, mixed and replaced, and the same operation is repeated. Find x when there are 17 gallons of brandy in P after the second replacement. (20 marks) 6. Two trains A and B are travelling on two railway tracks which are parallel to each other. Train A is 240 ft long and it is travelling at 22.5 miles per hour. Train B is 200 ft long and is travelling at the rate 15 miles per hour. Find the length of time in seconds from the instant when the ⟦illegible⟧ of the front cars of the two trains are together, to the instant when the P. T. O

**Traduction anglaise —**

- 2 - Mid-Year Exam. Cont., in Algebra ; 4th Year, Secondary, 17/2/1969. ⟦line⟧ two trains just clear away from each other in the two cases : (i) When the two trains are travelling in opposite directions. (ii) when the two trains are travelling in the same direction. (20 marks). ⟦line⟧ 3. (b) In the following equation, A, B, and C are constants and the equation is true for all values of x. Find the values of A, B and C. A(x² - 2x) + B(x + 4) + C = 3x² + x + 25 . (10 marks) (ii) Solve the following equations simultaneously : x² + xy + 2y² = 8 ⟦line⟧ (1) 2x² - 2xy - 3y² = 1 ⟦line⟧ (2) (10 marks) 4. (i) If x + 1/x = a and y + 1/y = b , find the value of the expression (x² + 1/x² + 2)(y² + 1/y² - 2) in terms of "a" and "b". Hence or otherwise find the value of (x + 1/x + y + 1/y) if a = 1 and b = 2 . (10 marks) (ii) Resolve the expression a² + b² - c² - d² + 2ab + 2cd into two factors one of which is (a + b - c + d). (10 marks) 5. A cask P is filled with 100 gallons of water, and a cask Q with 50 gallons of brandy; x gallons are drawn from each cask, mixed and replaced, and the same operation is repeated. Find x when there are 17 gallons of brandy in P after the second replacement. (20 marks) 6. Two trains A and B are travelling on two railway tracks which are parallel to each other. Train A is 240 ft long and it is travelling at 22.5 miles per hour. Train B is 200 ft long and is travelling at the rate 15 miles per hour. Find the length of time in seconds from the instant when the ⟦illegible⟧ of the front cars of the two trains are together, to the instant when the P. T. O

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### csp_3660ed7243fc50dd849023211f8520ad

B01 · header / latin 3

B02 · paragraph / latin New R. V = 32 X 27,750 ⟦line⟧ 24,000 = £ 37 Ans.

**Traduction anglaise —**

3 New R. V = 32 X 27,750 ⟦line⟧ 24,000 = £ 37 Ans.

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### csp_37845092986f5fe4a100b9c47f81fe32

[Marginalia] ⟦illegible⟧

-p.2- Algebra. 4th Year Scientific. 29/5/1966. -----

4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks). (ii) An invalid on a certain day was able to take a single step of 18 inches. If he was each day to walk twice as far as on the preceding day, how long would it be before he can take a walk of 512 yards ? (10 marks)

5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks) (ii) Using the same axes and scales, draw another graph to find the roots of the equation x³- 13/4 x - 3/2 = 0. ( 7 marks) (iii) From your diagram, find all values of x which make the expression [ x³-(13/4x+ 3/2) ] , positive. ( 7 marks).

-------

**Traduction anglaise —**

⟦illegible⟧ -p.2- Algebra. 4th Year Scientific. 29/5/1966. ⟦line⟧ 4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks). (ii) An invalid on a certain day was able to take a single step of 18 inches. If he was each day to walk twice as far as on the preceding day, how long would it be before he can take a walk of 512 yards ? (10 marks) 5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks) (ii) Using the same axes and scales, draw another graph to find the roots of the equation x³- 13/4 x - 3/2 = 0. ( 7 marks) (iii) From your diagram, find all values of x which make the expression [ x³-(13/4x+ 3/2) ] , positive. ( 7 marks). ⟦line⟧

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### csp_379d12f0123f551194f136a9f8b29267

Solutions to 2nd Quarter Exam. algebra, 4th year. page 2. 27/12/1966

1. (a) x = 3ay - 5bz / 3ay + 5bz ∴ 3axy + 5bxz = 3ay - 5bz ∴ 3ay - 3axy = 5bxz + 5bz ∴ 3ay(1-x) = 5bz(x+1) ∴ a = 5bz / 3y . 1+x / 1-x Ans. 1 also 5bxz + 5bz = 3ay - 3axy ∴ 5bz(x+1) = 3ay(1-x) (8 marks) ∴ b = 3ay / 5z . 1-x / 1+x Ans. 2

(b) a/b = k ∴ 4a - 5b / √18a² - 4b² = 4 a/b - 5 / b/b √18a² - 4b² = 4 a/b - 5 / √18 a²/b² - 4 = 4k - 5 / √18k² - 4 Ans. (8 marks)

2. (a) x = y / y+1 and y = a-2 / 2 . Prove that x(y+2) + x/y + 1/x = a x = (a-2)/2 / (a-2)/2 + 1 = a-2 / a-2+2 = a-2 / a . Now substitute x = a-2 / a , y = a-2 / 2 ∴ the expression x(y+2) + x/y + 1/x = a-2 / a ( a-2 / 2 + 2 ) + a-2 / a / a-2 / 2 + 1 / a-2 / a ∴ the expression = a-2 / a . a+2 / 2 + 2a-4 / a(a-2) + a²-2a / 2(a-2) = a²-4 / 2a + 2(a-2) / a(a-2) + a(a-2) / 2(a-2) = a²-4 / 2a + 2/a + a/2 = a²-4+4+a² / 2a = 2a² / 2a = a Q.E.D. (8 marks)

(b) 1 - 1.4x / 0.2 + x = 0.7(x-1) / 0.1 - 0.5x multiply the first fraction (both numerat. + Den.) by 5 + the second fraction by 10. , we get 5 - 7x / 1 + 5x = 7(x-1) / 1 - 5x ∴ (5-7x)(1-5x) = 7(1+5x)(x-1) ∴ 5 - 32x + 35x² = 7(5x² - 4x - 1) or 35x² - 32x + 5 = 35x² - 28x - 7 ∴ 4x = 12 ∴ x = 3 Ans. (8 marks)

________________________________________________________________________________

**Traduction anglaise —**

Solutions to 2nd Quarter Exam. algebra, 4th year. page 2. 27/12/1966 1. (a) x = 3ay - 5bz / 3ay + 5bz ∴ 3axy + 5bxz = 3ay - 5bz ∴ 3ay - 3axy = 5bxz + 5bz ∴ 3ay(1-x) = 5bz(x+1) ∴ a = 5bz / 3y . 1+x / 1-x Ans. 1 also 5bxz + 5bz = 3ay - 3axy ∴ 5bz(x+1) = 3ay(1-x) (8 marks) ∴ b = 3ay / 5z . 1-x / 1+x Ans. 2 (b) a/b = k ∴ 4a - 5b / √18a² - 4b² = 4 a/b - 5 / b/b √18a² - 4b² = 4 a/b - 5 / √18 a²/b² - 4 = 4k - 5 / √18k² - 4 Ans. (8 marks) 2. (a) x = y / y+1 and y = a-2 / 2 . Prove that x(y+2) + x/y + 1/x = a x = (a-2)/2 / (a-2)/2 + 1 = a-2 / a-2+2 = a-2 / a . Now substitute x = a-2 / a , y = a-2 / 2 ∴ the expression x(y+2) + x/y + 1/x = a-2 / a ( a-2 / 2 + 2 ) + a-2 / a / a-2 / 2 + 1 / a-2 / a ∴ the expression = a-2 / a . a+2 / 2 + 2a-4 / a(a-2) + a²-2a / 2(a-2) = a²-4 / 2a + 2(a-2) / a(a-2) + a(a-2) / 2(a-2) = a²-4 / 2a + 2/a + a/2 = a²-4+4+a² / 2a = 2a² / 2a = a Q.E.D. (8 marks) (b) 1 - 1.4x / 0.2 + x = 0.7(x-1) / 0.1 - 0.5x multiply the first fraction (both numerat. + Den.) by 5 + the second fraction by 10. , we get 5 - 7x / 1 + 5x = 7(x-1) / 1 - 5x ∴ (5-7x)(1-5x) = 7(1+5x)(x-1) ∴ 5 - 32x + 35x² = 7(5x² - 4x - 1) or 35x² - 32x + 5 = 35x² - 28x - 7 ∴ 4x = 12 ∴ x = 3 Ans. (8 marks) ⟦line⟧

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### csp_39d4f6016f5d598e8d17c00d524597af

2. (i)

tan θ = 6/20 = 0.3000 ∴ θ = 16° 42'

63. 60' 58 42 16 18" 47

x = 20 tan 47° 18' = 20 x 0.8920 = 19.84 ft 20 x 1.0837 19.84 + 6 = 25.84 ft 21.674 6 27.674

(ii) 8 x 50 = 400 ft² 400 cot 27 = 400 x 1.9626 = 785.04 ft²

⟦illegible⟧

**Traduction anglaise —**

2. (i) tan θ = 6/20 = 0.3000 ∴ θ = 16° 42' 63. 60' 58 42 16 18" 47 x = 20 tan 47° 18' = 20 x 0.8920 = 19.84 ft 20 x 1.0837 19.84 + 6 = 25.84 ft 21.674 6 27.674 (ii) 8 x 50 = 400 ft² 400 cot 27 = 400 x 1.9626 = 785.04 ft² ⟦illegible⟧

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### csp_3a3ff16235e357d4add23d8d4bcbd05c

B01 · paragraph / latin (4) (i) The 1st tap fills the bath in "a" minutes ∴ it fills 1/a of the bath in one minute The 2nd " " " " " "b" " ∴ it " 1/b " " " " " " ∴ both taps together fill (1/a + 1/b) of the bath in one minute " " " " " (a+b/ab) " " " " " " " " " " " (1/ab) " " " in (1/(a+b/ab)) minutes " " " " " (ab/ab) " " " i.e. all the bath in (ab/a+b) minutes. Ans.

B02 · paragraph / latin (ii) When running separately, the 1st tap fills the bath in 7 minutes less than the 2nd tap ⟦...⟧ i.e. b - a = 7 ⟦line⟧ ① also, when running together, they both fill the bath in (ab/a+b) minutes i.e. ab/a+b = 12 ⟦line⟧ ② from ① b = a + 7 and from ② ab = 12a + 12b ∴ a(a + 7) = 12a + 12(a + 7) or a² + 7a = 12a + 12a + 84 a² - 17a - 84 = 0 ∴ (a + 4)(a - 21) = 0 ∴ a = -4 to be discarded or a = 21 minutes ∴ b = a + 7 = 21 + 7 = 28 minutes or [and b = 28 minutes] Ans.

B03 · paragraph / latin 5(i) Let the number of boy candidates in any case be = b and the " " " girl " " " " " be = g 5b/8 + 7g/12 = 41/68 (b + g) or 5x17x3b + 7x17x2g = 6x41(b + g) 255b + 238g = 246b + 246g or 9b = 8g ∴ b/g = 8/9 No. of boy candidates = 8 : 9 Ans. No. of girl candidates

B04 · paragraph / latin (ii) When the no. of girl candidates who passed the examination is equal to 168, then in the above relation we get: 7/12 g = 168 ∴ g = 168 x 12 / 7 = 24 x 12 = 288 b = 8/9 g = 8/9 x 288 = 8 x 32 = 256 ∴ b + g = 256 + 288 = 544 ∴ Total number of candidates = 544 Ans.

**Traduction anglaise —**

(4) (i) The 1st tap fills the bath in "a" minutes ∴ it fills 1/a of the bath in one minute The 2nd " " " " " "b" " ∴ it " 1/b " " " " " " ∴ both taps together fill (1/a + 1/b) of the bath in one minute " " " " " (a+b/ab) " " " " " " " " " " " (1/ab) " " " in (1/(a+b/ab)) minutes " " " " " (ab/ab) " " " i.e. all the bath in (ab/a+b) minutes. Ans. (ii) When running separately, the 1st tap fills the bath in 7 minutes less than the 2nd tap ⟦...⟧ i.e. b - a = 7 ⟦line⟧ ① also, when running together, they both fill the bath in (ab/a+b) minutes i.e. ab/a+b = 12 ⟦line⟧ ② from ① b = a + 7 and from ② ab = 12a + 12b ∴ a(a + 7) = 12a + 12(a + 7) or a² + 7a = 12a + 12a + 84 a² - 17a - 84 = 0 ∴ (a + 4)(a - 21) = 0 ∴ a = -4 to be discarded or a = 21 minutes ∴ b = a + 7 = 21 + 7 = 28 minutes or [and b = 28 minutes] Ans. 5(i) Let the number of boy candidates in any case be = b and the " " " girl " " " " " be = g 5b/8 + 7g/12 = 41/68 (b + g) or 5x17x3b + 7x17x2g = 6x41(b + g) 255b + 238g = 246b + 246g or 9b = 8g ∴ b/g = 8/9 No. of boy candidates = 8 : 9 Ans. No. of girl candidates (ii) When the no. of girl candidates who passed the examination is equal to 168, then in the above relation we get: 7/12 g = 168 ∴ g = 168 x 12 / 7 = 24 x 12 = 288 b = 8/9 g = 8/9 x 288 = 8 x 32 = 256 ∴ b + g = 256 + 288 = 544 ∴ Total number of candidates = 544 Ans.

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### csp_3b2adcd036cb5a08a77c73818e3b85fa

(cont'd).. -2- Algebra 4th Year. Scientific 14/5/1969. ---

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**Traduction anglaise —**

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### csp_3d390b52ebe95d0f890940f7cc7bd171

Shamash Secondary School 4th Quarter Exam. May 5th, 1967

Subject: Mathematics Date: 7/5/1967 Class: 4th Year Secondary Time: 8:00 - 9:30

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1. Compute by logarithm the expression:

9 / (0.4007)³ x tan² 37° 19' / 50.72⁵ x Cos³ 14° 34' (30 marks)

2. Simplify: (i) (x^(1+q/p))^(p/(p+q)) ÷ p√(x^(2p) / (x⁻¹)^(-p)) (20 marks)

(ii) { (y^(1/2) + y^(-1/2)) / (y² - y + 1) - (y^(1/2) - y^(-1/2)) / (y² + y + 1) } ÷ { (y^(1/2) + 2y^(-1/2)) / (y³ - 1) - (y^(1/2) - 2y^(-1/2)) / (y³ + 1) } (20 marks)

3. Solve the equation: 2x3^(2x) = 4^(x-1) finding the answer correct to four significant figures. (30 marks)

**Traduction anglaise —**

Shamash Secondary School 4th Quarter Exam. May 5th, 1967 Subject: Mathematics Date: 7/5/1967 Class: 4th Year Secondary Time: 8:00 - 9:30 ⟦line⟧ 1. Compute by logarithm the expression: 9 / (0.4007)³ x tan² 37° 19' / 50.72⁵ x Cos³ 14° 34' (30 marks) 2. Simplify: (i) (x^(1+q/p))^(p/(p+q)) ÷ p√(x^(2p) / (x⁻¹)^(-p)) (20 marks) (ii) { (y^(1/2) + y^(-1/2)) / (y² - y + 1) - (y^(1/2) - y^(-1/2)) / (y² + y + 1) } ÷ { (y^(1/2) + 2y^(-1/2)) / (y³ - 1) - (y^(1/2) - 2y^(-1/2)) / (y³ + 1) } (20 marks) 3. Solve the equation: 2x3^(2x) = 4^(x-1) finding the answer correct to four significant figures. (30 marks)

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### csp_401a8eb12c3a5084b328cba1b48392a3

Shamash Secondary School Final Examination May 1969

Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m.

Answer all questions:

1. (i) In the expression x³+Ax²+31x+B, A and B are constant. Find the values of A and B which will make this expression divisible by (x-2) (x-3) and find the remaining factor. (10 marks) (ii) The wages of 12 men and 7 boys amount to £9 13s. If 3 men together receive 8s. more than 4 boys, what are the wages of each man and boy ? (10 marks)

2. (i) Solve the equation (log x)² = log x⁷ - ⟦10⟧, finding two values for x. (10 marks) (ii) Solve the two simultaneous equations: 3x²+xy-2y²+7=0 .......(1) x²-xy+y²-7=0 .........(2) (10 marks)

3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√ 8¹⁻ˣ) = (√ 2ˣ) (√ 50) (10 marks) (ii)Compute the value of ⁷√[(0.5002)² Sin³ 14° 25' / (4.003)³ Cos² 15° 27'] (10 marks)

4. (i) If (b+c)⁻¹, (c+a)⁻¹, (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks) (ii) A bouncing tennis ball rebounds each time to a height one half the height of the previous bounce. If it is dropped from a height of 10 ft., show: (a) that the total distance it has travelled when it hits the ground for the 10th time is equal to 29 123/128 ft. (b) Show also that the total distance it travels before coming to rest is 30 ft. (10 marks)

(cont'd.p.2)..

**Traduction anglaise —**

Shamash Secondary School Final Examination May 1969 Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m. Answer all questions: 1. (i) In the expression x³+Ax²+31x+B, A and B are constant. Find the values of A and B which will make this expression divisible by (x-2) (x-3) and find the remaining factor. (10 marks) (ii) The wages of 12 men and 7 boys amount to £9 13s. If 3 men together receive 8s. more than 4 boys, what are the wages of each man and boy ? (10 marks) 2. (i) Solve the equation (log x)² = log x⁷ - ⟦10⟧, finding two values for x. (10 marks) ii) Solve the two simultaneous equations: 3x²+xy-2y²+7=0 ⟦line⟧(1) x²-xy+y²-7=0 ⟦line⟧(2) (10 marks) 3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√ 8¹⁻ˣ) = (√ 2ˣ) (√ 50) (10 marks) (ii)Compute the value of ⁷√[(0.5002)² Sin³ 14° 25' / (4.003)³ Cos² 15° 27'] (10 marks) 4. (i) If (b+c)⁻¹, (c+a)⁻¹, (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks) (ii) A bouncing tennis ball rebounds each time to a height one half the height of the previous bounce. If it is dropped from a height of 10 ft., show: (a) that the total distance it has travelled when it hits the ground for the 10th time is equal to 29 123/128 ft. (b) Show also that the total distance it travels before coming to rest is 30 ft. (10 marks) (cont'd.p.2)..

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### csp_422261184c125975b75ab0db31507f40

Shamash Secondary School Final Examination, May, 1967

Subject: Algebra Date: 18/5/1967 Class: 4th Year Secondary Time: 8:00 - 11:00 a.m.

--- Answer all questions:

1. (i) If x = a + b/t and y = b + at, find an expression for y in terms of a, b, x. (6 marks) (ii) If 2^(x-y) = 8 and 3^(x-2y) = 9 find the values of x and y. (7 marks) (iii) If m/n = 5/4 and p/q = 3/4, find the value of (3m+5p)/(n+q) (7 marks)

2. A manufacturer calculated that, allowing for the initial outlay on equipment, the cost of production of n tables was £(300+8n). Write down the cost of production of 1 table when (a) n tables are made, (b) (50+n) tables are made. If the cost of production of one table decreases by £1 when the extra 50 tables are made, find the value of n and the cost of production of a table in each case. (20 marks)

3. (i) Compute by logarithms the following expression, arranging your work neatly: 7√ (Cos³ 42° 17' x tan⁵ 27° 34') / (1.009³ x 90.04⁵) (10 marks) (ii) If N = 2^4.136 and 10^0.301 = 2, find, without the use of tables, the logarithm of N to the base 10 correct to five significant figures. (10 marks)

4. (i) Prove that the sum of any number of terms of the progression 4, 12, 20, 28, ... is a perfect square. (ii) The first three of the four terms, 12, x, y, 4, are in arithmetical progression and the last three are in geometrical progression. Find x and y.

(p.2)..

**Traduction anglaise —**

Shamash Secondary School Final Examination, May, 1967 Subject: Algebra Date: 18/5/1967 Class: 4th Year Secondary Time: 8:00 - 11:00 a.m. ⟦line⟧ Answer all questions: 1. (i) If x = a + b/t and y = b + at, find an expression for y in terms of a, b, x. (6 marks) (ii) If 2^(x-y) = 8 and 3^(x-2y) = 9 find the values of x and y. (7 marks) (iii) If m/n = 5/4 and p/q = 3/4, find the value of (3m+5p)/(n+q) (7 marks) 2. A manufacturer calculated that, allowing for the initial outlay on equipment, the cost of production of n tables was £(300+8n). Write down the cost of production of 1 table when (a) n tables are made, (b) (50+n) tables are made. If the cost of production of one table decreases by £1 when the extra 50 tables are made, find the value of n and the cost of production of a table in each case. (20 marks) 3. (i) Compute by logarithms the following expression, arranging your work neatly: 7√ (Cos³ 42° 17' x tan⁵ 27° 34') / (1.009³ x 90.04⁵) (10 marks) (ii) If N = 2^4.136 and 10^0.301 = 2, find, without the use of tables, the logarithm of N to the base 10 correct to five significant figures. (10 marks) 4. (i) Prove that the sum of any number of terms of the progression 4, 12, 20, 28, ... is a perfect square. (ii) The first three of the four terms, 12, x, y, 4, are in arithmetical progression and the last three are in geometrical progression. Find x and y. (p.2)..

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### csp_43da8e6dc3f3530e91ec4a93e404083f

(ii) In a G.P., we have l_1 × l_4 = 1 also l_1 + l_4 = 9, S_7 = ? let the first term be a } ∴ (a)(ar³) = ar³ or a²r³=ar³ or ar = 1 ... ① let the Common Ratio be r } also a + ar³ = 9 ... ② ∴ a + 1/a = 9 ∴ a = 8 ∴ r³ = 1/8 ∴ r = 1/2 ∴ a = 8 } r = 1/2 } S_7 = a(1-rⁿ) / 1-r = 8(1-(1/2)⁷) / 1-1/2 = 16[1 - 1/128] = 16[128-1 / 128] S_7 = ? } ∴ S_7 = 16 × 127 / 128 = 127 / 8 = 15 7/8 Ans.

**Traduction anglaise —**

4 (i) A.P. 3, 7, 11, ... here a=3, d=4, no of terms = n, l_n = ? l_n = a+(n-1)d ∴ l_n = 3+(n-1)×4 = 4n-1 ∴ l_n = 4n-1 Ans. 1 ② S_n = bn+cn² let n=1 ∴ S_1 = 3 = b+c or b+c=3 ... ① also let n=2 ∴ S_2 = 3+7 = 10 = 2b+c(2)² or 2b+4c=10 ... ② b+2c=5 ... ③ Combining eq. ① + ③, we get c = 2 ∴ b = 1 Ans. ∴ S_30 = bn+cn² = 1×30 + 2(30)² = 30 + 1800 = 1830 Ans. 3 (ii) In a G.P., we have l_1 × l_4 = 1 also l_1 + l_4 = 9, S_7 = ? let the first term be a } ∴ (a)(ar³) = ar³ or a²r³=ar³ or ar = 1 ... ① let the Common Ratio be r } also a + ar³ = 9 ... ② ∴ a + 1/a = 9 ∴ a = 8 ∴ r³ = 1/8 ∴ r = 1/2 ∴ a = 8 } r = 1/2 } S_7 = a(1-rⁿ) / 1-r = 8(1-(1/2)⁷) / 1-1/2 = 16[1 - 1/128] = 16[128-1 / 128] S_7 = ? } ∴ S_7 = 16 × 127 / 128 = 127 / 8 = 15 7/8 Ans.

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### csp_44307023ab845f6b9d73b2328b093c35

Shamash Secondary School Conditional Examination, Sept. 63

Subject: Algebra Date: 12/9/1963 Class: 4th Year Secondary Time: 8.30-11.00 a.m.

---- Attempt all questions:

1. (i) The equation 7X + 2 / X² - 4 = A / X-2 + B / X+2 is true for all values of X. Find the values of A & B. (8 marks) (ii) The expression X³ + pX² + qX + 6 is factorable into (X-1) and (X+2). Find the values of p and q and find the third factor. (8 marks)

2. Solve for X the following equations, rejecting all extraneous roots: (i) 1 / 1 - X + 1 / √X +1 + 1 / √X -1 = 0 (6 marks) (ii) (X-7)^(1/2) = √X - 7 (6 marks) (iii) 3X^(-2/3) - 10X^(-1/3) + 3 = 0 (6 marks)

3. (i) Solve for X, using tables if necessary: 20^x = 2^(x+2) (ii) Compute by logarithms: ⁵√((0.0012)² X (1.003)³) / (7515 X 2.004) (8 marks)

4. A man travels 108 miles, and finds that he could have made the journey in 4 1/2 hours less, had he travelled 2 miles an hour faster. At what rate did he travel ? (16 marks)

5. (i) Find by series the value of the recurring fraction 0.3205 (8 marks) (ii) The three digits of a number are in arithmetical progression. The number itself divided by the sum of the digits is 48. The number formed by the same digits in reverse order is 396 less than the original number. What is the number ? (8 marks)

**Traduction anglaise —**

Shamash Secondary School Conditional Examination, Sept. 63 Subject: Algebra Date: 12/9/1963 Class: 4th Year Secondary Time: 8.30-11.00 a.m. ⟦line⟧ Attempt all questions: 1. (i) The equation 7X + 2 / X² - 4 = A / X-2 + B / X+2 is true for all values of X. Find the values of A & B. (8 marks) (ii) The expression X³ + pX² + qX + 6 is factorable into (X-1) and (X+2). Find the values of p and q and find the third factor. (8 marks) 2. Solve for X the following equations, rejecting all extraneous roots: (i) 1 / 1 - X + 1 / √X +1 + 1 / √X -1 = 0 (6 marks) (ii) (X-7)^(1/2) = √X - 7 (6 marks) (iii) 3X^(-2/3) - 10X^(-1/3) + 3 = 0 (6 marks) 3. (i) Solve for X, using tables if necessary: 20^x = 2^(x+2) (ii) Compute by logarithms: ⁵√((0.0012)² X (1.003)³) / (7515 X 2.004) (8 marks) 4. A man travels 108 miles, and finds that he could have made the journey in 4 1/2 hours less, had he travelled 2 miles an hour faster. At what rate did he travel ? (16 marks) 5. (i) Find by series the value of the recurring fraction 0.3205 (8 marks) (ii) The three digits of a number are in arithmetical progression. The number itself divided by the sum of the digits is 48. The number formed by the same digits in reverse order is 396 less than the original number. What is the number ? (8 marks)

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### csp_45eadb88f59e5a50b4f992e1d0a91883

Illustrative Examples:

In the financial year 1954-55 a man had an earned income of £ 1350 and a further unearned income of £ 200 from investments. In that year income-tax was levied according to the following rules: The first £ 210 of the man's income was free of tax and there was a further tax free allowance of 2/9 th. of the earned income. The remainder of the earned income, called "taxable income" was taxed as follows: The first £ 100 of the taxable income was taxed at 2s, 6d. in the £ 1, the next £ 150 at 5s, the next £ 150 at 7s. and the remainder at 9 s. in the £. How much tax did the man pay?

solution:

Total income = £ 1350 + £ 200 = £ 1550. Taxable income = £ 1550 - { £ 210 + £ 1350 x 2 / 9 = £ 1040 £ 100 @ 2 s 6 d £ 100 x 0.125 = £ 12.500 £ 150 @ 5 s £ 150 x 0.250 = £ 37.500 £ 150 @ 7 s £ 150 x 0.350 = £ 52.500 Remainder (£ 640) £ 640 x 0.450 = £ 288.000 Total tax = £ 390.500 or £ 390 10 s Ans.

P.T.O.

**Traduction anglaise —**

Illustrative Examples: In the financial year 1954-55 a man had an earned income of £ 1350 and a further unearned income of £ 200 from investments. In that year income-tax was levied according to the following rules: The first £ 210 of the man's income was free of tax and there was a further tax free allowance of 2/9 th. of the earned income. The remainder of the earned income, called "taxable income" was taxed as follows: The first £ 100 of the taxable income was taxed at 2s, 6d. in the £ 1, the next £ 150 at 5s, the next £ 150 at 7s. and the remainder at 9 s. in the £. How much tax did the man pay? solution: Total income = £ 1350 + £ 200 = £ 1550. Taxable income = £ 1550 - { £ 210 + £ 1350 x 2 / 9 = £ 1040 £ 100 @ 2 s 6 d £ 100 x 0.125 = £ 12.500 £ 150 @ 5 s £ 150 x 0.250 = £ 37.500 £ 150 @ 7 s £ 150 x 0.350 = £ 52.500 Remainder (£ 640) £ 640 x 0.450 = £ 288.000 Total tax = £ 390.500 or £ 390 10 s Ans. P.T.O.

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### csp_465ae0e23d2658da8c528121b0cf658c

[Marginalia] stencil

Taxation

(a) Income tax ضريبة الدخل (b) Rates = tax on <del>rateable</del> real property ضريبة الاملاك

(a) Income tax Income tax is levied in accordance with ascending scale (نسبة تصاعدية) i.e. the rate of tax increases with higher income.

Earned income = Income <del>as a</del> resulting from one's toil (اتعاب الشخص) e.g. salaries. Unearned income = Income from deposits, shares etc.

In the U.K income tax is <del>colle</del> levied at the rate of so many shillings in the £ e.g 10s 6d of £1 of income.

Total income = Total earnings Taxable income = Total income - tax- free allowance

_ _ _ _ _

(b) Rates Rates - ضريبة الاملاك Rent - الايجار Rateable value - assessed value مبلغ التقدير The rates are collected on

P T O

**Traduction anglaise —**

stencil Taxation (a) Income tax Income tax (b) Rates = tax on <del>rateable</del> real property Property tax (a) Income tax Income tax is levied in accordance with ascending scale (progressive rate) i.e. the rate of tax increases with higher income. Earned income = Income <del>as a</del> resulting from one's toil (person's labor) e.g. salaries. Unearned income = Income from deposits, shares etc. In the U.K income tax is <del>colle</del> levied at the rate of so many shillings in the £ e.g 10s 6d of £1 of income. Total income = Total earnings Taxable income = Total income - tax- free allowance ⟦line⟧ (b) Rates Rates - Property tax Rent - Rent Rateable value - assessed value Estimated amount The rates are collected on P T O

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### csp_4845775504a95041897ecdbf28b61b70

B01 · header / latin Shamash Secondary School 1st. Quarter Exam.

B02 · form / latin Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary. Date: 8/12/1957

B03 · paragraph / latin ⟦line⟧ All questions are to be attempted

B04 · paragraph / latin (1) State ⟦...⟧ how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m, by 25.00 m.

B05 · paragraph / latin (2) (a) Decimalise to 3 places the following:

B06 · table / latin £ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 3/4 d

B07 · paragraph / latin (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint.

B08 · paragraph / latin (3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ?

B09 · paragraph / latin (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together.

B10 · paragraph / latin (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation.

B11 · paragraph / latin (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

**Traduction anglaise —**

Shamash Secondary School 1st. Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary. Date: 8/12/1957 ⟦line⟧ All questions are to be attempted (1) State ⟦...⟧ how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m, by 25.00 m. (2) (a) Decimalise to 3 places the following: £ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 3/4 d (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint. (3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ? (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together. (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation. (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

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### csp_48b356d0413f540ebe24cd33db99b221

Solution to Conditional Exam - September 1962 4th year Secondary. Page 1

ax + by = c ---- ① bx + ay = ab ---- ② multiplying equ. ① by a, equ. ② by b, + you get a²x + aby = ac ---- ③ b²x + aby = ab² ---- ④ } subtracting ④ from ③, we get: (a² - b²)x = ac - ab² ∴ x = a(c - b²) / (a² - b²) again multiplying equ. ① by b + equ. ② by a, + you get: abx + b²y = bc ---- ⑤ abx + a²y = a²b ---- ⑥ } subtracting ⑤ from ⑥, we get: (a² - b²)y = a²b - bc ∴ y = b(a² - c) / (a² - b²) x = a(c - b²) / (a² - b²) } Ans. y = b(a² - c) / (a² - b²)

(ii) 4x² - 12x + 3 = 0 ∴ x = (12 ± √144 - 4x4x3) / (2x4) = (12 ± √96) / 8 ∴ x = (12 ± 4√6) / 8 = (3 ± √6) / 2 ∴ x₁ = (3 + √6) / 2 = (3 + 2.449) / 2 = 5.449 / 2 = 2.7245 = 2.72 Correct to 2 dec. x₂ = (3 - √6) / 2 = (3 - 2.449) / 2 = 0.551 / 2 = 0.2755 = 0.28 Cor. ⟦...⟧ Ans.

2. (i) x = 4.836, let √x² + x = y ∴ y = √x(x + 1) = √4.836 x 5.836 log 4.836 = 0.6844 log 5.836 = 0.7661 2 log y = 1.4505 log y = 0.72525 ∴ y = 5.311 or 5.31 Ans.

By actual multiplication: (ii) (1 - 2x + x²)(1 - kx + x²) ≡ 1 - (2 + k)x + 2(1 + k)x² - (2 + k)x³ + x⁴ ∴ the coefficient of x² is 2(1 + k) = 0 ∴ k = -1 Ans.

**Traduction anglaise —**

Solution to Conditional Exam - September 1962 4th year Secondary. Page 1 ax + by = c ⟦line⟧ ① bx + ay = ab ⟦line⟧ ② multiplying equ. ① by a, equ. ② by b, + you get a²x + aby = ac ⟦line⟧ ③ b²x + aby = ab² ⟦line⟧ ④ } subtracting ④ from ③, we get: (a² - b²)x = ac - ab² ∴ x = a(c - b²) / (a² - b²) again multiplying equ. ① by b + equ. ② by a, + you get: abx + b²y = bc ⟦line⟧ ⑤ abx + a²y = a²b ⟦line⟧ ⑥ } subtracting ⑤ from ⑥, we get: (a² - b²)y = a²b - bc ∴ y = b(a² - c) / (a² - b²) x = a(c - b²) / (a² - b²) } Ans. y = b(a² - c) / (a² - b²) (ii) 4x² - 12x + 3 = 0 ∴ x = (12 ± √144 - 4x4x3) / (2x4) = (12 ± √96) / 8 ∴ x = (12 ± 4√6) / 8 = (3 ± √6) / 2 ∴ x₁ = (3 + √6) / 2 = (3 + 2.449) / 2 = 5.449 / 2 = 2.7245 = 2.72 Correct to 2 dec. x₂ = (3 - √6) / 2 = (3 - 2.449) / 2 = 0.551 / 2 = 0.2755 = 0.28 Cor. ⟦to 2 dec.⟧ Ans. 2. (i) x = 4.836, let √x² + x = y ∴ y = √x(x + 1) = √4.836 x 5.836 log 4.836 = 0.6844 log 5.836 = 0.7661 2 log y = 1.4505 log y = 0.72525 ∴ y = 5.311 or 5.31 Ans. By actual multiplication: (ii) (1 - 2x + x²)(1 - kx + x²) ≡ 1 - (2 + k)x + 2(1 + k)x² - (2 + k)x³ + x⁴ ∴ the coefficient of x² is 2(1 + k) = 0 ∴ k = -1 Ans.

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### csp_492871890e1e5459a677262e5190f974

Shamash Secondary School 1st. Quarter Exam.

Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary Date: 8/12/1957

All questions are to be attempted.

(2) (a) Decimalise to 3 places the following:

£ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 1/4 d

(b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint.

(3) A watch which gains 5 sec.in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ?

(4) The average age of m boys is b years and of n girls is c years. Find the average age of all together.

(5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20°W., find the distance of the fort from the ship at the first observation.

(6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42°7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

**Traduction anglaise —**

Shamash Secondary School 1st. Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary Date: 8/12/1957 All questions are to be attempted. (1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m. by 25.00 m. (2) (a) Decimalise to 3 places the following: £ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 1/4 d (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint. (3) A watch which gains 5 sec.in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ? (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together. (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20°W., find the distance of the fort from the ship at the first observation. (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42°7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

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### csp_4aa02bc531a25c069b4b447ef0ce35ab

Solutions to Conditional Exam in Algebra Cont. Sept., 1965

4. (i) the first <del>number</del> which is divisible by 13 & greater than 60 is 65 = (5x13) and the last number divisible by 13 and less than 600 is 598 = (46x13) ∴ we have an Arith. Prog. in which a = 65 , l = 598 , d = 13 . to find n & S. l = a + (n-1) d ∴ 598 = 65 + (n-1) 13 ∴ 13(n-1) = 533 ∴ n-1 = 533/13 = 41 ∴ n = 42 ∴ S = n/2 (a + l) or S = 42/2 (65 + 598) or S = 21 x 663 or S = 13923 Ans. (10 marks)

(ii) ① total at the end of 5th time is = 256 + 64x2 + 16x2 + 4x2 + 1x2 S = 256 + 2 (64 + 16 + 4 + 1) = 256 + 2 [ 64 (1 - (1/4)⁴) / (1 - 1/4) ] = 256 + 2 [ (4 x 64)/3 (1 - 1/4⁴) ] = 256 + 512/3 (1 - 1/256) = 256 + 512/3 x 255/256 = 256 + 170 = 426 Ans. (5 marks)

256 256 64 16 64 16

② total dist. at the end of nth strike is S = 256 + 2 (64 + 16 + 4 + ... to (n-1) terms) S = 256 + 2 [ 64 { 1 - (1/4)ⁿ⁻¹ } / (1 - 1/4) ] Ans. = 256 + 2 [ (4 x 64)/3 (1 - 1/4ⁿ⁻¹) ] = 256 + 512/3 (1 - 1/4ⁿ⁻¹) = 256 + 512/3 - 512 / (3 x 4ⁿ⁻¹) = (768 + 512)/3 - 512 / (3 x 4ⁿ⁻¹) = 1280/3 - 512 / (3 x 4ⁿ⁻¹) = 1280/3 - 2⁹ / (3 x 2²ⁿ⁻²) = 1280/3 - 2¹¹⁻²ⁿ / 3 = (1280 - 2¹¹⁻²ⁿ) / 3 Ans (5 marks)

**Traduction anglaise —**

Solutions to Conditional Exam in Algebra Cont. Sept., 1965 4. (i) the first <del>number</del> which is divisible by 13 & greater than 60 is 65 = (5x13) and the last number divisible by 13 and less than 600 is 598 = (46x13) ∴ we have an Arith. Prog. in which a = 65 , l = 598 , d = 13 . to find n & S. l = a + (n-1) d ∴ 598 = 65 + (n-1) 13 ∴ 13(n-1) = 533 ∴ n-1 = 533/13 = 41 ∴ n = 42 ∴ S = n/2 (a + l) or S = 42/2 (65 + 598) or S = 21 x 663 or S = 13923 Ans. (10 marks) (ii) ① total at the end of 5th time is = 256 + 64x2 + 16x2 + 4x2 + 1x2 S = 256 + 2 (64 + 16 + 4 + 1) = 256 + 2 [ 64 (1 - (1/4)⁴) / (1 - 1/4) ] = 256 + 2 [ (4 x 64)/3 (1 - 1/4⁴) ] = 256 + 512/3 (1 - 1/256) = 256 + 512/3 x 255/256 = 256 + 170 = 426 Ans. (5 marks) 256 256 64 16 64 16 ② total dist. at the end of nth strike is S = 256 + 2 (64 + 16 + 4 + ... to (n-1) terms) S = 256 + 2 [ 64 { 1 - (1/4)ⁿ⁻¹ } / (1 - 1/4) ] Ans. = 256 + 2 [ (4 x 64)/3 (1 - 1/4ⁿ⁻¹) ] = 256 + 512/3 (1 - 1/4ⁿ⁻¹) = 256 + 512/3 - 512 / (3 x 4ⁿ⁻¹) = (768 + 512)/3 - 512 / (3 x 4ⁿ⁻¹) = 1280/3 - 512 / (3 x 4ⁿ⁻¹) = 1280/3 - 2⁹ / (3 x 2²ⁿ⁻²) = 1280/3 - 2¹¹⁻²ⁿ / 3 = (1280 - 2¹¹⁻²ⁿ) / 3 Ans (5 marks)

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### csp_4acc0d7197f95c9eb6b2e850cc6d49bd

SHAMASH SECONDARY SCHOOL 4th Quarter Examination, May, 1965.

Subject: Algebra Date: 2/5/1965 Class: 4th Secondary year Time: 8:00-9:30 a.m.

<del>A</del>ttempt all questions.

1. (a) Prove that: (a-a⁻¹)(a⁴/³ + a⁻²/³) = (a² - a⁻²) / a⁻¹/³ (13 marks) (b) Evaluate: (x³/⁴ + xy) / (xy - y³) -- √x / (√x - y) (1⟦2⟧ marks)

2. Solve the equation: (6√x - 7) / (√x - 1) - 5 = (7√x - 26) / (7√x - 21) (25 marks)

3. Find x from the equation: 3²ˣ = 5ˣ⁺¹ (25 marks)

4. Compute by logarithms the value of x, arranging your work neatly: ⁷√((1.001)² (0.0004061)²/³) / (Sin³ 24° 21' Cos² 41° 57') (25 marks)

--------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL 4th Quarter Examination, May, 1965. Subject: Algebra Date: 2/5/1965 Class: 4th Secondary year Time: 8:00-9:30 a.m. <del>A</del>ttempt all questions. 1. (a) Prove that: (a-a⁻¹)(a⁴/³ + a⁻²/³) = (a² - a⁻²) / a⁻¹/³ (13 marks) (b) Evaluate: (x³/⁴ + xy) / (xy - y³) -- √x / (√x - y) (1⟦2⟧ marks) 2. Solve the equation: (6√x - 7) / (√x - 1) - 5 = (7√x - 26) / (7√x - 21) (25 marks) 3. Find x from the equation: 3²ˣ = 5ˣ⁺¹ (25 marks) 4. Compute by logarithms the value of x, arranging your work neatly: ⁷√((1.001)² (0.0004061)²/³) / (Sin³ 24° 21' Cos² 41° 57') (25 marks) ⟦line⟧

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### csp_4cea4fdb1b06529db8a145caa33f1216

Solution to 3rd + 4th Quarter Exam - Cert. in Algebra 7/4/69 Page 2

√1+x + √1-x / √1+x - √1-x = (√1+x + √1-x)² / 1+x - (1-x) = 1+x + 1-x + 2√1-x² / 2x = 2 + 2√1-x² / 2x = = 2(1 + √1-x²) / 2x = 1 + √1-x² / x = 1 + √1 - 4b² / (b²+1)² / 2b / b²+1 = 1 + √((b²+1)² - 4b²) / b²+1 / 2b / b²+1 = b²+1 + √b⁴+2b²+1-4b² / 2b = b²+1 + √(b²-1)² / 2b = b²+1 + b²-1 / 2b = 2b² / 2b = b Ans.

3 (ii) solve for x : 2(log x)² - 5(log x) + 2 = 0 ∴ [2(log x) - 1][log x - 2] = 0 ∴ 2 log x = 1 or log x = 1/2 ∴ x = 10^(1/2) or x = √10 = 3.162 Correct to 3 dec. pl. or log x = 2 or x = 10² or x = 100 } Ans.

4.(i) a = 3 } Prove S_2n = 4 S_n d = 6 } S_2n = 2n/2 {2a + (2n-1)d} = n {6 + (2n-1)(6)} = n {6 + 12n - 6} = 12 n² S_n = n/2 {6 + (n-1)(6)} = n/2 {6n} = 3 n² but 12 n² = 4 (3 n²) ∴ S_2n = 4 S_n Q.E.D.

(ii) t_3 / t_6 = 11 / 26 and S_4 = 34 , S_8 = ? ∴ a + 2d / a + 5d = 11 / 26 or 26a + 52d = 11a + 55d or 15a - 3d = 0 .... ① also 34 = 4/2 {2a + 3d} or 34 = 2(2a + 3d) or 17 = 2a + 3d .... ② ∴ 15a - 3d = 0 .... ① } or <del>20 a + 7 d = 9</del> } <del>23 d = 170</del> <del>d = ⟦illegible⟧</del> 2a + 3d = 17 .... ② } <del>20 a + 30 d = 170</del> } 17a = 17 ∴ a = 1 ∴ 15 = 3d ∴ d = 5 ∴ a = 1 } Ans. 1 d = 5 } + the progression is 1, 6, 11, 16, ... S_8 = 8/2 {2 x 1 + 7 x 5} or S_8 = 4(2 + 35) = 4 x 37 = 148 Ans. 2

**Traduction anglaise —**

Solution to 3rd + 4th Quarter Exam - Cert. in Algebra 7/4/69 Page 2 √1+x + √1-x / √1+x - √1-x = (√1+x + √1-x)² / 1+x - (1-x) = 1+x + 1-x + 2√1-x² / 2x = 2 + 2√1-x² / 2x = = 2(1 + √1-x²) / 2x = 1 + √1-x² / x = 1 + √1 - 4b² / (b²+1)² / 2b / b²+1 = 1 + √((b²+1)² - 4b²) / b²+1 / 2b / b²+1 = b²+1 + √b⁴+2b²+1-4b² / 2b = b²+1 + √(b²-1)² / 2b = b²+1 + b²-1 / 2b = 2b² / 2b = b Ans. 3 (ii) solve for x : 2(log x)² - 5(log x) + 2 = 0 ∴ [2(log x) - 1][log x - 2] = 0 ∴ 2 log x = 1 or log x = 1/2 ∴ x = 10^(1/2) or x = √10 = 3.162 Correct to 3 dec. pl. or log x = 2 or x = 10² or x = 100 } Ans. 4.(i) a = 3 } Prove S_2n = 4 S_n d = 6 } S_2n = 2n/2 {2a + (2n-1)d} = n {6 + (2n-1)(6)} = n {6 + 12n - 6} = 12 n² S_n = n/2 {6 + (n-1)(6)} = n/2 {6n} = 3 n² but 12 n² = 4 (3 n²) ∴ S_2n = 4 S_n Q.E.D. (ii) t_3 / t_6 = 11 / 26 and S_4 = 34 , S_8 = ? ∴ a + 2d / a + 5d = 11 / 26 or 26a + 52d = 11a + 55d or 15a - 3d = 0 .... ① also 34 = 4/2 {2a + 3d} or 34 = 2(2a + 3d) or 17 = 2a + 3d .... ② ∴ 15a - 3d = 0 .... ① } or <del>20 a + 7 d = 9</del> } <del>23 d = 170</del> <del>d = ⟦illegible⟧</del> 2a + 3d = 17 .... ② } <del>20 a + 30 d = 170</del> } 17a = 17 ∴ a = 1 ∴ 15 = 3d ∴ d = 5 ∴ a = 1 } Ans. 1 d = 5 } + the progression is 1, 6, 11, 16, ... S_8 = 8/2 {2 x 1 + 7 x 5} or S_8 = 4(2 + 35) = 4 x 37 = 148 Ans. 2

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### csp_4d3df91170125c62a039709be6e17b18

⟦illegible⟧ to Delinquent Exam in Algebra ⟦illegible⟧ September 1961

S = n/2 (a+l) ∴ 35 = 7/2 (a+l) . also, l-a = 18 or l = a+18 ∴ 35 = 7/2 (a+a+18) or 10 = 2a+18 or 2a = -8 ∴ a = -4 ∴ l = a+18 or l = 14 . <del>Hence the seven numbers are:</del> but, l = a + (n-1)d or d = (l-a)/(n-1) = (14+4)/6 = 3 . Hence the seven numbers are: -4 , -1 , 2 , 5 , 8 , 11 , 14 Ans.

(ii) ar^3 + 6ar^4 = ar^2 Dividing by ar^2, we get: r + 6r^2 = 1 or 6r^2 + r - 1 = 0 or (3r-1)(2r+1) = 0 or r = 1/3 and r = -1/2 Ans. ar = 16 and r = -1/2 ∴ -a/2 = 16 ∴ a = -32 <del>⟦illegible⟧</del> S = a(1-r^n) / (1-r) = -32[1 - (-1/2)^6] / (1 + 1/2) = -32[1 - 1/64] / (3/2) = - (32 × 2 / 3) (63/64) = - 21 Ans.

5. Let the time be x minutes after 7: the Hour hand would have moved x/12 divisions beyond No. 7 ∴ x + 15 = 35 + x/12 ∴ x - x/12 = 20 ∴ 12x - x = 240 ∴ 11x = 240 ∴ x = 240/11 = 21 9/11 minutes ∴ The time is 21 9/11 minutes past 7. Ans.

⟦Diagram of a clock face showing hands between 7 and 8, and 4 and 5, with annotations: x minutes, x/12⟧

**Traduction anglaise —**

⟦illegible⟧ to Delinquent Exam in Algebra ⟦illegible⟧ September 1961 S = n/2 (a+l) ∴ 35 = 7/2 (a+l) . also, l-a = 18 or l = a+18 ∴ 35 = 7/2 (a+a+18) or 10 = 2a+18 or 2a = -8 ∴ a = -4 ∴ l = a+18 or l = 14 . <del>Hence the seven numbers are:</del> but, l = a + (n-1)d or d = (l-a)/(n-1) = (14+4)/6 = 3 . Hence the seven numbers are: -4 , -1 , 2 , 5 , 8 , 11 , 14 Ans. (ii) ar^3 + 6ar^4 = ar^2 Dividing by ar^2, we get: r + 6r^2 = 1 or 6r^2 + r - 1 = 0 or (3r-1)(2r+1) = 0 or r = 1/3 and r = -1/2 Ans. ar = 16 and r = -1/2 ∴ -a/2 = 16 ∴ a = -32 <del>⟦illegible⟧</del> S = a(1-r^n) / (1-r) = -32[1 - (-1/2)^6] / (1 + 1/2) = -32[1 - 1/64] / (3/2) = - (32 × 2 / 3) (63/64) = - 21 Ans. 5. Let the time be x minutes after 7: the Hour hand would have moved x/12 divisions beyond No. 7 ∴ x + 15 = 35 + x/12 ∴ x - x/12 = 20 ∴ 12x - x = 240 ∴ 11x = 240 ∴ x = 240/11 = 21 9/11 minutes ∴ The time is 21 9/11 minutes past 7. Ans. ⟦Diagram of a clock face showing hands between 7 and 8, and 4 and 5, with annotations: x minutes, x/12⟧

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### csp_4dec224fb6c95ecdb51f0c9e06d6027d

Shamash Secondary School Final Examination, 1955-1956

Subject: Algebra Date: 30/5/56 Class: 4th Year Time: 8:00 - 10:00 a.m.

All questions are to be attempted:

1. In 1955 a housewife could buy 7 more eggs for 10s. 6d. than she can buy for 14s. in 1956, when the price per egg has increased by one penny. Find the price of eggs, per dozen, in 1955. (Ans. 2s.)

2. A motorist travels a certain distance x at a certain uniform speed. If his speed had been 4 miles per hour greater, he would have saved 10 minutes on the journey; and if his speed had been 9 miles per hour greater he would have saved 20 minutes. Find the distance x. (Ans. 60 miles)

3. Using one pair of axes draw graphs of x² and of ½x + 2 between the values of x = -3 and x = +3, choosing your own scales. From your graphs read off the solutions of x² = ½x + 2. By drawing a further graph find the solutions of x² = ½x + 1. (Ans. 1.7 or -1.2, 1.3 or -0.8).

5. (i) Compute by logarithms ⁷√((0.00092)² x (4.006)³ / (0.006204)⁵) (ii) Find the value of x from the equation 4^(2-x) x 3^x = 243

4. (i) The sum of the first 29 terms of an Arithmetical progression, whose common difference is (-0.8), is zero. Find the first term. (Ans. 11.2) (ii) Find the tenth term + the sum of the first ten terms of the Arithmetical progression whose nth term is 3 - ½n (Ans. -2, 2.5).

[Marginalia] ⟦illegible arrow pointing to question 5⟧ [Marginalia] ⟦illegible arrow pointing to question 4⟧

**Traduction anglaise —**

Shamash Secondary School Final Examination, 1955-1956 Subject: Algebra Date: 30/5/56 Class: 4th Year Time: 8:00 - 10:00 a.m. All questions are to be attempted: 1. In 1955 a housewife could buy 7 more eggs for 10s. 6d. than she can buy for 14s. in 1956, when the price per egg has increased by one penny. Find the price of eggs, per dozen, in 1955. (Ans. 2s.) 2. A motorist travels a certain distance x at a certain uniform speed. If his speed had been 4 miles per hour greater, he would have saved 10 minutes on the journey; and if his speed had been 9 miles per hour greater he would have saved 20 minutes. Find the distance x. (Ans. 60 miles) 3. Using one pair of axes draw graphs of x² and of ½x + 2 between the values of x = -3 and x = +3, choosing your own scales. From your graphs read off the solutions of x² = ½x + 2. By drawing a further graph find the solutions of x² = ½x + 1. (Ans. 1.7 or -1.2, 1.3 or -0.8). 5. (i) Compute by logarithms ⁷√((0.00092)² x (4.006)³ / (0.006204)⁵) (ii) Find the value of x from the equation 4^(2-x) x 3^x = 243 4. (i) The sum of the first 29 terms of an Arithmetical progression, whose common difference is (-0.8), is zero. Find the first term. (Ans. 11.2) (ii) Find the tenth term + the sum of the first ten terms of the Arithmetical progression whose nth term is 3 - ½n (Ans. -2, 2.5). ⟦illegible arrow pointing to question 5⟧ ⟦illegible arrow pointing to question 4⟧

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### csp_4f542d613f3a5d358c5269947d225909

B01 · marginalia / latin ⟦illegible⟧ (21)

B02 · paragraph / latin A <del>⟦illegible⟧</del> can walk a mile in 2 minutes less time than B would take. In a walking race, B has a start of 1/4 mile and A overtakes B in 10 minutes. Assuming both men walk at a uniform rate, find their rates of walking in miles/hour. (10 marks.)

B03 · other / unknown ⟦line⟧ ⟦illegible blue ink bleed-through and faded text⟧

**Traduction anglaise —**

⟦illegible⟧ (21) A <del>⟦illegible⟧</del> can walk a mile in 2 minutes less time than B would take. In a walking race, B has a start of 1/4 mile and A overtakes B in 10 minutes. Assuming both men walk at a uniform rate, find their rates of walking in miles/hour. (10 marks.) ⟦line⟧ ⟦illegible blue ink bleed-through and faded text⟧

---

### csp_4f62e7d3965356bba76e2bb9ea8f2be0

Solution to Algebra paper Cont. (Final Exam 18/5/1967) Page 2

7 ⟦illegible⟧ x = ⟦illegible⟧ Cos³ 42° 17' x tan⁵ 27° 34' 1.009³ x 90.04⁵

(ii) N = 2⁴.¹³⁶ and 10⁰.³⁰¹ = 2 , log₁₀ N = ? log₁₀ N = 4.136 log₁₀ 2 but log₁₀ 2 = 0.301 ∴ log₁₀ N = 4.136 x 0.301 = 1.244936 = 1.2449 (Correct to Ans. 5 significant figures (10 marks)

⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

Solution to Algebra paper Cont. (Final Exam 18/5/1967) Page 2 7 ⟦illegible⟧ x = ⟦illegible⟧ Cos³ 42° 17' x tan⁵ 27° 34' 1.009³ x 90.04⁵ log cos 42° 17' = 1.8691 | 3 log cos 42° 17' = 1.6073 log tan 27° 34' = 1.7177 | 5 log tan 27° 34' = 2.5885 log 1.009 = 0.0037 | log Num. = 2.1958 log 90.04 = 1.9544 | log Den. = 9.7831 ⟦line⟧ 3 log 1.009 = 0.0111 | 7 log x = 12.4127 5 log 90.04 = 9.7720 | log x = 2.34467 log Den. = 9.7831 | or log x = 2.3447 | x = 0.02212 | or x = 2.212 x 10⁻² | Ans. | (10 marks) (ii) N = 2⁴.¹³⁶ and 10⁰.³⁰¹ = 2 , log₁₀ N = ? log₁₀ N = 4.136 log₁₀ 2 but log₁₀ 2 = 0.301 ∴ log₁₀ N = 4.136 x 0.301 = 1.244936 = 1.2449 (Correct to Ans. 5 significant figures (10 marks) ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_525fa49ca9ba57ec932f38c49805e527

⟦illegible⟧ 2 x 3^{2x} = 4^{x-1} log 2 + 2x log 3 = (x-1) log 4 2x log 3 - x log 4 = - log 4 - log 2 2x log 3 - 2x log 2 = - 2 log 2 - log 2 2x (log 3 - log 2) = - 3 log 2 x = - 3 log 2 / 2 (log 3 - log 2) = - 3 x 0.3010 / 2 (0.4771 - 0.3010)

∴ x = - 3 x 0.3010 / 2 x 0.1761 = - 0.9030 / 0.3522 = - 2.5638... ∴ x = - 2.564 Correct to four significant figures.

**Traduction anglaise —**

3 ⟦illegible⟧ 2 x 3^{2x} = 4^{x-1} log 2 + 2x log 3 = (x-1) log 4 2x log 3 - x log 4 = - log 4 - log 2 2x log 3 - 2x log 2 = - 2 log 2 - log 2 2x (log 3 - log 2) = - 3 log 2 x = - 3 log 2 / 2 (log 3 - log 2) = - 3 x 0.3010 / 2 (0.4771 - 0.3010) ∴ x = - 3 x 0.3010 / 2 x 0.1761 = - 0.9030 / 0.3522 = - 2.5638... ∴ x = - 2.564 Correct to four significant figures.

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### csp_52decbc67c595a6691bbb5a068b5cb6d

SHAMASH SECONDARY SCHOOL 4th Quarter Examination, May, 1965.

Subject: Algebra Date: 2/5/1965 Class: 4th Secondary year Time: 8:00-9:30 a.m.

Attempt all questions.

1. (a) Prove that: (a-a⁻¹)(a⁴/³ + a⁻²/³) = a² - a⁻² / a⁻¹/³ (13 marks) (b) Evaluate: x³/² + xy / xy - y³ - √x / √x-y (13 marks) 2. Solve the equation: 6√x - 7 / √x - 1 - 5 = 7√x - 26 / 7√x - 21 (25 marks) 3. Find x from the equation: 3²x = 5x+1 (25 marks) 4. Compute by logarithms the value of x, arranging your work neatly: ⁷√(1.001)² (0.0004061)¹/³ / Sin³ 24° 21' Cos² 41° 57' (25 marks)

--------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL 4th Quarter Examination, May, 1965. Subject: Algebra Date: 2/5/1965 Class: 4th Secondary year Time: 8:00-9:30 a.m. Attempt all questions. 1. (a) Prove that: (a-a⁻¹)(a⁴/³ + a⁻²/³) = a² - a⁻² / a⁻¹/³ (13 marks) (b) Evaluate: x³/² + xy / xy - y³ - √x / √x-y (13 marks) 2. Solve the equation: 6√x - 7 / √x - 1 - 5 = 7√x - 26 / 7√x - 21 (25 marks) 3. Find x from the equation: 3²x = 5x+1 (25 marks) 4. Compute by logarithms the value of x, arranging your work neatly: ⁷√(1.001)² (0.0004061)¹/³ / Sin³ 24° 21' Cos² 41° 57' (25 marks) ⟦line⟧

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### csp_542814b1471c5fac82e08f9dc8474d46

5. (i) Divide 3x³ - 2x² + Bx - 26 by x - 2 . Hence find the value of 'B' that makes the expression 3x³ - 2x² + Bx - 26 factorable into (x - 2), and find the other factor. (9 marks)

(ii) A man can swim at x m.p.h. in still water. His <del>speed</del> rate increases y m.p.h. when he swims with the current, and decreases y m.p.h. when swims against the current. The difference in his time to swim 2 miles with the current and 2 miles against the current is z hours. Find a formula for z in terms of x and y. (7 marks)

6. Give the English equivalent to the following: ١. اعداد زوجية واعداد فردية ٢. الجذر التربيعي لعدد كسري ٣. مقلوب العدد ٤. معامل حرفي ٥. اساس القوة وأس القوة ٦. الخطأ المطلق والخطأ النسبي ٧. نقل الحدود من طرف الى الطرف الاخر للمعادلة ٨. اوجد المقدار ٤١٩ر٤ صحيحاً لاقرب ثلاثة ارقام معنوية هي ٩. محيط المضلع ١٠. وتر القوس في دائرة

(20 marks)

**Traduction anglaise —**

5. (i) Divide 3x³ - 2x² + Bx - 26 by x - 2 . Hence find the value of 'B' that makes the expression 3x³ - 2x² + Bx - 26 factorable into (x - 2), and find the other factor. (9 marks) (ii) A man can swim at x m.p.h. in still water. His <del>speed</del> rate increases y m.p.h. when he swims with the current, and decreases y m.p.h. when swims against the current. The difference in his time to swim 2 miles with the current and 2 miles against the current is z hours. Find a formula for z in terms of x and y. (7 marks) 6. Give the English equivalent to the following: 1. Even numbers and odd numbers 2. The square root of a fractional number 3. Reciprocal of the number 4. Literal coefficient 5. The base of the power and the exponent of the power 6. Absolute error and relative error 7. Moving terms from one side to the other side of the equation 8. Find the value 4.419 correct to the nearest three significant figures 9. Perimeter of the polygon 10. Chord of an arc in a circle (20 marks)

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### csp_543989b94cbe53ab9ee6e95d473e7b0e

[Marginalia] Nadia Jacob.

-p.2- Algebra. 4th Year Scientific. 29/5/1966. -----

4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks).

5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks)

(ii) Using the same axes and scales, draw another graph to find the roots of the equation x³ - ⟦13/4⟧x - 3/2 = 0. ( 7 marks)

(iii) From your diagram, find all values of x which make the expression [x³ - (⟦13/4⟧x + 3/2)] , positive. ( 7 marks).

--------

**Traduction anglaise —**

Nadia Jacob. -p.2- Algebra. 4th Year Scientific. 29/5/1966. ⟦line⟧ 4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks). (ii) An invalid on a certain day was able to take a single step of 18 inches. If he was each day to walk twice as far as on the preceding day, how long would it be before he can take a walk of 512 yards ? (10 marks) 5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks) (ii) Using the same axes and scales, draw another graph to find the roots of the equation x³ - ⟦13/4⟧x - 3/2 = 0. ( 7 marks) (iii) From your diagram, find all values of x which make the expression [x³ - (⟦13/4⟧x + 3/2)] , positive. ( 7 marks). ⟦line⟧

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### csp_5467d47894bd541d8a8ab30faa2b0d5c

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965.

Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m.

----------------------- Attempt all questions :

1. (i) If m = 2x + y / x + 2y , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks).

3. (i) Compute by logarithms, arranging your work neatly : 7√ (cos² 18° 47) (sin³ 48° 21) / (10.09)³ (0.0002049) (6 marks). (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks).

------------------------------------------------- Look for question 2 at the back of this sheet. P.T.O.

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965. Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m. ⟦line⟧ Attempt all questions : 1. (i) If m = 2x + y / x + 2y , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks). 2. (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks). 3. (i) Compute by logarithms, arranging your work neatly : 7√ (cos² 18° 47) (sin³ 48° 21) / (10.09)³ (0.0002049) (6 marks). (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks). 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ...... (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks). 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axix and 0.2 inch for your unit on the y-axix. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks). ⟦line⟧ Look for question 2 at the back of this sheet. P.T.O.

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### csp_54df85b1a5045d67b0baf4851f2b8cda

6) 6000 : 4000 = 3 : 2.

At end of 1st year B's profit = 1400 / 5 X 2 = £ 560 At beginning of 2nd year B's capital = £ 4,560 At end " " " B's profit = 1584 X 4,560 / 10,560 = £ 684 At beginning of 3rd year B's capital = £ 4,560 + £ 684 = £ 5,244 At the end of 3rd year B's profit = (1639 15s) 5,244 / 11,244 = £ 764.75

At the end of 3rd year B's capital is £ 5,244 + £ 764.75 = £ 6008.75 or £ 6008 15s -

⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

6) 6000 : 4000 = 3 : 2. At end of 1st year B's profit = 1400 / 5 X 2 = £ 560 At beginning of 2nd year B's capital = £ 4,560 At end " " " B's profit = 1584 X 4,560 / 10,560 = £ 684 At beginning of 3rd year B's capital = £ 4,560 + £ 684 = £ 5,244 At the end of 3rd year B's profit = (1639 15s) 5,244 / 11,244 = £ 764.75 At the end of 3rd year B's capital is £ 5,244 + £ 764.75 = £ 6008.75 or £ 6008 15s - ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_54e8c3301c71589a81a680c2e1073b53

Shamash Secondary School 1st Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class : 4th year Secondary. Date: 8/12/1957 --------

All questions are to be attempted.

(2) (a) Decimalise to 3 places the following: £ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 1/4 d (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint.

(3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ?

(4) The average age of m boys is b years and of n girls is c years. Find the average age of all together.

(5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation.

(6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

**Traduction anglaise —**

Shamash Secondary School 1st Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class : 4th year Secondary. Date: 8/12/1957 ⟦line⟧ All questions are to be attempted. (1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m, by 25.00 m. (2) (a) Decimalise to 3 places the following: £ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 1/4 d (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint. (3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ? (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together. (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation. (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

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### csp_5678bb1e2ede536fad5071f1ab437d8e

4 (i) x³ + y³ + 1/x³ + 1/y³ = (x³ + 1/x³) + (y³ + 1/y³) = [(x + 1/x)³ - 3(x + 1/x)] + [(y + 1/y)³ - 3(y + 1/y)] = a³ - 3a + b³ - 3b = a³ + b³ - 3(a + b) = (a + b)(a² - ab + b² - 3) Ans. 1 = (1 + 2)(1² - 1x2 + 2² - 3) = (3)(1 - 2 + 4 - 3) = 3 x zero = 0 Ans. 2

(ii) a³ + a - 8b³ - 2b + c + 6abc + c³ = a³ - 8b³ + c³ + 6abc + a - 2b + c = [a³ + (-2b)³ + c³ - 3a(-2b)c] + [a - 2b + c] = (a - 2b + c)(a² + 4b² + c² + 2ab - ac + 2bc) + (a - 2b + c) = (a - 2b + c) [(a² + 4b² + c² + 2ab - ac + 2bc) + 1] = (a - 2b + c)(a² + 4b² + c² + 2ab - ac + 2bc + 1) Ans.

5. after the first replacement, there are x/2 gall. of Brandy in Cask P and (50 - x/2) gall. " " " Q At the beginning of the 2nd operation:

100 gall water P 50 gall Brandy Q

(x²/200) gall. of Brandy are removed from Cask P and x(50 - x/2) / 50 gall. " " " " Q

Mixture Brandy 100 x/2 gall. ? = x * x/2 / 100 x ? = x²/200 gall. 50 gall. (50 - x/2) gall. x ? ? = x(50 - x/2) / 50

x²/200 + x(50 - x/2) / 50 / 2 gall. of Brandy are deposited in P after 2nd replacement.

∴ (x/2 - x²/200) + (x²/200 + x(50 - x/2) / 50) / 2 = 17 or 100x - x² / 200 + x² + 4x(50 - x/2) / 2 = 17 ∴ 100x - x² / 200 + x² + 200x - 2x² / 2 = 17 ∴ 100x - x² / 200 + 200x - x² / 400 = 17 ∴ 200x - 2x² + 200x - x² = 6800 ∴ 3x² - 400x + 6800 = 0 ∴ (3x - 340)(x - 20) = 0 ∴ x = 340/3 = 113 1/3 inadmissible x = 20 gallons Ans.

**Traduction anglaise —**

4 4 (i) x³ + y³ + 1/x³ + 1/y³ = (x³ + 1/x³) + (y³ + 1/y³) = [(x + 1/x)³ - 3(x + 1/x)] + [(y + 1/y)³ - 3(y + 1/y)] = a³ - 3a + b³ - 3b = a³ + b³ - 3(a + b) = (a + b)(a² - ab + b² - 3) Ans. 1 = (1 + 2)(1² - 1x2 + 2² - 3) = (3)(1 - 2 + 4 - 3) = 3 x zero = 0 Ans. 2 (ii) a³ + a - 8b³ - 2b + c + 6abc + c³ = a³ - 8b³ + c³ + 6abc + a - 2b + c = [a³ + (-2b)³ + c³ - 3a(-2b)c] + [a - 2b + c] = (a - 2b + c)(a² + 4b² + c² + 2ab - ac + 2bc) + (a - 2b + c) = (a - 2b + c) [(a² + 4b² + c² + 2ab - ac + 2bc) + 1] = (a - 2b + c)(a² + 4b² + c² + 2ab - ac + 2bc + 1) Ans. 5. after the first replacement, there are x/2 gall. of Brandy in Cask P and (50 - x/2) gall. " " " Q At the beginning of the 2nd operation: 100 gall water P 50 gall Brandy Q (x²/200) gall. of Brandy are removed from Cask P and x(50 - x/2) / 50 gall. " " " " Q Mixture Brandy 100 x/2 gall. ? = x * x/2 / 100 x ? = x²/200 gall. 50 gall. (50 - x/2) gall. x ? ? = x(50 - x/2) / 50 x²/200 + x(50 - x/2) / 50 / 2 gall. of Brandy are deposited in P after 2nd replacement. ∴ (x/2 - x²/200) + (x²/200 + x(50 - x/2) / 50) / 2 = 17 or 100x - x² / 200 + x² + 4x(50 - x/2) / 2 = 17 ∴ 100x - x² / 200 + x² + 200x - 2x² / 2 = 17 ∴ 100x - x² / 200 + 200x - x² / 400 = 17 ∴ 200x - 2x² + 200x - x² = 6800 ∴ 3x² - 400x + 6800 = 0 ∴ (3x - 340)(x - 20) = 0 ∴ x = 340/3 = 113 1/3 inadmissible x = 20 gallons Ans.

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### csp_57034cd08d5055b88e22091606f81fa8

(cont'd).. -2- Algebra 4th Year. Scientific 14/5/1969.

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**Traduction anglaise —**

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### csp_5774009606d85a7f9ff9ff594ed895da

II. (a) sin 51° 40' = 0.7844 cos 33° 22' = 0.8352 or 0.8351 tan 88° 3' 9' = 42.495 (b) θ = 0° 14' Ans. 1 α = 30° Ans. 4 β = 82° 55' Ans. 2 γ = 60° Ans. 5 φ = 77° 56' Ans. 3 ε = 45° Ans. 6

III. y / 110 = sin 14° 12' ∴ y = 110 sin 14° 12' ∴ y = 110 x 0.2453 ∴ y = 26.9830 yds Ans. = 26.98 yds correct to two decimal places.

[Marginalia] 110 yds. [Marginalia] 14° 12'

**Traduction anglaise —**

I. (1) Inaccessible distances (5) The generation of an angle (2) Counterclockwise rotation (6) a homogeneous algebraic expression (3) Solution of oblique triangle (7) Transposing from one side of the equation (4) Circular measure of angles to the other & adding like terms (8) the base of a power, the index or exponent of a power, the nth power. II. (a) sin 51° 40' = 0.7844 cos 33° 22' = 0.8352 or 0.8351 tan 88° 3' 9' = 42.495 (b) θ = 0° 14' Ans. 1 α = 30° Ans. 4 β = 82° 55' Ans. 2 γ = 60° Ans. 5 φ = 77° 56' Ans. 3 ε = 45° Ans. 6 III. y / 110 = sin 14° 12' ∴ y = 110 sin 14° 12' ∴ y = 110 x 0.2453 ∴ y = 26.9830 yds Ans. = 26.98 yds correct to two decimal places. 110 yds. 14° 12'

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### csp_5ab91f0f26345a8ba4b4ca417a49ee17

[Marginalia] Audrey [Marginalia] Sam⟦...⟧

-p.2- Algebra. 4th Year Scientific. 29/5/1966. -----

4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks).

5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks)

(ii) Using the same axes and scales, draw another graph to find the roots of the equation x³ - 13/4x - 3/2 = 0. ( 7 marks)

(iii) From your diagram, find all values of x which make the expression [x³ - (13/4x + 3/2)] , positive. ( 7 marks). --------

**Traduction anglaise —**

Audrey Sam⟦...⟧ -p.2- Algebra. 4th Year Scientific. 29/5/1966. ⟦line⟧ 4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks). (ii) An invalid on a certain day was able to take a single step of 18 inches. If he was each day to walk twice as far as on the preceding day, how long would it be before he can take a walk of 512 yards ? (10 marks) 5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks) (ii) Using the same axes and scales, draw another graph to find the roots of the equation x³ - 13/4x - 3/2 = 0. ( 7 marks) (iii) From your diagram, find all values of x which make the expression [x³ - (13/4x + 3/2)] , positive. ( 7 marks). ⟦line⟧

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### csp_5dafe0a6199950319837cf3ece8a6e53

at 5 1/2 m/h 3 1/2 m/h A ------> B ------> C (2 1/2) (5 1/2) AB = 5/2 x 7/2 = 35/4 miles Let t hours = ⟦time taken A to⟧ reach C ∴ 5 1/2 t - 3 1/2 t = 35/4 ∴ t(5 1/2 - 3 1/2) = 35/4 ∴ 2t = 35/4 ∴ t = 35/8 hrs. ∴ t = 4 3/8 hours or 4 hours 22.5 minutes. Ans. (8 marks)

4. (i) boys men acres days 1 man = u boys { b y d { m a ? b boys = b/u men (9 marks) men acres days { b/u y d } ∴ No. of days = d.a.u / y.m { m a ? = adb / myu days Ans.

5 (i) x - 2 | 3x³ - 2x² + Bx - 26 | 3x² + 4x + 8 + B 3x³ - 6x² 4x² + Bx - 26 4x² - 8x (B + 8)x - 26 (B + 8)x - 16 - 2B 2B - 10 Hence 2B - 10 = 0 ∴ B = 5 Hence the other factor is the quotient 3x² + 4x + 8 + B or 3x² + 4x + 8 + 5 or (9 marks) 3x² + 4x + 13 Ans.

(ii) 3 / (x - y) - 2 / (x + y) = z or z = 3x + 3y - 2x + 2y / (x - y)(x + y) or z = x + 5y / x² - y² Ans. (7 marks)

**Traduction anglaise —**

at 5 1/2 m/h 3 1/2 m/h A ⟦line⟧ B ⟦line⟧ C (2 1/2) (5 1/2) AB = 5/2 x 7/2 = 35/4 miles Let t hours = ⟦time taken A to⟧ reach C ∴ 5 1/2 t - 3 1/2 t = 35/4 ∴ t(5 1/2 - 3 1/2) = 35/4 ∴ 2t = 35/4 ∴ t = 35/8 hrs. ∴ t = 4 3/8 hours or 4 hours 22.5 minutes. Ans. (8 marks) 4. (i) boys men acres days 1 man = u boys { b y d { m a ? b boys = b/u men (9 marks) men acres days { b/u y d } ∴ No. of days = d.a.u / y.m { m a ? = adb / myu days Ans. (ii) √16x⁴ + 8x² + 16/3 x²y + 4/3 y + 4/9 y² + 1 | 4x² + 1 + 2/3 y Ans. 16x⁴ 8x² + 1 | 8x² + 16/3 x²y + 4/3 y + 4/9 y² + 1 | 8x² + 1 8x² + 2 + 2/3 y | 16/3 x²y + 4/3 y + 4/9 y² + 2/3 y | 16/3 x²y + 4/3 y + 4/9 y² (8 marks) 5 (i) x - 2 | 3x³ - 2x² + Bx - 26 | 3x² + 4x + 8 + B 3x³ - 6x² 4x² + Bx - 26 4x² - 8x (B + 8)x - 26 (B + 8)x - 16 - 2B 2B - 10 Hence 2B - 10 = 0 ∴ B = 5 Hence the other factor is the quotient 3x² + 4x + 8 + B or 3x² + 4x + 8 + 5 or (9 marks) 3x² + 4x + 13 Ans. (ii) 3 / (x - y) - 2 / (x + y) = z or z = 3x + 3y - 2x + 2y / (x - y)(x + y) or z = x + 5y / x² - y² Ans. (7 marks)

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### csp_5efb0a2c70b85f978e51173c619658e3

- ٢ -

الرقم :: الاسم ::

٤٠- مقدار جبري متجانس ٤١- درجة المقدار الجبري ٤٢- المعامل الحرفي ٤٣- مقدار جبري من الدرجة الثانية ٤٤- ان حدي الكسر هما بسطه ومقامه

٤٥- في كل عملية قسمة يوجد مقسوم ومقسوم عليه وناتج قسمة وفي بعض الحالات باق للقسمة .

٤٦- ان الاعمدة المنصفة لاضلاع مثلث تلتقي في مركز الدائرة المرسومة ...........

٤٧- ان الخطوط المتوسطة في المثلث تلتقي في نقطة واحدة تقسم كلا منها الى ثلثين من جهة الرأس وثلث من جهة القاعدة . وتسمى هذه النقطة مركز ثقل المثلث.

٤٨- نقيس طول مستقيم فنجد انه يساوي ٥ر٦١ سم . ثم نجد فيما بعد ان طوله المضبوط ٦٠ سم . وفي هذه الحالة نقول ان الخطأ المطلق هو ........... والخطأ النسبي هو ........... والخطأ المئوي هو ...

٤٩- ان قيمة المقدار ٧٢ر٥٣٠٩ لاقرب اربعة ارقام معنوية هي ...........

٥٠- ان المعادلة ٣س٢ - ٢س ص + ص٢ = ٤٥ - س ع هي معادلة من الدرجة ....... في ....... مجاهيل .

(75 marks)

(II) Fill in the blanks in the following equations:-

1. one furlong = ( ) chains= ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds. = ( ) ft. 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre =( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. ⟦=⟧ ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones.

(25 marks).

**Traduction anglaise —**

- 2 - Number :: Name :: 40- Homogeneous algebraic expression 41- Degree of the algebraic expression 42- Literal coefficient 43- Second-degree algebraic expression 44- The two terms of a fraction are its numerator and denominator 45- In every division process there is a dividend, a divisor, and a quotient, and in some cases a remainder of the division. 46- The perpendicular bisectors of the sides of a triangle meet at the center of the circumscribed circle ⟦line⟧ 47- The medians of a triangle meet at a single point that divides each of them into two-thirds from the vertex side and one-third from the base side. This point is called the centroid of the triangle. 48- We measure the length of a straight line and find it equals 61.5 cm. Then we find later that its exact length is 60 cm. In this case we say that the absolute error is ⟦line⟧ And the relative error is ⟦line⟧ and the percentage error is ⟦line⟧ 49- The value of the quantity 5309.72 to the nearest four significant figures is ⟦line⟧ 50- The equation 3x2 - 2xy + y2 = 45 - xz is an equation of degree ⟦line⟧ in ⟦line⟧ unknowns. (75 marks) (II) Fill in the blanks in the following equations:- 1. one furlong = ( ) chains= ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds. = ( ) ft. 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre =( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. ⟦=⟧ ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones. (25 marks).

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### csp_5f9e013fa9a45716bdbd2160f5482b67

-p.2- Conditional Exam.in Algebra for the 4th Year, Sept.1964. (cont'd.) ------

V. (a) Sketch the curve of the function (x⁴-8x²) for values of x from -3 to 3 choosing ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (8 marks)

(b) From this curve find the values of x at which the function (x⁴-8x²) has minimum or maximum values. Find also these minimum and maximum values. (6 marks)

(c) Plot on the same diagram the graph of y= 3x-10 and from these two graphs find the solution of the equation x⁴-8x²+10 = 3x. (6 marks).

--------

**Traduction anglaise —**

-p.2- Conditional Exam.in Algebra for the 4th Year, Sept.1964. (cont'd.) ⟦line⟧ V. (a) Sketch the curve of the function (x⁴-8x²) for values of x from -3 to 3 choosing ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (8 marks) (b) From this curve find the values of x at which the function (x⁴-8x²) has minimum or maximum values. Find also these minimum and maximum values. (6 marks) (c) Plot on the same diagram the graph of y= 3x-10 and from these two graphs find the solution of the equation x⁴-8x²+10 = 3x. (6 marks). ⟦line⟧

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### csp_63e854d61d9656cbb6c755bb4130a524

⟦Conditional⟧ exams ⟦illegible⟧ 4th year secondary 2

_________________________________

⟦illegible⟧

**Traduction anglaise —**

⟦Conditional⟧ exams ⟦illegible⟧ 4th year secondary 2 5. (i) Taking 1 inch = 1 unit on the x-axis and 1 inch = 2 units on the y-axis draw the graphs of y = 4 - x² and 4y = 5x + 4 for values of x from -3 to +3. (8 marks) (ii) From your graph find: (a) the range of values of x for which 4 - x² is greater than 5/4 x + 1, (4 marks) (b) the values of x for which 4 - x² = 2.5, (4 marks) (c) the square root of 3.6. (4 marks) ⟦line⟧ ⟦illegible⟧

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### csp_6440e49d15c658009e4826c718b5fb00

Solutions to Algebra Final Exam. 4th Year, June 1952

(i) ⟦illegible⟧ Let x = 1, then 6 = 2A ∴ A = 3 Ans. Let x = -1, then 4 = -2B ∴ B = -2 Ans.

[Marginalia] 8

(ii) Method 1: By the remainder + factor theorem, when x = 2, then x - 2 = 0 and since x - 2 is a factor of the expression x³ + px² + qx + 42 then 2³ + p(2)² + q(2) + 42 = 0 ∴ 4p + 2q = -50 or 2p + q = -25 also 3³ + p(3)² + q(3) + 42 = 0 ∴ 9p + 3q = -69 or 3p + q = -23 Solving the two equations simultaneously, we get p = 2 and q = -27 Ans. Now the expression is x³ + 2x² - 27x + 42 But (x-2)(x-3) = x² - 5x + 6. The third factor is gotten by division as follows:

x³ + 2x² - 27x + 42 | x² - 5x + 6 x³ - 5x² + 6x | x + 7 ∴ the third factor is (x+7) Ans. II ----------------- 7x² - 33x + 42 7x² - 35x + 42 -----------------

Method 2: By Division: x³ + px² + qx + 42 | x - 2 x³ - 2x² | x² + (p+2)x + (2p+q+4) ----------------- (p+2)x² + qx + 42 (p+2)x² - 2(p+2)x ----------------- [q + 2(p+2)]x + 42 or (2p + q + 4)x + 42 (2p + q + 4)x - 2(2p + q + 4) ----------------- Remainder = 4p + 2q + 50 = 0 or 2p + q = -25 Also by actually dividing the same expression by x-3, we get 3p + q = -23 which leads to the same solution. Q.E.D.

2 (i) 3√x-1 + √2x-3 = 2√x+2 or 3√x-1 = 2√x+2 - √2x-3 squaring 9(x-1) = 4(x+2) + (2x-3) - 4√(x+2)(2x-3) or 4√2x²+x-6 = 14 - 3x squaring again, 16(2x²+x-6) = 196 - 84x + 9x² or 23x² + 100x - 292 = 0 ∴ (23x + 146)(x - 2) = 0 ∴ x = 2 Ans. 1 or x = -146/23 = Ans. 2 But Ans 2 should be rejected as an extraneous root since it does not verify the original equation.

[Marginalia] 8

**Traduction anglaise —**

Solutions to Algebra Final Exam. 4th Year, June 1952 (i) ⟦illegible⟧ Let x = 1, then 6 = 2A ∴ A = 3 Ans. Let x = -1, then 4 = -2B ∴ B = -2 Ans. 8 (ii) Method 1: By the remainder + factor theorem, when x = 2, then x - 2 = 0 and since x - 2 is a factor of the expression x³ + px² + qx + 42 then 2³ + p(2)² + q(2) + 42 = 0 ∴ 4p + 2q = -50 or 2p + q = -25 also 3³ + p(3)² + q(3) + 42 = 0 ∴ 9p + 3q = -69 or 3p + q = -23 Solving the two equations simultaneously, we get p = 2 and q = -27 Ans. Now the expression is x³ + 2x² - 27x + 42 But (x-2)(x-3) = x² - 5x + 6. The third factor is gotten by division as follows: x³ + 2x² - 27x + 42 | x² - 5x + 6 x³ - 5x² + 6x | x + 7 ∴ the third factor is (x+7) Ans. II ⟦line⟧ 7x² - 33x + 42 7x² - 35x + 42 ⟦line⟧ Method 2: By Division: x³ + px² + qx + 42 | x - 2 x³ - 2x² | x² + (p+2)x + (2p+q+4) ⟦line⟧ (p+2)x² + qx + 42 (p+2)x² - 2(p+2)x ⟦line⟧ [q + 2(p+2)]x + 42 or (2p + q + 4)x + 42 (2p + q + 4)x - 2(2p + q + 4) ⟦line⟧ Remainder = 4p + 2q + 50 = 0 or 2p + q = -25 Also by actually dividing the same expression by x-3, we get 3p + q = -23 which leads to the same solution. Q.E.D. 2 (i) 3√x-1 + √2x-3 = 2√x+2 or 3√x-1 = 2√x+2 - √2x-3 squaring 9(x-1) = 4(x+2) + (2x-3) - 4√(x+2)(2x-3) or 4√2x²+x-6 = 14 - 3x squaring again, 16(2x²+x-6) = 196 - 84x + 9x² or 23x² + 100x - 292 = 0 ∴ (23x + 146)(x - 2) = 0 ∴ x = 2 Ans. 1 or x = -146/23 = Ans. 2 But Ans 2 should be rejected as an extraneous root since it does not verify the original equation. 8

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### csp_66e48d9c019655b0b11d73c4cead3eeb

Solution to the 3rd + 4th Quarter Exam. in Algebra 7/4/1969

1. (i) The square root of (a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)² is equal: √[(a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)²] = √[(a³ - 1/a³) - 3(a - 1/a)]² = a³ - 1/a³ - 3(a - 1/a) = a³ - 3a + 3/a - 1/a³ also ³√[a³ - 3a + 3/a - 1/a³] = ³√(a - 1/a)³ = a - 1/a Ans.

(ii) Prove the identity: bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b) L.H.S. = bc(b-c) + ac² - a²c + a²b - ab² = bc(b-c) + a²(b-c) - a(b²-c²) = (b-c)[bc + a² - a(b+c)] = (b-c)(bc + a² - ab - ac) = (b-c)[a(a-b) - c(a-b)] = (b-c)(a-b)(a-c) = -(b-c)(c-a)(a-b) Q.E.D.

2. (i) solve: (x-1)/(√x - 1) = 3 + (√x + 1)/2 ∴ 2(x-1) = 6(√x - 1) + (√x - 1)(√x + 1) ∴ 2x - 2 = 6√x - 6 + x - 1 ∴ 6√x = x + 5 ∴ 36x = x² + 10x + 25 ∴ x² - 26x + 25 = 0 ∴ (x-1)(x-25) = 0 ∴ x = 1 and x = 25 } Ans. but x = 1 does not satisfy the original equation + should be rejected. Hence x = 25 Ans.

3 (i) x = ⁷√[(0.002001)³ (sin 16° 23')² / (1.003)⁵ (tan 41° 16')²] log 0.002001 = 3.3012 | 3 log 0.002001 = 9.9036 | 5 log 1.003 = 0.0060 log sin 16° 23' = 1.4504 | 2 log sin 16° 23' = 2.9008 | 2 log tan 41° 16' = 1.9466 log 1.003 = 0.0012 | log Num. = 10.8044 | log Den = 1.9526 log tan 41° 16' = 1.9433 | log Den. = 1.9526 | | 7 log x = 10.8518 | | log x = 2.7017 0.05031 | | x = 0.07975 | | or x = 7.975 X 10⁻² } Ans.

**Traduction anglaise —**

Solution to the 3rd + 4th Quarter Exam. in Algebra 7/4/1969 1. (i) The square root of (a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)² is equal: √[(a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)²] = √[(a³ - 1/a³) - 3(a - 1/a)]² = a³ - 1/a³ - 3(a - 1/a) = a³ - 3a + 3/a - 1/a³ also ³√[a³ - 3a + 3/a - 1/a³] = ³√(a - 1/a)³ = a - 1/a Ans. (ii) Prove the identity: bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b) L.H.S. = bc(b-c) + ac² - a²c + a²b - ab² = bc(b-c) + a²(b-c) - a(b²-c²) = (b-c)[bc + a² - a(b+c)] = (b-c)(bc + a² - ab - ac) = (b-c)[a(a-b) - c(a-b)] = (b-c)(a-b)(a-c) = -(b-c)(c-a)(a-b) Q.E.D. 2. (i) solve: (x-1)/(√x - 1) = 3 + (√x + 1)/2 ∴ 2(x-1) = 6(√x - 1) + (√x - 1)(√x + 1) ∴ 2x - 2 = 6√x - 6 + x - 1 ∴ 6√x = x + 5 ∴ 36x = x² + 10x + 25 ∴ x² - 26x + 25 = 0 ∴ (x-1)(x-25) = 0 ∴ x = 1 and x = 25 } Ans. but x = 1 does not satisfy the original equation + should be rejected. Hence x = 25 Ans. 3 (i) x = ⁷√[(0.002001)³ (sin 16° 23')² / (1.003)⁵ (tan 41° 16')²] log 0.002001 = 3.3012 | 3 log 0.002001 = 9.9036 | 5 log 1.003 = 0.0060 log sin 16° 23' = 1.4504 | 2 log sin 16° 23' = 2.9008 | 2 log tan 41° 16' = 1.9466 log 1.003 = 0.0012 | log Num. = 10.8044 | log Den = 1.9526 log tan 41° 16' = 1.9433 | log Den. = 1.9526 | | 7 log x = 10.8518 | | log x = 2.7017 0.05031 | | x = 0.07975 | | or x = 7.975 X 10⁻² } Ans.

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### csp_6745fb9cf0115c34aeb5ce8be00a166c

Final Exam. June, 1963 (cont'd.) - p. 2 - Algebra 4th Secondary . 9/6/63 -----

4. (a) Show that the sum of n terms of the series 1 + 1/3 + 1/9 + ..... is 1.5 - 1 / (2X3^(n-1)) How many terms of this series must be taken to make the sum equal to ( 3/2 - 1/13122 ) ? (10 marks)

(b) The first, second and fourth terms of an Arithmetical progression themselves form three successive terms of a Geometrical progression. Show that, if the common difference is not zero, it is equal to the first term. (10 marks).

5. Plot the graph of the function X² - 6X + 5 for values of X from -1 to 7 choosing ½ inch as one unit on the axis of X, and three tenths of an inch as one unit on the axis of y. From your graph, find: (a) the least value of the function. (b) the roots of the equation X² + 5 = 6X (c) by drawing another graph on the same figure, find the values of X between which the given function is less than (X-1). (d) find from the resulting figure the roots of the equation X² - 6X + 5 = X-1 (20 marks)

---- ⟦illegible carbon copy text repeating previous sections⟧

**Traduction anglaise —**

Final Exam. June, 1963 (cont'd.) - p. 2 - Algebra 4th Secondary . 9/6/63 ⟦line⟧ 4. (a) Show that the sum of n terms of the series 1 + 1/3 + 1/9 + ..... is 1.5 - 1 / (2X3^(n-1)) How many terms of this series must be taken to make the sum equal to ( 3/2 - 1/13122 ) ? (10 marks) (b) The first, second and fourth terms of an Arithmetical progression themselves form three successive terms of a Geometrical progression. Show that, if the common difference is not zero, it is equal to the first term. (10 marks). 5. Plot the graph of the function X² - 6X + 5 for values of X from -1 to 7 choosing ½ inch as one unit on the axis of X, and three tenths of an inch as one unit on the axis of y. From your graph, find: (a) the least value of the function. (b) the roots of the equation X² + 5 = 6X (c) by drawing another graph on the same figure, find the values of X between which the given function is less than (X-1). (d) find from the resulting figure the roots of the equation X² - 6X + 5 = X-1 (20 marks) ⟦line⟧ ⟦illegible⟧ carbon copy text repeating previous sections

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### csp_687103c842c65719973233c67da50e09

B01 · header / latin Final Examination in Algebra, ⟦illegible⟧ year 14/5/1968

B02 · paragraph / latin x⁹y³ - xy⁹ = x³(x⁶ - y⁶) - xy³(x⁶ - y⁶) (x³ + y³)(x³ - xy³) = (x³ + y³)(x² - x²y² + y⁴)(x - 2y)(x² + 2xy + 4y²) Ans.

B03 · paragraph / latin ⟦illegible⟧ ÷ ⟦illegible⟧ { ⟦illegible⟧ ÷ ⟦illegible⟧ } = ⟦illegible⟧ = ⟦illegible⟧ = x²/y³ Ans.

B04 · paragraph / latin 2x⁴ + 2x³ - 5x² - x + 3 + x + x + 3x² + B | x + x² - 2 ⟦line⟧ | 2x³ - x² + x + 3 Remainder = B + 6 = 0 ∴ B = -6 Ans.

B05 · table / latin x = ⁷√ (0.1023)³ cos 41° 28' / (1.007)⁴ tan 47° 51' log 0.1023 = 1.0098 | 3 log 0.1023 = 3.0294 log cos 41° 28' = 1.8747 | log cos 41° 28' = 1.7494 log 1.007 = 0.0029 | log Num. = 4.7788 log tan 47° 51' = 0.0433 | log Denom. = 0.2223 log x = 4.5565 log x = 1.50807 = 1.5081 x = 0.3222 Ans.

B06 · paragraph / latin (ii) (31.01)³ˣ = 104(2.003)²ˣ⁺¹ ∴ (3x-1) log 31.01 = log 104 + (2x+1) log 2.003 3x log 31.01 - log 31.01 = log 104 + 2x log 2.003 + log 2.003 x(3 log 31.01 - 2 log 2.003) = log 104 + log 2.003 + log 31.01 x = (log 104 + log 2.003 + log 31.01) / (3 log 31.01 - 2 log 2.003)

B07 · table / latin log 104 = 2.0170 | x = (2.0170 + 0.3016 + 1.4915) / (4.4745 - 0.6032) log 2.003 = 0.3016 | x = 3.8101 / 3.8713 = 0.98419 = 0.9842 log 31.01 = 1.4915

**Traduction anglaise —**

Final Examination in Algebra, ⟦illegible⟧ year 14/5/1968 x⁹y³ - xy⁹ = x³(x⁶ - y⁶) - xy³(x⁶ - y⁶) (x³ + y³)(x³ - xy³) = (x³ + y³)(x² - x²y² + y⁴)(x - 2y)(x² + 2xy + 4y²) Ans. ⟦illegible⟧ ÷ ⟦illegible⟧ { ⟦illegible⟧ ÷ ⟦illegible⟧ } = ⟦illegible⟧ = ⟦illegible⟧ = x²/y³ Ans. 2x⁴ + 2x³ - 5x² - x + 3 + x + x + 3x² + B | x + x² - 2 ⟦line⟧ | 2x³ - x² + x + 3 Remainder = B + 6 = 0 ∴ B = -6 Ans. x = ⁷√ (0.1023)³ cos 41° 28' / (1.007)⁴ tan 47° 51' log 0.1023 = 1.0098 | 3 log 0.1023 = 3.0294 log cos 41° 28' = 1.8747 | log cos 41° 28' = 1.7494 log 1.007 = 0.0029 | log Num. = 4.7788 log tan 47° 51' = 0.0433 | log Denom. = 0.2223 log x = 4.5565 log x = 1.50807 = 1.5081 x = 0.3222 Ans. (ii) (31.01)³ˣ = 104(2.003)²ˣ⁺¹ ∴ (3x-1) log 31.01 = log 104 + (2x+1) log 2.003 3x log 31.01 - log 31.01 = log 104 + 2x log 2.003 + log 2.003 x(3 log 31.01 - 2 log 2.003) = log 104 + log 2.003 + log 31.01 x = (log 104 + log 2.003 + log 31.01) / (3 log 31.01 - 2 log 2.003) log 104 = 2.0170 | x = (2.0170 + 0.3016 + 1.4915) / (4.4745 - 0.6032) log 2.003 = 0.3016 | x = 3.8101 / 3.8713 = 0.98419 = 0.9842 log 31.01 = 1.4915

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### csp_69417993aea2509b9a551e71ad485b3f

Let x be the amounted invested in £. 2 3/4% Stock purchased = £ (x * 100) / 95 3 1/2% Stock purchased = £ ( (100 x / 95) - 900).

Since his income remained the same (100 x / 95) * (2 3/4 / 100) = ( (100 x / 95) - 900) * (3 1/2 / 100) x / 95 * 11/4 = ( (x / 95) - 9) 7/2 = 7x / 2 * 95 - (7 * 9 * 100) / (2 * 95) ⟦line⟧ <del>⟦illegible⟧</del> <del>⟦illegible⟧</del> 7x / 2 * 95 - 11x / 4 * 95 = (7 * 900) / (2 * 95) 14x - 11x = 7 * 1800 3x = 7 * 1800 x = 7 * 600 = £ 4200

1.5 x = 7 * 855 x = (7 * 855 * 2) / 3 = £ 3990

[Marginalia] 285 [Marginalia] 14 [Marginalia] ⟦line⟧ [Marginalia] 1140 [Marginalia] 285 [Marginalia] ⟦line⟧ [Marginalia] 3990

⟦line⟧ ⟦line⟧ Now the correct clock would ⟦illegible⟧ after 120 ⟦illegible⟧ days of correct time The first clock will therefore read 1 hr ⟦illegible⟧ + 6 hours ⟦illegible⟧ The slower clock will read the

**Traduction anglaise —**

Let x be the amount invested in £. 2 3/4% Stock purchased = £ (x * 100) / 95 3 1/2% Stock purchased = £ ( (100 x / 95) - 900). Since his income remained the same (100 x / 95) * (2 3/4 / 100) = ( (100 x / 95) - 900) * (3 1/2 / 100) x / 95 * 11/4 = ( (x / 95) - 9) 7/2 = 7x / 2 * 95 - (7 * 9 * 100) / (2 * 95) ⟦line⟧ <del>⟦illegible⟧</del> <del>⟦illegible⟧</del> 7x / 2 * 95 - 11x / 4 * 95 = (7 * 900) / (2 * 95) 14x - 11x = 7 * 1800 3x = 7 * 1800 x = 7 * 600 = £ 4200 1.5 x = 7 * 855 x = (7 * 855 * 2) / 3 = £ 3990 285 14 ⟦line⟧ 1140 285 ⟦line⟧ 3990 ⟦line⟧ ⟦line⟧ Now the correct clock would ⟦illegible⟧ after 120 ⟦illegible⟧ days of correct time The first clock will therefore read 1 hr ⟦illegible⟧ + 6 hours ⟦illegible⟧ The slower clock will read the

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### csp_6b8db417dafe5019971dfbd16018c16d

B01 · paragraph / latin Let the 1st term be = a & the common difference be = d T₈ = a + 7d ∴ a + 7d = 6(a + 2d) ∴ a + 7d = 6a + 12d T₃ = a + 2d ∴ 5d = -5a ∴ d = -a ∴ Second term = a + d = a - a = 0 Ans.

B02 · paragraph / latin (ii) the 1ˢᵗ day the invalid walks 18 inches } and so on. " the 2nd " " " " 36 inches } ∴ we have a Geometric progression in which a = 18 inches , lₙ = 512 yards = 512 × 36 inches , and r = 2 Since lₙ = arⁿ⁻¹ ∴ 512 × 36 = 18 (2)ⁿ⁻¹ or 2ⁿ⁻¹ = ⟦512 × 36⟧ / 18 2ⁿ⁻¹ = 512 × 2 = 2¹⁰ ∴ 2ⁿ⁻¹ = 2¹⁰ ∴ n - 1 = 10 or n = 11 ∴ the invalid can take a walk of 512 yards on the 11ᵗʰ day (after 10 days) Ans.

**Traduction anglaise —**

Let the 1st term be = a & the common difference be = d T₈ = a + 7d ∴ a + 7d = 6(a + 2d) ∴ a + 7d = 6a + 12d T₃ = a + 2d ∴ 5d = -5a ∴ d = -a ∴ Second term = a + d = a - a = 0 Ans. (ii) the 1ˢᵗ day the invalid walks 18 inches } and so on. " the 2nd " " " " 36 inches } ∴ we have a Geometric progression in which a = 18 inches , lₙ = 512 yards = 512 × 36 inches , and r = 2 Since lₙ = arⁿ⁻¹ ∴ 512 × 36 = 18 (2)ⁿ⁻¹ or 2ⁿ⁻¹ = ⟦512 × 36⟧ / 18 2ⁿ⁻¹ = 512 × 2 = 2¹⁰ ∴ 2ⁿ⁻¹ = 2¹⁰ ∴ n - 1 = 10 or n = 11 ∴ the invalid can take a walk of 512 yards on the 11ᵗʰ day (after 10 days) Ans.

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### csp_6bf524b3d349512691b7176613af0473

⟦illegible⟧ Exam in Algebra September 1st, 1961 ⟦line⟧

4a²(a-1) - 9a + 9 = 4a²(a-1) - 9(a-1) = (a-1)(4a²-9) = (a-1)(2a-3)(2a+3) Ans. (ii) 7A - 3B / 5A + 6B = ? given B/A = 11 , 7A - 3B / 5A + 6B = 7 - 3(B/A) / 5 + 6(B/A) = 7 - 33 / 5 + 66 = -26 / 71 Ans. (iii) No. of apples = x - (2x/5 + x/6) = x - (12x + 5x / 30) is a fraction of the total = 30x - 17x / 30x = 13x / 30x = 13/30 Ans. ⟦line⟧

2. (i) Ax³ + Bx² + Cx + 2 = 0 , when x = 1, we get: A + B + C + 2 = 0 ... ① when x = -1 we get: -A + B - C + 2 = 0 ... ② when x = 2/3 we get: 8/27 A + 4/9 B + 2/3 C + 2 = 0 ... ③ from ① + ② we get: 2B + 4 = 0 or B = -2 Ans. I from ③ 8A + 12B + 18C + 54 = 0 or 4A + 6B + 9C + 27 = 0 ... ④ substitute B = -2 in equ. ② A + C = 0 ∴ A = -C substitute in eq. ④ we get -4C - 12 + 9C + 27 = 0 ∴ 5C = -15 ∴ C = -3 Ans. II ∴ A = -C = 3 Ans. III Hence A = 3, B = -2, C = -3 Ans. (ii) √Ay-1 / Ax+1 = y/x , A = ? ∴ Ay-1 / Ax+1 = y²/x² ∴ Ax²y + y² = Ax y² - x² ∴ Axy(x-y) = -x²-y² ∴ A = x²+y² / xy(y-x) Ans. ⟦line⟧

3. (i) (729)^(5/6) = (3⁶)^(5/6) = 3⁵ = 243 Ans. I (32/3125)^(-3/5) = (3125/32)^(3/5) = (5⁵/2⁵)^(3/5) = (5/2)³ = 125/8 = 15 5/8 = 15.625 Ans. II ⟦line⟧ (ii) ⁴√0.0001 = 10⁻⁴/⁴ = 10⁻¹ Ans. II ⟦line⟧ (iii) Let x = ⁷√(0.00561)² / 1.008 log 0.00561 = 3.7490 log 1.008 = 0.0033 2 log 0.00561 = 5.4980 log 1.008 = 0.0033 7 log x = 5.4947 log x = 1.35638... = 1.3564 Correct to 4 dec. pl. ∴ x = 0.2272 Ans.

**Traduction anglaise —**

⟦illegible⟧ Exam in Algebra September 1st, 1961 ⟦line⟧ 4a²(a-1) - 9a + 9 = 4a²(a-1) - 9(a-1) = (a-1)(4a²-9) = (a-1)(2a-3)(2a+3) Ans. (ii) 7A - 3B / 5A + 6B = ? given B/A = 11 , 7A - 3B / 5A + 6B = 7 - 3(B/A) / 5 + 6(B/A) = 7 - 33 / 5 + 66 = -26 / 71 Ans. (iii) No. of apples = x - (2x/5 + x/6) = x - (12x + 5x / 30) is a fraction of the total = 30x - 17x / 30x = 13x / 30x = 13/30 Ans. ⟦line⟧ 2. (i) Ax³ + Bx² + Cx + 2 = 0 , when x = 1, we get: A + B + C + 2 = 0 ... ① when x = -1 we get: -A + B - C + 2 = 0 ... ② when x = 2/3 we get: 8/27 A + 4/9 B + 2/3 C + 2 = 0 ... ③ from ① + ② we get: 2B + 4 = 0 or B = -2 Ans. I from ③ 8A + 12B + 18C + 54 = 0 or 4A + 6B + 9C + 27 = 0 ... ④ substitute B = -2 in equ. ② A + C = 0 ∴ A = -C substitute in eq. ④ we get -4C - 12 + 9C + 27 = 0 ∴ 5C = -15 ∴ C = -3 Ans. II ∴ A = -C = 3 Ans. III Hence A = 3, B = -2, C = -3 Ans. (ii) √Ay-1 / Ax+1 = y/x , A = ? ∴ Ay-1 / Ax+1 = y²/x² ∴ Ax²y + y² = Ax y² - x² ∴ Axy(x-y) = -x²-y² ∴ A = x²+y² / xy(y-x) Ans. ⟦line⟧ 3. (i) (729)^(5/6) = (3⁶)^(5/6) = 3⁵ = 243 Ans. I (32/3125)^(-3/5) = (3125/32)^(3/5) = (5⁵/2⁵)^(3/5) = (5/2)³ = 125/8 = 15 5/8 = 15.625 Ans. II ⟦line⟧ (ii) ⁴√0.0001 = 10⁻⁴/⁴ = 10⁻¹ Ans. II ⟦line⟧ (iii) Let x = ⁷√(0.00561)² / 1.008 log 0.00561 = 3.7490 log 1.008 = 0.0033 2 log 0.00561 = 5.4980 log 1.008 = 0.0033 7 log x = 5.4947 log x = 1.35638... = 1.3564 Correct to 4 dec. pl. ∴ x = 0.2272 Ans.

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### csp_71292120bc3e59b3b1ac2536d21a0909

[Marginalia] ⟦illegible⟧ [Marginalia] (3) [Marginalia] ⟦illegible⟧

original position of ladder A'B' later position

AA' = 2 ft find BB'

A A' 24' θ 52° B' B C

AC = 24 sin 52° = 24 X 0.7880 = 18.91 ft A'C = 18.91 - 2 = 16.91 ft sin θ = A'C / A'B' = 16.91 / 24 = 0.7046 ∠ θ = 44° 48' BC = 24 cos 52° = 24 X 0.6157 = 14.78 ft. B'C = 24 cos 44° 48' = 24 X 0.7096 = 17.03 ft. B'B = 17.03 - 14.78 = 2.25 ft

[Marginalia] 1 2 3 4 5 6 [Marginalia] 5 5 5 4 1

**Traduction anglaise —**

⟦illegible⟧ (3) ⟦illegible⟧ original position of ladder A'B' later position AA' = 2 ft find BB' A A' 24' θ 52° B' B C AC = 24 sin 52° = 24 X 0.7880 = 18.91 ft A'C = 18.91 - 2 = 16.91 ft sin θ = A'C / A'B' = 16.91 / 24 = 0.7046 ∠ θ = 44° 48' BC = 24 cos 52° = 24 X 0.6157 = 14.78 ft. B'C = 24 cos 44° 48' = 24 X 0.7096 = 17.03 ft. B'B = 17.03 - 14.78 = 2.25 ft 1 2 3 4 5 6 5 5 5 4 1

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### csp_71fbf3eff5a05e2ca2dbf06e864a4200

Shamash Secondary School 3rd Quarter Examination

Subject: Algebra Date: 17/3/1961. Class: 4th Secondary Time: 8:30-10:00 a.m.

----------------------- Attempt all Questions:

3. At what time between 12 o'clock and 1 o'clock are the two hands of a watch at right angles for the second time?

4. By lowering the price of eggs and selling them one penny per egg cheaper, a man finds that he can sell 5 more than he used to do for 5s. At what price per egg did he sell them at first?

5. A cistern can be filled by two pipes running together in 24 minutes. The larger pipe would fill the cistern in 20 minutes less than the smaller one. Find the time taken by each. -----------------------

**Traduction anglaise —**

Shamash Secondary School 3rd Quarter Examination Subject: Algebra Date: 17/3/1961. Class: 4th Secondary Time: 8:30-10:00 a.m. ⟦line⟧ Attempt all Questions: 1. A boy can swim at v m.p.h. in still water. When he swims downstream the current increases his speed by u m.p.h., and when he swims upstream the current decreases his speed by u m.p.h. The difference in his time to swim one mile downstream and one mile upstream is t hours. Find a formula for t in terms of v and u, and find v if u = 2, t = 4/5. 2. A dealer bought a horse, expecting to sell it again at a price that would have given him 10 per cent. profit on his purchase; but he had to sell it for £50 less than he expected, and he then found that he had lost 15 per cent. on what it cost him. What did he pay for the horse ? 3. At what time between 12 o'clock and 1 o'clock are the two hands of a watch at right angles for the second time? 4. By lowering the price of eggs and selling them one penny per egg cheaper, a man finds that he can sell 5 more than he used to do for 5s. At what price per egg did he sell them at first? 5. A cistern can be filled by two pipes running together in 24 minutes. The larger pipe would fill the cistern in 20 minutes less than the smaller one. Find the time taken by each. ⟦line⟧

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### csp_73394049a9585fba888391d2d02b784b

[Marginalia] Nour Basri 45

Shamash Secondary School Final Examination, May, 1966.

Subject: Algebra Date: 29/5/1966 Class: 4th Year, Scientific. Time: 8:00 - 11:00 a.m.

--- Answer all f i v e questions:

2. (a) Use the method of completing the square to show that the sum of the roots of the equation ax²+bx+c=0 is equal to (- b/a) and that their product is equal to ( c/a ). (10 marks) (b) Find the value of x from the following equation: 3.10²ˣ - 13.10ˣ + 4 = 0 (10 marks)

3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: log₃ (2x²-7x) = 2 (10 marks) (ii) Without using tables evaluate: (log₂ 9)(log₉ 32) (10 marks) (iii) Compute the value of y by logarithms, arranging your work neatly: y = ⁷√((tan 19°45')² x (cos 77°16')³) / ((3.004)⁵ x (50.06)³) (10 marks)

(cont'd.p.2)..

**Traduction anglaise —**

Nour Basri 45 Shamash Secondary School Final Examination, May, 1966. Subject: Algebra Date: 29/5/1966 Class: 4th Year, Scientific. Time: 8:00 - 11:00 a.m. ⟦line⟧ Answer all f i v e questions: 1. (a) Factor the following: (i) my⁴ + 16mx⁴ - 12mx²y² (4 marks) (ii) a²b²x² - a²b² - 2abx² + 2ab + x² - 1 (4 marks) (iii) 27x⁶y⁹ + 64y³ (4 marks) (b) Find the value of p and q which will make the expression 2x³ + px² + qx + 1 divisible by (x-1) and (x+1), and find the third factor. (8 marks) 2. (a) Use the method of completing the square to show that the sum of the roots of the equation ax²+bx+c=0 is equal to (- b/a) and that their product is equal to ( c/a ). (10 marks) (b) Find the value of x from the following equation: 3.10²ˣ - 13.10ˣ + 4 = 0 (10 marks) 3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: log₃ (2x²-7x) = 2 (10 marks) (ii) Without using tables evaluate: (log₂ 9)(log₉ 32) (10 marks) (iii) Compute the value of y by logarithms, arranging your work neatly: y = ⁷√((tan 19°45')² x (cos 77°16')³) / ((3.004)⁵ x (50.06)³) (10 marks) (cont'd.p.2)..

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### csp_7421ca98d5bf501ca345bc1d18b2ea8b

6. speed of train A = 22.5 mi/h = 22.5 x 22/15 ft/sec = 45/2 x 22/15 = 33 ft/sec. speed " " B = 15 mi/h = 15 x 22/15 ft/sec = 22 ft/sec. (i) when the two trains are travelling in opposite directions (see Fig. I), points C and D are separating at the rate of (33+22) ft/sec = 55 ft/sec. When the rear cars A and B just clear away from each other, (see Fig. II) points C and D have already separated by a distance = (240+200) ft = 440 ft. Time taken = total distance / rate of separation = 440 / 55 = 8 sec. Ans. 1

v₁ = 33 ft/sec. C A ————————→ 240 ft v₂ = 22 ft/sec Fig. I ←————— B 200 ft A ————————→ C D ←————— B Fig. II

(ii) When the two trains are travelling in the same direction, (See Fig. III), points C and D are separating at the rate of (33-22) ft/sec = 11 ft/sec. When the rear car A of the faster train and the front car D of the slower train just clear away from each other, (see Fig. IV), points C and D have already separated by a distance of 240 ft which is the length of the faster train.

240 ft A ————————→ C v₁ = 33 ft/sec 200 ft B —————→ D v₂ = 22 ft/sec Fig. III A 240 ft ———————→ C B 200 ft —————→ D Fig. IV

∴ Time taken = distance / rate of separation = 240 / 11 = 21 9/11 sec. Ans. 2

**Traduction anglaise —**

45 6. speed of train A = 22.5 mi/h = 22.5 x 22/15 ft/sec = 45/2 x 22/15 = 33 ft/sec. speed " " B = 15 mi/h = 15 x 22/15 ft/sec = 22 ft/sec. (i) when the two trains are travelling in opposite directions (see Fig. I), points C and D are separating at the rate of (33+22) ft/sec = 55 ft/sec. When the rear cars A and B just clear away from each other, (see Fig. II) points C and D have already separated by a distance = (240+200) ft = 440 ft. Time taken = total distance / rate of separation = 440 / 55 = 8 sec. Ans. 1 v₁ = 33 ft/sec. C A ⟦line⟧→ 240 ft v₂ = 22 ft/sec Fig. I ←⟦line⟧ B 200 ft A ⟦line⟧→ C D ←⟦line⟧ B Fig. II (ii) When the two trains are travelling in the same direction, (See Fig. III), points C and D are separating at the rate of (33-22) ft/sec = 11 ft/sec. When the rear car A of the faster train and the front car D of the slower train just clear away from each other, (see Fig. IV), points C and D have already separated by a distance of 240 ft which is the length of the faster train. 240 ft A ⟦line⟧→ C v₁ = 33 ft/sec 200 ft B ⟦line⟧→ D v₂ = 22 ft/sec Fig. III A 240 ft ⟦line⟧→ C B 200 ft ⟦line⟧→ D Fig. IV ∴ Time taken = distance / rate of separation = 240 / 11 = 21 9/11 sec. Ans. 2

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### csp_74d9118a7abd514ba577d1e35a899cb5

Shamash Secondary School 3rd Quarter + 4th Quarter Examination

Page. 1

Subject: Algebra Date: 7/4/1969 Class: 4th Year, scientific Time: 8:30 - 10:30 a.m.

Answer all questions:

1. (i) By first taking the square root and then the cube root, find the sixth root of: (a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)² (12 marks) (ii) Prove that the left hand side is always equal to the right hand side in the following equation: bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b) (13 marks)

2. (i) solve the equation : (x-1)/(√x -1) = 3 + (√x +1)/2 (12 marks) (ii) Rationalise the denominator and then find the value of: (√1+x + √1-x) / (√1+x - √1-x) , when x = 2b / (b²+1) (13 marks)

3. (i) Compute : ⁷√[ (6.002001)³ (sin 16° 23')² / (1.003)⁵ (tan 41° 16')² ] (12 marks) (ii) Solve the following equation for x: 2(log x)² - 5(log x) + 2 = 0 (13 marks)

4. (i) In an arithmetic Progression the first term is 3 and the common difference is 6 . Show that the sum of 2n terms is always equal to four times the sum of n terms . (12 marks) (ii) In an A.P. the ratio of the 3rd term to the 6th term is 11:26 and the sum of the first 4 terms is 34 . Find the progression and the sum of the first 8 terms . (13 marks)

**Traduction anglaise —**

Shamash Secondary School 3rd Quarter + 4th Quarter Examination Page. 1 Subject: Algebra Date: 7/4/1969 Class: 4th Year, scientific Time: 8:30 - 10:30 a.m. Answer all questions: 1. (i) By first taking the square root and then the cube root, find the sixth root of: (a³ - 1/a³)² - 6(a - 1/a)(a³ - 1/a³) + 9(a - 1/a)² (12 marks) (ii) Prove that the left hand side is always equal to the right hand side in the following equation: bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b) (13 marks) 2. (i) solve the equation : (x-1)/(√x -1) = 3 + (√x +1)/2 (12 marks) (ii) Rationalise the denominator and then find the value of: (√1+x + √1-x) / (√1+x - √1-x) , when x = 2b / (b²+1) (13 marks) 3. (i) Compute : ⁷√[ (6.002001)³ (sin 16° 23')² / (1.003)⁵ (tan 41° 16')² ] (12 marks) (ii) Solve the following equation for x: 2(log x)² - 5(log x) + 2 = 0 (13 marks) 4. (i) In an arithmetic Progression the first term is 3 and the common difference is 6 . Show that the sum of 2n terms is always equal to four times the sum of n terms . (12 marks) (ii) In an A.P. the ratio of the 3rd term to the 6th term is 11:26 and the sum of the first 4 terms is 34 . Find the progression and the sum of the first 8 terms . (13 marks)

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### csp_78792e8f6c2950e3babb727eb5e25d8b

B01 · header / latin Shamash Secondary School 1st. Quarter Exam. Subject : Arithmetic & Trigonometry Time : 12:00-1:30 p.m. Class : 4th year Secondary. Date : 8/12/1957 ⟦line⟧

B02 · paragraph / latin All questions are to be attempted.

B03 · paragraph / latin (1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m. by 25.00 m.

B04 · table / latin (2) (a) Decimalise to 3 places the following: £ 2 12s 2 3/4 d £ 8 10s 8 1/3 d £ 9 5s 10 3/4 d

B05 · paragraph / latin (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint.

B06 · paragraph / latin (3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ?

B07 · paragraph / latin (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together.

B08 · paragraph / latin (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation.

B09 · paragraph / latin (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

**Traduction anglaise —**

Shamash Secondary School 1st. Quarter Exam. Subject : Arithmetic & Trigonometry Time : 12:00-1:30 p.m. Class : 4th year Secondary. Date : 8/12/1957 ⟦line⟧ All questions are to be attempted. (1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m. by 25.00 m. (2) (a) Decimalise to 3 places the following: £ 2 12s 2 3/4 d £ 8 10s 8 1/3 d £ 9 5s 10 3/4 d (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint. (3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ? (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together. (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation. (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

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### csp_78dd28bf778952c08c309bd0862e4141

Shamash Secondary School 2nd Quarter Examination, December, 1966

Subject: Algebra Date: 27/12/1966 Class: 4th Scientific year Time: 11:00 - 12:30 <del>⟦illegible⟧</del>

All questions are to be attempted.

1. (a) Given that x = (3a² - 5b²) / (3a² + 5b²) ; make a, b respectively the subject of the formula. (9 marks) (b) If a/b = k, express (4a - 5b) / √(18a² - 4b²) in terms of k. (9 marks)

2. (a) Prove that x(y+z) + x/y + x/z is equal to x, if x = y / (y+1) and y = (z-2) / z (8 marks) (b) Solve the equation (1 - 0.4x) / (0.2 + x) = (0.7(x-1)) / (0.1 - 0.5x) (8 marks)

3. (a) Solve 5 / (6 - 5 / (6 - 5 / (6-x))) = x (8 marks) (b) Solve 3x³ + x² + 4 = 8x (10 marks) (c) Solve x²y² + 192 = 28xy ...... (1) x + y = 8 ...... (2) (10 marks)

4. A fast train travelling at x miles an hour takes t hours less to travel y miles than a slower one takes to travel z miles. Find the difference between their speeds in terms of (20 marks) x, t, y and z.

5. A man started for a walk when the hands of his watch were coincident between three and four o'clock. When he finished, the hands were again coincident between five & six o'clock. What was the time when he started, and how long did he walk? (20 marks)

**Traduction anglaise —**

Shamash Secondary School 2nd Quarter Examination, December, 1966 Subject: Algebra Date: 27/12/1966 Class: 4th Scientific year Time: 11:00 - 12:30 <del>⟦illegible⟧</del> All questions are to be attempted. 1. (a) Given that x = (3a² - 5b²) / (3a² + 5b²) ; make a, b respectively the subject of the formula. (9 marks) (b) If a/b = k, express (4a - 5b) / √(18a² - 4b²) in terms of k. (9 marks) 2. (a) Prove that x(y+z) + x/y + x/z is equal to x, if x = y / (y+1) and y = (z-2) / z (8 marks) (b) Solve the equation (1 - 0.4x) / (0.2 + x) = (0.7(x-1)) / (0.1 - 0.5x) (8 marks) 3. (a) Solve 5 / (6 - 5 / (6 - 5 / (6-x))) = x (8 marks) (b) Solve 3x³ + x² + 4 = 8x (10 marks) (c) Solve x²y² + 192 = 28xy ...... (1) x + y = 8 ...... (2) (10 marks) 4. A fast train travelling at x miles an hour takes t hours less to travel y miles than a slower one takes to travel z miles. Find the difference between their speeds in terms of (20 marks) x, t, y and z. 5. A man started for a walk when the hands of his watch were coincident between three and four o'clock. When he finished, the hands were again coincident between five & six o'clock. What was the time when he started, and how long did he walk? (20 marks)

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### csp_798523cd40c6545e9f8ea9453d7c70a3

⟦Delinquent⟧ Examination, September 1954 Algebra Date: 21st Sept., 1954 4th (Scientific) Time: 8:30 — 10:30 a.m.

All questions are to be attempted.

1. Find by how much (x+a)² + (x-a)² exceeds twice x². Hence find the difference between (4.3)² + (4.1)² and 2(4.2)².

2. Solve the equations: (i) 3x² + 1.7x - 2.6 = 0 (correct to two decimal places) (ii) 4x² - 8xy + 4y² - 3x + 3y - 1 = 0 ; x + 2y = 7.

3. Factorise: (i) 6x² + 7x - 20 (ii) (p² + pq + q²)² - (p² - pq + q²)² (iii) x³ - x² + 2x - 1

4. (i) What kind of series is 1/4, 3/10, 7/20, 2/5, ...? Find its nth term and the sum of the first ten terms. (ii) Three numbers in A.P. add up to 36. When they are increased by 1, 4, 43 respectively they form a G.P. What are the numbers?

5. A manufacturer produces a motor car at a cost of £ 440. He sells it to a dealer at a loss, and the dealer sells it to a customer for £ 480. Given that the dealer's percentage profit is double the manufacturer's percentage loss, find at what price the manufacturer sold the car to the dealer.

6. Draw the graph of y = 6 + 3x - x² for values of x from -2 to 5, taking 1 in. as unit on the x-axis and 1/2 in. as unit on the y-axis. (correct to one dec. pl.) From your graph find (i) the maximum value of y, and (ii) between what values of x the function is positive.

**Traduction anglaise —**

⟦Delinquent⟧ Examination, September 1954 Algebra Date: 21st Sept., 1954 4th (Scientific) Time: 8:30 — 10:30 a.m. All questions are to be attempted. 1. Find by how much (x+a)² + (x-a)² exceeds twice x². Hence find the difference between (4.3)² + (4.1)² and 2(4.2)². 2. Solve the equations: (i) 3x² + 1.7x - 2.6 = 0 (correct to two decimal places) (ii) 4x² - 8xy + 4y² - 3x + 3y - 1 = 0 ; x + 2y = 7. 3. Factorise: (i) 6x² + 7x - 20 (ii) (p² + pq + q²)² - (p² - pq + q²)² (iii) x³ - x² + 2x - 1 4. (i) What kind of series is 1/4, 3/10, 7/20, 2/5, ...? Find its nth term and the sum of the first ten terms. (ii) Three numbers in A.P. add up to 36. When they are increased by 1, 4, 43 respectively they form a G.P. What are the numbers? 5. A manufacturer produces a motor car at a cost of £ 440. He sells it to a dealer at a loss, and the dealer sells it to a customer for £ 480. Given that the dealer's percentage profit is double the manufacturer's percentage loss, find at what price the manufacturer sold the car to the dealer. 6. Draw the graph of y = 6 + 3x - x² for values of x from -2 to 5, taking 1 in. as unit on the x-axis and 1/2 in. as unit on the y-axis. (correct to one dec. pl.) From your graph find (i) the maximum value of y, and (ii) between what values of x the function is positive.

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### csp_7a09a3d86d9e57c88e3cd5b247072285

Solutions to Mid-year Exam., in Algebra 4th Secondary year, February, 1969. 1

1. Let the time now be x minutes after 5 o'clock. then x = 25 + x/12 + 30 (ACB = 30) ∴ x - x/12 = 55 ∴ 11x/12 = 55 ∴ x = 60 minutes after 5 ∴ the time now is exactly 6 o'clock Ans.

⟦Diagram of a clock with labels: B at 12, A at 6, C at 9, and an arrow at 5 labeled x min.⟧

2. (i) solve 6 x³+19 x²+x-6 = 0 By trial + error we discover that x = -3 satisfies the equation. Hence by the remainder + factor theorems, (x+3) is a factor. Factoring, we get (x+3) (6x² + x - 2) = 0 or (x+3) (3x+2) (2x-1) = 0 ∴ x = -3 x = - 2/3 } Ans. x = 1/2

(ii) (3x-7)/(x-2) + (2x-5)/(x-3) = (3x+7)/(x+2) + (2x+5)/(x+3) ∴ (3(x-2)-1)/(x-2) + (2(x-3)+1)/(x-3) = (3(x+2)+1)/(x+2) + (2(x+3)-1)/(x+3) ∴ 3 - 1/(x-2) + 2 + 1/(x-3) = 3 + 1/(x+2) + 2 - 1/(x+3) 1/(x-3) - 1/(x-2) = 1/(x+2) - 1/(x+3) ∴ (x-2-x+3)/((x-2)(x-3)) = (x+3-x-2)/((x+2)(x+3)) ∴ 1/((x-2)(x-3)) = 1/((x+2)(x+3)) ∴ (x+2)(x+3) = (x-2)(x-3) ∴ x²+5x+6 = x²-5x+6 ∴ 10x = 0 ∴ x = 0 Ans.

**Traduction anglaise —**

Solutions to Mid-year Exam., in Algebra 4th Secondary year, February, 1969. 1 1. Let the time now be x minutes after 5 o'clock. then x = 25 + x/12 + 30 (ACB = 30) ∴ x - x/12 = 55 ∴ 11x/12 = 55 ∴ x = 60 minutes after 5 ∴ the time now is exactly 6 o'clock Ans. ⟦Diagram of a clock with labels: B at 12, A at 6, C at 9, and an arrow at 5 labeled x min.⟧ 2. (i) solve 6 x³+19 x²+x-6 = 0 By trial + error we discover that x = -3 satisfies the equation. Hence by the remainder + factor theorems, (x+3) is a factor. Factoring, we get (x+3) (6x² + x - 2) = 0 or (x+3) (3x+2) (2x-1) = 0 ∴ x = -3 x = - 2/3 } Ans. x = 1/2 (ii) (3x-7)/(x-2) + (2x-5)/(x-3) = (3x+7)/(x+2) + (2x+5)/(x+3) ∴ (3(x-2)-1)/(x-2) + (2(x-3)+1)/(x-3) = (3(x+2)+1)/(x+2) + (2(x+3)-1)/(x+3) ∴ 3 - 1/(x-2) + 2 + 1/(x-3) = 3 + 1/(x+2) + 2 - 1/(x+3) 1/(x-3) - 1/(x-2) = 1/(x+2) - 1/(x+3) ∴ (x-2-x+3)/((x-2)(x-3)) = (x+3-x-2)/((x+2)(x+3)) ∴ 1/((x-2)(x-3)) = 1/((x+2)(x+3)) ∴ (x+2)(x+3) = (x-2)(x-3) ∴ x²+5x+6 = x²-5x+6 ∴ 10x = 0 ∴ x = 0 Ans.

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### csp_7b93f90f7cee54f981ff7c7c630c01d3

- p.2 - Shamash Secondary School Cond. Exam. September, 1963. Algebra 12/9/63 4th Year Secondary. -----

6. Draw the graph of Y = ½X³ for values of X between -3 and 4 taking ½ inch to represent one unit on the X-axis and two tenths of an inch to represent one unit on the Y-axis. By drawing other graphs on the same figure, solve the equations:

(i) ½X³ - 7/2 X - 3 = 0 (6 marks) (ii) ½X³ + 3/2 X - 2 = 0 (6 marks) (iii) From the graph find the range of values of X for which ½X³ is greater than 7/2 X + 3. (6 marks).

----

**Traduction anglaise —**

- p.2 - Shamash Secondary School Cond. Exam. September, 1963. Algebra 12/9/63 4th Year Secondary. ⟦line⟧ 6. Draw the graph of Y = ½X³ for values of X between -3 and 4 taking ½ inch to represent one unit on the X-axis and two tenths of an inch to represent one unit on the Y-axis. By drawing other graphs on the same figure, solve the equations: (i) ½X³ - 7/2 X - 3 = 0 (6 marks) (ii) ½X³ + 3/2 X - 2 = 0 (6 marks) (iii) From the graph find the range of values of X for which ½X³ is greater than 7/2 X + 3. (6 marks). ⟦line⟧

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### csp_7bd023af05fd57879198e94ce3181786

B01 · header / latin Solutions to 4th Quarter Exam. May 1967. 4th year scientific

B03 · paragraph / latin x = 9√((0.4007)^3 * tan^2 37° 19' / (50.72^5 * cos^3 14° 34'))

B04 · paragraph / latin 2. (i) simplify: (x^(1 + p/q))^(p/(p+q)) ÷ p√(x^(2p) / (x^-1)^-p) = x^(p/(p+q) + q/(p+q)) ÷ p√(x^(2p) * x^-p) = x^((p+q)/(p+q)) ÷ x^(p/p) = x ÷ x = 1 Ans.

**Traduction anglaise —**

Solutions to 4th Quarter Exam. May 1967. 4th year scientific log 0.4007 = 1.6029 | 3 log 0.4007 = 2.8087 log tan 37° 19' = 1.8821 | 2 log tan 37° 19' = 1.7642 log 50.72 = 1.7052 | log Num. = 2.5729 log cos 14° 34' = 1.9858 | log Den. = 8.4834 ⟦line⟧ 5 log 50.72 = 8.5260 | 9 log x = 10.0895 3 log cos 14° 34' = 1.9574 | log x = 2.8988 log Den. = 8.4834 | x = 0.07921 | = 7.921 x 10^-2 Ans. x = 9√((0.4007)^3 * tan^2 37° 19' / (50.72^5 * cos^3 14° 34')) 2. (i) simplify: (x^(1 + p/q))^(p/(p+q)) ÷ p√(x^(2p) / (x^-1)^-p) = x^(p/(p+q) + q/(p+q)) ÷ p√(x^(2p) * x^-p) = x^((p+q)/(p+q)) ÷ x^(p/p) = x ÷ x = 1 Ans.

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### csp_7c5cd0bc9621561a8007061bcd8de7fb

[Marginalia] الرقم : [Marginalia] الاسم :

Monthly Examination, August 1965

Subject: General Mathematics Date: 19/8/1965 Class: 4th Year Secondary Time: 7:30-8:15 a.m.

--- I. Give the English Equivalent of the following:

٦ـ المقسوم عليه ١ـ ارقام ٧ـ ناتج القسمة ٢ـ مراتب ٨ـ المقسوم ٣ـ الطرح ٩ـ باقي القسمة ٤ـ العوامل ١٠ـ مضاعف ٥ـ أس القوة

١١ـ اعداد زوجية متتالية ١٢ـ اعداد فردية متتالية ١٣ـ اعداد اولية ١٤ـ الجزء الصحيح من العدد ١٥ـ المقام المشترك الاصغر ١٦ـ كسر لفظي ١٧ـ مقلوب العدد ١٨ـ الكسور العشرية المنتهية ١٩ـ الكسور العشرية الدورية ٢٠ـ بسط الكسر ٢١ـ مقام الكسر ٢٢ـ الخطأ المئوي ٢٣ـ النسبة والتناسب ٢٤ـ الوسط المتناسب بين عددين ٢٥ـ ربح المساهم (ربح حامل الاسهم) ٢٦ـ البديهية ٢٧ـ الموضوعة ٢٨ـ زاوية حادة ٢٩ـ زاوية منفرجة ٣٠ـ زاوية منعكسة ٣١ـ قطعة دائرة ٣٢ـ قطاع دائرة ٣٣ـ المعاليم ٣٤ـ المجاهيل ٣٥ـ زاويتان متتامتان

ـ يتبع ـ

**Traduction anglaise —**

Number: Name: Monthly Examination, August 1965 Subject: General Mathematics Date: 19/8/1965 Class: 4th Year Secondary Time: 7:30-8:15 a.m. ⟦line⟧ I. Give the English Equivalent of the following: 6. Divisor | 1. Numbers 7. Quotient | 2. Places (Digits) 8. Dividend | 3. Subtraction 9. Remainder | 4. Factors 10. Multiple | 5. Exponent (Power) 11. Consecutive even numbers 12. Consecutive odd numbers 13. Prime numbers 14. Integer part of a number 15. Lowest Common Denominator 16. Verbal fraction 17. Reciprocal of a number 18. Terminating decimals 19. Recurring (Periodic) decimals 20. Numerator of a fraction 21. Denominator of a fraction 22. Percentage error 23. Ratio and Proportion 24. Geometric mean between two numbers 25. Shareholder's profit (Dividend) 26. Axiom 27. Postulate 28. Acute angle 29. Obtuse angle 30. Reflex angle 31. Segment of a circle 32. Sector of a circle 33. Knowns 34. Unknowns 35. Complementary angles - To be continued -

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### csp_7c5f4a1860615dc59109d86b6f66e49b

Shamash Secondary School Final Examination May 1969

Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m.

--- Answer all questions:

[Marginalia] (log x)² = log x⁷ - 10, finding two

2. (i) Solve the equation (log x)² = log x⟦⁷⟧ - 10, ⟦finding⟧ two values for x. (10 marks) (ii) Solve the two simultaneous equations: 3x² +xy-2y² +7=0 ....... (1) x² -xy+y² -7=0 ......... (2) (10 marks)

3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√8)¹⁻ˣ = (√2)ˣ (√50) (10 marks) (ii) Compute the value of ⁷√((0.5002)² sin³ 14° 25') / ((4.003)³ cos² 15° 27') (10 marks)

4. (i) If (b+c)⁻¹, (c+a)⁻¹, (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks)

[Marginalia] bounce

(ii) A bouncing tennis ball rebounds each time to a height one half the height of the previous ⟦bounce⟧. If it is dropped from a height of 10 ft., show: (a) that the total distance it has travelled when it hits the ground for the 10th time is equal to 29 123/128 ft. (b) Show also that the total distance it travels before coming to rest is 30 ft. (10 marks)

(cont'd.p.2)..

[Marginalia] (5)(4³ˣ⁻¹)(√8)¹⁻ˣ = (√2)ˣ (√50)

**Traduction anglaise —**

Shamash Secondary School Final Examination May 1969 Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m. ⟦line⟧ Answer all questions: 1. (i) In the expression x³+Ax²+31x+B, A and B are constant. Find the values of A and B which will make this expression divisible by (x-2) (x-3) and find the remaining factor. (10 marks) (ii) The wages of 12 men and 7 boys amount to £9 13s. If 3 men together receive 8s. more than 4 boys, what are the wages of each man and boy ? (10 marks) (log x)² = log x⁷ - 10, finding two 2. (i) Solve the equation (log x)² = log x⟦⁷⟧ - 10, ⟦finding⟧ two values for x. (10 marks) (ii) Solve the two simultaneous equations: 3x² +xy-2y² +7=0 ⟦line⟧ (1) x² -xy+y² -7=0 ⟦line⟧ (2) (10 marks) 3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√8)¹⁻ˣ = (√2)ˣ (√50) (10 marks) (ii) Compute the value of ⁷√((0.5002)² sin³ 14° 25') / ((4.003)³ cos² 15° 27') (10 marks) 4. (i) If (b+c)⁻¹, (c+a)⁻¹, (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks) bounce (ii) A bouncing tennis ball rebounds each time to a height one half the height of the previous ⟦bounce⟧. If it is dropped from a height of 10 ft., show: (a) that the total distance it has travelled when it hits the ground for the 10th time is equal to 29 123/128 ft. (b) Show also that the total distance it travels before coming to rest is 30 ft. (10 marks) (cont'd.p.2).. (5)(4³ˣ⁻¹)(√8)¹⁻ˣ = (√2)ˣ (√50)

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### csp_7ca433875d775fae9b4d580ca08839f4

Class: 4th year Secondary. 1st Quarter Exam. Subject: Arithmetic + Trigonometry January 25, 1957 TIME ALLOWED: 90 MINUTES

Attempt six questions only:

(1) a) Define sin A, cos A and tan A where A is an acute angle. b) Prove sin A = cos (90º-A) c) Find from first principles sin 30º and tan 45º.

(2) a) Find from tables sin 55º 57' and cos 60º 57' b) Given tan A = 2.2372 find A from tables.

(3) A B C D is a rectangle in which A B is 15 in. and A D is 8 in. Calculate the angle B A C. The rectangle is held in a vertical plane with A B inclined to the horizontal at 40º and with B C D lower than A. Calculate the depth of B below A.

(4) a) Divide £117 5s 3d by 49. b) Express the following as decimals of a pound

12s 3d 18s 9d 9s 8d

(5) The area of a rectangular field is 2 acres and its breadth is 88 yds. Find the perimeter of the field and the length of a diagonal correct to the nearest yard.

(6) What is the correct time when the two hands of a clock coincide between 8 and 9 o'clock.

(7) a) A class of 30 students; 12 of whom weigh 9 st. 2 lb each, 8 weigh 8 st. 4 lb. each, 7 weigh 8 st. 1 lb. each and the rest 10 st. 4 lb, 10 st. 6 lb. and 10 st. respectively. Find the average weight of a student to the nearest lb. b) Define the gallon and state how many pints it contains.

**Traduction anglaise —**

Class: 4th year Secondary. 1st Quarter Exam. Subject: Arithmetic + Trigonometry January 25, 1957 TIME ALLOWED: 90 MINUTES Attempt six questions only: (1) a) Define sin A, cos A and tan A where A is an acute angle. b) Prove sin A = cos (90º-A) c) Find from first principles sin 30º and tan 45º. (2) a) Find from tables sin 55º 57' and cos 60º 57' b) Given tan A = 2.2372 find A from tables. (3) A B C D is a rectangle in which A B is 15 in. and A D is 8 in. Calculate the angle B A C. The rectangle is held in a vertical plane with A B inclined to the horizontal at 40º and with B C D lower than A. Calculate the depth of B below A. (4) a) Divide £117 5s 3d by 49. b) Express the following as decimals of a pound 12s 3d 18s 9d 9s 8d (5) The area of a rectangular field is 2 acres and its breadth is 88 yds. Find the perimeter of the field and the length of a diagonal correct to the nearest yard. (6) What is the correct time when the two hands of a clock coincide between 8 and 9 o'clock. (7) a) A class of 30 students; 12 of whom weigh 9 st. 2 lb each, 8 weigh 8 st. 4 lb. each, 7 weigh 8 st. 1 lb. each and the rest 10 st. 4 lb, 10 st. 6 lb. and 10 st. respectively. Find the average weight of a student to the nearest lb. b) Define the gallon and state how many pints it contains.

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### csp_7cd57c423c9e54de8088ce70e9dc76fb

SHAMASH SECONDARY SCHOOL Final Examination, May 1966

Subject: Arithmetic & Trigonometry Date: 18.5.1966 Class: 4th Year Secondary Time: 8:00-10:30 a.m.

- - - Attempt five questions only including question (4).

1. A person, having bought a certain amount of 2¾% stock at 95, afterwards sold it, and with the proceeds bought 3½% stock. He obtained £900 less stock than before, but his income was unchanged. How much money did he originally invest? (20 marks)

2. A house holder owns his house which has a rateable value of £44 on which the annual rates are charged at 21s 10d in the £1. He also has to pay an annual property tax at the rate of 9s in the £ on an assessment of £44. Calculate, correct to the nearest penny, the average cost per week of the total of these charges, taking a year as 52 weeks. He subsequently sells his house for £3300, which sum he invests at the rate of 2½% per annum free of tax, and moves into a flat which he rents at £126 per annum. He has however to rent a garage for his car at 7s 6d per week. Find how much per annum he saves by the change. (20 marks)

3. (a) A watch was 5 minutes fast at 9 a.m. on Monday, and 10 minutes slow at 12 noon on the following Wednesday. Find when it was exactly right, assuming that it lost time uniformly. Note: (9 a.m. and 12 noon are correct time). (10 marks) (b) Two clocks sound the first stroke of 12 o'clock at the same instant; one clock allows an interval of 20 secs. between each stroke and the next, and the other allows 25 secs. How many strokes of the slower clock remain after the quicker one has finished striking, and what time will elapse between the 12th stroke of the quicker one and the following stroke of the slower one? (10 marks)

4. (a) A solid consists of a hemisphere, radius 8 cm., joined to a cone of the same base-radius and height 6 cm., so that the plane surfaces coincide. Find (i) the volume, (ii) the total area of the surface of the solid. (Give answer to 3 significant figures). (10 marks) (b) A sphere of radius 3 in. is filed down into the greatest possible cube; find the volume of the material removed. (Give answer to 4 significant figures). (10 marks)

5. Find the difference between the perimeters of a regular pentagon and a regular hexagon, each of which has an area of 24 square inches. (20 marks)

6. In response to an S O S call from a ship at A, another ship at B, 175 miles due east of A, starts toward A at a speed of 12 miles per hour. At the same time a third ship at C, which is 186 miles from B in a direction bearing ⟦30⟧° 15' west of north, also starts for A at a speed of 16 miles per hour. Which ship will reach A first, and how long will it take ? (20 marks) --------------------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Final Examination, May 1966 Subject: Arithmetic & Trigonometry Date: 18.5.1966 Class: 4th Year Secondary Time: 8:00-10:30 a.m. ⟦line⟧ Attempt five questions only including question (4). 1. A person, having bought a certain amount of 2¾% stock at 95, afterwards sold it, and with the proceeds bought 3½% stock. He obtained £900 less stock than before, but his income was unchanged. How much money did he originally invest? (20 marks) 2. A house holder owns his house which has a rateable value of £44 on which the annual rates are charged at 21s 10d in the £1. He also has to pay an annual property tax at the rate of 9s in the £ on an assessment of £44. Calculate, correct to the nearest penny, the average cost per week of the total of these charges, taking a year as 52 weeks. He subsequently sells his house for £3300, which sum he invests at the rate of 2½% per annum free of tax, and moves into a flat which he rents at £126 per annum. He has however to rent a garage for his car at 7s 6d per week. Find how much per annum he saves by the change. (20 marks) 3. (a) A watch was 5 minutes fast at 9 a.m. on Monday, and 10 minutes slow at 12 noon on the following Wednesday. Find when it was exactly right, assuming that it lost time uniformly. Note: (9 a.m. and 12 noon are correct time). (10 marks) (b) Two clocks sound the first stroke of 12 o'clock at the same instant; one clock allows an interval of 20 secs. between each stroke and the next, and the other allows 25 secs. How many strokes of the slower clock remain after the quicker one has finished striking, and what time will elapse between the 12th stroke of the quicker one and the following stroke of the slower one? (10 marks) 4. (a) A solid consists of a hemisphere, radius 8 cm., joined to a cone of the same base-radius and height 6 cm., so that the plane surfaces coincide. Find (i) the volume, (ii) the total area of the surface of the solid. (Give answer to 3 significant figures). (10 marks) (b) A sphere of radius 3 in. is filed down into the greatest possible cube; find the volume of the material removed. (Give answer to 4 significant figures). (10 marks) 5. Find the difference between the perimeters of a regular pentagon and a regular hexagon, each of which has an area of 24 square inches. (20 marks) 6. In response to an S O S call from a ship at A, another ship at B, 175 miles due east of A, starts toward A at a speed of 12 miles per hour. At the same time a third ship at C, which is 186 miles from B in a direction bearing ⟦30⟧° 15' west of north, also starts for A at a speed of 16 miles per hour. Which ship will reach A first, and how long will it take ? (20 marks) ⟦line⟧

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### csp_7ceea579fbf9599d9ad07394b2828f0b

Shamash Secondary School Conditional Examination, Sept. 1962.

Subject: Mathematics Date: 14/9/1962 Class: 4th Year Secondary Time: 8:00-10:30

--- Attempt all questions:

1. (i) Solve the following simultaneous equations for x and y : ax + by = c ...... (1) x/a + y/b = 1 ....... (2) (10 marks). (ii) Solve the equation 4x² - 12x + 3 = 0, giving the answer correct to two decimal places. (10 marks).

2. (i) Use logarithms to calculate by the shortest possible way, the value of √ x² + x, when x = 4.836. (10 marks) (ii) When (1-2x + x²) is multiplied by (1-kx+x²) the coefficient of x² is zero. Find the value of k. (10 marks)

3. (i) Find the 50th term and the sum of the first 100 terms of the arithmetical progression whose first term is 20 and whose 3rd term is 21. (10 marks). (ii) Find the first and sixth terms of a geometrical progression whose common ratio is ½ when the first four terms add up to 18¾. (10 marks).

4. Draw the graph of x²-2x between x =-2 and x=4. From your graph solve approximately the equation x²-2x = 1. With the help of a further graph, solve approximately the equation x²-2x = x + 1. (20 marks).

--------

⟦illegible blue ink bleed-through from reverse side⟧

**Traduction anglaise —**

Shamash Secondary School Conditional Examination, Sept. 1962. Subject: Mathematics Date: 14/9/1962 Class: 4th Year Secondary Time: 8:00-10:30 ⟦line⟧ Attempt all questions: 1. (i) Solve the following simultaneous equations for x and y : ax + by = c ...... (1) x/a + y/b = 1 ....... (2) (10 marks). (ii) Solve the equation 4x² - 12x + 3 = 0, giving the answer correct to two decimal places. (10 marks). 2. (i) Use logarithms to calculate by the shortest possible way, the value of √ x² + x, when x = 4.836. (10 marks) (ii) When (1-2x + x²) is multiplied by (1-kx+x²) the coefficient of x² is zero. Find the value of k. (10 marks) 3. (i) Find the 50th term and the sum of the first 100 terms of the arithmetical progression whose first term is 20 and whose 3rd term is 21. (10 marks). (ii) Find the first and sixth terms of a geometrical progression whose common ratio is ½ when the first four terms add up to 18¾. (10 marks). 4. The cost of turfing a piece of lawn for a tennis court at 10 shillings per square yard is £ 12 less than 12 times the cost of fencing it all round at 5 shillings per yard. If the lawn had been 12 ft. longer it would have been twice as long as it is wide. Find the dimensions of the lawn in yards. (20 marks). 4. Draw the graph of x²-2x between x =-2 and x=4. From your graph solve approximately the equation x²-2x = 1. With the help of a further graph, solve approximately the equation x²-2x = x + 1. (20 marks). ⟦line⟧ ⟦illegible blue ink bleed-through from reverse side⟧

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### csp_7d223f288ea754439b76f0997226acbb

B01 · header / latin 4th year Final Exam 18/5/1961 Algebra.

B02 · table / latin x | y₁ | y₂ | y₃ -2 | -8 | -4 | -8 -1 | -1 | -1 | -3 0 | 0 | 2 | 2 1 | 1 | 5 | ⟦illegible⟧ 2 | 8 | 8 | 12 3 | 27

B03 · paragraph / latin (6 marks) (i) The curve y = x³ is plotted as shown below (7 marks) (ii) From the equation x³ = 3x + 2 we get the same roots for x as the roots obtained from solving the two simultaneous equations: y₁ = x³ and y₂ = 3x + 2 the graph of y₂ = 3x + 2 is the st. line shown in the diagram. This line touches the curve y₁ = x³ at B(-1, -1) and intersects it at C(2, 8) ∴ the roots of the original equation are x = -1 and x = 2 Ans.

B04 · other / latin y₁ = x³ y₃ = 5x + 2 y₂ = 3x + 2 D(2.4, 14) C(2, 8) B(-1, -1) E(-0.4, -0.05) A(-2, -8)

B05 · paragraph / latin Again the roots of the equation x³ - 5x - 2 = 0 are the same as the roots of the two simultaneous equations y₁ = x³ and y₃ = 5x + 2 since x³ = 5x + 2 Thus the graph of y₃ = 5x + 2 is another st. line which intersects the curve y₁ = x³ at A(-2, -8) and at D(2.4, 14) and at E(-0.4, -0.05) ∴ the roots of x³ - 5x - 2 = 0 are x = -2 and x = 2.4 x = -0.4 Ans.

B06 · paragraph / latin (7 marks) (iii) From the diagram, since the st. line y₂ = 3x + 2 lies above the curve y₁ = x³ between x = -1 and x = 2 and also between x = -∞ and x = -1 ∴ 3x + 2 > x³ when x < -1 and also when -1 < x < 2 Ans.

**Traduction anglaise —**

4th year Final Exam 18/5/1961 Algebra. x | y₁ | y₂ | y₃ -2 | -8 | -4 | -8 -1 | -1 | -1 | -3 0 | 0 | 2 | 2 1 | 1 | 5 | ⟦illegible⟧ 2 | 8 | 8 | 12 3 | 27 | | (6 marks) (i) The curve y = x³ is plotted as shown below (7 marks) (ii) From the equation x³ = 3x + 2 we get the same roots for x as the roots obtained from solving the two simultaneous equations: y₁ = x³ and y₂ = 3x + 2 the graph of y₂ = 3x + 2 is the st. line shown in the diagram. This line touches the curve y₁ = x³ at B(-1, -1) and intersects it at C(2, 8) ∴ the roots of the original equation are x = -1 and x = 2 Ans. y₁ = x³ y₃ = 5x + 2 y₂ = 3x + 2 D(2.4, 14) C(2, 8) B(-1, -1) E(-0.4, -0.05) A(-2, -8) Again the roots of the equation x³ - 5x - 2 = 0 are the same as the roots of the two simultaneous equations y₁ = x³ and y₃ = 5x + 2 since x³ = 5x + 2 Thus the graph of y₃ = 5x + 2 is another st. line which intersects the curve y₁ = x³ at A(-2, -8) and at D(2.4, 14) and at E(-0.4, -0.05) ∴ the roots of x³ - 5x - 2 = 0 are x = -2 and x = 2.4 x = -0.4 Ans. (7 marks) (iii) From the diagram, since the st. line y₂ = 3x + 2 lies above the curve y₁ = x³ between x = -1 and x = 2 and also between x = -∞ and x = -1 ∴ 3x + 2 > x³ when x < -1 and also when -1 < x < 2 Ans.

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### csp_7df63fd1a88c568dac589382988f01ab

B01 · header / latin 1st Quarter Exam.

B02 · form / latin Subject: Geometry Date: 30/11/1958 Class: 4th Secondary Time: 90 minutes.

B03 · other / latin ⟦line⟧

B04 · paragraph / latin Attempt all questions.

B05 · paragraph / latin 1. Draw a triangle given two angles and the perimeter. Prove your construction. (30 marks).

B06 · paragraph / latin 2. Construct the quadrilateral A B C D given that AB=AD= 8cm. BC=CD = 6 cm. and the angle ABC = 75°. Construct the point ⟦K⟧ on AB produced such that triangle A⟦P⟧D is equal in area to the quadrilateral ABCD. Measure A⟦P⟧. (35 marks)

B07 · paragraph / latin 3. ABC is a triangle. Y & Z are the mid-points of AC and AB respectively. AC is produced to D so that CD = AY. DP is drawn parallel to BA to meet BC and ZY (both produced) at P and W respectively. Show that the triangles DCP and AYZ are congruent. Calculate the ratio of the area of the parallelogram ZBPW to that of the triangle ABC. (35 marks).

B08 · footer / latin ⟦line⟧

**Traduction anglaise —**

1st Quarter Exam. Subject: Geometry Date: 30/11/1958 Class: 4th Secondary Time: 90 minutes. ⟦line⟧ Attempt all questions. 1. Draw a triangle given two angles and the perimeter. Prove your construction. (30 marks). 2. Construct the quadrilateral A B C D given that AB=AD= 8cm. BC=CD = 6 cm. and the angle ABC = 75°. Construct the point ⟦K⟧ on AB produced such that triangle A⟦P⟧D is equal in area to the quadrilateral ABCD. Measure A⟦P⟧. (35 marks) 3. ABC is a triangle. Y & Z are the mid-points of AC and AB respectively. AC is produced to D so that CD = AY. DP is drawn parallel to BA to meet BC and ZY (both produced) at P and W respectively. Show that the triangles DCP and AYZ are congruent. Calculate the ratio of the area of the parallelogram ZBPW to that of the triangle ABC. (35 marks). ⟦line⟧

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### csp_7e50cae41bce549f82f0da018ed0bd5f

Shamash Secondary School Conditional Examination, Sept. 1966

Subject: Algebra Date: 8/9/1966 Class: 4th Year Secondary Time: 8:00 - 11:00 a.m.

Attempt all questions:-

1. (i) Resolve x¹⁶ - y¹⁶ into five factors. (8 marks) (ii) Divide (x² + y² + z²)(x + 1) + (2xy - 2xz)(x + 1) - 2xyz - 2yz by x + 1 and express the quotient as a perfect square. (8 marks) (iii) If a = (x² - 2x + 2) / x(2 - x) and b = (x - 1) / x(2 - x), find the numerical value of a² - 4b² (9 marks).

2. (a) A man sets out at 7 a.m. from a town A to drive his horse and cart to B, a distance of 20 miles. His average speed is 6 m.p.h. at 7:30 a.m. his son leaves A by bicycle on the same road, riding at an average speed of 15 m.p.h. Write down the distance from A of each (i) at 7:40 a.m., (ii) t hours after 7 a.m. Calculate the time when the boy overtakes his father. (10 marks) (b) When the boy reaches B, he spends half an hour there and then rides back along the same road at the same average speed. Calculate the time when he meets his father. (10 marks)

3. (i) Compute by logarithms the value of the following:- x = ⁸√((Sin 17° 14')² (Cos 9° 11')³ / (1.003)³ (15.04)⁵) (10 marks) (ii) Find the value of x from the following equation: (log x)² - 4 log x + 4 = 0 (10 marks)

4. (i) Find the sum of all the numbers between 92 and 4815 which are exactly divisible by 13. How many of these numbers are there? (10 marks) (ii) The product of the first and sixth terms of a geometric progression is equal to 99 times the fourth term and the sum of the first and fourth terms is (-286). Find the first term and the sum of the first seven terms. (10 marks)

5. (i) Draw the graph of y=x(x+1)(x-2) between x=-1.5 and x=+2.5, taking one inch as unit for x and one inch as unit for y. (9 marks) From your graph obtain:- (ii) Approximate solutions for the equation x³ - x² - 2x + 1 = 0 (8 marks) (iii) The positive value of x for which the expression x(x+1)(x-2) is a minimum. (8 marks).

[Marginalia] 6 [Marginalia] 6 [Marginalia] 8 [Marginalia] 7 [Marginalia] 7 [Marginalia] 6

**Traduction anglaise —**

Shamash Secondary School Conditional Examination, Sept. 1966 Subject: Algebra Date: 8/9/1966 Class: 4th Year Secondary Time: 8:00 - 11:00 a.m. Attempt all questions:- 1. (i) Resolve x¹⁶ - y¹⁶ into five factors. (8 marks) (ii) Divide (x² + y² + z²)(x + 1) + (2xy - 2xz)(x + 1) - 2xyz - 2yz by x + 1 and express the quotient as a perfect square. (8 marks) (iii) If a = (x² - 2x + 2) / x(2 - x) and b = (x - 1) / x(2 - x), find the numerical value of a² - 4b² (9 marks). 2. (a) A man sets out at 7 a.m. from a town A to drive his horse and cart to B, a distance of 20 miles. His average speed is 6 m.p.h. at 7:30 a.m. his son leaves A by bicycle on the same road, riding at an average speed of 15 m.p.h. Write down the distance from A of each (i) at 7:40 a.m., (ii) t hours after 7 a.m. Calculate the time when the boy overtakes his father. (10 marks) (b) When the boy reaches B, he spends half an hour there and then rides back along the same road at the same average speed. Calculate the time when he meets his father. (10 marks) 3. (i) Compute by logarithms the value of the following:- x = ⁸√((Sin 17° 14')² (Cos 9° 11')³ / (1.003)³ (15.04)⁵) (10 marks) (ii) Find the value of x from the following equation: (log x)² - 4 log x + 4 = 0 (10 marks) 4. (i) Find the sum of all the numbers between 92 and 4815 which are exactly divisible by 13. How many of these numbers are there? (10 marks) (ii) The product of the first and sixth terms of a geometric progression is equal to 99 times the fourth term and the sum of the first and fourth terms is (-286). Find the first term and the sum of the first seven terms. (10 marks) 5. (i) Draw the graph of y=x(x+1)(x-2) between x=-1.5 and x=+2.5, taking one inch as unit for x and one inch as unit for y. (9 marks) From your graph obtain:- (ii) Approximate solutions for the equation x³ - x² - 2x + 1 = 0 (8 marks) (iii) The positive value of x for which the expression x(x+1)(x-2) is a minimum. (8 marks). 6 6 8 7 7 6

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### csp_7e7b5dc5a9ec5ed2a0c14091a9d71fe3

B01 · marginalia / latin Solution to question :- Mid-Year Exam. 21/12/1947

B02 · other / latin ⟦graph showing a parabola and a straight line intersecting at points C and D⟧

B03 · other / latin y₁ = 4x - 3 y₂ = 4x² - 4x - 15

B04 · table / latin x | y₁ | y₂ -3 | 33 -2 | -11 | 9 -1 | -7 0 | -3 | -15 ½ | -16 1 | -15 2 | 5 | -7 3 | 9 4 | 33

B05 · paragraph / latin (i) The two graphs are drawn as shown above. (ii) The roots of the two simultaneous equations are the coordinates of the two points of intersection C(-1, -7) and D(3, 9) . (i.e.) x = -1 } Ans. 1 x = 3 } Ans. 2 y = -7 } y = 9 }

B06 · paragraph / latin (iii) The roots of the equation 4x² - 4x - 15 = 0 are the <del>⟦illegible⟧</del> abscissas of the points of intersection of the curve with the x-axis , i.e: x = -1.5 Ans. 1 x = 2.5 Ans. 2

**Traduction anglaise —**

Solution to question :- Mid-Year Exam. 21/12/1947 ⟦graph showing a parabola and a straight line intersecting at points C and D⟧ y₁ = 4x - 3 y₂ = 4x² - 4x - 15 x | y₁ | y₂ -3 | | 33 -2 | -11 | 9 -1 | | -7 0 | -3 | -15 ½ | | -16 1 | | -15 2 | 5 | -7 3 | | 9 4 | | 33 (i) The two graphs are drawn as shown above. (ii) The roots of the two simultaneous equations are the coordinates of the two points of intersection C(-1, -7) and D(3, 9) . (i.e.) x = -1 } Ans. 1 x = 3 } Ans. 2 y = -7 } y = 9 } (iii) The roots of the equation 4x² - 4x - 15 = 0 are the <del>⟦illegible⟧</del> abscissas of the points of intersection of the curve with the x-axis , i.e: x = -1.5 Ans. 1 x = 2.5 Ans. 2

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### csp_7fa614ba04705cd9b7c9c4b891910945

B01 · paragraph / latin S = a(r^n - 1) / (r - 1) = 11[(-3)^7 - 1] / (-3 - 1) = 11[1 - (-3)^7] / (1 + 3) = 11(1 - (-2187)) / 4 = 11(1 + 2187) / 4 = 11 x 2188 / 4 = 11 x 547 = 6017 Ans.

B02 · marginalia / latin (10 marks)

**Traduction anglaise —**

S = a(r^n - 1) / (r - 1) = 11[(-3)^7 - 1] / (-3 - 1) = 11[1 - (-3)^7] / (1 + 3) = 11(1 - (-2187)) / 4 = 11(1 + 2187) / 4 = 11 x 2188 / 4 = 11 x 547 = 6017 Ans. (10 marks)

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### csp_805783d0fde55353bfbc5ac08cea681f

Let the x mi/hr. = rate of rowing in still water + Let the y mi/hr. = rate of current.

∴ 2 / (x+y) + 2 / (x-y) = 1 2/3 or 2 / (x+y) + 2 / (x-y) = 5/3 or 12(x+2y) + 12(x-2y) = 5(x²-y²) or 24x = 5x² - 20y² ----- ①

also 3 / (x+y) + 3 / (x-y) = 2 or 3(x-y) + 3(x+y) = (x²-y²) or 6x = x² - y² ----- ②

Multiply equation ② by 20 + subtract, hence 120x = 20x² - 20y² ----- ③ 24x = 5x² - 20y² ----- ① 96x = 15x² ∴ 32 = 5x ∴ x = 0 + x = 32/5 Ans. I

Now y² = x² - 6x or y = √x²-6x or y = √ (1024/25 - 192/5) or y = √ (1024-960)/25 or y = ± √ 64/25 or y = 8/5 Ans. II

the rate of rowing in still water = 32/5 = 6.4 miles/hr. } Ans. and the " " current = 8/5 = 1.6 mils/hr. }

5. (i) To get the first stone in the box the boy has to travel 1+1 = 2 yds. To get the second stone in the box, the boy has to travel 2+2 = 4 yds. + so on. ∴ total distance travelled to get 10 stones in the box = 2 + 4 + 6 + .. ∴ S₁₀ = n/2 {2a + (n-1)d} = 10/2 {4 + 9x2} = 5 x 22 = 110 yards. Ans. I

To get the first n stones: S = n/2 {2x2 + (n-1)x2} = n/2 (2 + 2n) = n(n+1) Ans. II

To get the first (n-1) stones: S = (n-1)/2 {2x2 + (n-2)x2} = (n-1)/2 (2n) = n(n-1) Ans. III

[Marginalia] 8 [Marginalia] 8

**Traduction anglaise —**

13 Let the x mi/hr. = rate of rowing in still water + Let the y mi/hr. = rate of current. ∴ 2 / (x+y) + 2 / (x-y) = 1 2/3 or 2 / (x+y) + 2 / (x-y) = 5/3 or 12(x+2y) + 12(x-2y) = 5(x²-y²) or 24x = 5x² - 20y² ⟦line⟧ ① also 3 / (x+y) + 3 / (x-y) = 2 or 3(x-y) + 3(x+y) = (x²-y²) or 6x = x² - y² ⟦line⟧ ② Multiply equation ② by 20 + subtract, hence 120x = 20x² - 20y² ⟦line⟧ ③ 24x = 5x² - 20y² ⟦line⟧ ① 96x = 15x² ∴ 32 = 5x ∴ x = 0 + x = 32/5 Ans. I Now y² = x² - 6x or y = √x²-6x or y = √ (1024/25 - 192/5) or y = √ (1024-960)/25 or y = ± √ 64/25 or y = 8/5 Ans. II the rate of rowing in still water = 32/5 = 6.4 miles/hr. } Ans. and the " " current = 8/5 = 1.6 mils/hr. } 5. (i) To get the first stone in the box the boy has to travel 1+1 = 2 yds. To get the second stone in the box, the boy has to travel 2+2 = 4 yds. + so on. ∴ total distance travelled to get 10 stones in the box = 2 + 4 + 6 + .. ∴ S₁₀ = n/2 {2a + (n-1)d} = 10/2 {4 + 9x2} = 5 x 22 = 110 yards. Ans. I To get the first n stones: S = n/2 {2x2 + (n-1)x2} = n/2 (2 + 2n) = n(n+1) Ans. II To get the first (n-1) stones: S = (n-1)/2 {2x2 + (n-2)x2} = (n-1)/2 (2n) = n(n-1) Ans. III 8 8

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### csp_83b306c846c251c48e7e195316c323b3

[Marginalia] الرقم: [Marginalia] الاسم:

- ٢ -

٣٦ - زاويتان متكاملتان ٣٧ - مضلع متساوي الاضلاع ٣٨ - مثلث متساوي الساقين ٣٩ - الخطوط المتوسطة في المثلث ٤٠ - المعين ٤١ - المحل الهندسي ٤٢ - المستقيم القاطع للدائرة ٤٣ - ازالة وادخال الاقواس ٤٤ - نقل حدود المعادلة من جهة الى الجهة الاخرى ٤٥ - متطابقة ٤٦ - متباينة ٤٧ - مقدار جبري متجانس ٤٨ - درجة المقدار الجبري ٤٩ - المعامل الحرفي ٥٠ - مقدار جبري من الدرجة الثانية.

(75 marks)

II. Fill in the blanks in the following equations:-

1. one furlong = ( ) chains = ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds. = ( ) ft. 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre = ( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. ⟦~⟧ ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones.

(25 marks).

-----

**Traduction anglaise —**

Number: Name: - 2 - 36 - Supplementary angles 37 - Equilateral polygon 38 - Isosceles triangle 39 - Medians of a triangle 40 - Rhombus 41 - Locus 42 - Secant line of a circle 43 - Removing and inserting brackets 44 - Transferring terms of an equation from one side to the other 45 - Identity 46 - Inequality 47 - Homogeneous algebraic expression 48 - Degree of an algebraic expression 49 - Literal coefficient 50 - Second-degree algebraic expression. (75 marks) II. Fill in the blanks in the following equations:- 1. one furlong = ( ) chains = ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds. = ( ) ft. 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre = ( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. ⟦~⟧ ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones. (25 marks). ⟦line⟧

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### csp_8442c7f26dad5125ae60c5ac1b9ac507

Final Exam in Algebra Cont. 12/6/1961 June 1961

⟦In⟧ the equation y = 16 x (4-x) , 'y' represents the height in feet risen by a stone thrown vertically upward from a point A on the roof of a building 20 ft above the level of the ground; and 'x' represents the time in seconds taken by the stone to reach the height 'y' from that point.

Draw a graph between x = 0 and x = 5 showing the relationship between y and x. (Take 1 in. = 1 second and 20 ft respectively.)

From the graph, find (a) the maximum height above the level of the ground reached by the stone, (b) how long the stone remains at least 68 ft above the ground, (c) how many seconds elapse from the time the <del>ball</del> stone was thrown to the time the stone strikes the ground.

(20 marks)

**Traduction anglaise —**

Final Exam in Algebra Cont. 12/6/1961 June 1961 ⟦In⟧ the equation y = 16 x (4-x) , 'y' represents the height in feet risen by a stone thrown vertically upward from a point A on the roof of a building 20 ft above the level of the ground; and 'x' represents the time in seconds taken by the stone to reach the height 'y' from that point. Draw a graph between x = 0 and x = 5 showing the relationship between y and x. (Take 1 in. = 1 second and 20 ft respectively.) From the graph, find (a) the maximum height above the level of the ground reached by the stone, (b) how long the stone remains at least 68 ft above the ground, (c) how many seconds elapse from the time the <del>ball</del> stone was thrown to the time the stone strikes the ground. (20 marks)

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### csp_84d42b2612c1538e9b3d8fdd327aee1b

B01 · header / latin ⟦illegible⟧ ⟦illegible⟧

B02 · paragraph / latin 1. (i) Factorise ⟦illegible⟧ (ii) Find the value of k which will make the expression 6x³ + 8x² + x - k divisible by ⟦illegible⟧ the other factors. (10 marks)

B03 · paragraph / latin 2. Find ⟦illegible⟧ ⟦illegible⟧ x⁶ - 2x⁵ + ⟦illegible⟧ x⁴ - ⟦illegible⟧ x³ + ⟦illegible⟧ x² - ⟦illegible⟧ x + ⟦illegible⟧

B04 · paragraph / latin 3. (i) Reduce to simplest form : ⟦illegible⟧ (ii) Solve the equation : (5x - 9) / (x - 2) + (4x - 11) / (x - 3) = (10x - 8) / (x - 1) - (x - 2) / (x - 4) (10 marks)

B05 · paragraph / latin 4. Find the values of x, y and z from the following equations: 3x - 2y + 4z = 2y - 3x + 7 = 7x + 2z - 2 = 11

B06 · paragraph / latin 5. A basket of eggs is emptied by one person taking half of them and one more, a second person taking half of the remainder and one more, and a third person taking half of the remainder + one more. How many did the basket contain at first? (10 marks)

**Traduction anglaise —**

⟦illegible⟧ ⟦illegible⟧ 1. (i) Factorise ⟦illegible⟧ (ii) Find the value of k which will make the expression 6x³ + 8x² + x - k divisible by ⟦illegible⟧ the other factors. (10 marks) 2. Find ⟦illegible⟧ ⟦illegible⟧ x⁶ - 2x⁵ + ⟦illegible⟧ x⁴ - ⟦illegible⟧ x³ + ⟦illegible⟧ x² - ⟦illegible⟧ x + ⟦illegible⟧ 3. (i) Reduce to simplest form : ⟦illegible⟧ (ii) Solve the equation : (5x - 9) / (x - 2) + (4x - 11) / (x - 3) = (10x - 8) / (x - 1) - (x - 2) / (x - 4) (10 marks) 4. Find the values of x, y and z from the following equations: 3x - 2y + 4z = 2y - 3x + 7 = 7x + 2z - 2 = 11 5. A basket of eggs is emptied by one person taking half of them and one more, a second person taking half of the remainder and one more, and a third person taking half of the remainder + one more. How many did the basket contain at first? (10 marks)

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### csp_85c16266f74b50149d97449c685dd209

B01 · header / latin ⟦illegible⟧ 31/12/1967

B02 · paragraph / latin ⟦illegible⟧ (1) ⟦illegible⟧ (1 + (a-b)/(a+b)) / (1 + (a^2-b^2)/(a^2+b^2)) = (a+b+a-b)/(a+b) / (a^2+b^2+a^2-b^2)/(a^2+b^2) = 2a/(a+b) / 2a^2/(a^2+b^2) = a/(a+b) * (a^2+b^2)/a^2 = (a^2+b^2)/(a(a+b))

B03 · paragraph / latin (2) (5x-2)/(x-2) + (6x-44)/(x-7) - (10x-8)/(x-1) = (x-5)/(x-6) (5(x-2)+8)/(x-2) + (6(x-7)-2)/(x-7) - (10(x-1)+2)/(x-1) = (x-6+1)/(x-6) 5 + 8/(x-2) + 6 - 2/(x-7) - 10 - 2/(x-1) = 1 - 2/(x-6) 8/(x-2) - 2/(x-7) = 2/(x-1) - 2/(x-6) or 1/(x-2) - 1/(x-7) = 1/(x-1) - 1/(x-6) (x-7-(x-2))/((x-2)(x-7)) = (x-6-(x-1))/((x-1)(x-6)) or -5/((x-2)(x-7)) = -5/((x-1)(x-6)) 1/((x-2)(x-7)) = 1/((x-1)(x-6)) or (x-2)(x-7) = (x-1)(x-6) x^2 - 9x + 14 = x^2 - 7x + 6 or 2x = 8 or x = 4 Ans. (10 marks)

B04 · paragraph / latin ⟦line⟧ 3x - 2y + 4z = 11 ; 2y + 4x + 8z = 24 ; 7x + 2y - z = 11 3x - 2y + 4z = 11 ---- (1) ; 4x + 4y + 8z = 24 -7x + 2y - z = 11 ---- (2) ; -6x + 3y = 12 ⟦illegible⟧ + 8z = 24 2y - 2x + 7 = 11 ⟦illegible⟧ 24 7x + 2y = 13 ⟦illegible⟧ ⟦illegible⟧ = 54 25x + 8z = 34 ⟦illegible⟧ 21x + 4z = 22 ⟦illegible⟧ ⟦illegible⟧ = 174 ⟦illegible⟧ = 2 from eq (2) 14 + 2y - z = 11 or 2y - z = -3 from eq (1) -6x + 6(1) = 12 or -6x = 6 or x = -1

**Traduction anglaise —**

⟦illegible⟧ 31/12/1967 ⟦illegible⟧ (1) ⟦illegible⟧ (1 + (a-b)/(a+b)) / (1 + (a^2-b^2)/(a^2+b^2)) = (a+b+a-b)/(a+b) / (a^2+b^2+a^2-b^2)/(a^2+b^2) = 2a/(a+b) / 2a^2/(a^2+b^2) = a/(a+b) * (a^2+b^2)/a^2 = (a^2+b^2)/(a(a+b)) (2) (5x-2)/(x-2) + (6x-44)/(x-7) - (10x-8)/(x-1) = (x-5)/(x-6) (5(x-2)+8)/(x-2) + (6(x-7)-2)/(x-7) - (10(x-1)+2)/(x-1) = (x-6+1)/(x-6) 5 + 8/(x-2) + 6 - 2/(x-7) - 10 - 2/(x-1) = 1 - 2/(x-6) 8/(x-2) - 2/(x-7) = 2/(x-1) - 2/(x-6) or 1/(x-2) - 1/(x-7) = 1/(x-1) - 1/(x-6) (x-7-(x-2))/((x-2)(x-7)) = (x-6-(x-1))/((x-1)(x-6)) or -5/((x-2)(x-7)) = -5/((x-1)(x-6)) 1/((x-2)(x-7)) = 1/((x-1)(x-6)) or (x-2)(x-7) = (x-1)(x-6) x^2 - 9x + 14 = x^2 - 7x + 6 or 2x = 8 or x = 4 Ans. (10 marks) ⟦line⟧ 3x - 2y + 4z = 11 ; 2y + 4x + 8z = 24 ; 7x + 2y - z = 11 3x - 2y + 4z = 11 ---- (1) ; 4x + 4y + 8z = 24 -7x + 2y - z = 11 ---- (2) ; -6x + 3y = 12 ⟦illegible⟧ + 8z = 24 2y - 2x + 7 = 11 ⟦illegible⟧ 24 7x + 2y = 13 ⟦illegible⟧ ⟦illegible⟧ = 54 25x + 8z = 34 ⟦illegible⟧ 21x + 4z = 22 ⟦illegible⟧ ⟦illegible⟧ = 174 ⟦illegible⟧ = 2 from eq (2) 14 + 2y - z = 11 or 2y - z = -3 from eq (1) -6x + 6(1) = 12 or -6x = 6 or x = -1

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### csp_8759afd3b3d356c79f0ded63f7f63581

Solutions to Mid-Year Exam in Algebra Cont. page 2. 6/2/1967

(2 x³ + px² - 5x + 2) ÷ (x + 2) = 2x² + (p-4)x - 2p + 3 + (4p-4)/(x+2) ∴ 4p - 4 = 0 ∴ p = 1

2x³ + px² - 5x + 2 | 2x² + (p-4)x - 2p + 3 2x³ + 4x² (p-4)x² - 5x + 2 (p-4)x² + 2(p-4)x (-2p + 3)x + 2 (-2p + 3)x - 4p + 6 4p - 4 = Remainder = 0 ∴ p = 1 and the other two factors are (2x-1) and (x-1) Ans.

∴ Original expression = 2x³ + x² - 5x + 2 = = (x+2) [2x² + (p-4)x - 2p + 3] = (x+2) (2x² - 3x + 1) = = (x+2)(2x-1)(x-1)

(i) Rate of rowing upstream = a m.p.h. } let the distance rowed each way = x miles " " " downstream = b m.p.h. } ∴ x/a + x/b = 2x/s or 1/a + 1/b = 2/s ∴ bs + as = 2ab or 2ab - as = bs or a(2b-s) = bs or a = bs / (2b-s) Ans. 1 also 2ba - bs = as ∴ b(2a-s) = as ∴ b = as / (2a-s) Ans. 2 also bs + as = 2ab ∴ s(a+b) = 2ab ∴ s = 2ab / (a+b) Ans. 3. but b = as / (2a-s) = (2x3) / (2x2-3) = 6/1 = 6 Ans. 4

(ii) { x² + 4y² + 80 = 15x + 30y ... ① } from ② : 4xy = 24 ... ③ { xy = 6 ... ② } add eq. ① + ③ and you get: x² + 4xy + 4y² + 80 = 15x + 30y + 24 or (x+2y)² + 80 = 15(x+2y) + 24 or (x+2y)² - 15(x+2y) + 56 = 0 ∴ [(x+2y)-7][(x+2y)-8] = 0 or x + 2y - 7 = 0 or x = 7 - 2y ... ④ also x + 2y - 8 = 0 or x = 8 - 2y ... ⑤ from eq. ② + ④ (7-2y)y = 6 or 7y - 2y² = 6 ∴ 2y² - 7y + 6 = 0 ∴ (2y-3)(y-2) = 0 ∴ y = 2 and y = 3/2 ∴ from ④ : x = 7 - 4 = 3 and x = 7 - 3 = 4 x = 3 } Ans. x = 4 } Ans. || again from ② + ⑤ : (8-2y)y = 6 ∴ 8y - 2y² = 6 y = 2 } y = 3/2 } || ∴ y² - 4y + 3 = 0 ∴ (y-1)(y-3) = 0 ∴ y = 1, y = 3 from ⑤ : x = 8 - 2 = 6 and x = 8 - 6 = 2 ∴ x = 6 } Ans. x = 2 } Ans. y = 1 } y = 3 } x = 3 } Ans. 1 x = 4 } Ans. 2 x = 6 } Ans. 3 x = 2 } Ans. 4 y = 2 } y = 3/2 } y = 1 } y = 3 }

**Traduction anglaise —**

Solutions to Mid-Year Exam in Algebra Cont. page 2. 6/2/1967 (2 x³ + px² - 5x + 2) ÷ (x + 2) = 2x² + (p-4)x - 2p + 3 + (4p-4)/(x+2) ∴ 4p - 4 = 0 ∴ p = 1 2x³ + px² - 5x + 2 | 2x² + (p-4)x - 2p + 3 2x³ + 4x² (p-4)x² - 5x + 2 (p-4)x² + 2(p-4)x (-2p + 3)x + 2 (-2p + 3)x - 4p + 6 4p - 4 = Remainder = 0 ∴ p = 1 and the other two factors are (2x-1) and (x-1) Ans. ∴ Original expression = 2x³ + x² - 5x + 2 = = (x+2) [2x² + (p-4)x - 2p + 3] = (x+2) (2x² - 3x + 1) = = (x+2)(2x-1)(x-1) (i) Rate of rowing upstream = a m.p.h. } let the distance rowed each way = x miles " " " downstream = b m.p.h. } ∴ x/a + x/b = 2x/s or 1/a + 1/b = 2/s ∴ bs + as = 2ab or 2ab - as = bs or a(2b-s) = bs or a = bs / (2b-s) Ans. 1 also 2ba - bs = as ∴ b(2a-s) = as ∴ b = as / (2a-s) Ans. 2 also bs + as = 2ab ∴ s(a+b) = 2ab ∴ s = 2ab / (a+b) Ans. 3. but b = as / (2a-s) = (2x3) / (2x2-3) = 6/1 = 6 Ans. 4 (ii) { x² + 4y² + 80 = 15x + 30y ... ① } from ② : 4xy = 24 ... ③ { xy = 6 ... ② } add eq. ① + ③ and you get: x² + 4xy + 4y² + 80 = 15x + 30y + 24 or (x+2y)² + 80 = 15(x+2y) + 24 or (x+2y)² - 15(x+2y) + 56 = 0 ∴ [(x+2y)-7][(x+2y)-8] = 0 or x + 2y - 7 = 0 or x = 7 - 2y ... ④ also x + 2y - 8 = 0 or x = 8 - 2y ... ⑤ from eq. ② + ④ (7-2y)y = 6 or 7y - 2y² = 6 ∴ 2y² - 7y + 6 = 0 ∴ (2y-3)(y-2) = 0 ∴ y = 2 and y = 3/2 ∴ from ④ : x = 7 - 4 = 3 and x = 7 - 3 = 4 x = 3 } Ans. x = 4 } Ans. | again from ② + ⑤ : (8-2y)y = 6 ∴ 8y - 2y² = 6 y = 2 } y = 3/2 } | ∴ y² - 4y + 3 = 0 ∴ (y-1)(y-3) = 0 ∴ y = 1, y = 3 from ⑤ : x = 8 - 2 = 6 and x = 8 - 6 = 2 ∴ x = 6 } Ans. x = 2 } Ans. y = 1 } y = 3 } x = 3 } Ans. 1 x = 4 } Ans. 2 x = 6 } Ans. 3 x = 2 } Ans. 4 y = 2 } y = 3/2 } y = 1 } y = 3 }

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### csp_87963f4849f6532591bee3222f69605a

Shamash Secondary School Conditional Examination, Sept. 63

Subject: Algebra Date: 12/9/1963 Class: 4th Year Secondary Time: 8.30-11.00 a.m.

Attempt all questions:

2. Solve for X the following equations, rejecting all extraneous roots: (i) 1 / 1 - X + 1 / √X + 1 + 1 / √X - 1 = 0 (6 marks) (ii) (X-7)⅓ = √X - 7 (6 marks) (iii) 3X⁻⅔ - 10X⁻⅓ + 3 = 0 (6 marks)

3. (i) Solve for X, using tables if necessary: 20ˣ = 2ˣ⁺² (ii) Compute by logarithms: ⁵√ ( (0.0012)² X (1.003)³ ) / ( 7515 X 2.004 ) (8 marks)

4. A man travels 108 miles, and finds that he could have made the journey in 4½ hours less, had he travelled 2 miles an hour faster. At what rate did he travel ? (16 marks)

**Traduction anglaise —**

Shamash Secondary School Conditional Examination, Sept. 63 Subject: Algebra Date: 12/9/1963 Class: 4th Year Secondary Time: 8.30-11.00 a.m. Attempt all questions: 1. (i) The equation 7X + 2 / X² - 4 = A / X-2 + B / X+2 is true for all values of X. Find the values of A & B. (8 marks) (ii) The expression X³ + pX² + qX + 6 is factorable into (X-1) and (X+2). Find the values of p and q and find the third factor. (8 marks) 2. Solve for X the following equations, rejecting all extraneous roots: (i) 1 / 1 - X + 1 / √X + 1 + 1 / √X - 1 = 0 (6 marks) (ii) (X-7)⅓ = √X - 7 (6 marks) (iii) 3X⁻⅔ - 10X⁻⅓ + 3 = 0 (6 marks) 3. (i) Solve for X, using tables if necessary: 20ˣ = 2ˣ⁺² (ii) Compute by logarithms: ⁵√ ( (0.0012)² X (1.003)³ ) / ( 7515 X 2.004 ) (8 marks) 4. A man travels 108 miles, and finds that he could have made the journey in 4½ hours less, had he travelled 2 miles an hour faster. At what rate did he travel ? (16 marks) 5. (i) Find by series the value of the recurring fraction 0.3205 (8 marks) (ii) The three digits of a number are in arithmetical progression. The number itself divided by the sum of the digits is 48. The number formed by the same digits in reverse order is 396 less than the original number. What is the number ? (8 marks)

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### csp_88244496624d524d856140f662d875eb

SHAMASH SECONDARY SCHOOL: الرقم :: الاسم :: Monthly Examination, November 1968: Subject: General Mathematics: Date: 18/11/1968.: Class : 4th Year Secondary: Time: 8:30 - 10:00 a.m.:

1. Give the English Equivalent of the followinh, filling the blanks in this sheet and hand it over with your examination book.

١- ارقام ٢- مراتب ٣- الطرح ٤- العوامل ٥- اس القوة ٦- مضاعف ٧- اعداد زوجية متتالية ٨- اعداد فردية متتالية ٩- الجزء الصحيح من العدد ١٠- اعداد اولية ١١- المقام المشترك الاصغر ١٢- كسر لفظي ١٣- مقلوب العدد ١٤- الكسور العشرية المنتهية ١٥- <del>الكسور العشرية الدورية</del> ١٦- الخطأ المئوي ١٧- النسبة والتناسب ١٨- الوسط المتناسب بين عددين ١٩- ربح المساهم ( ربح حامل الاسهم ) ٢٠- البديهية ٢١- الموضوعة ٢٢- زاوية حادة ٢٣- زاوية منفرجة ٢٤- زاوية منعكسة ٢٥- قطعة دائرة ٢٦- قطاع دائرة ٢٧- المعاليم ٢٨- المجاهيل ٢٩- زاويتان متتامتان ٣٠- زاويتان متكاملتان ٣١- مضلع متساوي الاضلاع ٣٢- مثلث متساوي الساقين ٣٣- المعين ٣٤- المحل الهندسي ٣٥- المستقيم القاطع للدائرة ٣٦- ازالة وادخال الاقواس ٣٧- نقل حدود المعادلة من جهة الى الجهة الاخرى ٣٨- متطابقة ٣٩- متباينة

يتبع

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL: Number :: Name :: Monthly Examination, November 1968: Subject: General Mathematics: Date: 18/11/1968.: Class : 4th Year Secondary: Time: 8:30 - 10:00 a.m.: 1. Give the English Equivalent of the followinh, filling the blanks in this sheet and hand it over with your examination book. 1- Digits 2- Places 3- Subtraction 4- Factors 5- Exponent of power 6- Multiple 7- Consecutive even numbers 8- Consecutive odd numbers 9- The integer part of the number 10- Prime numbers 11- Lowest common denominator 12- Vulgar fraction 13- Reciprocal of the number 14- Terminating decimals 15- <del>Recurring decimals</del> 16- Percentage error 17- Ratio and proportion 18- Mean proportional between two numbers 19- Shareholder's profit (dividend) 20- Axiom 21- Postulate 22- Acute angle 23- Obtuse angle 24- Reflex angle 25- Segment of a circle 26- Sector of a circle 27- Knowns 28- Unknowns 29- Complementary angles 30- Supplementary angles 31- Equilateral polygon 32- Isosceles triangle 33- Rhombus 34- Locus 35- Secant line of a circle 36- Removing and inserting brackets 37- Transposing terms of the equation from one side to the other 38- Identity 39- Inequality To be continued

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### csp_8836ce52718d50cfa60ec3d47a6a42be

Exam, June 1964 4th Year Secondary

Let x m.p.h. = average speed of cyclist ∴ (x + 18) m.p.h. = " " " Motorist ∴ <del>44</del> / x = 44 / (x + 18) + 1 36/60 or 44 / x = 44 / (x + 18) + 1 3/5 or 44 / x = 44 / (x + 18) + 8/5 ∴ 5 × 44(x + 18) = 5 × 44x + 8x(x + 18) or 220x + 3960 = 220x + 8x² + 144x or 8x² + 144x - 3960 = 0 or x² + 18x - 495 = 0 or (x + 33)(x - 15) = 0 ∴ x = -33 (to be discarded) + x = 15 m.p.h. Ans. = av. sp. of cyclist ∴ x + 18 = 15 + 18 = 33 m.p.h. Ans. = av. sp. of motorist

II. (i) x/4 + y/4 = 4 = x/4 + y/2 ∴ x/2 + y/4 = 4 .... ① x/4 + y/2 = 4 .... ② ∴ 2x + y = 16 .... ①a ∴ { 4x + 2y = 32 .... ③ x + 2y = 16 .... ②a x + 2y = 16 .... ② ∴ 3x = 16 ∴ x = 16/3 from ①a, y = 16 - 2x = 16 - 32/3 = 16/3 ∴ x = 16/3 } Ans. y = 16/3

(ii) In order that the expression 3x² - 4x + 1 be expressed in the form a(x - 1)² + b(x - 1), the two forms should be identical. Hence the equation: 3x² - 4x + 1 = a(x - 1)² + b(x - 1) is an identity which is true for all values of x, Now let x = 0, then 1 = a(-1)² + b(-1) or a - b = 1 .... ① } Also let x = 2, then 5 = a + b or a + b = 5 .... ② ∴ 2a = 6 ∴ a = 3 and b = 2 ∴ { a = 3 } Ans. b = 2

III. (i) (a) Let the No. of terms be n. Now the 1st term = x and the common difference d = y - x ∴ the <del>⟦illegible⟧</del> n th term z = x + (n - 1)(y - x) ∴ (n - 1)(y - x) = z - x ∴ n - 1 = (z - x) / (y - x) ∴ n = (z - x) / (y - x) + 1 or n = (z + y - 2x) / (y - x) Ans.

[Marginalia] page 1

**Traduction anglaise —**

Exam, June 1964 4th Year Secondary Let x m.p.h. = average speed of cyclist ∴ (x + 18) m.p.h. = " " " Motorist ∴ <del>44</del> / x = 44 / (x + 18) + 1 36/60 or 44 / x = 44 / (x + 18) + 1 3/5 or 44 / x = 44 / (x + 18) + 8/5 ∴ 5 × 44(x + 18) = 5 × 44x + 8x(x + 18) or 220x + 3960 = 220x + 8x² + 144x or 8x² + 144x - 3960 = 0 or x² + 18x - 495 = 0 or (x + 33)(x - 15) = 0 ∴ x = -33 (to be discarded) + x = 15 m.p.h. Ans. = av. sp. of cyclist ∴ x + 18 = 15 + 18 = 33 m.p.h. Ans. = av. sp. of motorist II. (i) x/4 + y/4 = 4 = x/4 + y/2 ∴ x/2 + y/4 = 4 .... ① x/4 + y/2 = 4 .... ② ∴ 2x + y = 16 .... ①a ∴ { 4x + 2y = 32 .... ③ x + 2y = 16 .... ②a x + 2y = 16 .... ② ∴ 3x = 16 ∴ x = 16/3 from ①a, y = 16 - 2x = 16 - 32/3 = 16/3 ∴ x = 16/3 } Ans. y = 16/3 (ii) In order that the expression 3x² - 4x + 1 be expressed in the form a(x - 1)² + b(x - 1), the two forms should be identical. Hence the equation: 3x² - 4x + 1 = a(x - 1)² + b(x - 1) is an identity which is true for all values of x, Now let x = 0, then 1 = a(-1)² + b(-1) or a - b = 1 .... ① } Also let x = 2, then 5 = a + b or a + b = 5 .... ② ∴ 2a = 6 ∴ a = 3 and b = 2 ∴ { a = 3 } Ans. b = 2 III. (i) (a) Let the No. of terms be n. Now the 1st term = x and the common difference d = y - x ∴ the <del>⟦illegible⟧</del> n th term z = x + (n - 1)(y - x) ∴ (n - 1)(y - x) = z - x ∴ n - 1 = (z - x) / (y - x) ∴ n = (z - x) / (y - x) + 1 or n = (z + y - 2x) / (y - x) Ans. page 1

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### csp_89a1ee0a41335f1ab3d240ceccfdf106

[Marginalia] سلوى عزرا سويغ (١١)

Shamash Secondary School Mid-Year Exam. January, 1967

Subject: Arith. & Trig. Date: Jan. 29, 1967 Class: 4th Year Secondary. Time: 8:30 - 10:30 a.m.

---- Answer five questions which must include question 2.

1. A ladder, 24 ft. long, makes an angle of 52° with the ground and leans against a vertical wall. If the top of the ladder slips down 2 ft., how far will the foot of the ladder move ?

2. (i) The point C is 6 ft. above level ground and 20 ft. measured horizontally from a vertical pole AB where B is at ground level. If the angle ACB = 64°, calculate the length of AB. (ii) A vertical wall of length 50 ft. and height 8 ft. runs due N and S. Find the area of the shadow of the wall cast on level ground by the sun shining from the W at an elevation of 27°.

3. The total rateable value of a town is £540,000 and it is estimated that the necessary expenditure for 1959 will amount to £432,000. Calculate: (a) The rate in the £ which must be charged to meet the 1959 expenditure. (b) The rates to be paid on a house whose rateable value is £63. (c) The amount produced by a penny rate.

4. A length of 4800 ft. of paper is wrapped on a wooden cylinder of radius 3 in.; the thickness of the paper is 1/120 in. Find the radius of the whole roll to 1/100 in.

5. A man invested £990 in (£1) shares, paying 10%, at 27s. 6d.; he sold the shares at 32s. and invested the proceeds in (10s.) shares, paying 5%, at 9s. How many (10s.) shares did he buy and what was the change of his annual income ?

6. Two partners A, B started with capitals of £6000, £4000 respectively. The profits at the end of each year are divided in proportion to their capitals invested in the business at the beginning of the year. A withdrew his profits at the end of each year, while B left his in the business. The profits for the first 3 years were £1400, £1584, £1639 15s. respectively. What was B's capital at the end of 3 years?

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**Traduction anglaise —**

Salwa Ezra Sweigh (11) Shamash Secondary School Mid-Year Exam. January, 1967 Subject: Arith. & Trig. Date: Jan. 29, 1967 Class: 4th Year Secondary. Time: 8:30 - 10:30 a.m. ⟦line⟧ Answer five questions which must include question 2. 1. A ladder, 24 ft. long, makes an angle of 52° with the ground and leans against a vertical wall. If the top of the ladder slips down 2 ft., how far will the foot of the ladder move ? 2. (i) The point C is 6 ft. above level ground and 20 ft. measured horizontally from a vertical pole AB where B is at ground level. If the angle ACB = 64°, calculate the length of AB. (ii) A vertical wall of length 50 ft. and height 8 ft. runs due N and S. Find the area of the shadow of the wall cast on level ground by the sun shining from the W at an elevation of 27°. 3. The total rateable value of a town is £540,000 and it is estimated that the necessary expenditure for 1959 will amount to £432,000. Calculate: (a) The rate in the £ which must be charged to meet the 1959 expenditure. (b) The rates to be paid on a house whose rateable value is £63. (c) The amount produced by a penny rate. 4. A length of 4800 ft. of paper is wrapped on a wooden cylinder of radius 3 in.; the thickness of the paper is 1/120 in. Find the radius of the whole roll to 1/100 in. 5. A man invested £990 in (£1) shares, paying 10%, at 27s. 6d.; he sold the shares at 32s. and invested the proceeds in (10s.) shares, paying 5%, at 9s. How many (10s.) shares did he buy and what was the change of his annual income ? 6. Two partners A, B started with capitals of £6000, £4000 respectively. The profits at the end of each year are divided in proportion to their capitals invested in the business at the beginning of the year. A withdrew his profits at the end of each year, while B left his in the business. The profits for the first 3 years were £1400, £1584, £1639 15s. respectively. What was B's capital at the end of 3 years? ⟦line⟧

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### csp_8b6ae7cbc0fa5fc5b41aff1694762155

⟦illegible⟧ 1956 Solution to ⟦illegible⟧ Algebra ⟦illegible⟧

2. (b) 3 x 10^2x - 13 x 10^x + 4 = 0 let 10^x = y then 3y^2 - 13y + 4 = 0 ∴ (3y - 1)(y - 4) = 0 ∴ y = 4 or y = 1/3 ∴ 10^x = 4 hence x log 10 = log 4 ∴ x = log 4 = 0.6021 Ans. 1 10^x = 1/3 ∴ x log 10 = log 1 - log 3 or x = - log 3 ∴ x = - 0.4771 Ans. 2

3 (i) log_3 (2x^2 - 7x) = 2 ∴ 2x^2 - 7x = 3^2 or 2x^2 - 7x - 9 = 0 ∴ (2x - 9)(x + 1) = 0 ∴ x = 9/2 = 4.5 Ans. 1 or x = -1 Ans. 2

(ii) (log_2 9)(log_9 32) = ? Let log_2 9 = x ∴ 9 = 2^x or 3^2 = 2^x ∴ x log 2 = 2 log 3 ∴ x = 2 log 3 / log 2 also let log_9 32 = y ∴ 32 = 9^y or 2^5 = 3^2y ∴ 2y log 3 = 5 log 2 ∴ y = 5 log 2 / 2 log 3 Now (log_2 9)(log_9 32) = x . y = 2 log 3 / log 2 . 5 log 2 / 2 log 3 = 5 Ans.

[Marginalia] a more direct method would be as follows: [Marginalia] (log_2 9)(log_9 32) = log 9 / log 2 . log 32 / log 9 = 5 log 2 / log 2 = 5 Ans.

(iii) y = 7√((tan 19° 45')^2 x (cos 77° 16')^3 / (3.004)^5 x (50.06)^3)

2 log tan 19° 45' = 1.1102 3 log cos 77° 16' = 2.0296 log Num = 3.1398 log Den = 7.4870 7 log y = 11.6528 log y = 2.52183 y = 0.03325 = 3.325 x 10^-2 Ans.

**Traduction anglaise —**

⟦illegible⟧ 1956 Solution to ⟦illegible⟧ Algebra ⟦illegible⟧ 2. (b) 3 x 10^2x - 13 x 10^x + 4 = 0 let 10^x = y then 3y^2 - 13y + 4 = 0 ∴ (3y - 1)(y - 4) = 0 ∴ y = 4 or y = 1/3 ∴ 10^x = 4 hence x log 10 = log 4 ∴ x = log 4 = 0.6021 Ans. 1 10^x = 1/3 ∴ x log 10 = log 1 - log 3 or x = - log 3 ∴ x = - 0.4771 Ans. 2 3 (i) log_3 (2x^2 - 7x) = 2 ∴ 2x^2 - 7x = 3^2 or 2x^2 - 7x - 9 = 0 ∴ (2x - 9)(x + 1) = 0 ∴ x = 9/2 = 4.5 Ans. 1 or x = -1 Ans. 2 (ii) (log_2 9)(log_9 32) = ? Let log_2 9 = x ∴ 9 = 2^x or 3^2 = 2^x ∴ x log 2 = 2 log 3 ∴ x = 2 log 3 / log 2 also let log_9 32 = y ∴ 32 = 9^y or 2^5 = 3^2y ∴ 2y log 3 = 5 log 2 ∴ y = 5 log 2 / 2 log 3 Now (log_2 9)(log_9 32) = x . y = 2 log 3 / log 2 . 5 log 2 / 2 log 3 = 5 Ans. a more direct method would be as follows: (log_2 9)(log_9 32) = log 9 / log 2 . log 32 / log 9 = 5 log 2 / log 2 = 5 Ans. (iii) y = 7√((tan 19° 45')^2 x (cos 77° 16')^3 / (3.004)^5 x (50.06)^3) log tan 19° 45' = 1.5551 | 5 log 3.004 = 2.3885 | 2 log tan 19° 45' = 1.1102 log cos 77° 16' = <del>1.3432</del> 1.3431 | 3 log 50.06 = 5.0985 | 3 log cos 77° 16' = 2.0296 log 3.004 = 0.4777 | log Den = 7.4870 | log Num = 3.1398 log 50.06 = 1.6995 | | log Den = 7.4870 | | 7 log y = 11.6528 | | log y = 2.52183 2 log tan 19° 45' = 1.1102 3 log cos 77° 16' = 2.0296 log Num = 3.1398 log Den = 7.4870 7 log y = 11.6528 log y = 2.52183 y = 0.03325 = 3.325 x 10^-2 Ans. <del>y = 0.03437</del> <del>= 3.437 x 10^-2 Ans.</del>

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### csp_8bae47a6f1b452d883f38523578c26fa

B01 · header / mixed ⟦illegible⟧ في الرياضيات للصف ⟦illegible⟧ 97/10/⟦illegible⟧ للرابع العام

B02 · paragraph / latin No 1 Let the man has £x stock 4x / 100 his profit

B03 · paragraph / mixed x / 100 * 114 = £ 114x / 100 114x / 100 * 1/3 = £ 38x / 100 the first part 114x / 100 * 2/3 = £ 76x / 100 second part

B04 · paragraph / latin 38x / 100 * 100 / 95 * 5 / 100 = 2x / 100 profit 76x / 100 * 100 / 100 * 3 / 400 = 19x / 1200 profit 2x / 100 + 19x / 1200 = 43x / 1200 total profit

B05 · paragraph / latin 4x / 100 - 43x / 1200 = 25 / 2 (1200) <del>12x</del> 48x - 43x = ⟦illegible⟧ 5x = ⟦illegible⟧ x = ⟦illegible⟧ Stock

B06 · marginalia / unknown ⟦illegible diagram/sketch⟧

**Traduction anglaise —**

⟦illegible⟧ in Mathematics for the ⟦illegible⟧ grade 10/⟦illegible⟧/97 For the fourth general No 1 Let the man has £x stock 4x / 100 his profit x / 100 * 114 = £ 114x / 100 114x / 100 * 1/3 = £ 38x / 100 the first part 114x / 100 * 2/3 = £ 76x / 100 second part 38x / 100 * 100 / 95 * 5 / 100 = 2x / 100 profit 76x / 100 * 100 / 100 * 3 / 400 = 19x / 1200 profit 2x / 100 + 19x / 1200 = 43x / 1200 total profit 4x / 100 - 43x / 1200 = 25 / 2 (1200) <del>12x</del> 48x - 43x = ⟦illegible⟧ 5x = ⟦illegible⟧ x = ⟦illegible⟧ Stock ⟦illegible diagram/sketch⟧

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### csp_8c5e76da0e315dbfbd7178c599e68c5f

B01 · header / latin y = 1/4(3x² - 5x - 4) Final Exam in Algebra 14/5/1962

B02 · table / latin x | y -2 | 4 1/2 -1 | 1 0 | -1 1 | -1 1/2 2 | -1/2 3 | 2 4 | 6

B03 · other / latin y = 1/4(3x² - 5x - 4) (-0.81, 1/2) (2.48, 1/2) y = 1/2 (5/6, -1.5)

B04 · paragraph / latin (ii) From the curve the value of x for of y (i.e. for the root of the eq.) 1/4(3x² - 5x - 4) is (-0.54) is the least value of (3x² - 5x - 4) (i.e.) the least value of y is = 1/4(-1.54) = -6.08 or -6 1/2 Ans. II

B05 · paragraph / latin (iii) The equation 3x² - 5x + 6 = 0 is identical with the equation 3x² - 5x - 4 = 2 ≡ 1/4(3x² - 5x - 4) = 1/2 ∴ the roots of 3x² - 5x - 6 = 0 are the same as the roots of 1/4(3x² - 5x - 4) = 1/2 . Draw the curve y = 1/2, this line will intersect the original curve at (2.48, 1/2) and (-0.81, 1/2) ; the roots are: x = 2.48 x = -0.81 Ans.

**Traduction anglaise —**

y = 1/4(3x² - 5x - 4) Final Exam in Algebra 14/5/1962 x | y -2 | 4 1/2 -1 | 1 0 | -1 1 | -1 1/2 2 | -1/2 3 | 2 4 | 6 y = 1/4(3x² - 5x - 4) (-0.81, 1/2) (2.48, 1/2) y = 1/2 (5/6, -1.5) (ii) From the curve the value of x for of y (i.e. for the root of the eq.) 1/4(3x² - 5x - 4) is (-0.54) is the least value of (3x² - 5x - 4) (i.e.) the least value of y is = 1/4(-1.54) = -6.08 or -6 1/2 Ans. II (iii) The equation 3x² - 5x + 6 = 0 is identical with the equation 3x² - 5x - 4 = 2 ≡ 1/4(3x² - 5x - 4) = 1/2 ∴ the roots of 3x² - 5x - 6 = 0 are the same as the roots of 1/4(3x² - 5x - 4) = 1/2 . Draw the curve y = 1/2, this line will intersect the original curve at (2.48, 1/2) and (-0.81, 1/2) ; the roots are: x = 2.48 x = -0.81 Ans.

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### csp_8e69516e6297530aa2ce2ff7a3d3a223

(cont'd.).. -p.2-

Arith. & Trig. 17/5/67 4th Year Secondary.

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**Traduction anglaise —**

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### csp_9169731b0365525799b475f24b5230e1

B01 · header / latin Shamash Secondary School Final Examination, June 1960.

B02 · header / latin Subject: Algebra Class: 4th Year Secondary Date: 17/6/1960 Time: 8:00-11:00

B03 · paragraph / latin ⟦line⟧ All questions are to be attempted.

B04 · paragraph / latin 1. (a) If X = (2Y-1)/(3Y-4) , and Y = (Z + 1)/(Z - 1) , find Z in terms of X. (8 marks) (b) Prove that, if a + b = C, and none of these quantities is zero, the expression 1 / (a² + b² - C²) + 1 / (b² + C² - a²) + 1 / (C² + a² - b²) is equal to zero. (8 marks)

B05 · paragraph / latin 2. (i) Find the value of b for which the expression X³ - 2 - b (X-1) is equal to Zero when X = 2. (5 marks) (ii) Factorize the expression for this value of b, and find the other values of X for which the expression is Zero. ( 12 marks)

B06 · paragraph / latin 3. A train left station P at 10 a.m. on a non-stop run of 300 miles to station Q where it was due to arrive at 4:15 p.m. At a station B some miles from Q it was 3¾ minutes behind the Scheduled time. But by travelling from B to Q at 60 miles per hour the train arrived at its destination on time. How far is it from B to Q ? (17 marks)

B07 · paragraph / latin 4. (a) Compute by logarithms the following expression: ⁷√( (1.004² X 0.000491³) / (516.2 X 2.003²) ) (8 marks ) (b) Use logarithms to solve the equation 4²ˣ - 8 x 4ˣ + 12 = 0 (8 marks)

B08 · paragraph / latin 5. (a) The 21st term of an arithmetical progression is 2½ times the 8th term, and the arithmetic mean of the 5th and 13th terms is 29. Find the sum of the first 15 terms. (8 marks) (b) Prove that in any Geometric series the sum of the 4th, 5th, and 6th terms is the Geometric mean of the sum of the 1st, 2nd, and 3rd terms and the sum of the 7th, 8th, and 9th terms. (8 marks)

B09 · paragraph / latin 6. (a) Draw the graph of Y = X² - 3X + 2 for values of X between -2 and 5. (6 marks). (b) Use your graph to solve the equations: ( i) X² - 3X + 2 = 0 (6 marks) (ii) X² - 3X - 4 = 0 (6 marks)

B10 · footer / latin ⟦line⟧

**Traduction anglaise —**

Shamash Secondary School Final Examination, June 1960. Subject: Algebra Class: 4th Year Secondary Date: 17/6/1960 Time: 8:00-11:00 ⟦line⟧ All questions are to be attempted. 1. (a) If X = (2Y-1)/(3Y-4) , and Y = (Z + 1)/(Z - 1) , find Z in terms of X. (8 marks) (b) Prove that, if a + b = C, and none of these quantities is zero, the expression 1 / (a² + b² - C²) + 1 / (b² + C² - a²) + 1 / (C² + a² - b²) is equal to zero. (8 marks) 2. (i) Find the value of b for which the expression X³ - 2 - b (X-1) is equal to Zero when X = 2. (5 marks) (ii) Factorize the expression for this value of b, and find the other values of X for which the expression is Zero. ( 12 marks) 3. A train left station P at 10 a.m. on a non-stop run of 300 miles to station Q where it was due to arrive at 4:15 p.m. At a station B some miles from Q it was 3¾ minutes behind the Scheduled time. But by travelling from B to Q at 60 miles per hour the train arrived at its destination on time. How far is it from B to Q ? (17 marks) 4. (a) Compute by logarithms the following expression: ⁷√( (1.004² X 0.000491³) / (516.2 X 2.003²) ) (8 marks ) (b) Use logarithms to solve the equation 4²ˣ - 8 x 4ˣ + 12 = 0 (8 marks) 5. (a) The 21st term of an arithmetical progression is 2½ times the 8th term, and the arithmetic mean of the 5th and 13th terms is 29. Find the sum of the first 15 terms. (8 marks) (b) Prove that in any Geometric series the sum of the 4th, 5th, and 6th terms is the Geometric mean of the sum of the 1st, 2nd, and 3rd terms and the sum of the 7th, 8th, and 9th terms. (8 marks) 6. (a) Draw the graph of Y = X² - 3X + 2 for values of X between -2 and 5. (6 marks). (b) Use your graph to solve the equations: ( i) X² - 3X + 2 = 0 (6 marks) (ii) X² - 3X - 4 = 0 (6 marks) ⟦line⟧

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### csp_91fa71ace81259fabc13f98c5d334392

Shamash Secondary School Final Examination, June 1965

Subject: Algebra Date: 1/6/1965 Class: 4th Scientific, sections A + B Time: 8:00 - 11:00 a.m.

Attempt all questions

I. (i) If (2x+y)/(x+2y) = m, find an expression for y in terms of m and x. If also y = mx, find the values of m. (7 marks) (ii) Resolve into two factors: <del>⟦illegible⟧</del> a³ - 27b³ + a² + 3ab + c (7 marks) (iii) Resolve the expression 5x² - 14x + 8 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8 (6 marks)

II. A man can walk a mile in 2 minutes less time than B would take. In a walking race, B has a start of 1/4 mile and A overtakes B in 10 minutes. Assuming both men walk at a uniform rate, find their rates of walking in miles per hour. (20 marks)

III. (i) Compute by logarithms arranging your work neatly: 7√((cos² 18° 47' x sin³ 48° 21') / ((10.02)³ x 0.0002043)) (16 marks) (ii) If log a - 5 log b = 3 log c, find (a) in terms of b and c. (4 marks) (iii) Given log 4.41 = 2, calculate the value of a. (4 marks) (iv). Solve the equation 2^(3-x) = 3^(2x+1) giving your answer correct to 3 decimal places. (6 marks)

IV. (i) Write down and simplify an expression for the nth term of the Arithmetic progression 3, 7, 11, ... (4 marks) If the sum of n terms of this progression is kn + cn², find the values of k and c and the sum of the first thirty terms. (8 marks) (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term, and the sum of the first + fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks)

**Traduction anglaise —**

Shamash Secondary School Final Examination, June 1965 Subject: Algebra Date: 1/6/1965 Class: 4th Scientific, sections A + B Time: 8:00 - 11:00 a.m. Attempt all questions I. (i) If (2x+y)/(x+2y) = m, find an expression for y in terms of m and x. If also y = mx, find the values of m. (7 marks) (ii) Resolve into two factors: <del>⟦illegible⟧</del> a³ - 27b³ + a² + 3ab + c (7 marks) (iii) Resolve the expression 5x² - 14x + 8 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8 (6 marks) II. A man can walk a mile in 2 minutes less time than B would take. In a walking race, B has a start of 1/4 mile and A overtakes B in 10 minutes. Assuming both men walk at a uniform rate, find their rates of walking in miles per hour. (20 marks) III. (i) Compute by logarithms arranging your work neatly: 7√((cos² 18° 47' x sin³ 48° 21') / ((10.02)³ x 0.0002043)) (16 marks) (ii) If log a - 5 log b = 3 log c, find (a) in terms of b and c. (4 marks) (iii) Given log 4.41 = 2, calculate the value of a. (4 marks) (iv). Solve the equation 2^(3-x) = 3^(2x+1) giving your answer correct to 3 decimal places. (6 marks) IV. (i) Write down and simplify an expression for the nth term of the Arithmetic progression 3, 7, 11, ... (4 marks) If the sum of n terms of this progression is kn + cn², find the values of k and c and the sum of the first thirty terms. (8 marks) (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term, and the sum of the first + fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks)

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### csp_91fafd9426ec56e797c2130470333823

[Stamp] ⟦illegible⟧

Shamash Secondary School Final Examination May 1969 --------------------------

Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m.

Answer all questions:

1. (i) In the expression x³ +Ax² +31x+B, A and B are constant. Find the values of A and B which will make this expression divisible by (x-2) (x-3) and find the remaining factor. (10 marks) (ii) The wages of 12 men and 7 boys amount to £9 13s. If 3 men together receive 8s. more than 4 boys, what are the wages of each man and boy ? (10 marks)

2. (i) Solve the equation (log x)² = log x ⟦-10, finding two values for x.⟧ (10 marks) (ii) Solve the two simultaneous equations: 3x² +xy-2y² +7=0 ........(1) x² -xy+y² -7=0 .........(2) (10 marks)

3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√¹⁻ˣ 8) = (√ˣ 2)(√ 50) (10 marks) (ii)Compute the value of ⁷√ (0.5002)² Sin³ 14° 25' / (4.003)³ Cos² 15° 27' (10 marks)

4. (i) If (b+c)⁻¹ , (c+a)⁻¹ , (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks)

[Marginalia] bounce

(cont'd.p.2)..

**Traduction anglaise —**

⟦illegible⟧ Shamash Secondary School Final Examination May 1969 ⟦line⟧ Subject: Algebra Date: 14/5/1969 Class: 4th Year, Scientific Time: 8:00 - 11:00 a.m. Answer all questions: 1. (i) In the expression x³ +Ax² +31x+B, A and B are constant. Find the values of A and B which will make this expression divisible by (x-2) (x-3) and find the remaining factor. (10 marks) (ii) The wages of 12 men and 7 boys amount to £9 13s. If 3 men together receive 8s. more than 4 boys, what are the wages of each man and boy ? (10 marks) 2. (i) Solve the equation (log x)² = log x ⟦-10, finding two values for x.⟧ (10 marks) (ii) Solve the two simultaneous equations: 3x² +xy-2y² +7=0 ........(1) x² -xy+y² -7=0 .........(2) (10 marks) 3. (i) Find the value of x from the following equation without using the tables: (5)(4³ˣ⁻¹)(√¹⁻ˣ 8) = (√ˣ 2)(√ 50) (10 marks) (ii)Compute the value of ⁷√ (0.5002)² Sin³ 14° 25' / (4.003)³ Cos² 15° 27' (10 marks) 4. (i) If (b+c)⁻¹ , (c+a)⁻¹ , (a+b)⁻¹ are in arithmetical progression, prove that a², b², c² are also in arithmetical progression. (10 marks) bounce (ii) A bouncing tennis ball rebounds each time to a height one half the height of the previous ⟦bounce⟧. If it is dropped from a height of 10 ft., show: (a) that the total distance it has travelled when it hits the ground for the 10th time is equal to 29 123/128 ft. (b) Show also that the total distance it travels before coming to rest is 30 ft. (10 marks) (cont'd.p.2)..

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### csp_92254022b6a65d1c872f6700d0af51e1

Solutions to Conditional Exam in Algebra Sept., 1965, 4th year.

1. (i) 2(x²-2)² + 5(x²-2) - 12 = 0 or [2(x²-2)-3][(x²-2)+4] = 0 2(x²-2)-3 = 0 or 2x² = 7 ∴ x = ± √7/2 or x = ± 1/2 √14 Ans. I, II also (x²-2)+4 = 0 or x² = -2 ∴ x = ± √-2 = ± √2 i Ans. III + IV

(ii) (a) { (9ⁿ⁺¼)(√3.3ⁿ) / 3√3⁻ⁿ }¹/ⁿ = { (3²)ⁿ⁺¼ . (3)ⁿ⁺¹/² / 3 . 3⁻ⁿ/² }¹/ⁿ = { 3²ⁿ⁺¹/² . 3ⁿ⁺¹/² / 3¹⁻ⁿ/² }¹/ⁿ = = { 3 (4n+1+n+1+n-2)/2 }¹/ⁿ = (3³ⁿ)¹/ⁿ = 3³ = 27 Ans. (6 marks)

(b) 6x²y² / m+n ÷ [ 3(m-n)x / 7(r+s) ÷ { 4(r-s) / 21xy² ÷ r²-s² / 4(m²-n²) } ] = = 6x²y² / m+n ÷ [ 3(m-n)x / 7(r+s) ÷ { 4(r-s) } { 4(m²-n²) } / 21xy² (r²-s²) ] = = 6x²y² / m+n ÷ [ 3(m-n)x ] [ 21xy² (r²-s²) ] / [ 7(r+s) ] [ 16(r-s) (m²-n²) ] = 6x²y² / m+n × 7 × 16 (m²-n²) (r²-s²) / 3 × 21 (m-n) (r²+s²) (x³y²) = 6 × 7 × 16 / 3 × 21 = 32/3 Ans. (6 marks)

2 (i) log₁₀ y = a + b log₁₀ x } when x=1 , y=1000 } x=10 " x=0.1 , y=100 } y=? ∴ log₁₀ 1000 = a + b log₁₀ 1 or 3 = a or a = 3 log₁₀ 100 = a + b log₁₀ 0.1 or 2 = a - b or b = 1 ∴ log₁₀ y = 3 + log₁₀ x ∴ log₁₀ y = 3 + log₁₀ 10 or log₁₀ y = 4 ∴ y = 10⁴ Ans. (10 marks)

(ii) log₁₀ 2 = 0.301030 , log₁₀ 1.005 = 0.002166 , log₁₀ 402 = ? , log₁₀ 0.0804 ? 1.005 = 1005 / 1000 = 5 × 3 × 67 / 1000 = 3 × 67 / 200 ∴ 3 × 67 = 200 (1.005) 402 = 2 × 3 × 67 = 2 [ 200 (1.005) ] = 2² × 100 × 1.005 0.0804 = 804 / 10⁴ = 2 × 402 / 10⁴ = 2³ × 100 × 1.005 / 10⁴ = 2³ × 1.005 / 100

**Traduction anglaise —**

Solutions to Conditional Exam in Algebra Sept., 1965, 4th year. 1. (i) 2(x²-2)² + 5(x²-2) - 12 = 0 or [2(x²-2)-3][(x²-2)+4] = 0 2(x²-2)-3 = 0 or 2x² = 7 ∴ x = ± √7/2 or x = ± 1/2 √14 Ans. I, II also (x²-2)+4 = 0 or x² = -2 ∴ x = ± √-2 = ± √2 i Ans. III + IV (ii) (a) { (9ⁿ⁺¼)(√3.3ⁿ) / 3√3⁻ⁿ }¹/ⁿ = { (3²)ⁿ⁺¼ . (3)ⁿ⁺¹/² / 3 . 3⁻ⁿ/² }¹/ⁿ = { 3²ⁿ⁺¹/² . 3ⁿ⁺¹/² / 3¹⁻ⁿ/² }¹/ⁿ = = { 3 (4n+1+n+1+n-2)/2 }¹/ⁿ = (3³ⁿ)¹/ⁿ = 3³ = 27 Ans. (6 marks) (b) 6x²y² / m+n ÷ [ 3(m-n)x / 7(r+s) ÷ { 4(r-s) / 21xy² ÷ r²-s² / 4(m²-n²) } ] = = 6x²y² / m+n ÷ [ 3(m-n)x / 7(r+s) ÷ { 4(r-s) } { 4(m²-n²) } / 21xy² (r²-s²) ] = = 6x²y² / m+n ÷ [ 3(m-n)x ] [ 21xy² (r²-s²) ] / [ 7(r+s) ] [ 16(r-s) (m²-n²) ] = 6x²y² / m+n × 7 × 16 (m²-n²) (r²-s²) / 3 × 21 (m-n) (r²+s²) (x³y²) = 6 × 7 × 16 / 3 × 21 = 32/3 Ans. (6 marks) 2 (i) log₁₀ y = a + b log₁₀ x } when x=1 , y=1000 } x=10 " x=0.1 , y=100 } y=? ∴ log₁₀ 1000 = a + b log₁₀ 1 or 3 = a or a = 3 log₁₀ 100 = a + b log₁₀ 0.1 or 2 = a - b or b = 1 ∴ log₁₀ y = 3 + log₁₀ x ∴ log₁₀ y = 3 + log₁₀ 10 or log₁₀ y = 4 ∴ y = 10⁴ Ans. (10 marks) (ii) log₁₀ 2 = 0.301030 , log₁₀ 1.005 = 0.002166 , log₁₀ 402 = ? , log₁₀ 0.0804 ? 1.005 = 1005 / 1000 = 5 × 3 × 67 / 1000 = 3 × 67 / 200 ∴ 3 × 67 = 200 (1.005) 402 = 2 × 3 × 67 = 2 [ 200 (1.005) ] = 2² × 100 × 1.005 0.0804 = 804 / 10⁴ = 2 × 402 / 10⁴ = 2³ × 100 × 1.005 / 10⁴ = 2³ × 1.005 / 100

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### csp_927e49ba4fa65b7ebe3d0622364c156b

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965.

[Marginalia] ABE Daly [Marginalia] (5)

Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m.

Attempt all questions :

1. (i) If m = 2x + y / x + 2y , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks). (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks).

3. (i) Compute by logarithms, arranging your work neatly : ⁷√ (cos² 18° 47') (sin³ 48° 21') / (10.09)³ (0.0002049) (6 marks). ⟦log a - ⟦...⟧ log b + 3 log c⟧ (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks).

----------------------------------------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965. ABE Daly (5) Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m. Attempt all questions : 1. (i) If m = 2x + y / x + 2y , find an expression for y in terms of m and x. If also Y = mx , find the values of m. (7 marks). (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks). 3. (i) Compute by logarithms, arranging your work neatly : ⁷√ (cos² 18° 47') (sin³ 48° 21') / (10.09)³ (0.0002049) (6 marks). ⟦log a - ⟦uncertain⟧ log b + 3 log c⟧ (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks). 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ...... (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks). 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axix and 0.2 inch for your unit on the y-axix. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks). ⟦line⟧

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### csp_9366d27909285c6fa3ca39acaf6f57fb

2 ⟦illegible⟧ حلول الرياضيات للرابع العام ⟦illegible⟧ الامتحان النهائي

T A 22 34 B 200 yd C

5. AB = 200 Cos 34 = 200 x 0.829 = 165.8 yd. AT = 165.8 x tan 22 = 165.8 x 0.404 = 66.9832 yd. AC = 200 Sin 34 = 200 x 0.5592 = 111.84 tan θ = 66.9832 / 111.84 = 0.5985 ∴ θ = 30° 48'

[Marginalia] 5 [Marginalia] 6 [Marginalia] 5 [Marginalia] 5

⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

2 ⟦illegible⟧ Mathematics solutions for the fourth general year ⟦illegible⟧ Final exam T A 22 34 B 200 yd C 5. AB = 200 Cos 34 = 200 x 0.829 = 165.8 yd. AT = 165.8 x tan 22 = 165.8 x 0.404 = 66.9832 yd. AC = 200 Sin 34 = 200 x 0.5592 = 111.84 tan θ = 66.9832 / 111.84 = 0.5985 ∴ θ = 30° 48' 5 6 5 5 ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_93e99c13420d5b6b82e2d88a486e03fa

SHAMASH SECONDARY SCHOOL 4th Quarter Examination, May, 1965.

Subject: Algebra Date: 2/5/1965 Class: 4th Secondary year Time: 8:00-9:30 a.m.

Attempt all questions.

1. (a) Prove that: (a-a⁻¹)(a⁴/³ + a⁻²/³) = a² - a⁻² / a⁻¹/³ (13 marks) (b) Evaluate: x³/² + xy / xy - y³ - √x / √x-y (1⟦2⟧ marks)

2. Solve the equation: 6 √x - 7 / √x - 1 - 5 = 7 √x - 26 / 7 √x - 21 (25 marks)

3. Find x from the equation: 3²x = 5x+1 (25 marks)

4. Compute by logarithms the value of x, arranging your work neatly: ⁷√ (1.001)² (0.0004061)⅔ / Sin³ 24° 21' Cos² 41° 57' (25 marks)

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**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL 4th Quarter Examination, May, 1965. Subject: Algebra Date: 2/5/1965 Class: 4th Secondary year Time: 8:00-9:30 a.m. Attempt all questions. 1. (a) Prove that: (a-a⁻¹)(a⁴/³ + a⁻²/³) = a² - a⁻² / a⁻¹/³ (13 marks) (b) Evaluate: x³/² + xy / xy - y³ - √x / √x-y (1⟦2⟧ marks) 2. Solve the equation: 6 √x - 7 / √x - 1 - 5 = 7 √x - 26 / 7 √x - 21 (25 marks) 3. Find x from the equation: 3²x = 5x+1 (25 marks) 4. Compute by logarithms the value of x, arranging your work neatly: ⁷√ (1.001)² (0.0004061)⅔ / Sin³ 24° 21' Cos² 41° 57' (25 marks) ⟦line⟧

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### csp_96de23ecdb22538587e1477b9a5aaa7f

Shamash Secondary School Final Exams. June, 1963.

Subject: Algebra Class: 4th Year Secondary --- Date: 9/6/1963 Time: 8:00-10.30 a.m.

Attempt all questions.

1. (a) Resolve into three factors: 2X³ - 9X² + 7X + 6. (10 marks) (b) Resolve into four factors: 4(Xy + mn)² - (X² + y² - m² - n²)². (10 marks)

2. (a) Given √2 = 1.414, √3 = 1.732, √6 = 2.440, Find to two places of decimals the value of: (3 - √2) (7 + 4 √3) ÷ (2 √3 - 3), rationalising the denominator first. (10 marks) (b) Solve for X : 2 √X - 1 √X - 2 -------- = -------- (10 marks) 2 √X + 4 √X - 4 - - 3 3

3. (a) Solve for X without using the tables: (√3 √2)ˣ = 36 (10 marks) (b) Compute by logarithms, arranging your work neatly: ________________________ 7 / 2 3 / (0.0002003) --- (0.04031)--- / 3 5 V ----------------------------- (10 marks) 1.004 X 9.006

( cont'd.p.2)

**Traduction anglaise —**

Shamash Secondary School Final Exams. June, 1963. Subject: Algebra Class: 4th Year Secondary ⟦line⟧ Date: 9/6/1963 Time: 8:00-10.30 a.m. Attempt all questions. 1. (a) Resolve into three factors: 2X³ - 9X² + 7X + 6. (10 marks) (b) Resolve into four factors: 4(Xy + mn)² - (X² + y² - m² - n²)². (10 marks) 2. (a) Given √2 = 1.414, √3 = 1.732, √6 = 2.440, Find to two places of decimals the value of: (3 - √2) (7 + 4 √3) ÷ (2 √3 - 3), rationalising the denominator first. (10 marks) (b) Solve for X : 2 √X - 1 √X - 2 ⟦line⟧ = ⟦line⟧ (10 marks) 2 √X + 4 √X - 4 - - 3 3 3. (a) Solve for X without using the tables: (√3 √2)ˣ = 36 (10 marks) (b) Compute by logarithms, arranging your work neatly: ⟦line⟧ 7 / 2 3 / (0.0002003) --- (0.04031)--- / 3 5 V ⟦line⟧ (10 marks) 1.004 X 9.006 ( cont'd.p.2)

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### csp_993ce48b93dd5493a71ac50315bbd368

S = 2 - ⟦2^{n+1}/3^{n}⟧ Sum of 1st & 2nd terms = 2 - ⟦2^3/3^2 = 2 - 8/9 = 10/9⟧ ∴ Common ratio r = ⟦4/9 ÷ 2/3 = 4/9 × 3/2 = 2/3 Ans. 2⟧ ∴ the nth term l = ar^{n-1} = 2/3 (2/3)^{n-1} = (2/3)^n Ans.

(b) (i) In the Arithmetic Series, the common difference: d = b - a ∴ the nth term l = a + (n-1) (b - a) Ans. 1 (ii) In the Geometric Series, the common ratio: r = b/a ∴ the nth term: l = a (b/a)^{n-1} = b^{n-1}/a^{n-2} Ans. 2

5. (i) y = 3/4 x² ...... ① | x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | | y | 12 | 6 3/4 | 3 | 3/4 | 0 | 3/4 | 3 | 6 3/4 | 12 |

(ii) 3x + 2y = 12 ...... ② the st. line is determined by the two points: (0, 6) and (4, 0) | x | 0 | 4 | | y | 6 | 0 |

⟦Graph showing parabola and straight line intersection⟧ (-4, 12) 3x + 2y = 12 (2, 3) y = 3/4 x²

(iii) The points of intersections of the two graphs are: (-4, 12) and (2, 3) ∴ the solutions are: x = -4, y = 12 } Ans. 1 and x = 2, y = 3 } Ans. 2

To verify algebraically, from equation ② y = (12 - 3x)/2, substituting 12 - 3x / 2 = 3/4 x² ∴ 24 - 6x = 3x² ∴ 3x² + 6x - 24 = 0 ⟦x² + 2x - 8 = 0⟧ ⟦(x + 4)(x - 2) = 0⟧ ⟦x = -4 or x = 2⟧

**Traduction anglaise —**

S = 2 - ⟦2^{n+1}/3^{n}⟧ Sum of 1st & 2nd terms = 2 - ⟦2^3/3^2 = 2 - 8/9 = 10/9⟧ ∴ Common ratio r = ⟦4/9 ÷ 2/3 = 4/9 × 3/2 = 2/3 Ans. 2⟧ ∴ the nth term l = ar^{n-1} = 2/3 (2/3)^{n-1} = (2/3)^n Ans. (b) (i) In the Arithmetic Series, the common difference: d = b - a ∴ the nth term l = a + (n-1) (b - a) Ans. 1 (ii) In the Geometric Series, the common ratio: r = b/a ∴ the nth term: l = a (b/a)^{n-1} = b^{n-1}/a^{n-2} Ans. 2 5. (i) y = 3/4 x² ...... ① x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 y | 12 | 6 3/4 | 3 | 3/4 | 0 | 3/4 | 3 | 6 3/4 | 12 (ii) 3x + 2y = 12 ...... ② the st. line is determined by the two points: (0, 6) and (4, 0) x | 0 | 4 y | 6 | 0 ⟦Graph showing parabola and straight line intersection⟧ (-4, 12) 3x + 2y = 12 (2, 3) y = 3/4 x² (iii) The points of intersections of the two graphs are: (-4, 12) and (2, 3) ∴ the solutions are: x = -4, y = 12 } Ans. 1 and x = 2, y = 3 } Ans. 2 To verify algebraically, from equation ② y = (12 - 3x)/2, substituting 12 - 3x / 2 = 3/4 x² ∴ 24 - 6x = 3x² ∴ 3x² + 6x - 24 = 0 ⟦x² + 2x - 8 = 0⟧ ⟦(x + 4)(x - 2) = 0⟧ ⟦x = -4 or x = 2⟧

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### csp_99417916da355a35a431ea1c5d00f01f

Shamash Secondary School 3rd Quarter Examination, March 1964

Subject: Algebra Date: 26/3/1964 Class: 4th Year Secondary Time: 10:15-11.45

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Attempt all questions:

1. An education Committee spent £P in one year awarding n scholarships at Secondary schools. The school fees of the scholars, amounting to £F, were paid in each case; and in addition some of the scholars received a grant of £a and the remainder a grant of £b. How many received the grant of £a ? (20 marks)

2. Solve each of the following equations:- (i) (X²+2)² + 198 = 29(X²+2) (10 marks) (ii) X³+7X²+ 7X-15 = 0 (10 marks)

3. Solve the two simultaneous equations:- X²+4Y²+80 = 15X+30Y (20 marks) XY = 6

4. At what time between ten and eleven O'clock are the hands of a watch at right angles for the second time. (20 marks)

5. Find the value of X⁴-47X²Y²+Y⁴ in terms of p and q when X+Y = p and X-Y = q. (20 marks)

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**Traduction anglaise —**

Shamash Secondary School 3rd Quarter Examination, March 1964 Subject: Algebra Date: 26/3/1964 Class: 4th Year Secondary Time: 10:15-11.45 ⟦line⟧ Attempt all questions: 1. An education Committee spent £P in one year awarding n scholarships at Secondary schools. The school fees of the scholars, amounting to £F, were paid in each case; and in addition some of the scholars received a grant of £a and the remainder a grant of £b. How many received the grant of £a ? (20 marks) 2. Solve each of the following equations:- (i) (X²+2)² + 198 = 29(X²+2) (10 marks) (ii) X³+7X²+ 7X-15 = 0 (10 marks) 3. Solve the two simultaneous equations:- X²+4Y²+80 = 15X+30Y (20 marks) XY = 6 4. At what time between ten and eleven O'clock are the hands of a watch at right angles for the second time. (20 marks) 5. Find the value of X⁴-47X²Y²+Y⁴ in terms of p and q when X+Y = p and X-Y = q. (20 marks) ⟦line⟧

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### csp_9980f7560fa65be2b44edfc4b9fd7b3a

⟦...⟧ Algebra September, 1960 1

(1) (i) x^16 - y^16 = (x^8 - y^8)(x^8 + y^8) = (x^4 - y^4)(x^4 + y^4)(x^8 + y^8) = (x^2 - y^2)(x^2 + y^2)(x^4 + y^4)(x^8 + y^8) = (x - y)(x + y)(x^2 + y^2)(x^4 + y^4)(x^8 + y^8) Ans. (8 marks)

(ii) (x^2 + y^2 + z^2)(x + 1) + (2xy - xz)(x + 1) - 2xyz - 2yz ----------------------------------------------------------------- x + 1 = (x^2 + y^2 + z^2)(x + 1) + (2xy - xz)(x + 1) - 2yz(x + 1) ----------------------------------------------------------------- = x^2 + y^2 + z^2 + 2xy - 2xz - 2yz (x + 1) = (x + y - z)^2 Ans. (8 marks)

(iii) a = x^2 - 2x + 2 and b = x - 1 x(2 - x) x(2 - x) a^2 - 4b^2 = (a + 2b)(a - 2b) = ( x^2 - 2x + 2 + 2x - 2 ) ( x^2 - 2x + 2 - 2x - 2 ) x(2 - x) x(2 - x) x(2 - x) x(2 - x) = ( x^2 - 2x + 2 + 2x - 2 ) ( x^2 - 2x + 2 - 2x + 2 ) = ( x^2 ) ( x^2 - 4x + 4 ) = x^2(x - y)^2 x(2 - x) x(2 - x) ( x(2 - x) ) ( x(2 - x) ) x^2(2 - x)^2 = x^2(x - x)^2 = 1 Ans. (9 marks) x^2(2 - x)^2

2(a) (i) dist. from A of man at 7:40 = 40/60 x 6 = 4 miles Ans. (1 mark) distance of son at 7:40 = 10/60 x 15 = 2 1/2 miles Ans. (1 mark)

[Marginalia] 7:00 20 miles [Marginalia] A -> B [Marginalia] 7:30 -> 15 m.p.h. A C P B

(ii) dist. of man from A, t hours after 7:30 = 6t miles Ans. (1 mark) " " son " " , t - 1/2 " " = 15(t - 1/2) miles Ans. (1 mark) Let the time be t hrs. after 7:30 when the son overtake the father at pt. C .: 6t = 15(t - 1/2) or 6t = 15t - 15/2 or 9t = 15/2 .: t = 15/18 hrs. = 5/6 hrs. .: the time at the instant of overtaking is 7 5/6 or 7:50 a.m. Ans. (5 marks)

(b) the boy takes 20/15 hrs. to cover the distance AB or 4/3 hrs. ⟦...⟧ let the time taken by the boy to meet his father again on his way back from B be T hours .: ⟦...⟧ distance covered by the boy from B to the meeting point P is = 15T .: distance covered by father up till this instant = (20 - 15T) miles .: Time taken by the father to cover distance 15T = 20 - 15T hrs. 6 .: 20 - 15T = the no. of hours which have elapsed after seven 7:00 a.m. 6 .: 20 - 15T = 1/2 + 4/3 + T or 20 - 15T = 6 + 8 + 6T .: 21T = 6 .: T = 6/21 hrs. = 2/7 6 .: ⟦...⟧ time at this instant = 1/2 + 4/3 + 2/7 hrs. after 7:00 a.m. = 21 + 56 + 12 42

**Traduction anglaise —**

⟦...⟧ Algebra September, 1960 1 (1) (i) x^16 - y^16 = (x^8 - y^8)(x^8 + y^8) = (x^4 - y^4)(x^4 + y^4)(x^8 + y^8) = (x^2 - y^2)(x^2 + y^2)(x^4 + y^4)(x^8 + y^8) = (x - y)(x + y)(x^2 + y^2)(x^4 + y^4)(x^8 + y^8) Ans. (8 marks) (ii) (x^2 + y^2 + z^2)(x + 1) + (2xy - xz)(x + 1) - 2xyz - 2yz ⟦line⟧ x + 1 = (x^2 + y^2 + z^2)(x + 1) + (2xy - xz)(x + 1) - 2yz(x + 1) ⟦line⟧ = x^2 + y^2 + z^2 + 2xy - 2xz - 2yz (x + 1) = (x + y - z)^2 Ans. (8 marks) (iii) a = x^2 - 2x + 2 and b = x - 1 x(2 - x) x(2 - x) a^2 - 4b^2 = (a + 2b)(a - 2b) = ( x^2 - 2x + 2 + 2x - 2 ) ( x^2 - 2x + 2 - 2x - 2 ) x(2 - x) x(2 - x) x(2 - x) x(2 - x) = ( x^2 - 2x + 2 + 2x - 2 ) ( x^2 - 2x + 2 - 2x + 2 ) = ( x^2 ) ( x^2 - 4x + 4 ) = x^2(x - y)^2 x(2 - x) x(2 - x) ( x(2 - x) ) ( x(2 - x) ) x^2(2 - x)^2 = x^2(x - x)^2 = 1 Ans. (9 marks) x^2(2 - x)^2 2(a) (i) dist. from A of man at 7:40 = 40/60 x 6 = 4 miles Ans. (1 mark) distance of son at 7:40 = 10/60 x 15 = 2 1/2 miles Ans. (1 mark) 7:00 20 miles A -> B 7:30 -> 15 m.p.h. A C P B (ii) dist. of man from A, t hours after 7:30 = 6t miles Ans. (1 mark) " " son " " , t - 1/2 " " = 15(t - 1/2) miles Ans. (1 mark) Let the time be t hrs. after 7:30 when the son overtake the father at pt. C .: 6t = 15(t - 1/2) or 6t = 15t - 15/2 or 9t = 15/2 .: t = 15/18 hrs. = 5/6 hrs. .: the time at the instant of overtaking is 7 5/6 or 7:50 a.m. Ans. (5 marks) (b) the boy takes 20/15 hrs. to cover the distance AB or 4/3 hrs. ⟦...⟧ let the time taken by the boy to meet his father again on his way back from B be T hours .: ⟦...⟧ distance covered by the boy from B to the meeting point P is = 15T .: distance covered by father up till this instant = (20 - 15T) miles .: Time taken by the father to cover distance 15T = 20 - 15T hrs. 6 .: 20 - 15T = the no. of hours which have elapsed after seven 7:00 a.m. 6 .: 20 - 15T = 1/2 + 4/3 + T or 20 - 15T = 6 + 8 + 6T .: 21T = 6 .: T = 6/21 hrs. = 2/7 6 .: ⟦...⟧ time at this instant = 1/2 + 4/3 + 2/7 hrs. after 7:00 a.m. = 21 + 56 + 12 42

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### csp_99e2b3f35b92579c880b892295e6f5bc

SHAMASH SECONDARY SCHOOL 3rd & 4th quarter Examination.

Subject: Algebra Class: 4th Year Scientific Date: 7/4/1969. Time: 8:30-10:30 a.m.

⟦line⟧ Answer all Questions : 1. (i) By first taking the square root and then the cube root, find the sixth root of : ( a³ - 1/a³ )² - 6( a - 1/a )( a³ - 1/a³ ) + 9( a - 1/a )². (12 marks). (ii) Prove that the left-hand side is always equal to the right-hand side in the following equation : bc(b - c) + ca(c - a) + ab(a - b) = - (b - c)(c - a)(a - b). (13 marks).

2. (i) Solve the equation : (x - 1)/(√x - 1) = 3 + (√x + 1)/2. (12 marks). (ii) Rationalise the denominator and then find the value of : (√(1 + x) + √(1 - x)) / (√(1 + x) - √(1 - x)) , when x = 2b / (b² + 1) . (13 marks).

3. (i) Compute by logarithms : ⁷√[ (0.⟦002001⟧)³(sin16° 23')² / (1.003)⁵(tan41° 16')² ] (12 marks). (ii) Solve the following equation for x : 2(log x)² - 5(log x) + 2 = 0 . (13 marks).

4. (i) In an Arithmetic Progression the first term is 3 and the common difference is 6. Show that the sum of 2n terms is always equal to four times the sum of n terms. (12 marks). (ii) In an A. P. the ratio of the 3rd term to the 6th term is 11:26 and the sum of the first 4 terms is 34. Find the progression and the sum of the first 8 terms. (13 marks). ⟦line⟧

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL 3rd & 4th quarter Examination. Subject: Algebra Class: 4th Year Scientific Date: 7/4/1969. Time: 8:30-10:30 a.m. ⟦line⟧ Answer all Questions : 1. (i) By first taking the square root and then the cube root, find the sixth root of : ( a³ - 1/a³ )² - 6( a - 1/a )( a³ - 1/a³ ) + 9( a - 1/a )². (12 marks). (ii) Prove that the left-hand side is always equal to the right-hand side in the following equation : bc(b - c) + ca(c - a) + ab(a - b) = - (b - c)(c - a)(a - b). (13 marks). 2. (i) Solve the equation : (x - 1)/(√x - 1) = 3 + (√x + 1)/2. (12 marks). (ii) Rationalise the denominator and then find the value of : (√(1 + x) + √(1 - x)) / (√(1 + x) - √(1 - x)) , when x = 2b / (b² + 1) . (13 marks). 3. (i) Compute by logarithms : ⁷√[ (0.⟦002001⟧)³(sin16° 23')² / (1.003)⁵(tan41° 16')² ] (12 marks). (ii) Solve the following equation for x : 2(log x)² - 5(log x) + 2 = 0 . (13 marks). 4. (i) In an Arithmetic Progression the first term is 3 and the common difference is 6. Show that the sum of 2n terms is always equal to four times the sum of n terms. (12 marks). (ii) In an A. P. the ratio of the 3rd term to the 6th term is 11:26 and the sum of the first 4 terms is 34. Find the progression and the sum of the first 8 terms. (13 marks). ⟦line⟧

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### csp_9a0a669adb115a2caea8d98df5ec9061

SHAMASH SECONDARY SCHOOL Monthly Examination, November 1968 Subject: General Mathematics Date: 18/11/1968. Class : 4th Year Secondary Time: 8:30 - 10:00 a.m. الرقم: الاسم:

1. Give the English Equivalent of the following, filling the blanks in this sheet and hand it over with your examination book.

- يتبع -

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Monthly Examination, November 1968 Subject: General Mathematics Date: 18/11/1968. Class : 4th Year Secondary Time: 8:30 - 10:00 a.m. Number: Name: 1. Give the English Equivalent of the following, filling the blanks in this sheet and hand it over with your examination book. Numerals = figures | 1- Numerals Digits | 2- Digits Subtraction | 3- Subtraction Factors | 4- Factors The index or exponent of the power | 5- Exponent of the power Multiple | 6- Multiple Consecutive even numbers | 7- Consecutive even numbers ⟦Consecutive⟧ odd ⟦numbers⟧ | 8- Consecutive odd numbers The integral part of a number | 9- The integral part of the number Prime numbers | 10- Prime numbers The least common Denominator | 11- The least common denominator An improper fraction | 12- Improper fraction The reciprocal of a number | 13- The reciprocal of the number Terminating decimals | 14- Terminating decimal fractions Recurring or Repeating decimals | 15- Recurring decimal fractions The percentage error | 16- The percentage error Ratio + Proportion | 17- Ratio and Proportion The mean proportional between two numbers | 18- The mean proportional between two numbers The Dividend | 19- Shareholder's profit (stockholder's profit) Axiom | 20- Axiom Postulate | 21- Postulate an acute angle | 22- Acute angle an obtuse ⟦angle⟧ | 23- Obtuse angle a Reflex ⟦angle⟧ | 24- Reflex angle a segment of a circle | 25- Segment of a circle a Sector ⟦of a circle⟧ | 26- Sector of a circle The Data | 27- Knowns The unknowns | 28- Unknowns Two Complementary angles | 29- Two complementary angles ⟦Two⟧ Supplementary ⟦angles⟧ | 30- Two supplementary angles an equilateral polygon | 31- Equilateral polygon an isosceles triangle | 32- Isosceles triangle The rhombus | 33- The rhombus The Locus | 34- The geometric locus The secant to a circle | 35- The secant line to the circle The removal + insertion of brackets | 36- Removal and insertion of brackets Transposition from one side of an equation to the other | 37- Transposing equation terms from one side to the other Identity | 38- Identity Inequality | 39- Inequality - To be continued -

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### csp_9b1fb528b5d25ab3840f3b15f216b2dd

Shamash Secondary School 1st. Quarter Exam.

Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary. Date: 8/12/1957

----------------- All questions are to be attempted.

(2) (a) Decimalise to 3 places the following:

| £ 2 | 12s | 2 3/4 d | | £ 8 | 10s | 8 1/2 d | | £ 9 | 5s | 10 1/4 d |

(b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint.

(3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ?

(4) The average age of m boys is b years and of n girls is c years. Find the average age of all together.

(5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation.

(6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height. ------------------------

**Traduction anglaise —**

Shamash Secondary School 1st. Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary. Date: 8/12/1957 ⟦line⟧ All questions are to be attempted. (1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m, by 25.00 m. (2) (a) Decimalise to 3 places the following: £ 2 | 12s | 2 3/4 d £ 8 | 10s | 8 1/2 d £ 9 | 5s | 10 1/4 d (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint. (3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ? (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together. (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation. (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height. ⟦line⟧

---

### csp_9c21804660ac5da9ab5ba29fb1fb9ca7

B01 · stamp / latin ELBA RADO

B02 · marginalia / hebrew ⟦illegible⟧

B03 · stamp / latin RADO

B04 · other / latin C

**Traduction anglaise —**

ELBA RADO ⟦illegible⟧ RADO C

---

### csp_9d3dd6dd89c75f1bae403830516f4edf

Final Exams in Algebra June 12th 1961

Let x shillings be price of one gross. ∴ 12x/144 pence (= x/12 pence) = price of one pencil in pennies ∴ (20 x 12 x)/144 = 5/3 x pence price of one score in pennies in 1st case. ∴ (20 x 12)/x/12 = 2880/x = No. of pencils which can be bought for £1 in 1st case ∴ 2880/x + 120 = (2880 + 120x)/x " " " " " " 2nd case ∴ 240 / (2880 + 120x)/x = 240x / 120(24 + x) = 2x / 24 + x pence = price of one pencil in pennies in 2nd case 20 (2x / x + 24) = 40x / x + 24 pence = price of one score of pencils in 2nd case. ∴ 5/3 x - 40x / x + 24 = 2 ∴ 5x² + 120x - 120x = 6x + 144 or 5x² - 6x - 144 = 0 ∴ (5x + 24)(x - 6) = 0 ∴ x = - 24/5 to be discarded or x = 6 Ans. Hence Price of one gross = 6 shillings Ans.

⟦line⟧

2. (i) √a - x + √b + x = √a + √b squaring, we have: a - x + 2√(a - x)(b + x) + b + x = a + 2√ab + b or ∴ (a - x)(b + x) = a.b ∴ a.b + (a - b)x - x² = a.b. or x² - (a - b)x = 0 or x[x - (a - b)] = 0 ∴ x = 0 Ans. I or x = a - b Ans. II

(ii) { a^(p-q) / √a^(p-q) x a^(p-q) }^(n/(p-q)) = { a^(p-q) / a^((p-q)/2) x a^(p-q) }^n = { a^(p-q - (p-q)/2 + p-q) }^n = = [ a^(4(p-q)/2) ]^n = a^(2n(p-q)) Ans.

⟦line⟧

Ans. ii - ⟦illegible⟧ P. T. O.

**Traduction anglaise —**

Final Exams in Algebra June 12th 1961 Let x shillings be price of one gross. ∴ 12x/144 pence (= x/12 pence) = price of one pencil in pennies ∴ (20 x 12 x)/144 = 5/3 x pence price of one score in pennies in 1st case. ∴ (20 x 12)/x/12 = 2880/x = No. of pencils which can be bought for £1 in 1st case ∴ 2880/x + 120 = (2880 + 120x)/x " " " " " " 2nd case ∴ 240 / (2880 + 120x)/x = 240x / 120(24 + x) = 2x / 24 + x pence = price of one pencil in pennies in 2nd case 20 (2x / x + 24) = 40x / x + 24 pence = price of one score of pencils in 2nd case. ∴ 5/3 x - 40x / x + 24 = 2 ∴ 5x² + 120x - 120x = 6x + 144 or 5x² - 6x - 144 = 0 ∴ (5x + 24)(x - 6) = 0 ∴ x = - 24/5 to be discarded or x = 6 Ans. Hence Price of one gross = 6 shillings Ans. ⟦line⟧ 2. (i) √a - x + √b + x = √a + √b squaring, we have: a - x + 2√(a - x)(b + x) + b + x = a + 2√ab + b or ∴ (a - x)(b + x) = a.b ∴ a.b + (a - b)x - x² = a.b. or x² - (a - b)x = 0 or x[x - (a - b)] = 0 ∴ x = 0 Ans. I or x = a - b Ans. II (ii) { a^(p-q) / √a^(p-q) x a^(p-q) }^(n/(p-q)) = { a^(p-q) / a^((p-q)/2) x a^(p-q) }^n = { a^(p-q - (p-q)/2 + p-q) }^n = = [ a^(4(p-q)/2) ]^n = a^(2n(p-q)) Ans. ⟦line⟧ 3. (i) ⁷√[ (0.001021)² x (4.003)³ / (16.02)⁵ x (3.001)⁴ ] log 0.001021 = 3.0090 | 2 log 0.001021 = 6.0180 | 5 log 16.02 = 6.0230 log 4.003 = 0.6024 | 3 log 4.003 = 1.8072 | 4 log 3.001 = 1.9088 log 16.02 = 1.2046 | log Num. = 5.8252 | log Den. = 7.9318 log 3.001 = 0.4772 | log Den. = 7.9318 | | 7 log x = 13.8934 | | log x = 2.27048... = 2.2705 c. 4 d.p. | | x = 0.01864 Ans. | Ans. ii - ⟦illegible⟧ P. T. O.

---

### csp_9d731aa976f75256ac28f270f9ec64da

B01 · header / latin 27

B02 · paragraph / latin 2(a) The total rateable value of a town is £30,000. How much does the town receive in a year if the rate levied is 16s in the £, (b) If the town estimates that it will need an extra £3750 next year, by how much will the rate in the £ have to be increased if the total rateable value remains unchanged? (c) If the town decided to have the rate in the £ at 16s but to get the extra £3750 by increasing the rateable value of all property, what ought to be the new rateable value of a house which was formerly rated at £32 -

B03 · paragraph / latin solution:- (a) Total rates = £30,000 X 0.8 = £24,000 (b) 3750 / 30,000 = £0.125 = 2s 6d Increase in rate New Rate = 16s + 2s 6d = 18s 6d (c) New Rateable Value / Old Rateable Value = New tax (Rate) / Old tax (Rate) (provided the rate in the £ remains the same) New Rateable Value of the House / Old " " " " = 27,750 / 24,000 New R.V. = Old R.V. X 27,750 / 24,000

**Traduction anglaise —**

27 2(a) The total rateable value of a town is £30,000. How much does the town receive in a year if the rate levied is 16s in the £, (b) If the town estimates that it will need an extra £3750 next year, by how much will the rate in the £ have to be increased if the total rateable value remains unchanged? (c) If the town decided to have the rate in the £ at 16s but to get the extra £3750 by increasing the rateable value of all property, what ought to be the new rateable value of a house which was formerly rated at £32 - solution:- (a) Total rates = £30,000 X 0.8 = £24,000 (b) 3750 / 30,000 = £0.125 = 2s 6d Increase in rate New Rate = 16s + 2s 6d = 18s 6d (c) New Rateable Value / Old Rateable Value = New tax (Rate) / Old tax (Rate) (provided the rate in the £ remains the same) New Rateable Value of the House / Old " " " " = 27,750 / 24,000 New R.V. = Old R.V. X 27,750 / 24,000

---

### csp_9e2c8203c8b65a7482b9848268fc6f2e

Shamash Secondary School 3rd Quarter Examination, March, 1968

Subject: Algebra Date: Sunday 24/3/1968 Class: 4th Year Scientific Time: 10:15 - 11:45

----

1. Draw on the same diagram the graphs of the function 4x-3 and of the function 4x²-4x-15, taking ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (20 marks)

2. From your diagram, find the roots of the two simultaneous equations: y = 4x-3 .........(1) y = 4x²-4x-15 .....(2) (20 marks)

3. From hhe graph of the function 4x²-4x-15, find the roots of the equation 4x² = 4x + 15 (20 marks)

4. From your diagram find also the least value of 4x²-4x-15, and the value of x corresponding to the least value of the function. (20 marks)

5. By drawing an additional graph, find the values of x for which the expression 4x²-4x-15 is always less than 9. (20 marks).

------

**Traduction anglaise —**

Shamash Secondary School 3rd Quarter Examination, March, 1968 Subject: Algebra Date: Sunday 24/3/1968 Class: 4th Year Scientific Time: 10:15 - 11:45 ⟦line⟧ 1. Draw on the same diagram the graphs of the function 4x-3 and of the function 4x²-4x-15, taking ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (20 marks) 2. From your diagram, find the roots of the two simultaneous equations: y = 4x-3 ⟦line⟧ (1) y = 4x²-4x-15 ⟦line⟧ (2) (20 marks) 3. From ⟦the⟧ graph of the function 4x²-4x-15, find the roots of the equation 4x² = 4x + 15 (20 marks) 4. From your diagram find also the least value of 4x²-4x-15, and the value of x corresponding to the least value of the function. (20 marks) 5. By drawing an additional graph, find the values of x for which the expression 4x²-4x-15 is always less than 9. (20 marks). ⟦line⟧

---

### csp_9e52558c64005cc4a85b4c05ddbac38f

B01 · other / latin for two seconds at least 68 ft above the level of the ground

B02 · marginalia / latin 68 ft above the ground

B03 · other / latin level of ground

B04 · paragraph / latin (a) maximum height above ground = 84 ft Ans. (b) the stone remains 2 seconds above the ground level 68 ft Ans. (c) about 4.3 seconds elapse from the time the stone is thrown to the time it strikes the ground

**Traduction anglaise —**

for two seconds at least 68 ft above the level of the ground 68 ft above the ground level of ground (a) maximum height above ground = 84 ft Ans. (b) the stone remains 2 seconds above the ground level 68 ft Ans. (c) about 4.3 seconds elapse from the time the stone is thrown to the time it strikes the ground

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### csp_9ec23b8010b952e8b0cf82cd75332b5b

[Marginalia] Linda Meir Chitayat

Shamash Secondary School Final Examination, May, 1966.

Subject: Algebra Date: 29/5/1966 Class: 4th Year, Scientific. Time: 8:00 - 11:00 a.m.

--- Answer all f i v e questions:

1. (a) Factor the following: (i) my⁴ + 16mx⁴ - 12mx²y² (4 marks) (ii) a²b²x² - a²b² - 2abx² + 2ab + x²-1 (4 marks) (iii) 27x⁶y⁹ + 64y³ (4 marks) (b) Find the value of p and q which will make the expression 2x³ + px² + qx + 1 divisible by (x-1) and (x+1), and find the third factor. (8 marks)

3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: log₃ (2x²-7x) = 2 (10 marks) (ii) Without using tables evaluate: (log₂ 9)(log₉ 32) (10 marks) (iii) Compute the value of y by logarithms, arranging your work neatly: / ⁷/ (tan 19°45')² x (cos 77°16')³ y= V ---------------------------- (10 marks) (3.004)⁵ x (50.06)³

(cont'd.p.2)..

**Traduction anglaise —**

Linda Meir Chitayat Shamash Secondary School Final Examination, May, 1966. Subject: Algebra Date: 29/5/1966 Class: 4th Year, Scientific. Time: 8:00 - 11:00 a.m. ⟦line⟧ Answer all f i v e questions: 1. (a) Factor the following: (i) my⁴ + 16mx⁴ - 12mx²y² (4 marks) (ii) a²b²x² - a²b² - 2abx² + 2ab + x²-1 (4 marks) (iii) 27x⁶y⁹ + 64y³ (4 marks) (b) Find the value of p and q which will make the expression 2x³ + px² + qx + 1 divisible by (x-1) and (x+1), and find the third factor. (8 marks) 2. (a) Use the method of completing the square to show that the sum of the roots of the equation ax²+bx+c=0 is equal to (- b/a) and that their product is equal to ( c/a ). (10 marks) (b) Find the value of x from the following equation: 3.10²ˣ - 13.10ˣ + 4 = 0 (10 marks) ⟦3 x 10²ˣ - 13 x 10ˣ⟧ 3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: log₃ (2x²-7x) = 2 (10 marks) (ii) Without using tables evaluate: (log₂ 9)(log₉ 32) (10 marks) (iii) Compute the value of y by logarithms, arranging your work neatly: / ⁷/ (tan 19°45')² x (cos 77°16')³ y= V ⟦line⟧ (10 marks) (3.004)⁵ x (50.06)³ (cont'd.p.2)..

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### csp_9f7b592d8a7d5186842ab115a0f9a3fa

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION - MAY 1970

Subject: Mathematics Date: 27/5/1970 Class: 4th year Secondary (scientific section) Time: 2 hours

Q1- A chord AB of a circle is produced to T. From T a line TC is drawn to touch the circle at C . If BT = 9 cm. and TC = 12 cm. Find the length of and the ratio of the areas of the triangles BTC and ATC and then prove BC² : AC² = BT : AT .

Q2- Two ships leave the same port at the same time and steam at 10 and 16 m.p.h. respectively in the direction N. 55 W. and S. 75 W. Find their distance apart after 2 hours and the bearing of the first ship to the second .

Q3- PQ , CD are parallel chords of a circle , the tangent at D cuts PQ produ- ced at T , B is a point of contact of the other tangent from T ; prove that BC bisect PQ .

Q4- A man has a certain amount of 4 % stock . He sells it at 114 and invests one - third the proceed in 5% at 95 , and the rest in 2¼ % stock at 108. He then finds that his annual income is reduced by £ 12 10s. . Find the amount of the original stock had he .

Q5- ABC is a triangle in a horizotal plane $ BC = 120 yd. , angle BAC = 90 and angle ABC = 34 . At A is a vertical pole AT , the angle of elevation of T from B = 22 , calculate the height of AT , and the angle of elevation of T from C ,

[Marginalia] ⟦sketch⟧ [Marginalia] (TA) (TB) = (TC)² [Marginalia] 9 (TA) = 144 [Marginalia] TA = 144 / 9 = 16 [Marginalia] AB = 16 - 9 = 7 cm. [Marginalia] Δ ACT [Marginalia] Δ BCT

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION - MAY 1970 Subject: Mathematics Date: 27/5/1970 Class: 4th year Secondary (scientific section) Time: 2 hours Q1- A chord AB of a circle is produced to T. From T a line TC is drawn to touch the circle at C . If BT = 9 cm. and TC = 12 cm. Find the length of and the ratio of the areas of the triangles BTC and ATC and then prove BC² : AC² = BT : AT . Q2- Two ships leave the same port at the same time and steam at 10 and 16 m.p.h. respectively in the direction N. 55 W. and S. 75 W. Find their distance apart after 2 hours and the bearing of the first ship to the second . Q3- PQ , CD are parallel chords of a circle , the tangent at D cuts PQ produ- ced at T , B is a point of contact of the other tangent from T ; prove that BC bisect PQ . Q4- A man has a certain amount of 4 % stock . He sells it at 114 and invests one - third the proceed in 5% at 95 , and the rest in 2¼ % stock at 108. He then finds that his annual income is reduced by £ 12 10s. . Find the amount of the original stock had he . Q5- ABC is a triangle in a horizotal plane $ BC = 120 yd. , angle BAC = 90 and angle ABC = 34 . At A is a vertical pole AT , the angle of elevation of T from B = 22 , calculate the height of AT , and the angle of elevation of T from C , ⟦sketch⟧ (TA) (TB) = (TC)² 9 (TA) = 144 TA = 144 / 9 = 16 AB = 16 - 9 = 7 cm. Δ ACT Δ BCT

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### csp_9fcbe269d6fc57a1acf3ce07a2969184

SHAMASH SECONDARY SCHOOL Conditional Examination, September, 1969.

Subject: Algebra Date: 5/9/1969. Class: 4th Year, Secondary Time: 8:00-11:00 a⟦...⟧

----------------------------------------------------------------- Attempt all questions:

3. A can walk a mile in 2 minutes less time than B would take. In a walking race, B has a start of ¼ mile, and A overtakes B in 10 minutes. Assuming that both men walk at a uniform rate, find their rates of walking in miles per hour. (20 marks).

4. Find the value of x from the following equation : 10²ˣ - 11(10ˣ) + 10 = 0. (10 marks). (ii) Use logarithms to compute the value of the following expression : 7 / (0.002013)²(Sin 15° 12')³ √ --------------------------- (10 marks). (4.004)³(Cos 42° 13')²

5. (i) Plot the curve of the equation y = 2x² - x - 3 at half-unit inter- vals between x = -1.5 and x = 2, choosing one inch as one unit on each of the two axes. (8 marks). (ii) From your graph, find the roots of the equation x + 3 = 2x². (4 marks). (iii) Find the values of x for which the function 2x² - x - 3 is always positive. (4 marks). (iv) By drawing another straight line on your diagram, find the roots of the equation 2x² - x - 1 = 0. (4 marks).

-----------------------------------------------------------------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Conditional Examination, September, 1969. Subject: Algebra Date: 5/9/1969. Class: 4th Year, Secondary Time: 8:00-11:00 a⟦...⟧ ⟦line⟧ Attempt all questions: 1. (i) Find the values of 'a' and 'b' if 3x³ - ax + b is exactly divisible by (x+1)(x-2). If 'a' and 'b' have these values, factor the expre- ssion completely. (10 marks). (ii) If x³ + 3x + 5 = x(x+1)(x+2) + Ax(x+1) + Bx + C for all values of x, find the values of the constants A, B, and C. (10 marks). 2. (i) The 15th term of an arithmetic progression is 25 and the sum of the first 10 terms is 60. Find the first term of the progression, the common difference and the sum of the first 16 terms. (10 marks). (ii) p+ 3, p + 8, and p + 18, are the 3rd, 4th, and 5th terms of a geometric progression. Find the value of p. Find also the common ratio and the 9th term of the progression. (10 marks). 3. A can walk a mile in 2 minutes less time than B would take. In a walking race, B has a start of ¼ mile, and A overtakes B in 10 minutes. Assuming that both men walk at a uniform rate, find their rates of walking in miles per hour. (20 marks). 4. Find the value of x from the following equation : 10²ˣ - 11(10ˣ) + 10 = 0. (10 marks). (ii) Use logarithms to compute the value of the following expression : 7 / (0.002013)²(Sin 15° 12')³ √ ⟦line⟧ (10 marks). (4.004)³(Cos 42° 13')² 5. (i) Plot the curve of the equation y = 2x² - x - 3 at half-unit inter- vals between x = -1.5 and x = 2, choosing one inch as one unit on each of the two axes. (8 marks). (ii) From your graph, find the roots of the equation x + 3 = 2x². (4 marks). (iii) Find the values of x for which the function 2x² - x - 3 is always positive. (4 marks). (iv) By drawing another straight line on your diagram, find the roots of the equation 2x² - x - 1 = 0. (4 marks). ⟦line⟧

---

### csp_a16ce42c148c56c9b385e4260da11472

B01 · other / unknown ⟦illegible⟧

B02 · stamp / latin ELBA

**Traduction anglaise —**

⟦illegible⟧ ELBA

---

### csp_a3a8dcbb63975f5c8e02c6e0591950ac

Shamash Secondary School Final Examination, May, 1967.

Subject: Arithmetic & Trigonometry Date: 17/5/1967 Class: 4th Year Secondary. Time: 8:00 - 10:30 a.m.

--- Answer five questions which must include questions 2 & <del>5</del> 3.

1. (a) How much stock is obtained by investing £2,286 in a 4½ per cent stock at 95¼? After receiving the first annual dividend on this stock, it is immediately resold at 98. Calculate the total gain on the transaction. (b) I have a watch which gains six minutes in every true hour. I put the watch right at 8.30 a.m. What is the latest time indicated by the watch at which I must set out to catch a train which leaves at 10.25 a.m. if it takes me 15 minutes to walk to the station ?

2. Following a storm, water is pumped out of a flooded area through a pipe of 8 in. diameter at the rate of 1,000 gallons per minute. Taking 1 cu.ft. as 6¼ gallons and <del>⟦illegible⟧</del> π as 22/7 , calculate: (a) the speed in ft. per sec. at which the water is passing through the pipe. (b) how many tons of sediment will be pumped out in two days if it is known that the flood water contains ½ oz. of sediment in every cu.ft. of water.

3. A merchant bought 15 tons of potatoes from a farmer at £18 per ton. (a) He sold 4 ton 12 cwt of the potatoes in 1 cwt bags at £1 5s. per bag. The additional cost to the merchant in selling the potatoes in this way was 6s. 3d. per ton. (b) He sold 15 cwt retail at 4d. per lb for which he incurred additional labour costs of £5 10s. (c) He sold the remainder of the potatoes in bulk at £20 per ton. Calculate: (i) the merchant's total costs, including the initial cost of the potatoes, cost of selling the potatoes in bags, and additional labour cost for the retail sales. (ii) the total amount the merchant received from his sales. (iii) the merchant's profit calculated as a percentage, correct to 2 significant figures, of his total costs.

4. A borough is divided into two districts whose rateable values are respectively £52,320 and £127,460. The rate in the first district is 12s. 9d. in the £, and in the second district it is 19s. 3d. in the £. Find the average rate for the whole borough to the nearest farthing.

(cont'd.p.2)..

**Traduction anglaise —**

Shamash Secondary School Final Examination, May, 1967. Subject: Arithmetic & Trigonometry Date: 17/5/1967 Class: 4th Year Secondary. Time: 8:00 - 10:30 a.m. ⟦line⟧ Answer five questions which must include questions 2 & <del>5</del> 3. 1. (a) How much stock is obtained by investing £2,286 in a 4½ per cent stock at 95¼? After receiving the first annual dividend on this stock, it is immediately resold at 98. Calculate the total gain on the transaction. (b) I have a watch which gains six minutes in every true hour. I put the watch right at 8.30 a.m. What is the latest time indicated by the watch at which I must set out to catch a train which leaves at 10.25 a.m. if it takes me 15 minutes to walk to the station ? 2. Following a storm, water is pumped out of a flooded area through a pipe of 8 in. diameter at the rate of 1,000 gallons per minute. Taking 1 cu.ft. as 6¼ gallons and <del>⟦illegible⟧</del> π as 22/7 , calculate: (a) the speed in ft. per sec. at which the water is passing through the pipe. (b) how many tons of sediment will be pumped out in two days if it is known that the flood water contains ½ oz. of sediment in every cu.ft. of water. 3. A merchant bought 15 tons of potatoes from a farmer at £18 per ton. (a) He sold 4 ton 12 cwt of the potatoes in 1 cwt bags at £1 5s. per bag. The additional cost to the merchant in selling the potatoes in this way was 6s. 3d. per ton. (b) He sold 15 cwt retail at 4d. per lb for which he incurred additional labour costs of £5 10s. (c) He sold the remainder of the potatoes in bulk at £20 per ton. Calculate: (i) the merchant's total costs, including the initial cost of the potatoes, cost of selling the potatoes in bags, and additional labour cost for the retail sales. (ii) the total amount the merchant received from his sales. (iii) the merchant's profit calculated as a percentage, correct to 2 significant figures, of his total costs. 4. A borough is divided into two districts whose rateable values are respectively £52,320 and £127,460. The rate in the first district is 12s. 9d. in the £, and in the second district it is 19s. 3d. in the £. Find the average rate for the whole borough to the nearest farthing. (cont'd.p.2)..

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### csp_a3bb95fe24ef51cb89b889c7bcb6e164

y₁ = x³ y₂ = 3x² - 4

| x | y₁ = x³ | y₂ = 3x² - 4 | | -3 | -27 | 23 | | -2 | -8 | 8 | | -1 | -1 | -1 | | 0 | 0 | -4 | | 1 | 1 | -1 | | 2 | 8 | 8 | | 3 | 27 | 23 |

y₁ = x³ y₂ = 3x² - 4 y₂ = 3x² - 4 30 25 20 15 10 5 -1 A(-1, -1) B(2, 8) 5 10 15 20 25 30

(i) The two curves are plotted as they appear in the figure.

(ii) The roots of the equation x³ - 3x² + 4 = 0 are the same as the roots of the equation x³ = 3x² - 4 . These roots are the same as the abscissas of the points of intersections of the two curves y₁ = x³ and y₂ = 3x² - 4 , which are x = -1 and x = 2 . Since the two curves have points A(-1, -1) & B(2, 8) as common points between them.

(iii) ⟦as⟧ The expression x³ - 3x² + 4 is always negative when x³ < 3x² - 4 or when y₁ < y₂ . But y₁ < y₂ for all values of x < -1 since the curve y₁ = x³ lies below the curve y₂ = 3x² - 4 .

(b) Also the expression x³ - 3x² + 4 is always positive when y₁ > y₂ and this is the case for all values of x > -1 ⟦except x=2⟧, since for all these values of x, the curve y₁ lies above the curve y₂ . Q.E.D.

**Traduction anglaise —**

y₁ = x³ y₂ = 3x² - 4 x | y₁ = x³ | y₂ = 3x² - 4 -3 | -27 | 23 -2 | -8 | 8 -1 | -1 | -1 0 | 0 | -4 1 | 1 | -1 2 | 8 | 8 3 | 27 | 23 y₁ = x³ y₂ = 3x² - 4 y₂ = 3x² - 4 30 25 20 15 10 5 -1 A(-1, -1) B(2, 8) 5 10 15 20 25 30 (i) The two curves are plotted as they appear in the figure. (ii) The roots of the equation x³ - 3x² + 4 = 0 are the same as the roots of the equation x³ = 3x² - 4 . These roots are the same as the abscissas of the points of intersections of the two curves y₁ = x³ and y₂ = 3x² - 4 , which are x = -1 and x = 2 . Since the two curves have points A(-1, -1) & B(2, 8) as common points between them. (iii) ⟦as⟧ The expression x³ - 3x² + 4 is always negative when x³ < 3x² - 4 or when y₁ < y₂ . But y₁ < y₂ for all values of x < -1 since the curve y₁ = x³ lies below the curve y₂ = 3x² - 4 . (b) Also the expression x³ - 3x² + 4 is always positive when y₁ > y₂ and this is the case for all values of x > -1 ⟦except x=2⟧, since for all these values of x, the curve y₁ lies above the curve y₂ . Q.E.D.

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### csp_a444aae8fbd45476bd1dabe9213be71f

(3)

4. (i) Let a = 1st term , d = common difference " 3(a+2d) = 2(a+5d) or 3a+6d = 2a+10d or a-4d = 0 .... ① also a + (a+2d) + (a+4d) = 9 or 3a+6d = 9 or a+2d = 3 .... ② ∴ 6d = 3 ∴ d = 1/2 ∴ a = 2 (a) ∴ 9th term / 6th term = a+8d / a+5d = 2+4 / 2+2 1/2 = 6 / 4 1/2 = 12 / 9 = 4/3 Ans. 1 (b) S_13 = n/2 {2a + (n-1)d} = 13/2 {2x2 + 12x1/2} = 13/2 {10} = 65 Ans. 2 (ii) l_3 = 2/3 , l_1 + l_2 = 2 1/2 , a = ? , r = ? , l_4 = ? ar^2 = 2/3 } or ar^2 = 2/3 .... ① } by Division r^2 / 1+r = 2/3 x 2/5 or r^2 / 1+r = 4/15 a + ar = 5/2 } a(1+r) = 5/2 .... ② } ∴ 15r^2 = 4+4r or 15r^2 - 4r - 4 = 0 or (5r+2)(3r-2) = 0 or r = 2/3 and r = -2/5 {the latter value is to be discarded since} {all the terms of the progression are positive, given} ∴ from eq. (1) a(2/3)^2 = 2/3 ∴ 4/9 a = 2/3 ∴ a = 3/2 ∴ l_4 = ar^3 = 3/2 (2/3)^3 = 4/9 a = 3/2 , r = 2/3 , l_4 = 4/9 Ans.

________________________________________________________________ ⟦illegible⟧

**Traduction anglaise —**

(3) 4. (i) Let a = 1st term , d = common difference " 3(a+2d) = 2(a+5d) or 3a+6d = 2a+10d or a-4d = 0 .... ① also a + (a+2d) + (a+4d) = 9 or 3a+6d = 9 or a+2d = 3 .... ② ∴ 6d = 3 ∴ d = 1/2 ∴ a = 2 (a) ∴ 9th term / 6th term = a+8d / a+5d = 2+4 / 2+2 1/2 = 6 / 4 1/2 = 12 / 9 = 4/3 Ans. 1 (b) S_13 = n/2 {2a + (n-1)d} = 13/2 {2x2 + 12x1/2} = 13/2 {10} = 65 Ans. 2 (ii) l_3 = 2/3 , l_1 + l_2 = 2 1/2 , a = ? , r = ? , l_4 = ? ar^2 = 2/3 } or ar^2 = 2/3 .... ① } by Division r^2 / 1+r = 2/3 x 2/5 or r^2 / 1+r = 4/15 a + ar = 5/2 } a(1+r) = 5/2 .... ② } ∴ 15r^2 = 4+4r or 15r^2 - 4r - 4 = 0 or (5r+2)(3r-2) = 0 or r = 2/3 and r = -2/5 {the latter value is to be discarded since} {all the terms of the progression are positive, given} ∴ from eq. (1) a(2/3)^2 = 2/3 ∴ 4/9 a = 2/3 ∴ a = 3/2 ∴ l_4 = ar^3 = 3/2 (2/3)^3 = 4/9 a = 3/2 , r = 2/3 , l_4 = 4/9 Ans. ⟦line⟧ ⟦illegible⟧

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### csp_a70684c7632251718e07810c493d63c3

SHAMASH SCHOOL FINAL EXAMINATIONS 1953-1954

Subject: Trigonometry Date: 26/5/54 Class: Fourth year (secondary) - Time: 10:30-12:00

⟦line⟧ All questions are to be attempted 1. How far down a hill inclined at 7½° to the horizon must I walk in order to descend a distance of 70 ft. vertically? 2. A is 5 miles due South of a port O. A ship steaming at 10 miles an hour starts from O and steams in a straight line to B 1 mile due West of A. From B the ship steams 37° East of South. Calculate the ship's distance from O at the end of one hour after leaving O. 3. Solve the triangle ABC, having given: A = 43° 39', C = 17° 47', b = 4 ft. 4. Two points A and B are at sea level, B being due south of A and distant 2200 feet from it. A third point C, which is 200 feet above sea level, is due east of A and its bearing from B is 047° (N. 47° E.). Find the horizontal distance between B and C and the angle of elevation of C from B, correct to the nearest 10 feet.

θ = ? BD = ?

[Signature] Lecturer: Abdullah Obadiah

[Marginalia] 5.1 = x [Marginalia] 78 41' [Marginalia] 37 [Marginalia] z y [Marginalia] 4.9 [Marginalia] ⟦illegible⟧ [Marginalia] l = 4.952

**Traduction anglaise —**

SHAMASH SCHOOL FINAL EXAMINATIONS 1953-1954 Subject: Trigonometry Date: 26/5/54 Class: Fourth year (secondary) - Time: 10:30-12:00 ⟦line⟧ All questions are to be attempted 1. How far down a hill inclined at 7½° to the horizon must I walk in order to descend a distance of 70 ft. vertically? 2. A is 5 miles due South of a port O. A ship steaming at 10 miles an hour starts from O and steams in a straight line to B 1 mile due West of A. From B the ship steams 37° East of South. Calculate the ship's distance from O at the end of one hour after leaving O. 3. Solve the triangle ABC, having given: A = 43° 39', C = 17° 47', b = 4 ft. 4. Two points A and B are at sea level, B being due south of A and distant 2200 feet from it. A third point C, which is 200 feet above sea level, is due east of A and its bearing from B is 047° (N. 47° E.). Find the horizontal distance between B and C and the angle of elevation of C from B, correct to the nearest 10 feet. θ = ? BD = ? Lecturer: Abdullah Obadiah 5.1 = x 78 41' 37 z y 4.9 ⟦illegible⟧ l = 4.952

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### csp_a7c1e48b660a5adb899717efc95f45b3

Shamash Secondary School 3rd Quarter Examination, March, 1965

Subject: Algebra Date: 14/3/1965 Class: 4th Year, Section (B) Time: 10:15 - 11:45 a.m.

---- Attempt all questions: -

1. Solve simultaneously, the equations: (i) x² - 2xy + 8y² = 8 - - - (1) || (ii) (x-2)(y-1) = 3 - - - (1) 3xy - 2y² = 4 - - - (2) (13 marks) || (x+2)(2y-5) = 15 - (2) (12 marks)

2.(i) Show that 27 - 8x³ - 64y³ - 72xy is divisible by 3 - 2(x + 2y) and find the quotient in this way. (8 marks) (ii) Resolve into six factors x¹⁸ - y¹⁸ (8 marks) (iii) Find the value of x⁴ + x²y² + y⁴ in terms of a and b, having given: x + y = 2a and x - y = 2b (9 marks)

3. A man arrives by air at the airport of his city 3/4 of an hour earlier than the scheduled time, and sets out at once by a taxi, driving to his house at the rate of 20 miles per hour. At the same time, his driver who is supposed to leave his master's house to meet him at the airport in the scheduled time, did so according to plan and, instead, met him on the road to the airport after he has driven his master's private car for a distance of only 50 miles from his house. He immedi- ately picked his master and turned back to his home reaching it exactly 30 minutes earlier than was originally expected. How far is the man's house from the airport and at what rate was his private car driven? (25 marks)

4.(i) Determine the asymptotes and draw the curve of y = x / (x-1), for values of x from x = -2 to x = 4, taking 1 inch as the unit on the x-axis and 0.4 inch as the unit on the y-axis. (7 marks) (ii) Draw in the same diagram the graph of y = x(x - 1.5) for the same values of x. (7 marks) iii From your diagram find as accurately as possible (a) the value of 1.3 / (1.3 - ⟦illegible⟧) (5 marks). (b) two positive numbers differing by 1.5 whose product is 7. (6 marks). Show in your diagram how each answer is obtained.

**Traduction anglaise —**

Shamash Secondary School 3rd Quarter Examination, March, 1965 Subject: Algebra Date: 14/3/1965 Class: 4th Year, Section (B) Time: 10:15 - 11:45 a.m. ⟦line⟧ Attempt all questions: - 1. Solve simultaneously, the equations: (i) x² - 2xy + 8y² = 8 - - - (1) || (ii) (x-2)(y-1) = 3 - - - (1) 3xy - 2y² = 4 - - - (2) (13 marks) || (x+2)(2y-5) = 15 - (2) (12 marks) 2.(i) Show that 27 - 8x³ - 64y³ - 72xy is divisible by 3 - 2(x + 2y) and find the quotient in this way. (8 marks) (ii) Resolve into six factors x¹⁸ - y¹⁸ (8 marks) (iii) Find the value of x⁴ + x²y² + y⁴ in terms of a and b, having given: x + y = 2a and x - y = 2b (9 marks) 3. A man arrives by air at the airport of his city 3/4 of an hour earlier than the scheduled time, and sets out at once by a taxi, driving to his house at the rate of 20 miles per hour. At the same time, his driver who is supposed to leave his master's house to meet him at the airport in the scheduled time, did so according to plan and, instead, met him on the road to the airport after he has driven his master's private car for a distance of only 50 miles from his house. He immedi- ately picked his master and turned back to his home reaching it exactly 30 minutes earlier than was originally expected. How far is the man's house from the airport and at what rate was his private car driven? (25 marks) 4.(i) Determine the asymptotes and draw the curve of y = x / (x-1), for values of x from x = -2 to x = 4, taking 1 inch as the unit on the x-axis and 0.4 inch as the unit on the y-axis. (7 marks) (ii) Draw in the same diagram the graph of y = x(x - 1.5) for the same values of x. (7 marks) iii From your diagram find as accurately as possible (a) the value of 1.3 / (1.3 - ⟦illegible⟧) (5 marks). (b) two positive numbers differing by 1.5 whose product is 7. (6 marks). Show in your diagram how each answer is obtained.

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### csp_a834e93a9e0c5a5bbb6e82b22e73fc43

6. (a) Draw the graph of y = x² - 3x + 2 for values x between -2 and 5. (6 marks) (b) Use your graph to solve the equations: (i) x² - 3x + 2 = 0 (6 marks) (ii) x² - 3x - 4 = 0 (6 marks)

**Traduction anglaise —**

⟦(a)⟧ ⟦If⟧ ⟦the⟧ 2nd term of an arithmetic progression is 2 1/2 times the 5th term, and the arithmetic mean of the 5th and 12th terms is 27. Find the sum of the first 15 terms. (8 marks) (b) Prove that in any Geometric series the sum of the 4th, 5th, and 6th terms is the Geometric mean ⟦sum⟧ of the 1st, 2nd, and 3rd terms and the the 7th, 8th, and 9th terms. (8 marks) 6. (a) Draw the graph of y = x² - 3x + 2 for values x between -2 and 5. (6 marks) (b) Use your graph to solve the equations: (i) x² - 3x + 2 = 0 (6 marks) (ii) x² - 3x - 4 = 0 (6 marks)

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### csp_a8fe6870e0a45f96941a8e063cae8799

B01 · header / latin All questions are to be attempted

B02 · paragraph / latin 1. (a) Resolve into factors : (i) 3 (2x-5)² + 4(2x-5) - 32 ⟦illegible⟧ (ii) 5 (2y-4)² + 12(2y-4) - 9 ⟦illegible⟧ (iii) 4(x²-3y²)² + (x²-3y²)x - 3x²

B03 · marginalia / latin a = 5bk / 4(5b-k) b = 4ak / 5(4a-k) 4a / 5b

B04 · paragraph / latin ⟦illegible⟧ k = 20ab / 4a+5b , find: (i) "a" in terms of "b" and "k" (2 marks) (ii) "b" in terms of "a" and "k" (2 marks) (iii) The value of √((k-4a)/(k-5b)) in terms of a/b (6 marks)

B05 · paragraph / latin 2. (i) Using tables, compute by logarithms the value : ³√((0.002037² x 2.005) / 40.03²) (10 marks)

B06 · paragraph / latin (ii) Solve for x the equation 3²ˣ - 28 x 3ˣ⁻¹ + 3 = 0 (10 marks)

B07 · paragraph / latin 3. A man in a speed-boat sees the flash of a gun fired a ship directly towards him and hears the report 20 seconds later. The boat is travelling to meet the ship at 30 miles per hour. Find the distance between the two when the shot was fired, taking the velocity of sound to be 1100 ft. per second. (20 marks)

B08 · paragraph / latin 4. Draw the graph of y = 6 + 3x - x² for values of x from -2 to 5 taking 1 inch as unit on the x-axis and 1/2 inch as unit on the y-axis. From your graph find: (i) the maximum value of y (ii) the values of x between which the function 6 + 3x - x² is positive.

B09 · paragraph / latin 5. (i) What kind of series is 1/4, 3/10, 7/20, 2/5, ...? Find its nth term and the sum of the first ten terms. (10 marks) (ii) ⟦illegible⟧ 36

**Traduction anglaise —**

All questions are to be attempted 1. (a) Resolve into factors : (i) 3 (2x-5)² + 4(2x-5) - 32 ⟦illegible⟧ (ii) 5 (2y-4)² + 12(2y-4) - 9 ⟦illegible⟧ (iii) 4(x²-3y²)² + (x²-3y²)x - 3x² a = 5bk / 4(5b-k) b = 4ak / 5(4a-k) 4a / 5b ⟦illegible⟧ k = 20ab / 4a+5b , find: (i) "a" in terms of "b" and "k" (2 marks) (ii) "b" in terms of "a" and "k" (2 marks) (iii) The value of √((k-4a)/(k-5b)) in terms of a/b (6 marks) 2. (i) Using tables, compute by logarithms the value : ³√((0.002037² x 2.005) / 40.03²) (10 marks) (ii) Solve for x the equation 3²ˣ - 28 x 3ˣ⁻¹ + 3 = 0 (10 marks) 3. A man in a speed-boat sees the flash of a gun fired a ship directly towards him and hears the report 20 seconds later. The boat is travelling to meet the ship at 30 miles per hour. Find the distance between the two when the shot was fired, taking the velocity of sound to be 1100 ft. per second. (20 marks) 4. Draw the graph of y = 6 + 3x - x² for values of x from -2 to 5 taking 1 inch as unit on the x-axis and 1/2 inch as unit on the y-axis. From your graph find: (i) the maximum value of y (ii) the values of x between which the function 6 + 3x - x² is positive. 5. (i) What kind of series is 1/4, 3/10, 7/20, 2/5, ...? Find its nth term and the sum of the first ten terms. (10 marks) (ii) ⟦illegible⟧ 36

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### csp_aa1e7f9cb8cc587da04715ebacd0462e

Shamash Secondary School Mid-Year Examination, Feb. 1968

Subject: Algebra Date: 7/2/1968 Class: 4th Year, Scientific Section. Time: 8:30 - 11:00 a.m.

---- Attempt all questions:

1. (a) Resolve into factors: (i) (2a+b)x² - (a-b)x - (a + 2b) (4 marks) (ii) 201x² - 99x - 102 (4 " ) (iii) x⁹ - 64x³ - x⁶ + 64 (4 " ) (b) Show that any common factor of A and B is also a factor of mA & nB. (8 ma⟦rk⟧

2. (a) The following equation is true for all values of x: (2x-3)²-c = 2Ax² - 4Bx. Find the values of A,B and C. (8 " ) (b) Of the following three equations, one is always true, one is sometimes true and one is never true. Find which is which, giving your reasons: (i) 3x(x-4)+x = 5(x²-1)+13-11x (4 marks) (ii) x²(2x-5)+3(x-1) = 2x³-x(5x-3)-3 (4 " ) (iii) x(x²-1)+2(1+x)(1-x) = 0 (4 " )

3. (i) Solve the equation: 3x³-14x²+32 = 0 (10 " ) (ii) Solve the two simultaneous equations: x+y+2xy+x²+y² = 0 ...........(1) x-y-2xy+x²+y² = 6 ...........(2) (10 " )

4. (i) Running separately, two taps can fill a bath with water in "a" and "b" minutes respectively. Prove that they take ab/a+b minutes to fill it when running together. (10 marks) (ii) If, when they are running separately, the first tap can fill the bath in 7 minutes less time than the second, and when they are running together they fill it in 12 minutes, find the values of "a" and "b". (10 m⟦a⟧

5. (i) In an examination taken by both boys and girls, 41 candidates out of every 68 pass. Five boys out of every 8 pass and 7 girls out of every 12 pass. Find the ratio of boy candidates to girl candidates. (ii) If 168 girls passed the examination, find the total number of candidates.

-------

**Traduction anglaise —**

Shamash Secondary School Mid-Year Examination, Feb. 1968 Subject: Algebra Date: 7/2/1968 Class: 4th Year, Scientific Section. Time: 8:30 - 11:00 a.m. ⟦line⟧ Attempt all questions: 1. (a) Resolve into factors: (i) (2a+b)x² - (a-b)x - (a + 2b) (4 marks) (ii) 201x² - 99x - 102 (4 " ) (iii) x⁹ - 64x³ - x⁶ + 64 (4 " ) (b) Show that any common factor of A and B is also a factor of mA & nB. (8 ma⟦rk⟧ 2. (a) The following equation is true for all values of x: (2x-3)²-c = 2Ax² - 4Bx. Find the values of A,B and C. (8 " ) (b) Of the following three equations, one is always true, one is sometimes true and one is never true. Find which is which, giving your reasons: (i) 3x(x-4)+x = 5(x²-1)+13-11x (4 marks) (ii) x²(2x-5)+3(x-1) = 2x³-x(5x-3)-3 (4 " ) (iii) x(x²-1)+2(1+x)(1-x) = 0 (4 " ) 3. (i) Solve the equation: 3x³-14x²+32 = 0 (10 " ) (ii) Solve the two simultaneous equations: x+y+2xy+x²+y² = 0 ...........(1) x-y-2xy+x²+y² = 6 ...........(2) (10 " ) 4. (i) Running separately, two taps can fill a bath with water in "a" and "b" minutes respectively. Prove that they take ab/a+b minutes to fill it when running together. (10 marks) (ii) If, when they are running separately, the first tap can fill the bath in 7 minutes less time than the second, and when they are running together they fill it in 12 minutes, find the values of "a" and "b". (10 m⟦a⟧ 5. (i) In an examination taken by both boys and girls, 41 candidates out of every 68 pass. Five boys out of every 8 pass and 7 girls out of every 12 pass. Find the ratio of boy candidates to girl candidates. (ii) If 168 girls passed the examination, find the total number of candidates. ⟦line⟧

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### csp_aa4f57a5e1e95336b25e98811ada4cbc

∴ 3^{n-1} = 3^8 ∴ n-1 = 8 ∴ n = 9 Ans.

(b) Let the first term of the A.P. = a , the common diff = d ∴ 1st term = a ∴ a, (a+d), (a+3d) form a G.P. 2nd term = a+d 4th term = a+3d ∴ a+3d / a+d = a+d / a ∴ (a+d)^2 = a(a+3d) or a^2 + 2ad + d^2 = a^2 + 3ad ∴ d^2 = ad ∴ d = a Ans. Q.E.D.

5. f(x) = x^2 - 6x + 5

| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | | y | 12 | 5 | 0 | -3 | -4 | -3 | 0 | 5 | 12 |

(a) the least value of the function is (-4) at x = 3 (b) the roots of x^2 + 5 = 6x are the same as the roots of x^2 - 6x + 5 = 0 they are at A + B or x = 1 and x = 5 (c) Draw the graph of y = x - 1 it will intersect the first curve at A (1, 0) + C (6, 5) the function x^2 - 6x + 5 < x - 1 between x = 1 and x = 6 (d) the points A(1, 0) + C(6, 5) lie on both curves so they satisfy both equations y = x^2 - 6x + 5 and y = x - 1 Hence x = 1 and x = 6 are the roots of x^2 - 6x + 5 = x - 1

⟦Graph showing parabola and intersecting line with points A(1,0), B(5,0), C(6,5) and vertex (3,-4)⟧

**Traduction anglaise —**

4 (a) 1 + 1/3 + 1/9 + ... to n terms S_n = 1[1-(1/3)^n] / 1-1/3 = 3/2 (1 - 1/3^n) = 3/2 - 1 / 2x3^{n-1} ∴ S_n = 1.5 - 1 / 2x3^{n-1} Ans. 1 When S_n = 3/2 - 1/13122 , then 1.5 - 1 / 2x3^{n-1} = 3/2 - 1/13122 ∴ 1 / 2x3^{n-1} = 1 / 13122 ∴ 2x3^{n-1} = 13122 ∴ 3^{n-1} = 6561 but 6561 | 3 2187 | 3 729 | 3 243 | 3 81 | 3 ∴ 3^{n-1} = 3^8 ∴ n-1 = 8 ∴ n = 9 Ans. (b) Let the first term of the A.P. = a , the common diff = d ∴ 1st term = a ∴ a, (a+d), (a+3d) form a G.P. 2nd term = a+d 4th term = a+3d ∴ a+3d / a+d = a+d / a ∴ (a+d)^2 = a(a+3d) or a^2 + 2ad + d^2 = a^2 + 3ad ∴ d^2 = ad ∴ d = a Ans. Q.E.D. 5. f(x) = x^2 - 6x + 5 x | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 y | 12 | 5 | 0 | -3 | -4 | -3 | 0 | 5 | 12 (a) the least value of the function is (-4) at x = 3 (b) the roots of x^2 + 5 = 6x are the same as the roots of x^2 - 6x + 5 = 0 they are at A + B or x = 1 and x = 5 (c) Draw the graph of y = x - 1 it will intersect the first curve at A (1, 0) + C (6, 5) the function x^2 - 6x + 5 < x - 1 between x = 1 and x = 6 (d) the points A(1, 0) + C(6, 5) lie on both curves so they satisfy both equations y = x^2 - 6x + 5 and y = x - 1 Hence x = 1 and x = 6 are the roots of x^2 - 6x + 5 = x - 1 ⟦Graph showing parabola and intersecting line with points A(1,0), B(5,0), C(6,5) and vertex (3,-4)⟧

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### csp_aa80fa64d4455565af1a0c30d79da3a4

Shamash Secondary School 1st Quarter Examination, November, 1966.

Subject: Algebra Date: 8/11/1966 Class: 4th Secondary, Scientific Section. Time: 12:00 - 1:30 p.m.

all questions are to be attempted

1. (i) Given √xy + zx / √zx - yz = 3/4 solve each x, y, z in terms of the other two (6 marks) (ii) Simplify by removing brackets 35 [ 3x - 4y / 5 - 1/10 { 3x - 5/7 ( 7x - 4y ) } ] + 8 ( y - 2x ) ( 8 marks )

2. (i) Resolve into two factors (if possible) each of the following expressions ① a⁷ - b⁷ ② a⁶ + b⁶ ③ a⁶ - b⁶ ④ a⁹ + b⁹ (12 marks) (ii) Write down by inspection the quotient of (2a)⁵ - (3b)⁵ / 2a - 3b (4 marks)

3. (i) Solve the equation 2.4 = 0.24 / 0.6 - 0.16x - 7.6 / 0.8 (9 marks) (ii) Walking 5 1/2 miles an hour, I start 2 1/2 hours after a friend whose pace is 3 1/2 miles an hour. How long shall I be in overtaking him? (8 marks)

4. (i) How many days will "n" men take to mow "a" acres if "b" boys can mow "y" acres in "d" days and each man's work equals that of "u" boys? (9 marks) (ii) Find the square root of : 16 x⁴ + 16/3 x²y + 8x² + 4/9 y² + 4/3 y + 1 showing your steps neatly. (8 marks)

**Traduction anglaise —**

Shamash Secondary School 1st Quarter Examination, November, 1966. Subject: Algebra Date: 8/11/1966 Class: 4th Secondary, Scientific Section. Time: 12:00 - 1:30 p.m. all questions are to be attempted 1. (i) Given √xy + zx / √zx - yz = 3/4 solve each x, y, z in terms of the other two (6 marks) (ii) Simplify by removing brackets 35 [ 3x - 4y / 5 - 1/10 { 3x - 5/7 ( 7x - 4y ) } ] + 8 ( y - 2x ) ( 8 marks ) 2. (i) Resolve into two factors (if possible) each of the following expressions ① a⁷ - b⁷ ② a⁶ + b⁶ ③ a⁶ - b⁶ ④ a⁹ + b⁹ (12 marks) (ii) Write down by inspection the quotient of (2a)⁵ - (3b)⁵ / 2a - 3b (4 marks) 3. (i) Solve the equation 2.4 = 0.24 / 0.6 - 0.16x - 7.6 / 0.8 (9 marks) (ii) Walking 5 1/2 miles an hour, I start 2 1/2 hours after a friend whose pace is 3 1/2 miles an hour. How long shall I be in overtaking him? (8 marks) 4. (i) How many days will "n" men take to mow "a" acres if "b" boys can mow "y" acres in "d" days and each man's work equals that of "u" boys? (9 marks) (ii) Find the square root of : 16 x⁴ + 16/3 x²y + 8x² + 4/9 y² + 4/3 y + 1 showing your steps neatly. (8 marks)

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### csp_aab15b5ee3fb508687ebb60701e826f5

Shamash Secondary School 1st. Quarter Exam.

Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary. Date: 8/12/1957

-------- All questions are to be attempted.

(2) (a) Decimalise to 3 places the following:

£ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 1/4 d

(b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint.

(3) A watch which gains 5 sec.in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 0'clock ?

(4) The average age of m boys is b years and of n girls is c years. Find the average age of all together.

(5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20°W., find the distance of the fort from the ship at the first observation.

(6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42°7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

________

**Traduction anglaise —**

Shamash Secondary School 1st. Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary. Date: 8/12/1957 ⟦line⟧ All questions are to be attempted. (1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m, by 25.00 m. (2) (a) Decimalise to 3 places the following: £ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 1/4 d (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint. (3) A watch which gains 5 sec.in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 0'clock ? (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together. (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20°W., find the distance of the fort from the ship at the first observation. (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42°7', and the angle of depres- sion of its foot is 17° 59'. Find its height. ⟦line⟧

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### csp_ab2c5487f84e53debba7b6fa9731d044

Solutions to 2nd Quarter Exam in Algebra Cont. 27/12/1966 page 3

4. y/x hrs. = No. of hrs. taken by fast train to cover the distance y miles. Let v m.p.h. be the speed of the slower train. ∴ z/v hrs. = No of hrs taken by slower train to travel distance z ∴ z/v = y/x + t ∴ xz = vy + txv ∴ v(y + tx) = xz ∴ v = xz / (y + tx) & the difference between speeds = (x - v) m.p.h. ∴ diff. between the two speeds = x - xz / (y + tx) = (xy + tx² - xz) / (tx + y) = (tx² + x(y - z)) / (tx + y) Ans.

[Marginalia] (20 Marks)

5. Let the time when he started be x minutes after 3 ∴ x = 15 + x/12 ∴ 12x = 180 + x ∴ 11x = 180 ∴ x = 180/11 = 16 4/11 minutes

⟦Diagram of a clock face showing approx 3:16⟧ 1st case

Let the time when he finished = y minutes after five ∴ y = 25 + y/12 ∴ 12y = 300 + y ∴ 11y = 300 ∴ y = 300/11 = 27 3/11 minutes after five

⟦Diagram of a clock face showing approx 5:27⟧ 2nd case

he began at 16 4/11 min. after 3, and ended at 27 3/11 min past five. he walked for a period of 2 hrs 10 10/11 minutes Ans. = (5 hrs. 27 3/11 min - 3 hrs. 16 4/11 min)

[Marginalia] (20 marks)

**Traduction anglaise —**

Solutions to 2nd Quarter Exam in Algebra Cont. 27/12/1966 page 3 4. y/x hrs. = No. of hrs. taken by fast train to cover the distance y miles. Let v m.p.h. be the speed of the slower train. ∴ z/v hrs. = No of hrs taken by slower train to travel distance z ∴ z/v = y/x + t ∴ xz = vy + txv ∴ v(y + tx) = xz ∴ v = xz / (y + tx) & the difference between speeds = (x - v) m.p.h. ∴ diff. between the two speeds = x - xz / (y + tx) = (xy + tx² - xz) / (tx + y) = (tx² + x(y - z)) / (tx + y) Ans. (20 Marks) 5. Let the time when he started be x minutes after 3 ∴ x = 15 + x/12 ∴ 12x = 180 + x ∴ 11x = 180 ∴ x = 180/11 = 16 4/11 minutes ⟦Diagram of a clock face showing approx 3:16⟧ 1st case Let the time when he finished = y minutes after five ∴ y = 25 + y/12 ∴ 12y = 300 + y ∴ 11y = 300 ∴ y = 300/11 = 27 3/11 minutes after five ⟦Diagram of a clock face showing approx 5:27⟧ 2nd case he began at 16 4/11 min. after 3, and ended at 27 3/11 min past five. he walked for a period of 2 hrs 10 10/11 minutes Ans. = (5 hrs. 27 3/11 min - 3 hrs. 16 4/11 min) (20 marks)

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### csp_ac40813199f75b6ea9b418a384003bec

Shamash Secondary School Final Exams. June, 1964.

Subject: Algebra Date: 3/6/1964 Class: 4th Secondary (Scientific Section) Time: 8:00-10:45 a.m.

------

1. (a) Draw the graph of Y = 2(2X-1) / 3(X-1) for values of X from X=-2 to X =4, choosing one inch to represent one unit on the X-axis and 9 tenths of an inch to represent one unit on the Y-axis. (8 marks) (b) Plot another graph on the same axes to find a value of X for which 2(2X-1) / 3(X-1) = X and verify the result by solving this equation algebraically. ( 8 marks)

2. (a) Solve simultaneously: X³ + Y³ = 9 X²Y + XY² = 6 ( 8 marks) (b) Find all values of X which satisfy the equation: X⁴ + 4 = 5X² ( 8 marks)

3. (a) Find the factors of: i- X³ + Y³ + 5X²Y + 5XY² ( 4 marks) ii- (X+Y+Z)² + X² - Y² - Z² ( 4 marks) iii- (a² - b²)(x² - y²) + 4abxy ( 4 marks) (b) If X + 1/X = √3, prove that X³ + 1/X³ = 0 ( 8 marks)

4. (a) One boat in a race was rowed over the course at an average pace of 4 yards a second; the other moved over the first half of the course at the rate of 3½ yards a second, and over the last half at the rate of 4½ yards a second. Which of them won and by how many seconds ? ( 8 marks) (b) A person has 'a' hours free. How far can he ride at 'b' miles an hour so that walking back at 'c' miles an hour he may reach home in time ? ( 8 marks)

(cont'd.p.2)

**Traduction anglaise —**

Shamash Secondary School Final Exams. June, 1964. Subject: Algebra Date: 3/6/1964 Class: 4th Secondary (Scientific Section) Time: 8:00-10:45 a.m. ⟦line⟧ 1. (a) Draw the graph of Y = 2(2X-1) / 3(X-1) for values of X from X=-2 to X =4, choosing one inch to represent one unit on the X-axis and 9 tenths of an inch to represent one unit on the Y-axis. (8 marks) (b) Plot another graph on the same axes to find a value of X for which 2(2X-1) / 3(X-1) = X and verify the result by solving this equation algebraically. ( 8 marks) 2. (a) Solve simultaneously: X³ + Y³ = 9 X²Y + XY² = 6 ( 8 marks) (b) Find all values of X which satisfy the equation: X⁴ + 4 = 5X² ( 8 marks) 3. (a) Find the factors of: i- X³ + Y³ + 5X²Y + 5XY² ( 4 marks) ii- (X+Y+Z)² + X² - Y² - Z² ( 4 marks) iii- (a² - b²)(x² - y²) + 4abxy ( 4 marks) (b) If X + 1/X = √3, prove that X³ + 1/X³ = 0 ( 8 marks) 4. (a) One boat in a race was rowed over the course at an average pace of 4 yards a second; the other moved over the first half of the course at the rate of 3½ yards a second, and over the last half at the rate of 4½ yards a second. Which of them won and by how many seconds ? ( 8 marks) (b) A person has 'a' hours free. How far can he ride at 'b' miles an hour so that walking back at 'c' miles an hour he may reach home in time ? ( 8 marks) (cont'd.p.2)

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### csp_ac53affedf8350369336dc9500c3ab19

B01 · header / latin Shamash Secondary School Final Examination, May, 1967.

B02 · form / latin Subject:: Arithmetic & Trigonometry Date:: 17/5/1967 Class:: 4th Year Secondary. Time:: 8:00 - 10:30 a.m.

B03 · other / latin ⟦line⟧

B04 · marginalia / latin 3

B05 · paragraph / latin Answer five questions which must include questions 2 & 5.

B06 · paragraph / latin 1. (a) How much stock is obtained by investing £2,286 in a 4½ per cent stock at 95¼? After receiving the first annual dividend on this stock, it is immediately resold at 98. Calculate the total gain on the transaction. (b) I have a watch which gains six minutes in every true hour. I put the watch right at 8.30 a.m. What is the latest time indicated by the watch at which I must set out to catch a train which leaves at 10.25 a.m. if it takes me 15 minutes to walk to the station ?

B07 · paragraph / latin 2. Following a storm, water is pumped out of a flooded area through a pipe of 8 in. diameter at the rate of 1,000 gallons per minute. Taking 1 cu.ft as 6¼ gallons and π as 22/7, calculate: (a) the speed in ft. per sec. at which the water is passing through the pipe. (b) how many tons of sediment will be pumped out in two days if it is known that the flood water contains ½ oz. of sediment in every cu.ft of water.

B08 · paragraph / latin 3. A merchant bought 15 tons of potatoes from a farmer at £18 per ton. (a) He sold 4 ton 12 cwt of the potatoes in 1 cwt bags at £1 5s. per bag. The additional cost to the merchant in selling the potatoes in this way was 6s. 3d. per ton. (b) He sold 15 cwt retail at 4d. per lb for which he incurred additional labour costs of £5 10s. (c) He sold the remainder of the potatoes in bulk at £20 per ton. Calculate: (i) the merchant's total costs, including the initial cost of the potatoes, cost of selling the potatoes in bags, and additional labour cost for the retail sales. (ii) the total amount the merchant received from his sales. (iii) the merchant's profit calculated as a percentage, correct to 2 significant figures, of his total costs.

B09 · paragraph / latin 4. A borough is divided into two districts whose rateable values are respectively £52,320 and £127,460. The rate in the first district is 12s. 9d. in the £, and in the second district it is 19s. 3d. in the £. Find the average rate for the whole borough to the nearest farthing.

B10 · footer / latin (cont'd.p.2)..

**Traduction anglaise —**

Shamash Secondary School Final Examination, May, 1967. Subject:: Arithmetic & Trigonometry Date:: 17/5/1967 Class:: 4th Year Secondary. Time:: 8:00 - 10:30 a.m. ⟦line⟧ 3 Answer five questions which must include questions 2 & 5. 1. (a) How much stock is obtained by investing £2,286 in a 4½ per cent stock at 95¼? After receiving the first annual dividend on this stock, it is immediately resold at 98. Calculate the total gain on the transaction. (b) I have a watch which gains six minutes in every true hour. I put the watch right at 8.30 a.m. What is the latest time indicated by the watch at which I must set out to catch a train which leaves at 10.25 a.m. if it takes me 15 minutes to walk to the station ? 2. Following a storm, water is pumped out of a flooded area through a pipe of 8 in. diameter at the rate of 1,000 gallons per minute. Taking 1 cu.ft as 6¼ gallons and π as 22/7, calculate: (a) the speed in ft. per sec. at which the water is passing through the pipe. (b) how many tons of sediment will be pumped out in two days if it is known that the flood water contains ½ oz. of sediment in every cu.ft of water. 3. A merchant bought 15 tons of potatoes from a farmer at £18 per ton. (a) He sold 4 ton 12 cwt of the potatoes in 1 cwt bags at £1 5s. per bag. The additional cost to the merchant in selling the potatoes in this way was 6s. 3d. per ton. (b) He sold 15 cwt retail at 4d. per lb for which he incurred additional labour costs of £5 10s. (c) He sold the remainder of the potatoes in bulk at £20 per ton. Calculate: (i) the merchant's total costs, including the initial cost of the potatoes, cost of selling the potatoes in bags, and additional labour cost for the retail sales. (ii) the total amount the merchant received from his sales. (iii) the merchant's profit calculated as a percentage, correct to 2 significant figures, of his total costs. 4. A borough is divided into two districts whose rateable values are respectively £52,320 and £127,460. The rate in the first district is 12s. 9d. in the £, and in the second district it is 19s. 3d. in the £. Find the average rate for the whole borough to the nearest farthing. (cont'd.p.2)..

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### csp_ac69fbd33ae45a24a7cbe80164b96b0f

SHAMASH SECONDARY SCHOOL Final Examination, May 1966

Subject: Arithmetic & Trigonometry Date: 18.5.1966 Class: 4th Year Secondary Time: 8:00-10:30 a.m.

- - - Attempt five questions only including question (4).

⟦قليل⟧ 1. A person, having bought a certain amount of 2¾% stock at 95, afterwards sold it, and with the proceeds bought 3½% stock. He obtained £900 less stock than before, but his income was unchanged. How much money did he originally invest? (20 marks)

⟦مستقل⟧ 3. (a) A watch was 5 minutes fast at 9 a.m. on Monday, and 10 minutes slow at 12 noon on the following Wednesday. Find when it was exactly right, assuming that it lost time uniformly. Note:(9 a.m. and 12 noon are correct time). (10 marks) (b) Two clocks sound the first stroke of 12 o'clock at the same instant; one clock allows an interval of 20 secs. between each stroke and the next, and the other allows 25 secs. How many strokes of the slower clock remain after the quicker one has finished striking, and what time will elapse between the 12th stroke of the quicker one and the following stroke of the slower one? (10 marks)

⟦جيد⟧ 4. (a) A solid consists of a hemisphere, radius 8 cm., joined to a cone of the same base-radius and height 6 cm., so that the plane surfaces coincide. Find (i) the volume, (ii) the total area of the surface of the solid. (Give answer to 3 significant figures). (10 marks) (b) A sphere of radius 3 in. is filed down into the greatest possible cube; find the volume of the material removed. (Give answer to 4 significant figures). (10 marks)

5. Find the difference between the perimeters of a regular pentagon and a regular hexagon, each of which has an area of 24 square inches. ⟦سهلة⟧ (20 marks)

------------------------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Final Examination, May 1966 Subject: Arithmetic & Trigonometry Date: 18.5.1966 Class: 4th Year Secondary Time: 8:00-10:30 a.m. ⟦line⟧ Attempt five questions only including question (4). ⟦little⟧ 1. A person, having bought a certain amount of 2¾% stock at 95, afterwards sold it, and with the proceeds bought 3½% stock. He obtained £900 less stock than before, but his income was unchanged. How much money did he originally invest? (20 marks) 2. A house holder owns his house which has a rateable value of £44 on which the annual rates are charged at 21s 10d in the £1. He also has to pay an annual property tax at the rate of 9s in the £ on an assessment of £44. Calculate, correct to the nearest penny, the average cost per week of the total of these charges, taking a year as 52 weeks. He subsequently sells his house for £3300, which sum he invests at the rate of 2½% per annum free of tax, and moves into a flat which he rents at £126 per annum. He has however to rent a garage for his car at 7s 6d per week. Find how much per annum he saves by the change. (20 marks) ⟦independent⟧ 3. (a) A watch was 5 minutes fast at 9 a.m. on Monday, and 10 minutes slow at 12 noon on the following Wednesday. Find when it was exactly right, assuming that it lost time uniformly. Note:(9 a.m. and 12 noon are correct time). (10 marks) (b) Two clocks sound the first stroke of 12 o'clock at the same instant; one clock allows an interval of 20 secs. between each stroke and the next, and the other allows 25 secs. How many strokes of the slower clock remain after the quicker one has finished striking, and what time will elapse between the 12th stroke of the quicker one and the following stroke of the slower one? (10 marks) ⟦good⟧ 4. (a) A solid consists of a hemisphere, radius 8 cm., joined to a cone of the same base-radius and height 6 cm., so that the plane surfaces coincide. Find (i) the volume, (ii) the total area of the surface of the solid. (Give answer to 3 significant figures). (10 marks) (b) A sphere of radius 3 in. is filed down into the greatest possible cube; find the volume of the material removed. (Give answer to 4 significant figures). (10 marks) 5. Find the difference between the perimeters of a regular pentagon and a regular hexagon, each of which has an area of 24 square inches. ⟦easy⟧ (20 marks) 6. In response to an S O S call from a ship at A, another ship at B, 175 miles due east of A, starts toward A at a speed of 12 miles per hour. At the same time a third ship at C, which is 186 miles from B in a direction bearing ⟦30⟧° 15' west of north, also starts for A at a speed of 16 miles per hour. Which ship will reach A first, and how long will it take ? (20 marks) ⟦line⟧

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### csp_ad03482cb07f583c8979ebd23e380e9d

6. (a) X = ⁵√[ (1.005)³ × (0.0004007)⁵ / (0.06109)² × (10.71)⅓ ] log 1.005 = 0.0021 / 3 log 1.005 = 0.0063 | 2 log 0.06109 = 3̅.5718 log 0.0004007 = 4̅.6029 / 5 log 0.0004007 = 1̅7̅.0145 | ⅓ log 10.71 = 0.3433 log 0.06109 = 2̅.7859 / | + log 10.71 = 1.0298 / log Num. = 1̅7̅.0208 | log Den. = 3̅.9151 log Den. = 3̅.9151 | ③ 5 log x = 1̅5̅.1057 log x = 3̅.02114 ② x = 1.050 X 10⁻³ = 0.001050 Ans.

(b) log₁₀ 2 = 0.30103 , log₁₀ 3 = 0.47712 , log₃ 648 = ? log₃ 648 = log₃ 2³ × 3⁴ = log₃ 2³ + log₃ 3⁴ = 3 log₃ 2 + 4 log₃ 3 = = 4 + 3 ( log₁₀ 2 / log₁₀ 3 ) = 4 + 3 × 0.30103 / 0.47712 = 4 + .90309 / .47712 = 4 + 1.892794 = 5.892794 = 5.89279 Correct to five decimal places Ans.

[Marginalia] ② ⑥ [Marginalia] 3⁴ = (648)

3 log 2 : 3 × 0.30103 = .90309 4 log 3 : 4 × 0.47712 = 1.90848 2.81157 log₃ 648 = log₁₀ 648 / log₁₀ 3 = 2.81157 / 0.47712 = 5.892794 = 5.89279 Correct to 5 dec. pl. Ans

**Traduction anglaise —**

3 6. (a) X = ⁵√[ (1.005)³ × (0.0004007)⁵ / (0.06109)² × (10.71)⅓ ] log 1.005 = 0.0021 / 3 log 1.005 = 0.0063 | 2 log 0.06109 = 3̅.5718 log 0.0004007 = 4̅.6029 / 5 log 0.0004007 = 1̅7̅.0145 | ⅓ log 10.71 = 0.3433 log 0.06109 = 2̅.7859 / | + log 10.71 = 1.0298 / log Num. = 1̅7̅.0208 | log Den. = 3̅.9151 log Den. = 3̅.9151 | ③ 5 log x = 1̅5̅.1057 log x = 3̅.02114 ② x = 1.050 X 10⁻³ = 0.001050 Ans. (b) log₁₀ 2 = 0.30103 , log₁₀ 3 = 0.47712 , log₃ 648 = ? log₃ 648 = log₃ 2³ × 3⁴ = log₃ 2³ + log₃ 3⁴ = 3 log₃ 2 + 4 log₃ 3 = = 4 + 3 ( log₁₀ 2 / log₁₀ 3 ) = 4 + 3 × 0.30103 / 0.47712 = 4 + .90309 / .47712 = 4 + 1.892794 = 5.892794 = 5.89279 Correct to five decimal places Ans. ② ⑥ 3⁴ = (648) 3 log 2 : 3 × 0.30103 = .90309 4 log 3 : 4 × 0.47712 = 1.90848 2.81157 log₃ 648 = log₁₀ 648 / log₁₀ 3 = 2.81157 / 0.47712 = 5.892794 = 5.89279 Correct to 5 dec. pl. Ans

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### csp_ad0c453f43ad5dbc8de8212a8c857d66

B01 · header / latin Solution to Arithmetic + Trig Exam. Mid-year Session., Jan., 29th 1967 4th Year.

B02 · other / latin ⟦line⟧ 24 ft 52 a b

B03 · paragraph / latin a = 24 sin 52 = 24 x 0.7880 = 18.912 ft b = 24 cos 52 = 24 x 0.6157 = 14.7768 ft.

B04 · paragraph / latin 18.912 - 2 = 16.912 ft

B05 · paragraph / latin sin θ = 16.912 / 24 = 0.7047 ∴ θ = 44° 48'

B06 · other / latin 24 θ b' 16.912

B07 · paragraph / latin b' : 24 cos 44° 48' = 24 x 0.7096 = 17.0304

B08 · table / latin 17.0304 14.7768 ⟦line⟧ 2.2536 ft ⟦line⟧ 2.25 ft.

**Traduction anglaise —**

Solution to Arithmetic + Trig Exam. Mid-year Session., Jan., 29th 1967 4th Year. ⟦line⟧ 24 ft 52 a b a = 24 sin 52 = 24 x 0.7880 = 18.912 ft b = 24 cos 52 = 24 x 0.6157 = 14.7768 ft. 18.912 - 2 = 16.912 ft sin θ = 16.912 / 24 = 0.7047 ∴ θ = 44° 48' 24 θ b' 16.912 b' : 24 cos 44° 48' = 24 x 0.7096 = 17.0304 17.0304 14.7768 ⟦line⟧ 2.2536 ft ⟦line⟧ 2.25 ft.

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### csp_ae757d41bf3854b9a1cea726d691205e

[Marginalia] 4/

A E B A₁ θ B₁ 42° 55° O

AA₁ = OA₁ tan 42° = OB₁ tan θ (a) OB₁ = OA₁ sec 55° (b) tan θ = OA₁ / OB₁ tan 42° from (a) = (OA₁ tan 42°) / (OA₁ sec 55°) from (b) = tan 42° cos 55° = <del>0.9004 X 0.7071</del> = 0.9004 X 0.5736 = 0.5165 θ = angle of elevation = 27° 19'

**Traduction anglaise —**

4/ A E B A₁ θ B₁ 42° 55° O AA₁ = OA₁ tan 42° = OB₁ tan θ (a) OB₁ = OA₁ sec 55° (b) tan θ = OA₁ / OB₁ tan 42° from (a) = (OA₁ tan 42°) / (OA₁ sec 55°) from (b) = tan 42° cos 55° = <del>0.9004 X 0.7071</del> = 0.9004 X 0.5736 = 0.5165 θ = angle of elevation = 27° 19'

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### csp_aed37107c05a5003a5933c3dbf0df68b

B01 · paragraph / latin ⟦illegible⟧ ⟦illegible⟧ (b) Solve the triangle ABC, given that: a = 30 ; B = 11° 15' ; c = 35 a = 35, b = ⟦illegible⟧, c = ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ (b) If tanα + tanβ = 1 + tanα tanβ, find A ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ (c) Find the height of the tower ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

⟦illegible⟧ ⟦illegible⟧ (b) Solve the triangle ABC, given that: a = 30 ; B = 11° 15' ; c = 35 a = 35, b = ⟦illegible⟧, c = ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ (b) If tanα + tanβ = 1 + tanα tanβ, find A ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ (c) Find the height of the tower ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_af7ca4bc789b5ed5bbaedf41725f70c8

B01 · table / latin 1 | 1 2 | 4 3 | 9 4 | 16 1 1/2 | 2 1/4 2 1/2 | 6 1/4 1/2 | 1/4 1 1/4 | 1 9/16 2 1/4 | 5 1/16 3 1/2 | 12 1/4 4 1/2 | 20 1/4

B02 · other / latin approximate max (1 1/2, 1 1/8) (3, 0) Exact Minimum y x A (0, -9)

B03 · paragraph / latin Answer: (i) The graph is plotted as above. ⟦line⟧ (ii) When x = 1 1/2 y = 1 1/8 an approximate maximum When x = 3 , y = 0 an exact minimum. (iii) The roots of the equation (x-1)(x-3)² = 5x - 9 are the abscissae of the points ⟦illegible⟧, ⟦illegible⟧, ⟦illegible⟧ at which x = 0 x = 2 ⟦line⟧ x = 5 (iv) The function (x-1)(x-3)² is positive when ⟦illegible⟧ < x < ⟦illegible⟧ also when x > 5 Since for these values of x the curve of the function (x-1)(x-3)² is above the curve of 5x - 9

**Traduction anglaise —**

1 | 1 2 | 4 3 | 9 4 | 16 1 1/2 | 2 1/4 2 1/2 | 6 1/4 1/2 | 1/4 1 1/4 | 1 9/16 2 1/4 | 5 1/16 3 1/2 | 12 1/4 4 1/2 | 20 1/4 approximate max (1 1/2, 1 1/8) (3, 0) Exact Minimum y x A (0, -9) Answer: (i) The graph is plotted as above. ⟦line⟧ (ii) When x = 1 1/2 y = 1 1/8 an approximate maximum When x = 3 , y = 0 an exact minimum. (iii) The roots of the equation (x-1)(x-3)² = 5x - 9 are the abscissae of the points ⟦illegible⟧, ⟦illegible⟧, ⟦illegible⟧ at which x = 0 x = 2 ⟦line⟧ x = 5 (iv) The function (x-1)(x-3)² is positive when ⟦illegible⟧ < x < ⟦illegible⟧ also when x > 5 Since for these values of x the curve of the function (x-1)(x-3)² is above the curve of 5x - 9

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### csp_afc22d4079865422874cba19aa1cf973

Conditional examination, September 1967 1

subject: Algebra Date: 8/9/1967 class: 4th Year Secondary Time: 8:00 - 11:00 a.m.

Attempt all questions:

1. (i) Find the value of k if the expression 6x³-13x²+12x+k is exactly divisible by 2x²-3x+4 (10 marks)

(ii) If the n'th term of a series is ⟦(2n+1)/(2n+1)⟧ write down the first three terms and express the difference between the n'th and (n+1)th term as a single fraction in its simplest form. (10 marks)

2. (i) If 3x²-4x+5 = a(x-1)² + ⟦b(x-1)⟧ + c for all values of x, find the values of 'a', 'b', and 'c'. (10 marks) Hence, or otherwise, find the least value of 3x²-4x+5.

(ii) Find the lapse of time in minutes between the two instants when the two hands of a watch are at right angles for the 1st and the 2nd time between <del>four o'clock and five o'clock</del> four o'clock and five o'clock. (10 marks)

3. (i) Compute by logarithms the following expression: ⟦√[ (sin 15° 04' * cos³ 21° 31') / ((5.127)² * (4.007)³) ]⟧ (10 marks)

(ii) Find the value of x from the following equation correct to four significant figures. <del>⟦illegible⟧</del> 3^(2x-1) = 6.4 * 40 (10 marks)

(ii) The third term of a geometric progression, in which all the terms are positive, is 2 and the ⟦illegible⟧

**Traduction anglaise —**

Conditional examination, September 1967 1 subject: Algebra Date: 8/9/1967 class: 4th Year Secondary Time: 8:00 - 11:00 a.m. Attempt all questions: 1. (i) Find the value of k if the expression 6x³-13x²+12x+k is exactly divisible by 2x²-3x+4 (10 marks) (ii) If the n'th term of a series is ⟦(2n+1)/(2n+1)⟧ write down the first three terms and express the difference between the n'th and (n+1)th term as a single fraction in its simplest form. (10 marks) 2. (i) If 3x²-4x+5 = a(x-1)² + ⟦b(x-1)⟧ + c for all values of x, find the values of 'a', 'b', and 'c'. (10 marks) Hence, or otherwise, find the least value of 3x²-4x+5. (ii) Find the lapse of time in minutes between the two instants when the two hands of a watch are at right angles for the 1st and the 2nd time between <del>four o'clock and five o'clock</del> four o'clock and five o'clock. (10 marks) 3. (i) Compute by logarithms the following expression: ⟦√[ (sin 15° 04' * cos³ 21° 31') / ((5.127)² * (4.007)³) ]⟧ (10 marks) (ii) Find the value of x from the following equation correct to four significant figures. <del>⟦illegible⟧</del> 3^(2x-1) = 6.4 * 40 (10 marks) 4. (i) Three times the third term of an arithmetic progression is twice the sixth term. The sum of the first, third and fifth terms is 9. Find: (a) the ratio of the ninth term to the sixth term, (b) the sum of the first thirteen terms of the progression. (10 marks) (ii) The third term of a geometric progression, in which all the terms are positive, is 2 and the ⟦illegible⟧

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### csp_b00ba4559ab45be58744d1a937c14f60

SHAMASH SECONDARY SCHOOL Final Examination, 1958-1959.

Subject: Algebra Date: 28/5/1959 Class: 4th Year Secondary Time: 8:00-10:30 a.m.

All questions are to be attempted.

1. (a) Resolve into factors: (i) (7x + 8)² - 2(7x + 8) - 15. (3 marks) (ii) 2(x - y)² - 3x + 3y - 5. (4 marks) (iii) a(a - 4) - b(b - 4). (3 marks) (b) If 15(2x² - y²) = 7xy, and if x and y are both positive, find the ratio of x to y. Use the shortest possible way. (10 marks)

2(i) Using tables, compute by logarithms the value of : ⁵√ (0.004678)² x 1.002 (10 marks). ------------------- (30.04)³ (ii) Given : log70 = 1.8451, log110 = 2.0414, log34.62 = 1.5394, compute, without using tables, the value of : ³√ 41503 , correct to four sugnificant figures. (10 marks).

3. A certain alloy contains 6 parts by weight of a metal A and 5 parts by weight of a metal B; another alloy contains 7 parts by weight of A and 13 parts by weight of B. If these alloys are melted and mixed together, how many pounds of the second alloy must be mixed with 11 pounds of the first alloy to make a mixture which contains 40 per cent. of A ? (20 marks).

4. (a) The expression 2 - (2ⁿ⁺¹ / 3ⁿ) is a formula for the sum of 'n' terms of a certain geometric series, n being any positive integer. Find the first term of the series, the common ratio, and the formula for the n-th term. (10 marks). (b) The first and second terms of a series are 'a' and 'b' respectively. Find the n th term (i) if the series is an arithmetic series; (ii) if it is a geometric one. (10 marks).

5. (i) Taking ½ in. as one unit on the x-axis and on the y-axis, plot the curve y = ¾x² for values of x between x = -4 and x = 4. (7 marks). (ii) On the same axes of coordinates draw the graph of the equation 3x + 2y = 12. (6 marks). (iii) From the above graphs find two roots for the simultaneous equations y = ¾x² and 3x + 2y = 12. Verify your graphical answers by solving algebraically. (7 marks).

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Final Examination, 1958-1959. Subject: Algebra Date: 28/5/1959 Class: 4th Year Secondary Time: 8:00-10:30 a.m. All questions are to be attempted. 1. (a) Resolve into factors: (i) (7x + 8)² - 2(7x + 8) - 15. (3 marks) (ii) 2(x - y)² - 3x + 3y - 5. (4 marks) (iii) a(a - 4) - b(b - 4). (3 marks) (b) If 15(2x² - y²) = 7xy, and if x and y are both positive, find the ratio of x to y. Use the shortest possible way. (10 marks) 2(i) Using tables, compute by logarithms the value of : ⁵√ (0.004678)² x 1.002 (10 marks). ⟦line⟧ (30.04)³ (ii) Given : log70 = 1.8451, log110 = 2.0414, log34.62 = 1.5394, compute, without using tables, the value of : ³√ 41503 , correct to four sugnificant figures. (10 marks). 3. A certain alloy contains 6 parts by weight of a metal A and 5 parts by weight of a metal B; another alloy contains 7 parts by weight of A and 13 parts by weight of B. If these alloys are melted and mixed together, how many pounds of the second alloy must be mixed with 11 pounds of the first alloy to make a mixture which contains 40 per cent. of A ? (20 marks). 4. (a) The expression 2 - (2ⁿ⁺¹ / 3ⁿ) is a formula for the sum of 'n' terms of a certain geometric series, n being any positive integer. Find the first term of the series, the common ratio, and the formula for the n-th term. (10 marks). (b) The first and second terms of a series are 'a' and 'b' respectively. Find the n th term (i) if the series is an arithmetic series; (ii) if it is a geometric one. (10 marks). 5. (i) Taking ½ in. as one unit on the x-axis and on the y-axis, plot the curve y = ¾x² for values of x between x = -4 and x = 4. (7 marks). (ii) On the same axes of coordinates draw the graph of the equation 3x + 2y = 12. (6 marks). (iii) From the above graphs find two roots for the simultaneous equations y = ¾x² and 3x + 2y = 12. Verify your graphical answers by solving algebraically. (7 marks).

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### csp_b1aaf2a068d8593f987df9fb0726f181

- p.2 - Algebra. 4th Year 16/6/1964 (cont'd.) ----

V. A ball is thrown vertically upwards into the air from a point which is 20 feet above the sea. After x seconds the height y feet of the ball above the point from which it is thrown is given by y = 16x (4-x).

Draw a graph between x=0 and x=5 showing the relationship between y and x. (Take 1 in. = 1 second and 20 feet respectively). (5 marks)

From the graph, find: (a) the maximum height above the sea reached by the ball. (5 marks) (b) how long the ball remains at least 30 feet above (5 marks) the sea, (c) how many seconds elapse from the time the ball was (5 marks) thrown to the time the ball strikes the sea.

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**Traduction anglaise —**

- p.2 - Algebra. 4th Year 16/6/1964 (cont'd.) ⟦line⟧ V. A ball is thrown vertically upwards into the air from a point which is 20 feet above the sea. After x seconds the height y feet of the ball above the point from which it is thrown is given by y = 16x (4-x). Draw a graph between x=0 and x=5 showing the relationship between y and x. (Take 1 in. = 1 second and 20 feet respectively). (5 marks) From the graph, find: (a) the maximum height above the sea reached by the ball. (5 marks) (b) how long the ball remains at least 30 feet above (5 marks) the sea, (c) how many seconds elapse from the time the ball was (5 marks) thrown to the time the ball strikes the sea. ⟦line⟧

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### csp_b416f6df60f557398139d6ddcc895c27

Shamash Secondary School 4th Quarter Exam. May 5th, 1967

Subject: Mathematics Date: 7/5/1967 Class: 4th Year Secondary Time: 8:00 - 9:30

-----

1. Compute by logarithm the expression:

9 / (0.4007)³ x tan² 37° 19' / ----------------------- \/ 50.72⁵ x Cos³ 14° 34' (30 marks)

2. Simplify: (i) (x^(1+q/p))^(p/(p+q)) ÷ p\/ (x^2p / (x^-1)^-p) (20 marks)

{ y^1/2 + y^-1/2 y^1/2 - y^-1/2 } { y^1/2 + 2y^-1/2 y^1/2 - 2y^-1/2 } (ii) { --------------- - --------------- } ÷ { ---------------- - ---------------- } { y^2 - y + 1 y^2 + y + 1 } { y^3 - 1 y^3 + 1 } (20 marks)

3. Solve the equation: 2x3^2x = 4^x-1 finding the answer correct to four significant figures. (30 marks)

**Traduction anglaise —**

Shamash Secondary School 4th Quarter Exam. May 5th, 1967 Subject: Mathematics Date: 7/5/1967 Class: 4th Year Secondary Time: 8:00 - 9:30 ⟦line⟧ 1. Compute by logarithm the expression: 9 / (0.4007)³ x tan² 37° 19' / ⟦line⟧ \/ 50.72⁵ x Cos³ 14° 34' (30 marks) 2. Simplify: (i) (x^(1+q/p))^(p/(p+q)) ÷ p\/ (x^2p / (x^-1)^-p) (20 marks) { y^1/2 + y^-1/2 y^1/2 - y^-1/2 } { y^1/2 + 2y^-1/2 y^1/2 - 2y^-1/2 } (ii) { ⟦line⟧ - ⟦line⟧ } ÷ { ⟦line⟧ - ⟦line⟧ } { y^2 - y + 1 y^2 + y + 1 } { y^3 - 1 y^3 + 1 } (20 marks) 3. Solve the equation: 2x3^2x = 4^x-1 finding the answer correct to four significant figures. (30 marks)

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### csp_b6c33e85d9d05e4b8d6bb0ebdef35627

Shamash Secondary School 2nd Quarter Examination, December, 1966 Subject: Algebra Date: 27/12/1966 Class: 4th Scientific Year Time: 11:00 - 12:30 morning. ------

All questions are to be attempted.

1. (a) Given that x = 3ay - 5bz ; make a, b respectively the subject of --------- 3ay + 5bz the formula. (8 marks) (b) If a/b = k, express 4a - 5b in terms of k. --------------- √ 18a² - 4b² (8 marks)

2. (a) Prove that x(y+2) + x/y + y/x is equal to a, if x= y/(y+1) and y= a - 2 ----- 2 (8 marks) (b) Solve the equation 1-1.4x = 0.7(x-1) ------ -------- 0.2+x 0.1-0.5x (8 marks)

3. (a) Solve 5 = x (8 marks) ------------- 6 - 5 --------- 6 - 5 ----- 6 - x (b) Solve 3x³ + x² + 4 = 8x (10marks) (c) Solve x²y² + 192 = 28xy .......(1) (10 marks) x + y = 8 .......(2)

4. A fast train travelling at x miles an hour takes t hours less to travel y miles than a slower one takes to travel z miles. Find the difference between their speeds in terms of ⟦x, t,⟧ y and z. (20 marks)

5. A man started for a walk when the hands of his watch were coincident between three and four o'clock. When he finished, the hands were again coincidents between five and six o'clock. What was the time when he started, and how long did he walk ? (20 marks)

**Traduction anglaise —**

Shamash Secondary School 2nd Quarter Examination, December, 1966 Subject: Algebra Date: 27/12/1966 Class: 4th Scientific Year Time: 11:00 - 12:30 morning. ⟦line⟧ All questions are to be attempted. 1. (a) Given that x = 3ay - 5bz ; make a, b respectively the subject of ⟦line⟧ 3ay + 5bz the formula. (8 marks) (b) If a/b = k, express 4a - 5b in terms of k. ⟦line⟧ √ 18a² - 4b² (8 marks) 2. (a) Prove that x(y+2) + x/y + y/x is equal to a, if x= y/(y+1) and y= a - 2 ⟦line⟧ 2 (8 marks) (b) Solve the equation 1-1.4x = 0.7(x-1) ⟦line⟧ ⟦line⟧ 0.2+x 0.1-0.5x (8 marks) 3. (a) Solve 5 = x (8 marks) ⟦line⟧ 6 - 5 ⟦line⟧ 6 - 5 ⟦line⟧ 6 - x (b) Solve 3x³ + x² + 4 = 8x (10marks) (c) Solve x²y² + 192 = 28xy .......(1) (10 marks) x + y = 8 .......(2) 4. A fast train travelling at x miles an hour takes t hours less to travel y miles than a slower one takes to travel z miles. Find the difference between their speeds in terms of ⟦x, t,⟧ y and z. (20 marks) 5. A man started for a walk when the hands of his watch were coincident between three and four o'clock. When he finished, the hands were again coincidents between five and six o'clock. What was the time when he started, and how long did he walk ? (20 marks)

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### csp_b7fe1f7ee71e5b28a40220987b092f9f

-p.2- Algebra. 4th Secondary. 6/2/1967 -- --

5. (i) Draw on the same diagram the graphs of the function 4x-3, and of the function 4x²-4x-15, taking ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (8 marks)

(ii) From your diagram, find the roots of the two simultaneous equations y₁ = 4x-3 ........(1) y₂ = 4x²-4x-15 ....(2) (7 marks)

(iii) From the graph of the function 4x²-4x-15, find the roots of the equation 4x²-4x-15=0. (5 marks)

-----

| x | y₁ | y₂ | | - 2 | - 11 | 9 | | - 1 | | - 7 | | 0 | - 3 | - 15 | | ½ | | - 16 | | 1 | | - 15 | | 2 | 5 | - 7 | | 3 | | 9 | | 4 | | 33 |

**Traduction anglaise —**

-p.2- Algebra. 4th Secondary. 6/2/1967 ⟦line⟧ 5. (i) Draw on the same diagram the graphs of the function 4x-3, and of the function 4x²-4x-15, taking ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (8 marks) (ii) From your diagram, find the roots of the two simultaneous equations y₁ = 4x-3 ........(1) y₂ = 4x²-4x-15 ....(2) (7 marks) (iii) From the graph of the function 4x²-4x-15, find the roots of the equation 4x²-4x-15=0. (5 marks) ⟦line⟧ x | y₁ | y₂ - 2 | - 11 | 9 - 1 | | - 7 0 | - 3 | - 15 ½ | | - 16 1 | | - 15 2 | 5 | - 7 3 | | 9 4 | | 33

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### csp_b8330279740953cfbb812e69a89569ff

B01 · header / latin ⟦illegible⟧ Examination ⟦illegible⟧ ⟦illegible⟧ 1958

B02 · paragraph / latin 1. (i) Find the value of ⟦illegible⟧ (ii) Find the values of a and b which will make the expression x⁴ + ax³ + bx - 6 divisible by (x-2) and (x+3), and find the ⟦illegible⟧ ⟦line⟧

B03 · paragraph / latin 2. Find the square root of: x⁶/25 - 2/5 x⁵ + 244/225 x⁴ - 122/105 x³ + 136/49 x² - 2/3 x + 1/4 (20 marks)

B04 · paragraph / latin 3. (i) Reduce to simplest form: x - 4/x ⟦line⟧ x + 2 - 4x / (1 + x/ (2x-1 / (1 + 1/x-1))) (10 marks)

B05 · paragraph / latin (ii) Solve the equation: 1 - 1.4x / 0.2 + x = 0.7(x-1) / 0.1 - 0.5x (10 marks)

B06 · paragraph / latin 4. Find the values of x, y and z from the following equations: 4x - y + 2z = 15; y + 2z = 3x - 2; y + 4z = ⟦illegible⟧ (20 marks)

B07 · paragraph / latin 5. A football match a charge of 2s is made for sitting in the ground, and an extra charge of 2s 6d for a reserved seat. If N people are admitted to the ground and x people take seats, show that the total amount received, £P, is given by 40P = 4N + 5x. Write the formula so that x is the subject. Then find the number of people taking seats if 2000 enter the ground and the amount received is £600. (20 marks) ⟦line⟧

**Traduction anglaise —**

⟦illegible⟧ Examination ⟦illegible⟧ ⟦illegible⟧ 1958 1. (i) Find the value of ⟦illegible⟧ (ii) Find the values of a and b which will make the expression x⁴ + ax³ + bx - 6 divisible by (x-2) and (x+3), and find the ⟦illegible⟧ ⟦line⟧ 2. Find the square root of: x⁶/25 - 2/5 x⁵ + 244/225 x⁴ - 122/105 x³ + 136/49 x² - 2/3 x + 1/4 (20 marks) 3. (i) Reduce to simplest form: x - 4/x ⟦line⟧ x + 2 - 4x / (1 + x/ (2x-1 / (1 + 1/x-1))) (10 marks) (ii) Solve the equation: 1 - 1.4x / 0.2 + x = 0.7(x-1) / 0.1 - 0.5x (10 marks) 4. Find the values of x, y and z from the following equations: 4x - y + 2z = 15; y + 2z = 3x - 2; y + 4z = ⟦illegible⟧ (20 marks) 5. A football match a charge of 2s is made for sitting in the ground, and an extra charge of 2s 6d for a reserved seat. If N people are admitted to the ground and x people take seats, show that the total amount received, £P, is given by 40P = 4N + 5x. Write the formula so that x is the subject. Then find the number of people taking seats if 2000 enter the ground and the amount received is £600. (20 marks) ⟦line⟧

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### csp_b84f7db8f810511cbc01379cd6153aba

Shamash Secondary School 1st. Quarter Exam.

Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary. Date: 8/12/1957

---------- All questions are to be attempted.

(2) (a) Decimalise to 3 places the following:

£ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 3/4 d

(b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint.

(3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ?

(4) The average age of m boys is b years and of n girls is c years. Find the average age of all together.

(5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation.

(6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

**Traduction anglaise —**

Shamash Secondary School 1st. Quarter Exam. Subject: Arithmetic & Trigonometry Time: 12:00-1:30 p.m. Class: 4th year Secondary. Date: 8/12/1957 ⟦line⟧ All questions are to be attempted. (1) State to how many significant digits are the following underlined numbers given ? My expected profit from my business in the year 1960 is £ 8500. I have to pay my landlord with whom I have just concluded a 10 years agreement £ 1250 per annum. I have to pay my assistant a fixed sum of £ 500 per annum plus a commission of 0.5 per cent on my turnover. His earning from commission may amount to £ 650 per annum. My business premises measures 19.10m, by 25.00 m. (2) (a) Decimalise to 3 places the following: £ 2 12s 2 3/4 d £ 8 10s 8 1/2 d £ 9 5s 10 3/4 d (b) Convert into shillings and pence to nearest 1/4 d the following: £ 0.509 , £ 0.620, £ 0.945. (c) Express 4.316 gallons into gallons, quarts and pints to the nearest pint. (3) A watch which gains 5 sec. in every 3 min. of true time was set right at 6 a.m. What was the true time in the afternoon of the same day when the watch indicated a quarter-past 3 O'clock ? (4) The average age of m boys is b years and of n girls is c years. Find the average age of all together. (5) At 9 a.m. a ship which is sailing in a direction E.37° S. at the rate of 8 miles an hour observes a fort in a direction 53° North of East. At 11 a.m. the fort is observed to bear N.20° W., find the distance of the fort from the ship at the first observation. (6) From the roof of a house 30 feet high the angle of elevation of the top of a monument is 42° 7', and the angle of depres- sion of its foot is 17° 59'. Find its height.

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### csp_b855c8f3282a53a9ba161bd7afec0513

B01 · header / mixed حلول اسئلة الحساب والرياضيات ⟦illegible⟧ الامتحان النهائي الذي جرى في 17/5/1967 Arith. + ⟦illegible⟧ ⟦illegible⟧, 4th year, 1967

B02 · paragraph / latin 1 (a) Amount of stock obtained = £ 2286 / 95 1/4 X 100 = £ 2,400 Dividend = 2400 / 100 X 4 1/2 = £ 108 sale of Amount realized from stock = £ 2400 / 100 X 98 = £ 2,352 Total gain = £ 2,352 + £ 108 - £ 2,286 = £ 174

B03 · paragraph / latin (b) I must set out at 10.25 - 00.15 = 10.10 a.m (true time) There are 100 minutes between 8.30 a.m and 10.10 a.m. My watch gains in 100 minutes of true time 6 / 60 X 100 = 10 minutes Latest time indicated by watch at which I must set out is 10.20 a.m

**Traduction anglaise —**

Solutions to Arithmetic and Mathematics questions ⟦illegible⟧ The final exam that took place on 17/5/1967 Arith. + ⟦illegible⟧ ⟦illegible⟧, 4th year, 1967 1 (a) Amount of stock obtained = £ 2286 / 95 1/4 X 100 = £ 2,400 Dividend = 2400 / 100 X 4 1/2 = £ 108 Amount realized from sale of stock = £ 2400 / 100 X 98 = £ 2,352 Total gain = £ 2,352 + £ 108 - £ 2,286 = £ 174 (b) I must set out at 10.25 - 00.15 = 10.10 a.m (true time) There are 100 minutes between 8.30 a.m and 10.10 a.m. My watch gains in 100 minutes of true time 6 / 60 X 100 = 10 minutes Latest time indicated by watch at which I must set out is 10.20 a.m

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### csp_b860b97fbe4258bf86303419198bc2fb

B01 · paragraph / latin ⟦illegible⟧ ⟦illegible⟧ x³ - 7x - 6 = 0 ⟦illegible⟧ x³ = 7x + 6 the curve y = x³ and y = 7x + 6 intersect at (-2, -8) (-1, 0) and (3, 27) ∴ the roots of ⟦illegible⟧ are -2, -1 and 3 (ii) the equation x³ + 3x - 4 = 0 can be written in the form x³ = 4 - 3x ⟦illegible⟧ the curves y = x³ and y = 4 - 3x intersect at (1, 1). Therefore the roots of equation ⟦illegible⟧ are 1 ⟦illegible⟧

B02 · table / latin x | y = 7x + 6 0 | 6 -2 | -8 x | y = 4 - 3x 0 | 4 -1 | 7 2 | -2

B03 · marginalia / latin (iii) From the curve of x³ and 7x + 6, we see that x³ > 7x + 6 when -2 < x < -1 Ans. also when x > 3 Ans.

B04 · other / latin 13 y = x³ y = 7x + 6 (3, 27) (1, 1) (-1, 0) (-2, -8) y = 4 - 3x x y

**Traduction anglaise —**

⟦illegible⟧ ⟦illegible⟧ x³ - 7x - 6 = 0 ⟦illegible⟧ x³ = 7x + 6 the curve y = x³ and y = 7x + 6 intersect at (-2, -8) (-1, 0) and (3, 27) ∴ the roots of ⟦illegible⟧ are -2, -1 and 3 (ii) the equation x³ + 3x - 4 = 0 can be written in the form x³ = 4 - 3x ⟦illegible⟧ the curves y = x³ and y = 4 - 3x intersect at (1, 1). Therefore the roots of equation ⟦illegible⟧ are 1 ⟦illegible⟧ x | y = 7x + 6 0 | 6 -2 | -8 x | y = 4 - 3x 0 | 4 -1 | 7 2 | -2 (iii) From the curve of x³ and 7x + 6, we see that x³ > 7x + 6 when -2 < x < -1 Ans. also when x > 3 Ans. 13 y = x³ y = 7x + 6 (3, 27) (1, 1) (-1, 0) (-2, -8) y = 4 - 3x x y

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### csp_bb1fb2526da75ce2a44df5f259ad5f83

⟦Final Examination, June 1960⟧ Subject: Algebra Date: 17/6/1960 Class: 4th year Secondary Time: 8:00 - 11:00

All questions are to be attempted.

1. (a) If x = (2y-1)/(3y-4), and y = (z+1)/(z-1), find z in terms of x. (8 marks) (b) Prove that, if a + b = c, and none of these quantities is zero, the expression 1/(a² + b² - c²) + 1/(b² + c² - a²) + 1/(c² + a² - b²) is equal to zero. (8 marks)

2. (i) Find the value of b for which the expression x³ - 2 - b(x - 1) is to zero when x = 2. (5 marks) (ii) Factorize the expression for this value of b, and find the values of x for which the expression is zero. (12 marks)

3. A train left station P at 10 a.m. on a non-stop run 300 miles to station Q where it was due to arrive at 4:15 p.m. At a station B some miles from Q it was 3 3/4 minutes behind the scheduled time. But by travelling from B to Q at 60 miles per hour the train arrived at its destination on time. How far is it from B to Q? (17 marks)

4. (a) Compute by logarithms the following expression: ⁷√((1.004² × 0.000401³)/(516.2 × 2.003⁵)) (8 marks) (b) Use logarithms to solve the equation 4²ˣ - 8 × 4ˣ + 12 = 0 ⟦(8 marks)⟧

**Traduction anglaise —**

⟦Final Examination, June 1960⟧ Subject: Algebra Date: 17/6/1960 Class: 4th year Secondary Time: 8:00 - 11:00 All questions are to be attempted. 1. (a) If x = (2y-1)/(3y-4), and y = (z+1)/(z-1), find z in terms of x. (8 marks) (b) Prove that, if a + b = c, and none of these quantities is zero, the expression 1/(a² + b² - c²) + 1/(b² + c² - a²) + 1/(c² + a² - b²) is equal to zero. (8 marks) 2. (i) Find the value of b for which the expression x³ - 2 - b(x - 1) is to zero when x = 2. (5 marks) (ii) Factorize the expression for this value of b, and find the values of x for which the expression is zero. (12 marks) 3. A train left station P at 10 a.m. on a non-stop run 300 miles to station Q where it was due to arrive at 4:15 p.m. At a station B some miles from Q it was 3 3/4 minutes behind the scheduled time. But by travelling from B to Q at 60 miles per hour the train arrived at its destination on time. How far is it from B to Q? (17 marks) 4. (a) Compute by logarithms the following expression: ⁷√((1.004² × 0.000401³)/(516.2 × 2.003⁵)) (8 marks) (b) Use logarithms to solve the equation 4²ˣ - 8 × 4ˣ + 12 = 0 ⟦(8 marks)⟧

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### csp_bcd57cdcddc856dc9a68a8c5656c59e9

Two clocks are set right simultaneously at 12 noon; one of which loses 6 sec. in hour, and the other gains 3 sec. in 50 min. (a) How long will it be before the minute hands are again in the same direction? (b) What will then be the time am. or P.m. by each clock? (c) When will both clocks simultaneously indicate correct time?

[Marginalia] 240

**Traduction anglaise —**

The are two clocks, one of which gains 2 minutes, while the other loses 2 1/2 minutes each day. They are set right at 4 o'clock on Friday afternoon. What is the difference between them at noon on the following Wednesday? In how many days from the time they are set <del>at noon on the</del> right will they both show ⟦the same⟧ time? ⟦What⟧ what ⟦will⟧ that time be? Two clocks are set right simultaneously at 12 noon; one of which loses 6 sec. in hour, and the other gains 3 sec. in 50 min. (a) How long will it be before the minute hands are again in the same direction? (b) What will then be the time am. or P.m. by each clock? (c) When will both clocks simultaneously indicate correct time? 240

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### csp_be55265dd39152d2bbed6ca9a7e42726

B01 · header / latin ⟦illegible⟧ Exam 14/5/1968

B02 · paragraph / latin Let distance ridden be x miles also ½ minute = 1/120 hours. x/3x10 + x/3x8 + x/3x8 = x/2x10 + x/2x8 - 1/120 x/30 + x/24 + x/24 = x/20 + x/16 - 1/120 the least common denominator is 2³x3x5 or x/2x3x5 + x/2³x3 + x/2³x3 = x/2²x5 + x/2⁴ - 1/2³x3x5 multiplying all the equation by ⟦illegible⟧ 72x + 90x + 90x = 108x + 135x - 18 x = 18 Ans.

B03 · paragraph / latin 4(i) the sum Sₙ = ⅓ n (4n²-1) when n=1 S₁ = ⅓(1)(4-1) = ⅓(3) = 1 = 1st term when n=2 S₂ = ⅓ ⋅ 2 (4x4-1) = 2/3 x 15 = 10 = sum of 1st + 2nd terms ∴ 1st term = 1 Ans. 2nd " = 10 - 1 = 9

B04 · paragraph / latin (ii) 2s. 3d. = 27d. ∴ the boring of the 1st foot costs 27d. " " 2nd " " 28d. + 29d. ∴ we have an A.P. in which a = 27, d = 1, n = 400 ∴ l = a + (n-1)d = 27 + 399 = 426 d. = £1. 15s. 6d. cost of boring the last foot S = n/2 (a + l) = 400/2 (27 + 426) = 200 x 453 = 90600 d. (be used) ∴ S = 90600 / 240 = £377 ½ = £377. 10s. cost of boring the entire well Ans.

B05 · paragraph / latin (iii) l₃ = 18 a = ? 40.5 = ar⁴ ∴ r² = 81/2 / 18 = 81/36 l₅ = 40.5 S₆ = ? 18 = ar² ∴ r = ± 9/6 = ± 3/2 Since all the terms of the G.P. are positive ∴ r = 3/2 ∴ 18 = a(3/2)² ∴ a = 18x4 / 9 = 8 Ans. 1 S₆ = a(r⁶-1) / r-1 = 8[(3/2)⁶-1] / 3/2-1 = 8[729/64 - 1] / 1/2 = 16(729-64 / 64) ∴ S₆ = 16 x 665 / 64 = 665 / 4 = 166 ¼ Ans. 2

B06 · paragraph / latin 5.(i) Draw the graph of y = ¼(3x²-5x-4) from x = -2 to x = 4 using 1 inch = 1 unit on both axes (ii) Find the least value of 3x²-5x-4 (iii) solve the equation 3x²-5x-6 = 0

B07 · table / latin x | y -2 | 4.5 -1 | 1 0 | -1 1 | -1.5 2 | -0.5 3 | 2 4 | 6

**Traduction anglaise —**

⟦illegible⟧ Exam 14/5/1968 Let distance ridden be x miles also ½ minute = 1/120 hours. x/3x10 + x/3x8 + x/3x8 = x/2x10 + x/2x8 - 1/120 x/30 + x/24 + x/24 = x/20 + x/16 - 1/120 the least common denominator is 2³x3x5 or x/2x3x5 + x/2³x3 + x/2³x3 = x/2²x5 + x/2⁴ - 1/2³x3x5 multiplying all the equation by ⟦illegible⟧ 72x + 90x + 90x = 108x + 135x - 18 x = 18 Ans. 4(i) the sum Sₙ = ⅓ n (4n²-1) when n=1 S₁ = ⅓(1)(4-1) = ⅓(3) = 1 = 1st term when n=2 S₂ = ⅓ ⋅ 2 (4x4-1) = 2/3 x 15 = 10 = sum of 1st + 2nd terms ∴ 1st term = 1 Ans. 2nd " = 10 - 1 = 9 (ii) 2s. 3d. = 27d. ∴ the boring of the 1st foot costs 27d. " " 2nd " " 28d. + 29d. ∴ we have an A.P. in which a = 27, d = 1, n = 400 ∴ l = a + (n-1)d = 27 + 399 = 426 d. = £1. 15s. 6d. cost of boring the last foot S = n/2 (a + l) = 400/2 (27 + 426) = 200 x 453 = 90600 d. (be used) ∴ S = 90600 / 240 = £377 ½ = £377. 10s. cost of boring the entire well Ans. (iii) l₃ = 18 a = ? 40.5 = ar⁴ ∴ r² = 81/2 / 18 = 81/36 l₅ = 40.5 S₆ = ? 18 = ar² ∴ r = ± 9/6 = ± 3/2 Since all the terms of the G.P. are positive ∴ r = 3/2 ∴ 18 = a(3/2)² ∴ a = 18x4 / 9 = 8 Ans. 1 S₆ = a(r⁶-1) / r-1 = 8[(3/2)⁶-1] / 3/2-1 = 8[729/64 - 1] / 1/2 = 16(729-64 / 64) ∴ S₆ = 16 x 665 / 64 = 665 / 4 = 166 ¼ Ans. 2 5.(i) Draw the graph of y = ¼(3x²-5x-4) from x = -2 to x = 4 using 1 inch = 1 unit on both axes (ii) Find the least value of 3x²-5x-4 (iii) solve the equation 3x²-5x-6 = 0 x | y -2 | 4.5 -1 | 1 0 | -1 1 | -1.5 2 | -0.5 3 | 2 4 | 6

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### csp_be67e852ca67561a903a7a55d3215c78

3. (i) A (x² - 2x) + B (x + 4) + C = 3x² + x + 25 ∴ Ax² - 2Ax + Bx + 4B + C = 3x² + x + 25 ∴ Ax² + (B - 2A) x + 4B + C = 3x² + x + 25 . Equating coefficients of like terms, ∴ A = 3 , B - 2A = 1 ∴ B - 6 = 1 or B = 7 4B + C = 25 ∴ 4 x 7 + C = 25 or C = -3 ∴ A = 3 , B = 7 and C = -3 Ans. Another method : In the original identity, let x = 0 , then 4B + C = 25 ... ① again let x = 2 , then 6B + C = 12 + 2 + 25 = 39 ... ② 6B + C = 39 ... ② 4B + C = 25 ... ① } subtract 2B = 14 ∴ B = 7 and C = -3 now let x = 1 , then A (1 - 2) + B (1 + 4) + C = 3 + 1 + 25 or -A + 5 x 7 - 3 = 29 ∴ A = 32 - 29 ∴ A = 3 , B = 7 and C = -3 Ans.

(ii) Solve : x² + xy + 2y² = 8 ... ① 2x² - 2xy - 3y² = 1 ... ② Let y = mx ∴ x² + mx² + 2m²x² = 8 from ① ..... (1.a) 2x² - 2m x² - 3m²x² = 1 from ② ..... (2.a) x² (1 + m + 2m²) = 8 } dividing, (1 + m + 2m²) / (2 - 2m - 3m²) = 8 / 1 x² (2 - 2m - 3m²) = 1 } ∴ 1 + m + 2m² = 16 - 16m - 24m² ∴ 26m² + 17m - 15 = 0 ∴ (13m + 15) (2m - 1) = 0 ∴ m = 1/2 or m = -15/13 when m = 1/2 , from (1.a) , we get: x² + x²/2 + x²/2 = 8 or 2x² = 8 ∴ x² = 4 ∴ x = ± 2 when x = 2 , y = mx = 1/2 x 2 = 1 and when x = -2 , y = 1/2 (-2) = -1 x = 2 } Ans. 1 x = -2 } Ans. 2 y = 1 } y = -1 } when m = -15/13 then x² (1 - 15/13 + 2 x 225/169) = 8 or x² ( (169 - 13 x 15 + 450) / 169 ) = 8 ∴ x² ( (169 - 195 + 450) / 169 ) = 8 ∴ x² ( 424 / 169 ) = 8 ∴ x² = (8 x 169) / 424 = 169 / 53 ∴ x = ± 13 / √53 when x = 13 / √53 , y = -15/13 . 13 / √53 = -15 / √53 and when x = -13 / √53 , y = (-15/13) (-13 / √53) = 15 / √53 x = 13 / √53 } Ans. 3 x = -13 / √53 } Ans. 4 y = -15 / √53 } y = 15 / √53 }

**Traduction anglaise —**

2 3. (i) A (x² - 2x) + B (x + 4) + C = 3x² + x + 25 ∴ Ax² - 2Ax + Bx + 4B + C = 3x² + x + 25 ∴ Ax² + (B - 2A) x + 4B + C = 3x² + x + 25 . Equating coefficients of like terms, ∴ A = 3 , B - 2A = 1 ∴ B - 6 = 1 or B = 7 4B + C = 25 ∴ 4 x 7 + C = 25 or C = -3 ∴ A = 3 , B = 7 and C = -3 Ans. Another method : In the original identity, let x = 0 , then 4B + C = 25 ... ① again let x = 2 , then 6B + C = 12 + 2 + 25 = 39 ... ② 6B + C = 39 ... ② 4B + C = 25 ... ① } subtract 2B = 14 ∴ B = 7 and C = -3 now let x = 1 , then A (1 - 2) + B (1 + 4) + C = 3 + 1 + 25 or -A + 5 x 7 - 3 = 29 ∴ A = 32 - 29 ∴ A = 3 , B = 7 and C = -3 Ans. (ii) Solve : x² + xy + 2y² = 8 ... ① 2x² - 2xy - 3y² = 1 ... ② Let y = mx ∴ x² + mx² + 2m²x² = 8 from ① ..... (1.a) 2x² - 2m x² - 3m²x² = 1 from ② ..... (2.a) x² (1 + m + 2m²) = 8 } dividing, (1 + m + 2m²) / (2 - 2m - 3m²) = 8 / 1 x² (2 - 2m - 3m²) = 1 } ∴ 1 + m + 2m² = 16 - 16m - 24m² ∴ 26m² + 17m - 15 = 0 ∴ (13m + 15) (2m - 1) = 0 ∴ m = 1/2 or m = -15/13 when m = 1/2 , from (1.a) , we get: x² + x²/2 + x²/2 = 8 or 2x² = 8 ∴ x² = 4 ∴ x = ± 2 when x = 2 , y = mx = 1/2 x 2 = 1 and when x = -2 , y = 1/2 (-2) = -1 x = 2 } Ans. 1 x = -2 } Ans. 2 y = 1 } y = -1 } when m = -15/13 then x² (1 - 15/13 + 2 x 225/169) = 8 or x² ( (169 - 13 x 15 + 450) / 169 ) = 8 ∴ x² ( (169 - 195 + 450) / 169 ) = 8 ∴ x² ( 424 / 169 ) = 8 ∴ x² = (8 x 169) / 424 = 169 / 53 ∴ x = ± 13 / √53 when x = 13 / √53 , y = -15/13 . 13 / √53 = -15 / √53 and when x = -13 / √53 , y = (-15/13) (-13 / √53) = 15 / √53 x = 13 / √53 } Ans. 3 x = -13 / √53 } Ans. 4 y = -15 / √53 } y = 15 / √53 }

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### csp_bf5eb92289555857976bf148c65e94d9

[Marginalia] متروك [Marginalia] ③

21 s 10 d + 9 s = 30 s 10 d or 30 5/6 s = 185/6 s Cost per year = £ 44 x 185 / 6 x 20 Average cost per week = £ 44 x 185 / 6 x 20 x 52 = 11 x 37 / 6 x 52 = 1.3045 = £ 1 6 s 1 d Income from investment = £ 3300 x 2.5 / 100 = £ 82.5 Garage rent = 7.5 x 52 / 20 = £ 19.5 Total expenditure = £ 126 + £ 19.5 = £ 145.5 £ 145.5 - £ 82.5 = £ 63 net expenditure He saves £ 44 x 185 / 6 x 20 = £ 67 5/6 = £ 67 5/6 - £ 63 = £ 4 5/6 = £ 4 100/6 s = £ 4 16 2/3 s = £ 4 16 s 8 d ⟦illegible⟧ ⟦illegible⟧ after 150 days of ⟦illegible⟧ ⟦illegible⟧ The first ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

**Traduction anglaise —**

Left blank ③ 21 s 10 d + 9 s = 30 s 10 d or 30 5/6 s = 185/6 s Cost per year = £ 44 x 185 / 6 x 20 Average cost per week = £ 44 x 185 / 6 x 20 x 52 = 11 x 37 / 6 x 52 = 1.3045 = £ 1 6 s 1 d Income from investment = £ 3300 x 2.5 / 100 = £ 82.5 Garage rent = 7.5 x 52 / 20 = £ 19.5 Total expenditure = £ 126 + £ 19.5 = £ 145.5 £ 145.5 - £ 82.5 = £ 63 net expenditure He saves £ 44 x 185 / 6 x 20 = £ 67 5/6 = £ 67 5/6 - £ 63 = £ 4 5/6 = £ 4 100/6 s = £ 4 16 2/3 s = £ 4 16 s 8 d ⟦illegible⟧ ⟦illegible⟧ after 150 days of ⟦illegible⟧ ⟦illegible⟧ The first ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_bf626c5ad74b5dc19565e7c763a192f1

B01 · header / latin ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ in Algebra July 8th 1966 The Year Scientific

B02 · paragraph / latin ⟦illegible⟧ (xy + 8z) / (2x - yz) = 5/4 ∴ 16x + 16z = 50x - 25y x(16y + 16z - 50) x = -25y / (16y + 16z - 50) Ans. I (2 marks) y(16x + 25) = 50x - 16z x y = (50x - 16z) / (16x + 25) Ans. II (2 marks) z(16x + 5) = 50x - 16xy ∴ z = (50x - 16xy) / (16x + 25) Ans. III (2 marks)

B03 · paragraph / latin (ii) Simplify by removing brackets 35 [ (3x - 4y) / 5 - 1/10 { 2x - 5/7 (7x - 4y) } ] + 8(y - x) = = 35 [ (3x - 4y) / 5 - 1/10 { 2x - 5x + 20/7 y } ] + 8y - 16x = 7(3x - 4y) - 7/2 (3x - 5x + 20/7 y) + 8y - 16x = 21x - 28y - 21/2 x + 35/2 x - 10y + 8y - 16x = = x(21 - 21/2 + 35/2 - 16) + y(-28 - 10 + 8) = = 12x - 30y Ans. (8 marks)

B04 · paragraph / latin 2 (i) ① a^7 - b^7 = (a - b)(a^6 + a^5b + a^4b^2 + a^3b^3 + a^2b^4 + ab^5 + b^6) Ans. I 3 ② a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4) Ans. II ③ a^6 - b^6 = (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5) Ans. III ④ a^4 + b^4 = not factorable

B05 · paragraph / latin (ii) (2a)^5 - (3b)^5 / (2a - 3b) = (2a)^4 + (2a)^3(3b) + (2a)^2(3b)^2 + (2a)(3b)^3 + (3b)^4 = 16a^4 + 24a^3b + 36a^2b^2 + 54ab^3 + 81b^4

B06 · paragraph / latin 3 (i) 2.4 = 0.24 / 0.6 - (0.16x - 7.6) / 0.8 ∴ 2.4 = 24/60 - (16x - 760) / 80 2.4 = 0.4 - (x/5 - 7.6) / 0.8 ∴ 2.0 = - (x/5 - 7.6) / 0.8 1.6 = - (x/5 - 7.6) ∴ 8 = -x + 38 ∴ x = 30 ⟦illegible⟧ 48 = -5x + 228 ∴ 5x = 180 x = 36 Ans.

**Traduction anglaise —**

⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ in Algebra July 8th 1966 The Scientific Year ⟦illegible⟧ (xy + 8z) / (2x - yz) = 5/4 ∴ 16x + 16z = 50x - 25y x(16y + 16z - 50) x = -25y / (16y + 16z - 50) Ans. I (2 marks) y(16x + 25) = 50x - 16z x y = (50x - 16z) / (16x + 25) Ans. II (2 marks) z(16x + 5) = 50x - 16xy ∴ z = (50x - 16xy) / (16x + 25) Ans. III (2 marks) (ii) Simplify by removing brackets 35 [ (3x - 4y) / 5 - 1/10 { 2x - 5/7 (7x - 4y) } ] + 8(y - x) = = 35 [ (3x - 4y) / 5 - 1/10 { 2x - 5x + 20/7 y } ] + 8y - 16x = 7(3x - 4y) - 7/2 (3x - 5x + 20/7 y) + 8y - 16x = 21x - 28y - 21/2 x + 35/2 x - 10y + 8y - 16x = = x(21 - 21/2 + 35/2 - 16) + y(-28 - 10 + 8) = = 12x - 30y Ans. (8 marks) 2 (i) ① a^7 - b^7 = (a - b)(a^6 + a^5b + a^4b^2 + a^3b^3 + a^2b^4 + ab^5 + b^6) Ans. I 3 ② a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4) Ans. II ③ a^6 - b^6 = (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5) Ans. III ④ a^4 + b^4 = not factorable (ii) (2a)^5 - (3b)^5 / (2a - 3b) = (2a)^4 + (2a)^3(3b) + (2a)^2(3b)^2 + (2a)(3b)^3 + (3b)^4 = 16a^4 + 24a^3b + 36a^2b^2 + 54ab^3 + 81b^4 3 (i) 2.4 = 0.24 / 0.6 - (0.16x - 7.6) / 0.8 ∴ 2.4 = 24/60 - (16x - 760) / 80 2.4 = 0.4 - (x/5 - 7.6) / 0.8 ∴ 2.0 = - (x/5 - 7.6) / 0.8 1.6 = - (x/5 - 7.6) ∴ 8 = -x + 38 ∴ x = 30 ⟦illegible⟧ 48 = -5x + 228 ∴ 5x = 180 x = 36 Ans.

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### csp_bf68ca66082c573882beaa0f145d9b14

B01 · table / latin x | y = x^4 - 8x^2 + 3x - 10 -3 | 9 -2

-16 | -4 3

B02 · paragraph / latin Graph of the function f(x) = x^4 - 8x^2 is the curve shown below Ans. 1 In the graph it is clear that: when x = -2 , x^4 - 8x^2 = -16 minimum value when x = 0 , x^4 - 8x^2 = 0 maximum value when x = 2 , x^4 - 8x^2 = -16 a second minimum value

B03 · paragraph / latin 2) The solution of the equation x^4 - 8x^2 + 10 = 3x is the same as the solution of equation x^4 - 8x^2 = 3x - 10 But the L.H.S. of the last equation is f(x) and the R.H.S. is y The two graphs intersect at x = -2 (they touch) x = 1 and x = 2.8 approximately Ans. 2

B04 · other / latin ⟦graph with axes and curves⟧ f(x) = x^4 - 8x^2 y = 3x - 10 (0,0) (-2, -16) (2, -16) (1, -7) (2.8, -1.3) approximately

**Traduction anglaise —**

x | y = x^4 - 8x^2 + 3x - 10 -3 | 9 -2 | | -16 | -4 | 3 Graph of the function f(x) = x^4 - 8x^2 is the curve shown below Ans. 1 In the graph it is clear that: when x = -2 , x^4 - 8x^2 = -16 minimum value when x = 0 , x^4 - 8x^2 = 0 maximum value when x = 2 , x^4 - 8x^2 = -16 a second minimum value 2) The solution of the equation x^4 - 8x^2 + 10 = 3x is the same as the solution of equation x^4 - 8x^2 = 3x - 10 But the L.H.S. of the last equation is f(x) and the R.H.S. is y The two graphs intersect at x = -2 (they touch) x = 1 and x = 2.8 approximately Ans. 2 ⟦graph with axes and curves⟧ f(x) = x^4 - 8x^2 y = 3x - 10 (0,0) (-2, -16) (2, -16) (1, -7) (2.8, -1.3) approximately

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### csp_c072603de6ae51af8bc4a0182cb695f4

B01 · header / latin (cont'd.).. -p.2- Arith. & Trig. 17/5/67 4th Year Secondary.

B02 · other / latin ⟦line⟧

B03 · paragraph / latin 5. ABC is a triangle with AB = AC = 100 ft. and the angle BAC is 70°. At A is a vertical pole AO = 80 ft. high. Calculate: (i) the length of BC, (ii) the angle of elevation of the top of the pole from B, (iii) the angle of elevation of the top of the pole from the mid-point of BC.

B04 · paragraph / latin 6. A, B and C are three points on a coastline which runs from north to south; B is south of A and C is 1000 yards south of B. A boat is moving in a straight line towards C and when it is at a point P which is 2000 yd. from A on a bearing of (N.60.E.) its bearing from B is ( N.38E.). Calculate: (a) the distance from P to the nearest point X on the coastline. (b) the distance AB. (c) the bearing of P from C.

B05 · other / latin ⟦line⟧

**Traduction anglaise —**

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### csp_c1c1a1372dfe566e9e2b28af2dae2c84

B01 · paragraph / latin 3 (I) Merchant's total Costs 15 X 18 + (6s 3d) 4.6 + £ 5 10s = £ 270 + £ 1 8s 9d + £ 5 10s = £ 276 18s 9d

B02 · paragraph / latin (II) Total amount merchant received from sale 92 X (£ 1 5s) + 15 X 112 X 4 / 240 + {15 - (4.6 + 0.75)} 20 = £ 115 + £ 28 + £ 193 = £ 336

B03 · paragraph / latin (III) Profit 336 - 276 18s 9d = £ 59 1s 3d Percentage profit = 59.0625 X 100 / 276.9375 = 59062500 / 2769375 Since to two significant figures we need only take four in calculations Percentage profit = 59060000 / 2769000 = 21 %

**Traduction anglaise —**

3 (I) Merchant's total Costs 15 X 18 + (6s 3d) 4.6 + £ 5 10s = £ 270 + £ 1 8s 9d + £ 5 10s = £ 276 18s 9d (II) Total amount merchant received from sale 92 X (£ 1 5s) + 15 X 112 X 4 / 240 + {15 - (4.6 + 0.75)} 20 = £ 115 + £ 28 + £ 193 = £ 336 (III) Profit 336 - 276 18s 9d = £ 59 1s 3d Percentage profit = 59.0625 X 100 / 276.9375 = 59062500 / 2769375 Since to two significant figures we need only take four in calculations Percentage profit = 59060000 / 2769000 = 21 %

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### csp_c221cb56ab485795a3a50e28c52f2112

B01 · paragraph / latin (i) (7x+8) - 2 (7x+8) - 15 = (7x+8) - 5 (7x+8) + 3 (7x+8) (ii) 2(x-y)^2 - 3x + 3y - 2 = 2(x-y)^2 - 3(x-y) - 2 = (2x-2y+1)(x-y-2) (iii) a(a-4) - b(b-4) = a^2 - 4a - b^2 + 4b = (a^2 - b^2) - 4(a-b) = (a-b)(a+b) - 4(a-b) = (a-b)(a+b-4)

B02 · paragraph / latin 15(x^2-y^2) = 7xy ∴ 15x^2 - 7xy - 15y^2 = 0 ∴ (3x-5y)(5x+3y) = 0 ∴ 3x-5y = 0 ∴ x/y = 5/3 Ans. or 5x+3y = 0 or x/y = -3/5 to be discarded or divide 15(x^2-y^2) = 7xy by y^2 and we get 15((x/y)^2 - 1) = 7(x/y) ∴ 15(x/y)^2 - 7(x/y) - 15 = 0 ∴ (3x/y - 5)(5x/y + 3) = 0 ∴ x/y = 5/3 or x/y = -3/5 to be discarded

B05 · paragraph / latin 3. In the 11 lbs of the first alloy there are 6 lbs of metal A + 5 lbs of metal B In x lbs of the second alloy there are (7/(7+13) x) lbs of metal A + (13/20 x) lbs of metal B ∴ (6 + 7/20 x) lbs = 40/100 (11+x) ∴ 6 + 7x/20 = 2/5 (11+x) or 120 + 7x = 8(11+x) or x = 32 lbs. Ans.

**Traduction anglaise —**

(i) (7x+8) - 2 (7x+8) - 15 = (7x+8) - 5 (7x+8) + 3 (7x+8) (ii) 2(x-y)^2 - 3x + 3y - 2 = 2(x-y)^2 - 3(x-y) - 2 = (2x-2y+1)(x-y-2) (iii) a(a-4) - b(b-4) = a^2 - 4a - b^2 + 4b = (a^2 - b^2) - 4(a-b) = (a-b)(a+b) - 4(a-b) = (a-b)(a+b-4) 15(x^2-y^2) = 7xy ∴ 15x^2 - 7xy - 15y^2 = 0 ∴ (3x-5y)(5x+3y) = 0 ∴ 3x-5y = 0 ∴ x/y = 5/3 Ans. or 5x+3y = 0 or x/y = -3/5 to be discarded or divide 15(x^2-y^2) = 7xy by y^2 and we get 15((x/y)^2 - 1) = 7(x/y) ∴ 15(x/y)^2 - 7(x/y) - 15 = 0 ∴ (3x/y - 5)(5x/y + 3) = 0 ∴ x/y = 5/3 or x/y = -3/5 to be discarded 5√((0.004678)^2 × 1.002) / (30.04)^3 = x log 0.004678 = ̅3.6700 | 2 log 0.004678 = ̅5.3400 log 1.002 = 0.0008 | log 1.002 = 0.0008 log 30.04 = 1.4777 | log Num. = ̅5.3408 | log Den. = 4.4331 | 5 log x = ̅10.9077 | log x = ̅2.18154 | x = 0.01519 Ans. (ii) log 70 = 1.8451 | ∛√41503 = ∛√7^3 × 11^2 = x log 110 = 2.0414 | log 7 = 0.8451 ∴ 3 log 7 = 2.5353 log 34.62 = 1.5394 | log 11 = 1.0414 ∴ 2 log 11 = 2.0828 | ∴ 3 log x = 4.6181 | ∴ log x = 1.5394 | ∴ x = 34.62 Ans. Since log 34.62 = 1.5394 3. In the 11 lbs of the first alloy there are 6 lbs of metal A + 5 lbs of metal B In x lbs of the second alloy there are (7/(7+13) x) lbs of metal A + (13/20 x) lbs of metal B ∴ (6 + 7/20 x) lbs = 40/100 (11+x) ∴ 6 + 7x/20 = 2/5 (11+x) or 120 + 7x = 8(11+x) or x = 32 lbs. Ans.

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### csp_c3fc7abe6573541e927197ecf57b6895

B01 · table / latin x | y 0 | 0 1 | 48 2 | 64 3 | 48 4 | 0 5 | -80

B02 · other / latin y = 10 x 4.3 sec. 5 -20 sea level

B03 · paragraph / latin ⟦illegible⟧ height above the sea of the ball = 64 + 20 = 84 ft. Ans. b) The ball remains at least 30 ft above the sea when it is always above the line y = 10. Therefore, the ball remains in this height between t = 0.1 seconds + t = 3.9 seconds or time = 3.9 - 0.1 = 3.8 seconds approximately Ans. c) The ball strikes the sea after 4.3 seconds Approximate Ans.

**Traduction anglaise —**

x | y 0 | 0 1 | 48 2 | 64 3 | 48 4 | 0 5 | -80 y = 10 x 4.3 sec. 5 -20 sea level ⟦illegible⟧ height above the sea of the ball = 64 + 20 = 84 ft. Ans. b) The ball remains at least 30 ft above the sea when it is always above the line y = 10. Therefore, the ball remains in this height between t = 0.1 seconds + t = 3.9 seconds or time = 3.9 - 0.1 = 3.8 seconds approximately Ans. c) The ball strikes the sea after 4.3 seconds Approximate Ans.

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### csp_c454863c279c5c1f9188f81e168290bc

B01 · header / latin SHAMASH SECONDARY SCHOOL Subject: Algebra Date: 7/1/1969 Class : 4th Year Secondary Time: 10:15-11:45

B02 · paragraph / latin Attempt all questions:

B03 · paragraph / latin 1. The expression 2 x³ + Ax² + Bx - 4 is exactly divisible by x² - 4. Find the values of A and B and find the remaining factor. (20 marks)

B04 · paragraph / latin 2. Find the square root of: 4x⁴ - 3x⁵ - 3x³ + 9/4 x⁶ + 5/3 x² - 2/3 x + 1/9 (20 marks)

B05 · paragraph / latin 3. Which of the following equations is always true, which is sometimes true and which is never true? Find the values of x which satisfy the equation which is sometimes true. (a) 4(x² - 1) + 2(x + 3) = 2 + 2x(1 + 2x) (b) x(6x + 1) = 2x + 1 (c) x(x + 2) = 2(x - 2) (20 marks)

B06 · paragraph / latin (i) Solve simultaneously the following equations: 1/x + 1/y + 3/z = 2½ ⟦line⟧ (1) 2/x + 4/y - 6/z = 2 ⟦line⟧ (2) 3/x - 5/y + 7/z = 2 5/6 ⟦line⟧ (3)

B07 · paragraph / latin (ii) Simplify the following expression to simplest form: ((x/y + y/x - 1) / (x²/y² + x/y + 1)) . ((1 + y/x) / (x - y)) ÷ ((1 + y³/x³) / (x²/y - y²/x)) (20 marks)

B08 · paragraph / latin 5. A man bought "A" lbs of coffee for a certain sum of money. He kept "B" lbs to himself and sold the remainder at "C" shillings a pound more than he paid for it. He found that he received for this portion an amount equal to the original sum of money which he paid for the whole. Find the original sum of money which he paid for the whole. (20 marks)

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Subject: Algebra Date: 7/1/1969 Class : 4th Year Secondary Time: 10:15-11:45 Attempt all questions: 1. The expression 2 x³ + Ax² + Bx - 4 is exactly divisible by x² - 4. Find the values of A and B and find the remaining factor. (20 marks) 2. Find the square root of: 4x⁴ - 3x⁵ - 3x³ + 9/4 x⁶ + 5/3 x² - 2/3 x + 1/9 (20 marks) 3. Which of the following equations is always true, which is sometimes true and which is never true? Find the values of x which satisfy the equation which is sometimes true. (a) 4(x² - 1) + 2(x + 3) = 2 + 2x(1 + 2x) (b) x(6x + 1) = 2x + 1 (c) x(x + 2) = 2(x - 2) (20 marks) (i) Solve simultaneously the following equations: 1/x + 1/y + 3/z = 2½ ⟦line⟧ (1) 2/x + 4/y - 6/z = 2 ⟦line⟧ (2) 3/x - 5/y + 7/z = 2 5/6 ⟦line⟧ (3) (ii) Simplify the following expression to simplest form: ((x/y + y/x - 1) / (x²/y² + x/y + 1)) . ((1 + y/x) / (x - y)) ÷ ((1 + y³/x³) / (x²/y - y²/x)) (20 marks) 5. A man bought "A" lbs of coffee for a certain sum of money. He kept "B" lbs to himself and sold the remainder at "C" shillings a pound more than he paid for it. He found that he received for this portion an amount equal to the original sum of money which he paid for the whole. Find the original sum of money which he paid for the whole. (20 marks)

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### csp_c49e1a5d865d525c8aabbc405b3f892e

[Marginalia] الرقم: [Marginalia] الاسم:

- ص ٢ -

(75 marks)

II. Fill in the blanks in the following equations:-

1. one furlong = ( ) chains = ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds.=( ) ft. 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre = ( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. ⟦~⟧ ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones.

(25 marks). -----

**Traduction anglaise —**

Number: Name: - p. 2 - 36 - Supplementary angles 37 - Equilateral polygon 38 - Isosceles triangle 39 - Medians of a triangle 40 - Rhombus 41 - Locus 42 - Secant line of a circle 43 - Removing and inserting parentheses 44 - Moving equation terms from one side to the other 45 - Identity 46 - Inequality 47 - Homogeneous algebraic expression 48 - Degree of an algebraic expression 49 - Literal coefficient 50 - Second-degree algebraic expression. (75 marks) II. Fill in the blanks in the following equations:- 1. one furlong = ( ) chains = ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds.=( ) ft. 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre = ( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. ⟦~⟧ ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones. (25 marks). ⟦line⟧

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### csp_c4acf5444cc652e8aaec528149c91961

[Marginalia] (5)

Alternative I 68 X 7 + 196 X 2 3/4 = 476 + 196 X 11/4 = 476 + 539 = 1015 d,

Alternative II (68 + 196) x 1 1/2 d + £ 2 5 s 6 d = 264 X 3/2 d + 546 d = 942 d Method II is cheaper by 1015 - 942 = 73 d or 6 s 1 d

117 X 7 = 819 d or £ 3 8 s 3 d for light £ 5 3 s 1 d - £ 3 8 s 3 d = £ 1 14 s 10 d for power no of units for power is: { £ 1 14 s 10 d } ÷ 2 3/4 d = 418 X 4/11 = 152 units for power

**Traduction anglaise —**

(5) Alternative I 68 X 7 + 196 X 2 3/4 = 476 + 196 X 11/4 = 476 + 539 = 1015 d, Alternative II (68 + 196) x 1 1/2 d + £ 2 5 s 6 d = 264 X 3/2 d + 546 d = 942 d Method II is cheaper by 1015 - 942 = 73 d or 6 s 1 d 117 X 7 = 819 d or £ 3 8 s 3 d for light £ 5 3 s 1 d - £ 3 8 s 3 d = £ 1 14 s 10 d for power no of units for power is: { £ 1 14 s 10 d } ÷ 2 3/4 d = 418 X 4/11 = 152 units for power

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### csp_c4c2f399c8ea5837a8ce29d2ab0dd4bf

⟦illegible⟧ 1/4/59

⟦Diagram of a geometric figure with labels E, D, C, B, A, H, and angles 75°, 30°⟧ z/30 = sin 75° ∴ z = 30 sin 75° l/20 = sin 30° ∴ l = 20 sin 30° ∴ z + l = 30 sin 75 + 20 sin 30° = 10 (3 sin 75 + 2 sin 30°) ∴ z + l = 10 (3 x 0.9659 + 2 x 1/2) = 10 (2.8977 + 1) = 38.977 ⟦illegible⟧

x/30 = cos 75° ∴ x = 30 cos 75° = 30 x 0.2588 y/20 = cos 30° ∴ y = 20 cos 30° = 20 x 0.8660 x + y = 30 cos 75 + 20 cos 30 = 10 (3 x cos 75 + 2 cos 30) = 10 (3 x 0.2588 + 2 x 0.866) ∴ x + y = 10 (0.7764 + 1.7320) = 10 x 2.5084 = 25.084 ⟦illegible⟧ ∴ tan θ = (x + y) / (z + l) = 25.084 / 38.977 = 0.6436 θ = 32° 46' Ans. ∴ bearing = N 32° 46' E Ans.

∠ACD = 20° ⟦Diagram of a circle segment with labels A, B, C, D, and angle 20°⟧ x/14 = sin 20° ∴ x = 14 sin 20° ∴ AB = 14 + x = 14 + 14 sin 20° = 14 (1 + sin 20°) = 14 (1 + 0.3420) = 14 x 1.342 = 18.788 ⟦illegible⟧ = 18.79 in. correct to the nearest 0.01 in. Ans.

**Traduction anglaise —**

⟦illegible⟧ 1/4/59 ⟦Diagram of a geometric figure with labels E, D, C, B, A, H, and angles 75°, 30°⟧ z/30 = sin 75° ∴ z = 30 sin 75° l/20 = sin 30° ∴ l = 20 sin 30° ∴ z + l = 30 sin 75 + 20 sin 30° = 10 (3 sin 75 + 2 sin 30°) ∴ z + l = 10 (3 x 0.9659 + 2 x 1/2) = 10 (2.8977 + 1) = 38.977 ⟦illegible⟧ x/30 = cos 75° ∴ x = 30 cos 75° = 30 x 0.2588 y/20 = cos 30° ∴ y = 20 cos 30° = 20 x 0.8660 x + y = 30 cos 75 + 20 cos 30 = 10 (3 x cos 75 + 2 cos 30) = 10 (3 x 0.2588 + 2 x 0.866) ∴ x + y = 10 (0.7764 + 1.7320) = 10 x 2.5084 = 25.084 ⟦illegible⟧ ∴ tan θ = (x + y) / (z + l) = 25.084 / 38.977 = 0.6436 θ = 32° 46' Ans. ∴ bearing = N 32° 46' E Ans. ∠ACD = 20° ⟦Diagram of a circle segment with labels A, B, C, D, and angle 20°⟧ x/14 = sin 20° ∴ x = 14 sin 20° ∴ AB = 14 + x = 14 + 14 sin 20° = 14 (1 + sin 20°) = 14 (1 + 0.3420) = 14 x 1.342 = 18.788 ⟦illegible⟧ = 18.79 in. correct to the nearest 0.01 in. Ans.

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### csp_c772ceb3ff2a5322a7a83bdba9f0c34d

Solution to Conditional Exam in Algebra, Sept. 1967 Cont. 2

3 (i) 9 √ [ (sin² 15° 04' × Cos³ 31° 31') / (510.7)² × (4.007)³ ] = x

(ii) 32^(2x-1) = 64^x · 40 or 2^(5(2x-1)) = 2^(6x) · 40 or 2^(5(2x-1)-6x) = 40

<del>⟦illegible⟧</del> <del>∴ (4x+5) log 2 = log 40</del> <del>or 2^(4x+5) = 2² × 10 or 2^(4x+5-2) = 10 or 2^(4x+3) = 10</del> <del>(4x+3) log 2 = log 10 ∴ (4x+3) log 2 = 1 ∴ 4x log 2 + 3 log 2 = 1</del> <del>∴ 4x log 2 = 1 - 3 log 2 ∴ x = (1 - 3 log 2) / (4 log 2) = (1 - 3 × 0.3010) / (4 × 0.3010)</del> <del>∴ x = (1 - 0.9030) / 1.2040 = 0.0970 / 1.2040 = 97 / 1204 = 0.0805647...</del> <del>i.e. x = 0.08056 Correct to 4 significant figures.</del>

or 2^(4x-5) = 2² × 10 or 2^(4x-5-2) = 10 or 2^(4x-7) = 10 ∴ (4x-7) log 2 = 1 or 4x log 2 - 7 log 2 = 1 or 4x log 2 = 1 + 7 log 2 or x = (1 + 7 log 2) / (4 log 2) = (1 + 7 × 0.3010) / (4 × 0.3010) = (1 + 2.1070) / 1.2040 = 3.1070 / 1.2040 = 3.107 / 1.204 ∴ x = 2.5772... = 2.577 Correct to 4 significant figures Ans.

**Traduction anglaise —**

Solution to Conditional Exam in Algebra, Sept. 1967 Cont. 2 3 (i) 9 √ [ (sin² 15° 04' × Cos³ 31° 31') / (510.7)² × (4.007)³ ] = x log sin 15° 04' = Ī.4148 | 2 log sin 15° 04' = Ī.8296 log cos 31° 31' = Ī.9307 | 3 log cos 31° 31' = Ī.7921 log 510.7 = 2.7082 | log Num. = Ī.6217 log 4.007 = 0.6029 | log Den. = 7.2251 ⟦line⟧ 2 log 510.7 = 5.4164 | 9 log x = Ī.3966 3 log 4.007 = 1.8087 | log x = Ī.04407 log Den. = 7.2251 | = Ī.0441 x = 0.1107 Ans. (ii) 32^(2x-1) = 64^x · 40 or 2^(5(2x-1)) = 2^(6x) · 40 or 2^(5(2x-1)-6x) = 40 <del>⟦illegible⟧</del> <del>∴ (4x+5) log 2 = log 40</del> <del>or 2^(4x+5) = 2² × 10 or 2^(4x+5-2) = 10 or 2^(4x+3) = 10</del> <del>(4x+3) log 2 = log 10 ∴ (4x+3) log 2 = 1 ∴ 4x log 2 + 3 log 2 = 1</del> <del>∴ 4x log 2 = 1 - 3 log 2 ∴ x = (1 - 3 log 2) / (4 log 2) = (1 - 3 × 0.3010) / (4 × 0.3010)</del> <del>∴ x = (1 - 0.9030) / 1.2040 = 0.0970 / 1.2040 = 97 / 1204 = 0.0805647...</del> <del>i.e. x = 0.08056 Correct to 4 significant figures.</del> or 2^(4x-5) = 2² × 10 or 2^(4x-5-2) = 10 or 2^(4x-7) = 10 ∴ (4x-7) log 2 = 1 or 4x log 2 - 7 log 2 = 1 or 4x log 2 = 1 + 7 log 2 or x = (1 + 7 log 2) / (4 log 2) = (1 + 7 × 0.3010) / (4 × 0.3010) = (1 + 2.1070) / 1.2040 = 3.1070 / 1.2040 = 3.107 / 1.204 ∴ x = 2.5772... = 2.577 Correct to 4 significant figures Ans.

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### csp_c7b4fb96bcb450a0b0172f409aeebe92

SHAMASH SCHOOL FINAL EXAMINATIONS 1953-1954

Subject: Trigonometry Date: 26/5/54 Class: Fourth year (secondary) Time: 10:30-12:00

All questions are to be attempted 1. How far down a hill inclined at 7½° to the horizon must I walk in order to descend a distance of 70 ft. vertically? 2. A is 5 miles due South of a port O. A ship steaming at 10 miles an hour starts from O and steams in a straight line to B 1 mile due West of A. From B the ship steams 37° East of South. Calculate the ship's distance from O at the end of one hour after leaving O. 3. Solve the triangle ABC, having given: A = 43° 39', C = 17° 47', b = 4 ft. 4. Two points A and B are at sea level, B being due south of A and distant 2200 feet from it. A third point C, which is 200 feet above sea level, is due east of A and its bearing from B is 047° (N. 47° E.). Find the horizontal distance between B and C and the angle of elevation of C from B, correct to the nearest 10 feet.

θ = ? BD = ? ⟦Diagram showing a 3D geometric figure with points A, B, C, D and labels 2200, 200, 47°, θ⟧

[Signature] Lecturer: Abdullah Obadiah

**Traduction anglaise —**

SHAMASH SCHOOL FINAL EXAMINATIONS 1953-1954 Subject: Trigonometry Date: 26/5/54 Class: Fourth year (secondary) Time: 10:30-12:00 All questions are to be attempted 1. How far down a hill inclined at 7½° to the horizon must I walk in order to descend a distance of 70 ft. vertically? 2. A is 5 miles due South of a port O. A ship steaming at 10 miles an hour starts from O and steams in a straight line to B 1 mile due West of A. From B the ship steams 37° East of South. Calculate the ship's distance from O at the end of one hour after leaving O. 3. Solve the triangle ABC, having given: A = 43° 39', C = 17° 47', b = 4 ft. 4. Two points A and B are at sea level, B being due south of A and distant 2200 feet from it. A third point C, which is 200 feet above sea level, is due east of A and its bearing from B is 047° (N. 47° E.). Find the horizontal distance between B and C and the angle of elevation of C from B, correct to the nearest 10 feet. θ = ? BD = ? ⟦Diagram showing a 3D geometric figure with points A, B, C, D and labels 2200, 200, 47°, θ⟧ Lecturer: Abdullah Obadiah

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### csp_c822f3d4ab2458d7b125390808b0227d

[Marginalia] الرقم : [Marginalia] الاسم :

Monthly Examination, August 1965

Subject: General Mathematics Date: 19/8/1965 Class: 4th Year Secondary Time: 7:30-8:15 a.m.

--- I. Give the English Equivalent of the following:

١١- اعداد زوجية متتالية ١٢- اعداد فردية متتالية ١٣- اعداد اولية ١٤- الجزء الصحيح من العدد ١٥- المقام المشترك الاصغر ١٦- كسر لفظي ١٧- مقلوب العدد ١٨- الكسور العشرية المنتهية ١٩- الكسور العشرية الدورية ٢٠- بسط الكسر ٢١- مقام الكسر ٢٢- الخطأ المئوي ٢٣- النسبة والتناسب ٢٤- الوسط المتناسب بين عددين ٢٥- ربح المساهم (ربح حامل الاسهم) ٢٦- البديهية ٢٧- الموضوعة ٢٨- زاوية حادة ٢٩- زاوية منفرجة ٣٠- زاوية منعكسة ٣١- قطعة دائرة ٣٢- قطاع دائرة ٣٣- المعاليم ٣٤- المجاهيل ٣٥- زاويتان متتامتان

- يتبع -

**Traduction anglaise —**

Number: Name: Monthly Examination, August 1965 Subject: General Mathematics Date: 19/8/1965 Class: 4th Year Secondary Time: 7:30-8:15 a.m. ⟦line⟧ I. Give the English Equivalent of the following: 6- Divisor | 1- Numbers 7- Quotient | 2- Positions (Places) 8- Dividend | 3- Subtraction 9- Remainder | 4- Factors 10- Multiple | 5- Exponent (Power index) 11- Consecutive even numbers 12- Consecutive odd numbers 13- Prime numbers 14- Integer part of a number 15- Least common denominator 16- Verbal fraction 17- Reciprocal of a number 18- Terminating decimals 19- Recurring (periodic) decimals 20- Numerator of a fraction 21- Denominator of a fraction 22- Percentage error 23- Ratio and proportion 24- Mean proportional between two numbers 25- Shareholder's profit (Dividend) 26- Axiom 27- Postulate 28- Acute angle 29- Obtuse angle 30- Reflex angle 31- Segment of a circle 32- Sector of a circle 33- Knowns 34- Unknowns 35- Two complementary angles - To be continued -

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### csp_caa25351c6fc571ba42a885f70e17e03

(ii) √x-5 = ∛x-5 or (x-5)½ = (x-5)⅓ Raising both sides to the 6th power, we get (x-5)³ = (x-5)² or (x-5)³ - (x-5)² = 0 or (x-5)² [(x-5)-1] = 0 or (x-5)² (x-6) = 0 ∴ x = 5 Ans. I both roots x = 6 Ans. II satisfy the original equation

[Marginalia] 8

(iii) 3x⁻¹ - 10x⁻½ + 3 = 0 Let x⁻½ = y ∴ x⁻¹ = y² ∴ 3y² - 10y + 3 = 0 or (3y-1)(y-3) = 0 ∴ y = 3 or y = ⅓ when y = 3, then x⁻½ = 3 or x⁻¹ = 9 or 1/x = 9 or x = 1/9 Ans. I when y = ⅓, then x⁻½ = ⅓ or x⁻¹ = 1/9 or 1/x = 1/9 or x = 9 Ans. II Both roots satisfy the original equation

3. (i) (a) Nⁱ/(ₓ²⁻³ₓ₊₉) = ⁷√N or Nⁱ/(ₓ²⁻³ₓ₊₉) = N⅑ ∴ 1/(x²-3x+9) = 1/7 ∴ x²-3x+9 = 7 or x²-3x+2 = 0 or (x-2)(x-1) = 0 or x = 2 Ans. both roots satisfy the original x = 1 equation

[Marginalia] 3

(b) √5²-4² = ⃣√81 or 3 = 3⁴/ₓ ∴ 4/x = 1 ∴ x = 4 Ans.

[Marginalia] 3

(ii) y²ₓ - 5yₓ + 6 = 0 ∴ (yₓ - 2)(yₓ - 3) = 0 ∴ yₓ = 2 ∴ x log y = log 2 ∴ x = log 2 / log y = log 2 / log 100 = ½ log 2 Ans. 1 or yₓ = 3 ∴ x log y = log 3 ∴ x = log 3 / log y = log 3 / log 100 = ½ log 3 Ans. 2

(iii) x = ⁷√[(0.0104)² × (0.00003012)³] / [4020 × (3019)²]

[Marginalia] 6

log 0.0104 = ̄2.0170 | 2 log 0.0104 = ̄4.0340 | log 4020 = 3.6042 log 0.00003012 = ̄5.4789 | 3 log 0.00003012 = ̄14.4367 | 2 log 3019 = 6.9598 log 4020 = 3.6042 | log Numerator = ̄18.4707 | log Denomin = 10.5640 log 3019 = 3.4799 | log Denominator = 10.5640 | | 7 log x = ̄29.9067 | | log x = ̄5.98667 = ̄5.9867 Correct to 4 dec. | ∴ x = 9.699 × 10⁻⁵ or 0.00009699 | Ans.

**Traduction anglaise —**

8 (ii) √x-5 = ∛x-5 or (x-5)½ = (x-5)⅓ Raising both sides to the 6th power, we get (x-5)³ = (x-5)² or (x-5)³ - (x-5)² = 0 or (x-5)² [(x-5)-1] = 0 or (x-5)² (x-6) = 0 ∴ x = 5 Ans. I both roots x = 6 Ans. II satisfy the original equation 8 (iii) 3x⁻¹ - 10x⁻½ + 3 = 0 Let x⁻½ = y ∴ x⁻¹ = y² ∴ 3y² - 10y + 3 = 0 or (3y-1)(y-3) = 0 ∴ y = 3 or y = ⅓ when y = 3, then x⁻½ = 3 or x⁻¹ = 9 or 1/x = 9 or x = 1/9 Ans. I when y = ⅓, then x⁻½ = ⅓ or x⁻¹ = 1/9 or 1/x = 1/9 or x = 9 Ans. II Both roots satisfy the original equation 3. (i) (a) Nⁱ/(ₓ²⁻³ₓ₊₉) = ⁷√N or Nⁱ/(ₓ²⁻³ₓ₊₉) = N⅙ ∴ 1/(x²-3x+9) = 1/7 ∴ x²-3x+9 = 7 or x²-3x+2 = 0 or (x-2)(x-1) = 0 or x = 2 Ans. both roots satisfy the original x = 1 equation 3 (b) √5²-4² = ⃻√81 or 3 = 3⁴/ₓ ∴ 4/x = 1 ∴ x = 4 Ans. 3 (ii) y²ₓ - 5yₓ + 6 = 0 ∴ (yₓ - 2)(yₓ - 3) = 0 ∴ yₓ = 2 ∴ x log y = log 2 ∴ x = log 2 / log y = log 2 / log 100 = ½ log 2 Ans. 1 or yₓ = 3 ∴ x log y = log 3 ∴ x = log 3 / log y = log 3 / log 100 = ½ log 3 Ans. 2 (iii) x = ⁷√[(0.0104)² × (0.00003012)³] / [4020 × (3019)²] 6 log 0.0104 = ̄2.0170 | 2 log 0.0104 = ̄4.0340 | log 4020 = 3.6042 log 0.00003012 = ̄5.4789 | 3 log 0.00003012 = ̄14.4367 | 2 log 3019 = 6.9598 log 4020 = 3.6042 | log Numerator = ̄18.4707 | log Denomin = 10.5640 log 3019 = 3.4799 | log Denominator = 10.5640 | | 7 log x = ̄29.9067 | | log x = ̄5.98667 = ̄5.9867 Correct to 4 dec. | ∴ x = 9.699 × 10⁻⁵ or 0.00009699 | Ans.

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### csp_cbcf1f5520db5981899d604dbd3ce679

SHAMASH SECONDARY SCHOOL Final Examination, May 1966

Subject: Arithmetic & Trigonometry Date: 18.5.1966 Class: 4th Year Secondary Time: 8:00-10:30 a.m.

Attempt five questions only including question (4).

2. A house holder owns his house which has a rateable value of £44 on which the annual rates are charged at 21s 10d in the £1. He also has to pay an annual property tax at the rate of 9s in the £ on an assessment of £44. Calculate, correct to the nearest penny, the average cost per week of the total of these charges, taking a year as 52 weeks. He subsequently sells his house for £3300, which sum he invests at the rate of 2½% per annum free of tax, and moves into a flat which he rents at £128 per annum. He has however to rent a garage for his car at 7s 6d per week. Find how much per annum he saves by the change. (20 marks)

4. (a) A solid consists of a hemisphere, radius 8 cm., joined to a cone of the same base-radius and height 6 cm., so that the plane surfaces coincide. Find (i) the volume, (ii) the total area of the surface of the solid. (Give answer to 3 significant figures). (10 marks) بردت (b) A sphere of radius 3 in. is filed down into the greatest possible cube; find the volume of the material removed. (Give answer to 4 significant figures). (10 marks)

5. Find the difference between the perimeters of a regular pentagon and a regular hexagon, each of which has an area of 24 square inches. (20 marks)

[Marginalia] طليقة

6. In response to an S O S call from a ship at A, another ship at B, 175 miles due east of A, starts toward A at a speed of 12 miles per hour. At the same time a third ship at C, which is 186 miles from B in a direction bearing ⟦3⟧°15' west of north, also starts for A at a speed of 16 miles per hour. Which ship will reach A first, and how long will it take ? (20 marks)

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Final Examination, May 1966 Subject: Arithmetic & Trigonometry Date: 18.5.1966 Class: 4th Year Secondary Time: 8:00-10:30 a.m. Attempt five questions only including question (4). 1. A person, having bought a certain amount of 2¾% stock at 95, afterwards sold it, and with the proceeds bought 3½% stock. He obtained £900 less stock than before, but his income was unchanged. How much money did he originally invest? (20 marks) 2. A house holder owns his house which has a rateable value of £44 on which the annual rates are charged at 21s 10d in the £1. He also has to pay an annual property tax at the rate of 9s in the £ on an assessment of £44. Calculate, correct to the nearest penny, the average cost per week of the total of these charges, taking a year as 52 weeks. He subsequently sells his house for £3300, which sum he invests at the rate of 2½% per annum free of tax, and moves into a flat which he rents at £128 per annum. He has however to rent a garage for his car at 7s 6d per week. Find how much per annum he saves by the change. (20 marks) 3. (a) A watch was 5 minutes fast at 9 a.m. on Monday, and 10 minutes slow at 12 noon on the following Wednesday. Find when it was exactly right, assuming that it lost time uniformly. Note: (9 a.m. and 12 noon are correct time). (10 marks) (b) Two clocks sound the first stroke of 12 o'clock at the same instant; one clock allows an interval of 20 secs. between each stroke and the next, and the other allows 25 secs. How many strokes of the slower clock remain after the quicker one has finished striking, and what time will elapse between the 12th stroke of the quicker one and the following stroke of the slower one? (10 marks) 4. (a) A solid consists of a hemisphere, radius 8 cm., joined to a cone of the same base-radius and height 6 cm., so that the plane surfaces coincide. Find (i) the volume, (ii) the total area of the surface of the solid. (Give answer to 3 significant figures). (10 marks) ⟦cold⟧ (b) A sphere of radius 3 in. is filed down into the greatest possible cube; find the volume of the material removed. (Give answer to 4 significant figures). (10 marks) 5. Find the difference between the perimeters of a regular pentagon and a regular hexagon, each of which has an area of 24 square inches. (20 marks) ⟦free⟧ 6. In response to an S O S call from a ship at A, another ship at B, 175 miles due east of A, starts toward A at a speed of 12 miles per hour. At the same time a third ship at C, which is 186 miles from B in a direction bearing ⟦3⟧°15' west of north, also starts for A at a speed of 16 miles per hour. Which ship will reach A first, and how long will it take ? (20 marks)

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### csp_cc0cc761180d53c7a70ae4f099849a6d

Solution to Monthly Quiz 12/11/1967 4th year secondary, 1967

II: (a) k = 20 ab / (4a + 5b) (i) 20ab = 4ak + 5bk ∴ 20ab - 4ak = 5bk ∴ 4a(5b - k) = 5bk ∴ a = 5bk / 4(5b - k) Ans. 1 5 (ii) 20ab - 5bk = 4ak ∴ 5b(4a - k) = 4ak ∴ b = 4ak / 5(4a - k) Ans. 2 5

√((k - 4a) / (k - 5b)) = √((20ab / (4a + 5b) - 4a) / (20ab / (4a + 5b) - 5b)) = √((20ab - 16a² - 20ab) / (20ab - 20ab - 25b²)) = √(16a² / 25b²) = 4a / 5b Ans. 3. 5

(b) a = 0, b = 1, c = -2, d = 2 10 (3abc - abcd) ∛(a³bc - c³bd + 3) = [(0) - (0)] ∛(0 - (-2)³(1)(2) + 3) = 12 ∛(24 + 3) = 12 ∛(27) Ans. = 36 Ans.

(c) 3/2 x² - ax - 2/3 a² 3/4 x² - 1/2 ax + 1/3 a² x ________________________ 9/8 x⁴ - 3/4 ax³ - 1/2 a²x² 15 - 3/4 ax³ + 1/2 a²x² + 1/3 a³x 1/2 a²x² - 1/3 a³x - 2/9 a⁴ ________________________________ 9/8 x⁴ - 3/2 ax³ + 1/2 a²x² - 2/9 a⁴

[Stamp] 40

**Traduction anglaise —**

Solution to Monthly Quiz 12/11/1967 4th year secondary, 1967 II: (a) k = 20 ab / (4a + 5b) (i) 20ab = 4ak + 5bk ∴ 20ab - 4ak = 5bk ∴ 4a(5b - k) = 5bk ∴ a = 5bk / 4(5b - k) Ans. 1 5 (ii) 20ab - 5bk = 4ak ∴ 5b(4a - k) = 4ak ∴ b = 4ak / 5(4a - k) Ans. 2 5 √((k - 4a) / (k - 5b)) = √((20ab / (4a + 5b) - 4a) / (20ab / (4a + 5b) - 5b)) = √((20ab - 16a² - 20ab) / (20ab - 20ab - 25b²)) = √(16a² / 25b²) = 4a / 5b Ans. 3. 5 (b) a = 0, b = 1, c = -2, d = 2 10 (3abc - abcd) ∛(a³bc - c³bd + 3) = [(0) - (0)] ∛(0 - (-2)³(1)(2) + 3) = 12 ∛(24 + 3) = 12 ∛(27) Ans. = 36 Ans. (c) 3/2 x² - ax - 2/3 a² 3/4 x² - 1/2 ax + 1/3 a² x ⟦line⟧ 9/8 x⁴ - 3/4 ax³ - 1/2 a²x² 15 - 3/4 ax³ + 1/2 a²x² + 1/3 a³x 1/2 a²x² - 1/3 a³x - 2/9 a⁴ ⟦line⟧ 9/8 x⁴ - 3/2 ax³ + 1/2 a²x² - 2/9 a⁴ 40

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### csp_ccb5462d1aa255f88d0553f21c9c9948

[Marginalia] ⟦Naim Shahrabani⟧ [Marginalia] ⟦Linda Meir Chitayat⟧

-p.2- Algebra. 4th Year Scientific. 29/5/1966. -----

4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks).

5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks)

(ii) Using the same axes and scales, draw another graph to find the roots of the equation x³- 13/4x - 3/2 = 0. ( 7 marks)

(iii) From your diagram, find all values of x which make the expression [ x³-(13/4x+ 3/2)] , positive. ( 7 marks).

--------

**Traduction anglaise —**

⟦Naim Shahrabani⟧ ⟦Linda Meir Chitayat⟧ -p.2- Algebra. 4th Year Scientific. 29/5/1966. ⟦line⟧ 4. (i) The eighth term of an arithmetical progression is six times the third term. Find the second term of the progression. (10 marks). (ii) An invalid on a certain day was able to take a single step of 18 inches. If he was each day to walk twice as far as on the preceding day, how long would it be before he can take a walk of 512 yards ? (10 marks) 5. (i) Draw the graph of y=x³ for values of x at half-unit intervals from -2 to 2.2, taking one inch as one unit on the axis of x and 0.4 inch as one unit on the axis of y. ( 6 marks) (ii) Using the same axes and scales, draw another graph to find the roots of the equation x³- 13/4x - 3/2 = 0. ( 7 marks) (iii) From your diagram, find all values of x which make the expression [ x³-(13/4x+ 3/2)] , positive. ( 7 marks). ⟦line⟧

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### csp_cd9cf70ee5925daeb9976ba283af36d4

Shamash Secondary School Conditional Examination, Sept. 1965

Subject: Algebra Date: 16/9/1965 Class: 4th Year, Scientific Time: 8:00-11:00

Attempt all questions:

1. (i) Solve by the shortest possible way the following equation: 2(x²-2)² + 5(x²-2) - 12 = 0 ( 8 marks) Give your answers correct to two decimal places using tables if necessary. (ii) Reduce to simplest form the expressions: (a) ⟦[(9ⁿ⁺¼)(√3.3ⁿ)] / [3√3⁻ⁿ]⟧¹/ⁿ (6 marks) (b) 6x²y² / m+n ÷ [ 3(m-n)x / 7(r+s) ÷ { 4(r-s) / 21xy² ÷ r²-s² / 4(m²-n²) } ] (6 marks)

2. (i) The variables x and y are related by the equation log₁₀ y = a + b log₁₀ x where a and b are constants. If y = 1000 when x=1 and y=100 when x =0.1, find the value of y when x=10 (10 marks) (ii) Given that log₁₀ 2 = 0.301030 and log₁₀ 1.005=0.002166. Calculate, without the use of tables: (a) log₁₀ 402 , (b) log₁₀ 0.0804 (10 marks)

3. (i) A car travelling steadily at 48 m.p.h. is 550 yd. behind a car travelling steadily at 32 m.p.h. Find, to the nearest second, the time taken by the faster car to overtake the slower. (ii) If the faster car is travelling at x m.p.h. and the slower at y m.p.h. and D yds. is the distance between them, find a formula for the time, t sec., taken to overtake. (iii) From your formula express y in terms of the other letters. (20 marks)

(p.2..)

**Traduction anglaise —**

Shamash Secondary School Conditional Examination, Sept. 1965 Subject: Algebra Date: 16/9/1965 Class: 4th Year, Scientific Time: 8:00-11:00 Attempt all questions: 1. (i) Solve by the shortest possible way the following equation: 2(x²-2)² + 5(x²-2) - 12 = 0 ( 8 marks) Give your answers correct to two decimal places using tables if necessary. (ii) Reduce to simplest form the expressions: (a) ⟦[(9ⁿ⁺¼)(√3.3ⁿ)] / [3√3⁻ⁿ]⟧¹/ⁿ (6 marks) (b) 6x²y² / m+n ÷ [ 3(m-n)x / 7(r+s) ÷ { 4(r-s) / 21xy² ÷ r²-s² / 4(m²-n²) } ] (6 marks) 2. (i) The variables x and y are related by the equation log₁₀ y = a + b log₁₀ x where a and b are constants. If y = 1000 when x=1 and y=100 when x =0.1, find the value of y when x=10 (10 marks) (ii) Given that log₁₀ 2 = 0.301030 and log₁₀ 1.005=0.002166. Calculate, without the use of tables: (a) log₁₀ 402 , (b) log₁₀ 0.0804 (10 marks) 3. (i) A car travelling steadily at 48 m.p.h. is 550 yd. behind a car travelling steadily at 32 m.p.h. Find, to the nearest second, the time taken by the faster car to overtake the slower. (ii) If the faster car is travelling at x m.p.h. and the slower at y m.p.h. and D yds. is the distance between them, find a formula for the time, t sec., taken to overtake. (iii) From your formula express y in terms of the other letters. (20 marks) (p.2..)

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### csp_cdb5729cf0e95e4f834e9e6187b6fad8

[Marginalia] ⟦...⟧ taken total from G.C.E. past papers [Marginalia] not considered as an exam.

Shamash Secondary School Final Exams. June, 1 9 6 4. Paper II

Subject: Algebra Date: 16/6/1964 Class: 4th Year (Scientific Section) Time: 8:00-10:00 a.m.

All questions are to be attempted.

I. A and B are two towns 44 miles apart. A cyclist and a motorist travel from A to B. The motorist leaves A 1 hr. 36 min. later than the cyclist but they reach B at exactly the same time. If the average speed of the motorist is 18 m.p.h. greater than that of the cyclist, find the average speed of each. (20 marks)

II. (i) Find the values of x and y if x y x y - + - = 4 = - + - . (10 marks) 2 4 4 2 (ii) Find the values of a and b so that 3x²-4x+1 can be expressed in the form a(x-1)²+b(x-1). (10 marks)

III. (i) The first, second and last terms of an arithmetic progression are x, y and z respectively. a- Express the number of terms of the progression in terms of x, y and z. (5 marks) b- Show that the sum of the progression is (x+z)(y+z-2x) -------------- (5 marks) 2(y - x) (ii) Four positive numbers are in geometric progression. The product of the first and third is 36 and the product of the second and fourth is 324. Find the numbers. (10 marks)

IV. (i) If S denotes the sum 1+2+3+....+n, and T denotes the sum 1+2+3+....+(n-1), find in terms of n the value of S²-T². Give your answer in its lowest terms. (10 marks) (ii) What number must be added to each of the numbers 3, 6, 10½ to form the first three terms of a geometric progression? Find the sum of the first six terms of this progression. (10 marks)

(cont'd.p.2)

**Traduction anglaise —**

⟦...⟧ taken total from G.C.E. past papers not considered as an exam. Shamash Secondary School Final Exams. June, 1 9 6 4. Paper II Subject: Algebra Date: 16/6/1964 Class: 4th Year (Scientific Section) Time: 8:00-10:00 a.m. All questions are to be attempted. I. A and B are two towns 44 miles apart. A cyclist and a motorist travel from A to B. The motorist leaves A 1 hr. 36 min. later than the cyclist but they reach B at exactly the same time. If the average speed of the motorist is 18 m.p.h. greater than that of the cyclist, find the average speed of each. (20 marks) II. (i) Find the values of x and y if x y x y - + - = 4 = - + - . (10 marks) 2 4 4 2 (ii) Find the values of a and b so that 3x²-4x+1 can be expressed in the form a(x-1)²+b(x-1). (10 marks) III. (i) The first, second and last terms of an arithmetic progression are x, y and z respectively. a- Express the number of terms of the progression in terms of x, y and z. (5 marks) b- Show that the sum of the progression is (x+z)(y+z-2x) ⟦line⟧ (5 marks) 2(y - x) (ii) Four positive numbers are in geometric progression. The product of the first and third is 36 and the product of the second and fourth is 324. Find the numbers. (10 marks) IV. (i) If S denotes the sum 1+2+3+....+n, and T denotes the sum 1+2+3+....+(n-1), find in terms of n the value of S²-T². Give your answer in its lowest terms. (10 marks) (ii) What number must be added to each of the numbers 3, 6, 10½ to form the first three terms of a geometric progression? Find the sum of the first six terms of this progression. (10 marks) V. (cont'd.p.2)

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### csp_ce1fcd0d4ae753dfaa09ad8fb6ce77f5

B01 · header / latin Solutions to Algebra Exam. (2nd quart.) 7/2/1969 page 1

B02 · paragraph / latin 1. 2x³ + Ax² + Bx - 4 | x - 2 2x³ - 4x² | 2x² + (4+A)x + (B+2A+8) | x+2 (4+A)x² + Bx - 4 | 2x² + 4x | 2x+A (4+A)x² - 2(4+A)x | [(4+A)-4]x + B+2A+8 [2(4+A)+B]x - 4 or Bx + 2A+B+8 (B+2A+8)x - 2(B+2A+8) | Ax + 2A 2(B+2A+8) - 4 = 0 B+8 = 0 ----② ∴ 4+12B = -12 ∴ B = -8 from ① 2A-8 = -6 ∴ A = 1 or 2A + B = -6 ----① B = -8 Ans. 1 ∴ the remaining factor is the last quotient, namely: 2x+A or 2x+1 Ans. 2

B03 · paragraph / latin an alternative method: the factors of x²-4 are (x-2) & (x+2). By the remainder theorem, when x=2 the expression 2x³+Ax²+Bx-4 becomes zero ∴ 2x8 + 4A + 2B - 4 = 0 or 4A + 2B = -12 or 2A + B = -6 ----① also, when x = -2, then 2(-2)³ + 4A - 2B - 4 = 0 ∴ 4A - 2B = 20 ∴ 2A - B = 10 ----② Now 2A + B = -6 ----① ∴ 4A = 4 ∴ A = 1 } Ans. 1 2A - B = 10 ----② ∴ B = -8 ∴ 2x³ + x² - 8x - 4 = (x²-4)(2x+1) ∴ (2x+1) is the remaining factor Ans. 2

B04 · paragraph / latin 2. 9/4 x⁶ - 3x⁵ + 4x⁴ - 5x³ + 5/3 x² - 2/3 x + 1/9 (arranging according to descending powers) 9/4 x⁶ - 3x⁵ + 4x⁴ - 3x³ + 5/3 x² - 2/3 x + 1/9 | 3/2 x³ - x² + x - 1/3 9/4 x⁶ 3x³ - x² | -3x⁵ + 4x⁴ - 5x³ x | -3x⁵ + x⁴ 3x³ - 2x² + x | +3x⁴ - 5x³ + 5/3 x² x | +3x⁴ - 2x³ + x² 3x³ - 2x² + 2x - 1/3 | -x³ + 2/3 x² - 2/3 x + 1/9 | -x³ + 2/3 x² - 2/3 x + 1/9 ∴ the square root is; 3/2 x³ - x² + x - 1/3 Ans.

B05 · marginalia / latin (20 marks) (20 marks)

B06 · paragraph / latin 3. ⟦illegible⟧ ⟦illegible⟧ = 1 Ans.

**Traduction anglaise —**

Solutions to Algebra Exam. (2nd quart.) 7/2/1969 page 1 1. 2x³ + Ax² + Bx - 4 | x - 2 2x³ - 4x² | 2x² + (4+A)x + (B+2A+8) | x+2 (4+A)x² + Bx - 4 | 2x² + 4x | 2x+A (4+A)x² - 2(4+A)x | [(4+A)-4]x + B+2A+8 [2(4+A)+B]x - 4 or Bx + 2A+B+8 (B+2A+8)x - 2(B+2A+8) | Ax + 2A 2(B+2A+8) - 4 = 0 B+8 = 0 ⟦line⟧② ∴ 4+12B = -12 ∴ B = -8 from ① 2A-8 = -6 ∴ A = 1 or 2A + B = -6 ⟦line⟧① B = -8 Ans. 1 ∴ the remaining factor is the last quotient, namely: 2x+A or 2x+1 Ans. 2 an alternative method: the factors of x²-4 are (x-2) & (x+2). By the remainder theorem, when x=2 the expression 2x³+Ax²+Bx-4 becomes zero ∴ 2x8 + 4A + 2B - 4 = 0 or 4A + 2B = -12 or 2A + B = -6 ⟦line⟧① also, when x = -2, then 2(-2)³ + 4A - 2B - 4 = 0 ∴ 4A - 2B = 20 ∴ 2A - B = 10 ⟦line⟧② Now 2A + B = -6 ⟦line⟧① ∴ 4A = 4 ∴ A = 1 } Ans. 1 2A - B = 10 ⟦line⟧② ∴ B = -8 ∴ 2x³ + x² - 8x - 4 = (x²-4)(2x+1) ∴ (2x+1) is the remaining factor Ans. 2 2. 9/4 x⁶ - 3x⁵ + 4x⁴ - 5x³ + 5/3 x² - 2/3 x + 1/9 (arranging according to descending powers) 9/4 x⁶ - 3x⁵ + 4x⁴ - 3x³ + 5/3 x² - 2/3 x + 1/9 | 3/2 x³ - x² + x - 1/3 9/4 x⁶ 3x³ - x² | -3x⁵ + 4x⁴ - 5x³ x | -3x⁵ + x⁴ 3x³ - 2x² + x | +3x⁴ - 5x³ + 5/3 x² x | +3x⁴ - 2x³ + x² 3x³ - 2x² + 2x - 1/3 | -x³ + 2/3 x² - 2/3 x + 1/9 | -x³ + 2/3 x² - 2/3 x + 1/9 ∴ the square root is; 3/2 x³ - x² + x - 1/3 Ans. (20 marks) (20 marks) 3. ⟦illegible⟧ ⟦illegible⟧ = 1 Ans.

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### csp_ce5598f585a95c74a3729563713f0119

the rateable value which may or may not be equal to the rent. Rates are <del>⟦illegible⟧</del> levied on the basis of so many shillings in the £ 1 of rateable value e.g the rate of a house is 5s in the £ If the rateable value is £ 100, the rates are £ 25. The rates may sometimes exceed the rateable value; Reason: the assessment was made many years ago when rents were low. Instead of making reassessment, it is sometimes decided to raise the rate or the tax per £ 1. We thus find in certain boroughs the rate is 25s or more in the £. Penny rate = a rate of 1d in the £ 1 of rateable value.

(a) Tax

**Traduction anglaise —**

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### csp_cf4638dfdd935961ba2edf3e55712ec7

Shamash Secondary School Final Examination May 1969 Solution to Algebra Questions, 4th Year 14/5/1969.

1. (i) Divide by (x-2) | x-2 | x³ + Ax² + 31x + B | x² + (A+2)x + 2A + 35 x³ - 2x² ⟦line⟧ (A+2)x² + 31x + B (A+2)x² - 2(A+2)x ⟦line⟧ or (2A + 4 + 31)x + B (2A + 35)x + B (2A + 35)x - 4A - 70 ⟦line⟧ 4A + B + 70 = 0 ∴ B = -4A - 70 .... ①

Now Divide the Quotient by (x-3) ∴ x-3 | x² + (A+2)x + 2A + 35 | x + A + 5 x² - 3x ⟦line⟧ (A+5)x + 2A + 35 (A+5)x - 3A - 15 ⟦line⟧ 5A + 50 ∴ 5A + 50 = 0 .... ② ∴ 5A + 50 = 0 ∴ A = -50/5 = -10 ∴ the 3rd Factor is = x + A + 5 = ∴ B = -4(-10) - 70 = -30 = x - 10 + 5 = x - 5

∴ A = -10 } Ans. | and the expression is: x³ - 10x² + 31x - 30 = (x-2)(x-3)(x-5) B = -30 } 3rd factor = x-5 }

An alternative method: <del>By</del> the remainder theorem, when x=2 then 2³ + 2²A + 2x31 + B = 0 ∴ 4A + B = -70 ... ① also when x=3, then also 3³ + 3²A + 3x31 + B = 0 ∴ 9A + B = -120 ... ② 5A = -50 ∴ A = -10 ∴ from ①: 4(-10) + B = -70 ∴ B = -30 ∴ A = -10 } Ans. Factoring, we get x³ - 10x² + 31x - 30 = (x-2)(x-3)(x-5) B = -30 } Ans. 2

⟦line⟧

(ii) 12 men + 7 boys = £ 9 13s. also 3 men = 4 boys + 8s. Let x shillings be the wages of one boy y " " " " " " man ∴ 12y + 7x = 9 x 20 + 13 or 12y + 7x = 193 ... ① also 3y = 4x + 8 or 3y - 4x = 8 ... ② 12y + 7x = 193 ... ① ∴ 23x = 161 ∴ x = 161/23 = 7 shillings 12y - 16x = 32 ... ② ∴ y = 4x+8/3 = 28+8/3 = 36/3 = 12s. ∴ wages of one boy = x shill. = 7s. } Ans. " " " man = y shill = 12s. }

**Traduction anglaise —**

Shamash Secondary School Final Examination May 1969 Solution to Algebra Questions, 4th Year 14/5/1969. 1. (i) Divide by (x-2) | x-2 | x³ + Ax² + 31x + B | x² + (A+2)x + 2A + 35 x³ - 2x² ⟦line⟧ (A+2)x² + 31x + B (A+2)x² - 2(A+2)x ⟦line⟧ or (2A + 4 + 31)x + B (2A + 35)x + B (2A + 35)x - 4A - 70 ⟦line⟧ 4A + B + 70 = 0 ∴ B = -4A - 70 .... ① Now Divide the Quotient by (x-3) ∴ x-3 | x² + (A+2)x + 2A + 35 | x + A + 5 x² - 3x ⟦line⟧ (A+5)x + 2A + 35 (A+5)x - 3A - 15 ⟦line⟧ 5A + 50 ∴ 5A + 50 = 0 .... ② ∴ 5A + 50 = 0 ∴ A = -50/5 = -10 ∴ the 3rd Factor is = x + A + 5 = ∴ B = -4(-10) - 70 = -30 = x - 10 + 5 = x - 5 ∴ A = -10 } Ans. | and the expression is: x³ - 10x² + 31x - 30 = (x-2)(x-3)(x-5) B = -30 } 3rd factor = x-5 } An alternative method: <del>By</del> the remainder theorem, when x=2 then 2³ + 2²A + 2x31 + B = 0 ∴ 4A + B = -70 ... ① also when x=3, then also 3³ + 3²A + 3x31 + B = 0 ∴ 9A + B = -120 ... ② 5A = -50 ∴ A = -10 ∴ from ①: 4(-10) + B = -70 ∴ B = -30 ∴ A = -10 } Ans. Factoring, we get x³ - 10x² + 31x - 30 = (x-2)(x-3)(x-5) B = -30 } Ans. 2 ⟦line⟧ (ii) 12 men + 7 boys = £ 9 13s. also 3 men = 4 boys + 8s. Let x shillings be the wages of one boy y " " " " " " man ∴ 12y + 7x = 9 x 20 + 13 or 12y + 7x = 193 ... ① also 3y = 4x + 8 or 3y - 4x = 8 ... ② 12y + 7x = 193 ... ① ∴ 23x = 161 ∴ x = 161/23 = 7 shillings 12y - 16x = 32 ... ② ∴ y = 4x+8/3 = 28+8/3 = 36/3 = 12s. ∴ wages of one boy = x shill. = 7s. } Ans. " " " man = y shill = 12s. }

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### csp_cf62e2015d445f7db6f20d4fe2450a3b

Ⅳ (b) Suppose that the G.P. is a, ar, ar², ... Then S₅ = a (r⁵-1) / r-1 and the sixth term = ar⁵. The sum of the terms from the sixth to the tenth inclusive = ar⁵ (r⁵-1) / r-1 ∴ S₁₀ = S₅ + ar⁵ (r⁵-1) / r-1 or S₁₀ = a (r⁵-1) / r-1 + ar⁵ (r⁵-1) / r-1 But 1023 S₅ = 31 S₁₀ or 1023 [a (r⁵-1) / r-1] = 31 [a (r⁵-1) / r-1 + ar⁵ (r⁵-1) / r-1] or 1023 [a (r⁵-1) / r-1] = 31 (a (r⁵-1) / r-1) [1 + r⁵] or 1023 = 31 (1 + r⁵) or 1 + r⁵ = 1023 / 31 or 1 + r⁵ = 33 or r⁵ = 32 or r⁵ = 2⁵ ∴ r = 2 Ans.

**Traduction anglaise —**

3 Ⅳ (b) Suppose that the G.P. is a, ar, ar², ... Then S₅ = a (r⁵-1) / r-1 and the sixth term = ar⁵. The sum of the terms from the sixth to the tenth inclusive = ar⁵ (r⁵-1) / r-1 ∴ S₁₀ = S₅ + ar⁵ (r⁵-1) / r-1 or S₁₀ = a (r⁵-1) / r-1 + ar⁵ (r⁵-1) / r-1 But 1023 S₅ = 31 S₁₀ or 1023 [a (r⁵-1) / r-1] = 31 [a (r⁵-1) / r-1 + ar⁵ (r⁵-1) / r-1] or 1023 [a (r⁵-1) / r-1] = 31 (a (r⁵-1) / r-1) [1 + r⁵] or 1023 = 31 (1 + r⁵) or 1 + r⁵ = 1023 / 31 or 1 + r⁵ = 33 or r⁵ = 32 or r⁵ = 2⁵ ∴ r = 2 Ans.

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### csp_d2472466dd765ba2988bfd1f58f34cac

B01 · paragraph / latin y = 1/4 (3x² - 5x - 4)

B02 · table / latin x | y -2 | 4 1/2 -1 | 1 0 | -1 1 | -1 1/2 2 | -1/2 3 | 2

B03 · paragraph / latin (i) The curve and the table are as shown. Ans. (6 marks) (ii) The least value of (3x² - 5x - 4) is the same as the least value of 4y = 4(least value of y) = 4(-1.52) = -6.08 Ans. (6 marks)

B04 · paragraph / latin (iii) The roots of the equation 3x² - 5x - 6 = 0 are the same as the roots of 3x² - 5x - 4 - 2 = 0 also " " " " " 3x² - 5x - 4 = 2 also " " " " " 1/4(3x² - 5x - 4) = 1/4(2) " " " " " 1/4(3x² - 5x - 4) = 1/2 ∴ Draw the str. line y = 1/2 and the abscissas of the pts. of intersections give the roots x = -0.8 and x = 2.47 Ans. (8 marks)

B05 · other / latin A B y = 1/2 x = -0.8 x = 2.47

B06 · footer / latin Solution to Conditional Exam in Algebra Sept, 1965

**Traduction anglaise —**

y = 1/4 (3x² - 5x - 4) x | y -2 | 4 1/2 -1 | 1 0 | -1 1 | -1 1/2 2 | -1/2 3 | 2 (i) The curve and the table are as shown. Ans. (6 marks) (ii) The least value of (3x² - 5x - 4) is the same as the least value of 4y = 4(least value of y) = 4(-1.52) = -6.08 Ans. (6 marks) (iii) The roots of the equation 3x² - 5x - 6 = 0 are the same as the roots of 3x² - 5x - 4 - 2 = 0 also " " " " " 3x² - 5x - 4 = 2 also " " " " " 1/4(3x² - 5x - 4) = 1/4(2) " " " " " 1/4(3x² - 5x - 4) = 1/2 ∴ Draw the str. line y = 1/2 and the abscissas of the pts. of intersections give the roots x = -0.8 and x = 2.47 Ans. (8 marks) A B y = 1/2 x = -0.8 x = 2.47 Solution to Conditional Exam in Algebra Sept, 1965

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### csp_d4035d1e1346505da90b48a4585161d5

⟦illegible⟧ SCHOOL ⟦illegible⟧ EDIATE & PRIMARY Baghdad Telephone No. 91693

مدرسة ⟦illegible⟧ متوسطة وابتدائية بغداد رقم الهاتف ٩١٦٩٣

No: Date: الرقم: التاريخ:

١- ترجم الى الانكليزية : حكى أبو جعفر محمد بن الفضل الصغيري قال : كان في بلدنا عجوز صالحة كثيرة الصيام والصلاة وكان لها ابن صبي قد انهمك على الشرب واللعب . وكان يتشاغل بدكانه أكثر نهاره ثم يعود الى منزله فيخبئ كيسه عند والدته ويمضي فيبيت في مواضع يشرب فيها فبلغ بعض اللصوص على كيسه ليأخذه فجاء وراءه ودخل الى الدار وهو لا يعلم فاختبأ فيها وسلم هو كيسه الى أمه وخرج وبقيت هي وحدها في الدار وكان لها في دارها بيت مؤزر بالساج عليه باب من حديد تجعل قماشها فيه والكيس فخبأت الكيس فيه خلف الباب وحلبت فأفطرت بين يديه فقال اللص الساعة تغفله وتنام وانزل واقلع الباب وآخذ الكيس فلما أفطرت قامت تصلي ومرت الصلاة ومضى نصف الليل وتحير اللص وخاف أن يدركه الصبح فطاف في الدار فوجد إزاراً جديداً وبخوراً فاتزر بالإزار وأوقد البخور

٢- اكتب معاني هذه الكلمات بالانكليزية :- الطفيلية - الزنادقة - وليمة - الورطة - المقلى - أمة - محدث - كرهها - الأمان - الإسناد -

3. Translate into Arabic: Summer Holidays Then comes July, and with it examinations, but these are soon finished, and with them ends the school year. Boys and girls have nearly two months holiday before them ⟦illegible⟧ school by train and car to return home ⟦illegible⟧ and mothers. The ⟦illegible⟧ holidays

**Traduction anglaise —**

⟦illegible⟧ SCHOOL ⟦illegible⟧ EDIATE & PRIMARY Baghdad Telephone No. 91693 ⟦illegible⟧ School Intermediate & Primary Baghdad Telephone No. 91693 No: Date: Number: Date: 1- Translate into English: Abu Ja'far Muhammad ibn al-Fadl al-Saghiri narrated, saying: There was in our town a righteous old woman who fasted and prayed much, and she had a young son who was engrossed in drinking and play. He would busy himself at his shop most of the day, then return to his home, hide his pouch with his mother, and go out to spend the night in places where he would drink. One of the thieves learned about his pouch in order to take it, so he followed him and entered the house without him knowing, and hid inside. He (the son) handed his pouch to his mother and left, and she remained alone in the house. She had a room in her house paneled with teak with an iron door where she kept her cloth and the pouch, so she hid the pouch in it behind the door, milked (the animal), and broke her fast before him. The thief said: In a moment she will overlook it and sleep, then I will go down, tear off the door, and take the pouch. When she broke her fast, she stood up to pray, and the prayer continued, and half the night passed. The thief became confused and feared that morning would catch him. He circled the house and found a new waist-wrap and incense, so he put on the wrap and lit the incense. 2- Write the meanings of these words in English: Parasitism - Heretics - Banquet - Predicament - Frying pan - Nation - Hadith narrator - Hated her - Safety - Chain of transmission - 3. Translate into Arabic: Summer Holidays Then comes July, and with it examinations, but these are soon finished, and with them ends the school year. Boys and girls have nearly two months holiday before them ⟦illegible⟧ school by train and car to return home ⟦illegible⟧ and mothers. The ⟦illegible⟧ holidays

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### csp_d76b840caca15359a5854450ac89f17f

Shamash Secondary School Conditional Examination, Sept. 1967

Subject: Algebra Date: 8/9/1967 Class: 4th Year Secondary Time: 8:00 - 11:00 a.m.

--- Attempt all questions:

1. (i) Find the value of k if the expression 6x³-13x²+18x+k is exactly divisible by 2x²-3x+4 (10 marks) (ii) If the n th term of a series is (2n+1)/(2n+3) write down the first three terms and express the difference between the n th and (n+1)th terms as a single fraction in its simplest form. (10 marks)

2. (i) If 3x²-4x+5 = a(x-b)²+c for all values of x, find the values of a, b, and c. (10 marks) Hence, or otherwise, find the least value of 3x²-4x+5. (ii) Find the lapse of time in minutes between the two instants when the two hands of a watch are at right angles for the 1st and the 2nd time between four O'clock and five O'clock. (10 marks)

3. (i) Compute by logarithms the following expression : ⁹√[ (Sin² 15° 04' x Cos³ 31° 31') / ((510.7)² x (4.007)³) ] (10 marks) (ii) Find the value of x from the following equation correct to four significant figures: 32^(2x-1) = 64^x * 40 (10 marks)

4. (i) Three times the third term of an arithmetic progression is twice the sixth term. The sum of the first, third and fifth terms is 9. Find: (a) the ratio of the ninth term to the sixth term, (b) the sum of the first thirteen terms of the progression. (10 marks) (ii) The third term of a geometric progression, in which all the terms are positive, is 2/3 and the sum of the first two terms is 2½. Find the first term, the common ratio and the fourth term of the progression. (10 marks)

5. (i) Taking 1 inch = 1 unit on the x-axis and 1 inch = 2 units on the y-axis draw the graphs of y = 4-x² and 4y = 5x + 4 for values of x from -3 to +3. ( 8 marks) (ii) From your graph, find: a- the range of values of x for which 4-x² is greater than 5/4x+1, (4 marks) b- the values of x for which 4-x²=2.5, (4 marks) c- the square root of 5.6. (4 marks)

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**Traduction anglaise —**

Shamash Secondary School Conditional Examination, Sept. 1967 Subject: Algebra Date: 8/9/1967 Class: 4th Year Secondary Time: 8:00 - 11:00 a.m. ⟦line⟧ Attempt all questions: 1. (i) Find the value of k if the expression 6x³-13x²+18x+k is exactly divisible by 2x²-3x+4 (10 marks) (ii) If the n th term of a series is (2n+1)/(2n+3) write down the first three terms and express the difference between the n th and (n+1)th terms as a single fraction in its simplest form. (10 marks) 2. (i) If 3x²-4x+5 = a(x-b)²+c for all values of x, find the values of a, b, and c. (10 marks) Hence, or otherwise, find the least value of 3x²-4x+5. (ii) Find the lapse of time in minutes between the two instants when the two hands of a watch are at right angles for the 1st and the 2nd time between four O'clock and five O'clock. (10 marks) 3. (i) Compute by logarithms the following expression : ⁹√[ (Sin² 15° 04' x Cos³ 31° 31') / ((510.7)² x (4.007)³) ] (10 marks) (ii) Find the value of x from the following equation correct to four significant figures: 32^(2x-1) = 64^x * 40 (10 marks) 4. (i) Three times the third term of an arithmetic progression is twice the sixth term. The sum of the first, third and fifth terms is 9. Find: (a) the ratio of the ninth term to the sixth term, (b) the sum of the first thirteen terms of the progression. (10 marks) (ii) The third term of a geometric progression, in which all the terms are positive, is 2/3 and the sum of the first two terms is 2½. Find the first term, the common ratio and the fourth term of the progression. (10 marks) 5. (i) Taking 1 inch = 1 unit on the x-axis and 1 inch = 2 units on the y-axis draw the graphs of y = 4-x² and 4y = 5x + 4 for values of x from -3 to +3. ( 8 marks) (ii) From your graph, find: a- the range of values of x for which 4-x² is greater than 5/4x+1, (4 marks) b- the values of x for which 4-x²=2.5, (4 marks) c- the square root of 5.6. (4 marks) ⟦line⟧

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### csp_d7daca81edba5bad8239a3334865d3fa

- ٢ - الرقم: الاسم:

٤٠- مقدار جبرى متجانس ⟦line⟧ 1 mark A homogeneous algebraic expression ٤١- درجة المقدار الجبرى ⟦line⟧ " The degree or the dimension of an algebraic expression ٤٢- المعامل الحرفى ⟦line⟧ " The literal coefficient ٤٣- مقدار جبرى من الدرجة الثانية ⟦line⟧ " An algebraic expression of the second degree or a quadratic expression ٤٤- ان حدى الكسر هما بسطه ومقامه ⟦line⟧ 2 marks The two terms of a fraction are its numerator and denominator ٤٥- في كل عملية قسمة يوجد مقسوم ومقسوم عليه وناتج قسمة وفي بعض الحالات باق للقسمة . 5 marks In every process of division there is a dividend, a divisor, a quotient + in some cases a remainder ٤٦- ان الاعمدة المنصفة لاضلاع مثلث تلتقي في مركز الدائرة المرسومة ⟦line⟧ " The perpendicular bisectors of the sides of a triangle meet at the centre of the circumscribed circle. ٤٧- ان الخطوط المتوسطة في المثلث تلتقي في نقطة واحدة تقسم كلا منها الى ثلثين من جهة الرأس " وثلث من جهة القاعدة . وتسمى هذه النقطة مركز ثقل المثلث. The medians of a triangle meet at a point which divides each of them two thirds from the vertex and one third from the base. This point is called the centroid of the triangle. ٤٨- نقيس طول مستقيم فنجد انه يساوى ٦١,٥ سم . ثم نجد فيما بعد ان طوله المضبوط ٦٠ سم . " وفي هذه الحالة نقول ان الخطأ المطلق هو ⟦line⟧ والخطأ النسبي هو ⟦line⟧ والخطأ المئوى هو ⟦line⟧ We measure the length of a st. line + we find that it is equal 61.5 cms. We then find that its exact length is 60 cms. In this case we say that the absolute error is 1.5 cm, the relative error is 1.5/60 and the percentage error is 2.5% ٤٩- ان قيمة المقدار ٥٣٠٩,٧٢ لاقرب اربعة ارقام معنوية هي ⟦line⟧ 5 marks The value of 5309.72 correct to 4 significant figures is 5310.00 ٥٠- ان المعادلة ٣س٢ - ٢س ص + ص = ٥ع - ٣ع هي معادلة من الدرجة ⟦line⟧ في ⟦line⟧ " مجاهيل ⟦line⟧ The equation 3x² - 2xy + y = 5z - 3z is a quadratic equation in three unknowns. (75 marks)

(II) Fill in the blanks in the following equations:-

| (2.5 marks) | 1. | one furlong = | ( 10 ) | chains= | ( 1/8 ) | mile | | " | 2. | one chain = | ( 22 ) | yards = | ( 100 ) | links | | " | 3. | one statute mile = | ( 1760 ) | yds. = | ( 5280 ) | ft. | | " | 4. | one nautical mile = | ( 6080 ) | ft. | | " | 5. | one sq. chain = | ( 484 ) | sq. yds. | | " | 6. | one acre = | ( 10 ) | sq. ch. = | ( 4840 ) | sq. yds. | | " | 7. | one gallon = | ( 8 ) | pints | | " | 8. | one bushel = | ( 8 ) | gallons = | ( 4 ) | pecks | | " | 9. | one English ton = | ( 2240 ) | lbs. = | ( 1016 ) | kilograms | | " | 10. | one English ton = | ( 20 ) | cwt. = | ( 80 ) | qr. = | ( 160 ) | stones. |

[Marginalia] 1 quarter = 1/4 of one cwt = 28 lbs = 2 stones [Marginalia] 1 stone = 14 lbs

(25 marks).

**Traduction anglaise —**

- 2 - Number: Name: 40- A homogeneous algebraic expression ⟦line⟧ 1 mark A homogeneous algebraic expression 41- The degree of an algebraic expression ⟦line⟧ " The degree or the dimension of an algebraic expression 42- The literal coefficient ⟦line⟧ " The literal coefficient 43- An algebraic expression of the second degree ⟦line⟧ " An algebraic expression of the second degree or a quadratic expression 44- The two terms of a fraction are its numerator and denominator ⟦line⟧ 2 marks The two terms of a fraction are its numerator and denominator 45- In every process of division there is a dividend, a divisor, a quotient, and in some cases a remainder. 5 marks In every process of division there is a dividend, a divisor, a quotient + in some cases a remainder 46- The perpendicular bisectors of the sides of a triangle meet at the center of the circumscribed circle ⟦line⟧ " The perpendicular bisectors of the sides of a triangle meet at the centre of the circumscribed circle. 47- The medians of a triangle meet at one point which divides each of them into two-thirds from the vertex " and one-third from the base. This point is called the centroid of the triangle. The medians of a triangle meet at a point which divides each of them two thirds from the vertex and one third from the base. This point is called the centroid of the triangle. 48- We measure the length of a straight line and find it equals 61.5 cm. Then we later find its exact length is 60 cm. " In this case we say the absolute error is ⟦line⟧ and the relative error is ⟦line⟧ and the percentage error is ⟦line⟧ We measure the length of a st. line + we find that it is equal 61.5 cms. We then find that its exact length is 60 cms. In this case we say that the absolute error is 1.5 cm, the relative error is 1.5/60 and the percentage error is 2.5% 49- The value of the expression 5309.72 to the nearest four significant figures is ⟦line⟧ 5 marks The value of 5309.72 correct to 4 significant figures is 5310.00 50- The equation 3x² - 2xy + y = 5z - 3z is an equation of degree ⟦line⟧ in ⟦line⟧ " unknowns ⟦line⟧ The equation 3x² - 2xy + y = 5z - 3z is a quadratic equation in three unknowns. (75 marks) (II) Fill in the blanks in the following equations:- (2.5 marks) | 1. | one furlong = | ( 10 ) | chains= | ( 1/8 ) | mile " | 2. | one chain = | ( 22 ) | yards = | ( 100 ) | links " | 3. | one statute mile = | ( 1760 ) | yds. = | ( 5280 ) | ft. " | 4. | one nautical mile = | ( 6080 ) | ft. " | 5. | one sq. chain = | ( 484 ) | sq. yds. " | 6. | one acre = | ( 10 ) | sq. ch. = | ( 4840 ) | sq. yds. " | 7. | one gallon = | ( 8 ) | pints " | 8. | one bushel = | ( 8 ) | gallons = | ( 4 ) | pecks " | 9. | one English ton = | ( 2240 ) | lbs. = | ( 1016 ) | kilograms " | 10. | one English ton = | ( 20 ) | cwt. = | ( 80 ) | qr. = | ( 160 ) | stones. 1 quarter = 1/4 of one cwt = 28 lbs = 2 stones 1 stone = 14 lbs (25 marks).

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### csp_d805800e30de5cb5a9550806730e3dea

B01 · header / latin ⟦illegible⟧ Algebra Exam Mid-Year Exam

B02 · paragraph / latin (i) (x+2)x - 3(x+2) = (x+2)⟦illegible⟧ = (x+2)(x-3) Ans. (ii) 201x² - 99x - 102 = 3(67x² - 33x - 34) = 3(67x+34)(x-1) Ans. (iii) x⁹ - 64x³ - x⁶ + 64 = (x⁹ - x⁶ - 64x³ + 64) = x⁶(x³-1) - 64(x³-1) = (x⁶-64)(x³-1) = (x³-8)(x³+8)(x³-1) = (x-2)(x²+2x+4)(x+2)(x²-2x+4)(x-1)(x²+x+1) Ans.

B03 · paragraph / latin (b) Let x and y be factors of A and B in F, and let A = aF and B = bF. Then ⟦illegible⟧ F ⊆ (a+b)F if F is also a common factor of ⟦illegible⟧

B04 · paragraph / latin 2(a) (A+B)x² + (B+C)x + (A+C) = 5x² + 3x + 4 is true for all values of x. Letting x=0, we get A+C=4; x=1, A+B+C=5 ... ① Letting x=-1, A-B+C=3 ... ② From ① & ② by subtraction, 2B = 2 or B = 1 From ①, A+1+C=5 or A+C=4 From ②, A-1+C=3 or A+C=4 ⟦illegible⟧ A=2, B=1, C=2 Ans.

B05 · paragraph / latin ⟦illegible⟧ x² + ax + b = (x-3)Q + 0 ∴ (x-3) is a factor ⟦illegible⟧ (x-2) is a factor ⟦illegible⟧ A=2, B=3 Ans.

**Traduction anglaise —**

⟦illegible⟧ Algebra Exam Mid-Year Exam (i) (x+2)x - 3(x+2) = (x+2)⟦illegible⟧ = (x+2)(x-3) Ans. (ii) 201x² - 99x - 102 = 3(67x² - 33x - 34) = 3(67x+34)(x-1) Ans. (iii) x⁹ - 64x³ - x⁶ + 64 = (x⁹ - x⁶ - 64x³ + 64) = x⁶(x³-1) - 64(x³-1) = (x⁶-64)(x³-1) = (x³-8)(x³+8)(x³-1) = (x-2)(x²+2x+4)(x+2)(x²-2x+4)(x-1)(x²+x+1) Ans. (b) Let x and y be factors of A and B in F, and let A = aF and B = bF. Then ⟦illegible⟧ F ⊆ (a+b)F if F is also a common factor of ⟦illegible⟧ 2(a) (A+B)x² + (B+C)x + (A+C) = 5x² + 3x + 4 is true for all values of x. Letting x=0, we get A+C=4; x=1, A+B+C=5 ... ① Letting x=-1, A-B+C=3 ... ② From ① & ② by subtraction, 2B = 2 or B = 1 From ①, A+1+C=5 or A+C=4 From ②, A-1+C=3 or A+C=4 ⟦illegible⟧ A=2, B=1, C=2 Ans. ⟦illegible⟧ x² + ax + b = (x-3)Q + 0 ∴ (x-3) is a factor ⟦illegible⟧ (x-2) is a factor ⟦illegible⟧ A=2, B=3 Ans.

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### csp_d875dc7887dd5987a8f7630760a06630

- ٢ -

الرقم:: الاسم::

٤٠ - مقدار جبري متجانس ٤١ - درجة المقدار الجبري ٤٢ - المعامل الحرفي ٤٣ - مقدار جبري من الدرجة الثانية ٤٤ - ان حدي الكسر هما بسطه ومقامه ٤٥ - في كل عملية قسمة يوجد مقسوم ومقسوم عليه وناتج قسمة وفي بعض الحالات باق للقسمة . ٤٦ - ان الاعمدة المنصفة لاضلاع مثلث تلتقي في مركز الدائرة المرسومة ........... ٤٧ - ان الخطوط المتوسطة في المثلث تلتقي في نقطة واحدة تقسم كلا منها الى ثلثين من جهة الراس وثلث من جهة القاعدة . وتسمى هذه النقطة مركز ثقل المثلث. ٤٨ - نقيس طول مستقيم فنجد انه يساوي ٦١,٥ سم . ثم نجد فيما بعد ان طوله المضبوط ٦٠ سم . وفي هذه الحالة نقول ان الخطأ المطلق هو ........... والخطأ النسبي هو ........... والخطأ المئوي هو ... ٤٩ - ان قيمة المقدار ٥٣,٠٧٢ لاقرب اربعة ارقام معنوية هي ........... ٥٠ - ان المعادلة ٣س٢ - ٢س ص + ص٢ = ٤٥ - ٤ع هي معادلة من الدرجة ........ في ...... مجاهيل .

(75 marks)

(II) Fill in the blanks in the following equations:-

1. one furling = ( ) chains = ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds. = ( ) 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre = ( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. = ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones.

**Traduction anglaise —**

- 2 - Number:: Name:: 40 - Homogeneous algebraic expression 41 - Degree of the algebraic expression 42 - Literal coefficient 43 - Algebraic expression of the second degree 44 - The terms of a fraction are its numerator and its denominator 45 - In every division process, there is a dividend, a divisor, a quotient, and in some cases, a remainder. 46 - The perpendicular bisectors of the sides of a triangle meet at the center of the drawn circle ⟦line⟧ 47 - The medians of a triangle meet at a single point that divides each of them into two-thirds from the vertex side and one-third from the base side. This point is called the center of gravity (centroid) of the triangle. 48 - We measure the length of a straight line and find it equals 61.5 cm. Then we later find that its exact length is 60 cm. In this case, we say the absolute error is ⟦line⟧ and the relative error is ⟦line⟧ and the percentage error is ⟦line⟧ 49 - The value of the expression 53.072 to the nearest four significant figures is ⟦line⟧ 50 - The equation 3x^2 - 2xy + y^2 = 45 - 4z is an equation of the ⟦line⟧ degree in ⟦line⟧ unknowns. (75 marks) (II) Fill in the blanks in the following equations:- 1. one furling = ( ) chains = ( ) mile 2. one chain = ( ) yards = ( ) links 3. one statute mile = ( ) yds. = ( ) 4. one nautical mile = ( ) ft. 5. one sq. chain = ( ) sq. yds. 6. one acre = ( ) sq. ch. = ( ) sq. yds. 7. one gallon = ( ) pints 8. one bushel = ( ) gallons = ( ) pecks 9. one English ton = ( ) lbs. = ( ) kilograms 10. one English ton = ( ) cwt. = ( ) qr. = ( ) stones.

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### csp_d88068af0efd5ac2953284c421ebd6b0

Shamash Secondary School 3rd Quarter Examination, March 1967.

Subject:: Algebra Date:: 20/3/1967 Class:: 4th Secondary Time:: 8:30 - 10:00 a.m.

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1- (i) Draw the graphs of y = x³ and y = 3x²-4 on the same axes, for values of x from x = -3 to x = 3 (20 marks) (ii) From your graphs find the roots of the equation x³-3x²+4=0 (15 marks). (iii) For what values of x is the expression x³-3x²+4 always negative? always positive ? Use your graphs to explain why. (15 marks)

2- Simplify (i) / √a . ∛/ac . √/c³ / √b . √b⁻¹ / a⁻¹/6 (15 marks) (ii) [ (9ⁿ⁺¼) . √((3)(3ⁿ)) / 3√3⁻ⁿ ]¹/ⁿ (15 marks)

3- Divide (∛x² + 2x⅓ - 16x⁻⅔ - 32/x) by (xℙ + 4x⁻ℙ + 4/√x) (20 marks)

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**Traduction anglaise —**

Shamash Secondary School 3rd Quarter Examination, March 1967. Subject:: Algebra Date:: 20/3/1967 Class:: 4th Secondary Time:: 8:30 - 10:00 a.m. ⟦line⟧ 1- (i) Draw the graphs of y = x³ and y = 3x²-4 on the same axes, for values of x from x = -3 to x = 3 (20 marks) (ii) From your graphs find the roots of the equation x³-3x²+4=0 (15 marks). (iii) For what values of x is the expression x³-3x²+4 always negative? always positive ? Use your graphs to explain why. (15 marks) 2- Simplify (i) √a . ∛ac . √c³ / √b . √b⁻¹ / a⁻¹/6 (15 marks) (ii) [ (9ⁿ⁺¼) . √((3)(3ⁿ)) / 3√3⁻ⁿ ]¹/ⁿ (15 marks) 3- Divide (∛x² + 2x⅓ - 16x⁻⅔ - 32/x) by (xℙ + 4x⁻ℙ + 4/√x) (20 marks) ⟦line⟧

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### csp_d8af73bfe24b5bf7994f514d2ac10bc4

| x | y | | -2 | 1 1/9 | | -1 | 1 1/2 | | 0 | 2/3 | | 1/2 | 0 | | 2/3 | - 2/3 | | 1 1/3 | 3 1/3 | | 2 | 2 | | 4 | 2/3 | 1 1/2 |

y 4 3 x=1 y=x 2 A(2,2) 1 y₂ = 2(2x-1)/3(x-1) B(1/3, 1/3) -2 -1 0 1 2 3 4 x -1 x=1 is a vertical asymptote -2

**Traduction anglaise —**

⟦illegible⟧ simultaneously, we have: ⟦illegible⟧ = 3x(x-1) ∴ 4x-2 = 3x²-3x ∴ 3x²-7x+2 = 0 or (3x-1)(x-2) = 0 ∴ x=1/3 or x=2 the points of intersection A(2,2) + B(1/3, 1/3) the two functions 2(2x-1)/3(x-1) and x are equal ∴ x=1/3 } Ans. x=2 } x | y -2 | 1 1/9 -1 | 1 1/2 0 | 2/3 1/2 | 0 2/3 | - 2/3 1 1/3 | 3 1/3 2 | 2 4 | 2/3 | 1 1/2 y 4 3 x=1 y=x 2 A(2,2) 1 y₂ = 2(2x-1)/3(x-1) B(1/3, 1/3) -2 -1 0 1 2 3 4 x -1 x=1 is a vertical asymptote -2

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### csp_d91f68231ad75205a03206ab46333029

B01 · header / latin ⟦illegible⟧ Secondary School Final Examination, May 1968

B02 · form / latin Subject: Mathematics Date: 14/5/1968 Class: 4th Scientific Year Time: 8:00-11:00 a.m.

B03 · paragraph / latin ⟦line⟧ All questions are to be attempted.

B04 · paragraph / latin 1. (i) Resolve into four factors: x⁹+x³y⁶-8x⁶y³-8y⁹ (6 marks) (ii) Simplify: { (a⁴-x⁴)/(a²-2ax+x²) ÷ (a²+ax)/(a-x) } x { (a⁵-a³x²)/(a³+x³) ÷ (a⁴-2a³x+a²x²)/(a²-ax+x²) } (7 marks) (iii) Find the value of B if 2x⁴+2x^(7/2)-5x³-x^(5/2)+3x²-x^(3/2)+x^(1/2)+x+3 +B is exactly divisible by x+x^(1/2)-2 (7 marks)

B05 · paragraph / latin 2. (i) Compute by logarithms the value of x, if x= ⁷√[ (0.1023)³.Cos²41°28' / (1.007)²(tan⁵47°51') ] arranging your work neatly. (10 marks) (ii) Solve for x the equation: (31.01)^(3x-1) = 104(2.003)^(2x+1) (10 marks)

B06 · paragraph / latin 3. I rode one third of a journey at 10 miles an hour, one third more at 9, and the rest at 8 miles an hour. If I had ridden half the journey at 10, and the other half at 8 miles per hour, I should have been half a minute longer on the way. What distance did I ride ? (20 marks)

B07 · paragraph / latin 4. (i) The sum of n terms of a series is 1/3n(4n²-1). Find the first two terms. (6 marks) (ii) In boring a well 400 ft deep the cost is 2s. 3d. for the first foot and an additional penny for each subsequent foot. What is the cost of boring the last foot, and also of boring the entire well ? (7 marks) (iii) The third term of a geometric series, in which all the terms are positive, is 18 and the fifth term is 40.5. Find the first term and the sum of the first six terms. (7 marks)

B08 · paragraph / latin 5. (i) Draw the graph of y=1/4(3x²-5x-4) for values of x from -2 to +3, using a scale of 1 inch to 1 unit on each axis. (7 marks) (ii) Use your graph to find the least value of 3x²-5x-4. (6 marks) (iii) By drawing the appropriate straight line on your graph solve the equation 3x²-5x-6=0. (7 marks)

B09 · footer / latin ⟦line⟧

**Traduction anglaise —**

⟦illegible⟧ Secondary School Final Examination, May 1968 Subject: Mathematics Date: 14/5/1968 Class: 4th Scientific Year Time: 8:00-11:00 a.m. ⟦line⟧ All questions are to be attempted. 1. (i) Resolve into four factors: x⁹+x³y⁶-8x⁶y³-8y⁹ (6 marks) (ii) Simplify: { (a⁴-x⁴)/(a²-2ax+x²) ÷ (a²+ax)/(a-x) } x { (a⁵-a³x²)/(a³+x³) ÷ (a⁴-2a³x+a²x²)/(a²-ax+x²) } (7 marks) (iii) Find the value of B if 2x⁴+2x^(7/2)-5x³-x^(5/2)+3x²-x^(3/2)+x^(1/2)+x+3 +B is exactly divisible by x+x^(1/2)-2 (7 marks) 2. (i) Compute by logarithms the value of x, if x= ⁷√[ (0.1023)³.Cos²41°28' / (1.007)²(tan⁵47°51') ] arranging your work neatly. (10 marks) (ii) Solve for x the equation: (31.01)^(3x-1) = 104(2.003)^(2x+1) (10 marks) 3. I rode one third of a journey at 10 miles an hour, one third more at 9, and the rest at 8 miles an hour. If I had ridden half the journey at 10, and the other half at 8 miles per hour, I should have been half a minute longer on the way. What distance did I ride ? (20 marks) 4. (i) The sum of n terms of a series is 1/3n(4n²-1). Find the first two terms. (6 marks) (ii) In boring a well 400 ft deep the cost is 2s. 3d. for the first foot and an additional penny for each subsequent foot. What is the cost of boring the last foot, and also of boring the entire well ? (7 marks) (iii) The third term of a geometric series, in which all the terms are positive, is 18 and the fifth term is 40.5. Find the first term and the sum of the first six terms. (7 marks) 5. (i) Draw the graph of y=1/4(3x²-5x-4) for values of x from -2 to +3, using a scale of 1 inch to 1 unit on each axis. (7 marks) (ii) Use your graph to find the least value of 3x²-5x-4. (6 marks) (iii) By drawing the appropriate straight line on your graph solve the equation 3x²-5x-6=0. (7 marks) ⟦line⟧

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### csp_db7071ff575e55c7afaf2d5e910c8e76

- p. 2- Algebra. 4th Second. 3/6/1964 (cont'd.)

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5. (i) What is the sum of the first (n+1) even numbers ? ( 4 marks) (ii) What is the sum of the first ten numbers beginning with 22 that are divisible by 11 ? ( 4 marks) (iii)A body falling freely falls approximately 16 ft., in the first second, and in each succeeding second 32 ft. more than in the second immediately preceding. If a stone dropped from a stationary balloon reaches the ground in 12 seconds, how far does it fall in the last second ? How high is the balloon ? ( 4 marks) (iv) If it were possible for a rubber ball to fall 10 ft., and bound back 5 ft., then to fall 5 ft., and bound back 2½ ft., and to continue this forever, what is the limit of the total distance through which the ball would pass ? ( 4 marks)

6. (a) Compute by logarithms the value of the following, arranging your work neatly: _________________________ 5 / (1.005)³ x (0.0004007)⁵ / ----------------------- V (0.06109)² x (10.71)⟦⅕⟧ ( 8 marks) (b) Find, correct to five decimal places, the value of log 648, 3 having given: log 2 = 0.30103 and log 3 = 0.47712 10 10 ( 8 marks)

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**Traduction anglaise —**

- p. 2- Algebra. 4th Second. 3/6/1964 (cont'd.) ⟦line⟧ 5. (i) What is the sum of the first (n+1) even numbers ? ( 4 marks) (ii) What is the sum of the first ten numbers beginning with 22 that are divisible by 11 ? ( 4 marks) (iii)A body falling freely falls approximately 16 ft., in the first second, and in each succeeding second 32 ft. more than in the second immediately preceding. If a stone dropped from a stationary balloon reaches the ground in 12 seconds, how far does it fall in the last second ? How high is the balloon ? ( 4 marks) (iv) If it were possible for a rubber ball to fall 10 ft., and bound back 5 ft., then to fall 5 ft., and bound back 2½ ft., and to continue this forever, what is the limit of the total distance through which the ball would pass ? ( 4 marks) 6. (a) Compute by logarithms the value of the following, arranging your work neatly: ⟦line⟧ 5 / (1.005)³ x (0.0004007)⁵ / ⟦line⟧ V (0.06109)² x (10.71)⟦⅕⟧ ( 8 marks) (b) Find, correct to five decimal places, the value of log 648, 3 having given: log 2 = 0.30103 and log 3 = 0.47712 10 10 ( 8 marks) ⟦line⟧

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### csp_dc6db62042785c2492ac9ab141de2395

-p.2-

Algebra. 4th Secondary, 6/2/1967 --

5. (i) Draw on the same diagram the graphs of the function 4x-3, and of the function 4x²-4x-15, taking ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (8 marks)

(ii) From your diagram, find the roots of the two simultaneous equations y = 4x-3 ........(1) y = 4x²-4x-15 ....(2) (7 marks)

(iii) From the graph of the function 4x²-4x-15, find the roots of the equation 4x²-4x-15=0. (5 marks)

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**Traduction anglaise —**

-p.2- Algebra. 4th Secondary, 6/2/1967 ⟦line⟧ 5. (i) Draw on the same diagram the graphs of the function 4x-3, and of the function 4x²-4x-15, taking ½ inch as one unit on the x-axis and one tenth of an inch as one unit on the y-axis. (8 marks) (ii) From your diagram, find the roots of the two simultaneous equations y = 4x-3 ........(1) y = 4x²-4x-15 ....(2) (7 marks) (iii) From the graph of the function 4x²-4x-15, find the roots of the equation 4x²-4x-15=0. (5 marks) ⟦line⟧

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### csp_dd470cc3459b55b398fbca82cd6feb32

II. (b) Let x kilometres = the distance between the two cities on the right Bank ∴ x/5 k.p.hr. is the average speed of the first car ∴ x+50/7.5 k.p.h. " " " " " second car } ∴ x/5 - x+50/7.5 = 20 Multiplying the terms of the equation by 15, we get: 3x - 2(x+50) = 300 or 3x - 2x - 100 = 300 or x = 400 km. Ans. 1 = the length of the first course x+50 = 400+50=450 km. Ans. 2 = " " " 2nd "

III (a) If log₂(4x-4) = 2 , then log₄x = ? From the equation we get 4x-4 = 2² or 4x-4 = 4 ∴ x = 2 Now log₄x = log₄2 = y ∴ 2 = 4ʸ or 2 = 2²ʸ ∴ 2y = 1 ∴ y = 1/2 ∴ log₄x = log₄2 = 1/2 Ans.

(b) Let x = ⁷√[(0.1062)²(0.0071)³ / (1.005)(3.007)⁵]

log x = 2̄.45907 = 2̄.4591 Correct to 4 dec. pl. ∴ x = 0.02878 or = 2.878 x 10⁻² Ans.

IV (a) The distances fallen during the consecutive seconds beginning with the first are : 16 + 48 + 80 + 112 + ... which is the sum of an A.P. in which a = 16 , d = 32 , s = 1936 , n = ? From the formula s = n/2 {2a + (n-1)d} we get: 1936 = n/2 {2x16 + (n-1)x32} or 2 x 1936 = n {32 + 32n - 32} or 32n² = 3872 or n² = 121 or n = 11 ∴ the number of seconds which will take the stone to reach the bottom = 11 seconds Ans.

**Traduction anglaise —**

2 II. (b) Let x kilometres = the distance between the two cities on the right Bank ∴ x/5 k.p.hr. is the average speed of the first car ∴ x+50/7.5 k.p.h. " " " " " second car } ∴ x/5 - x+50/7.5 = 20 Multiplying the terms of the equation by 15, we get: 3x - 2(x+50) = 300 or 3x - 2x - 100 = 300 or x = 400 km. Ans. 1 = the length of the first course x+50 = 400+50=450 km. Ans. 2 = " " " 2nd " III (a) If log₂(4x-4) = 2 , then log₄x = ? From the equation we get 4x-4 = 2² or 4x-4 = 4 ∴ x = 2 Now log₄x = log₄2 = y ∴ 2 = 4ʸ or 2 = 2²ʸ ∴ 2y = 1 ∴ y = 1/2 ∴ log₄x = log₄2 = 1/2 Ans. (b) Let x = ⁷√[(0.1062)²(0.0071)³ / (1.005)(3.007)⁵] log 0.1062 = 1̄.0261 | 2 log 0.1062 = 2̄.0522 | log 1.005 = 0.0021 log 0.0071 = 3̄.8513 | 3 log 0.0071 = 7̄.5539 | 5 log 3.007 = 2.3905 log 1.005 = 0.0021 | log Num. = 9̄.6061 | log Den. = 2.3926 log 3.007 = 0.4781 | log Den. = 2.3926 | | 7 log x = 11̄.2135 | log x = 2̄.45907 = 2̄.4591 Correct to 4 dec. pl. ∴ x = 0.02878 or = 2.878 x 10⁻² Ans. IV (a) The distances fallen during the consecutive seconds beginning with the first are : 16 + 48 + 80 + 112 + ... which is the sum of an A.P. in which a = 16 , d = 32 , s = 1936 , n = ? From the formula s = n/2 {2a + (n-1)d} we get: 1936 = n/2 {2x16 + (n-1)x32} or 2 x 1936 = n {32 + 32n - 32} or 32n² = 3872 or n² = 121 or n = 11 ∴ the number of seconds which will take the stone to reach the bottom = 11 seconds Ans.

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### csp_deced1f4c9d358298533bc21412779b3

Shamash Secondary Mid-Year Exam. 1965-1966

Subject:: Arithmetic Date:: 1.2.1966 Class :: 4th Year, Scientific Time:: 8:30 - 10:30

----- Answer all questions:

1. A dealer sells 2640 articles for £ 341 his profit being 24% of his outlay. Find the cost price of each article. If the cost price to the dealer increased by 8% and he does not change his selling price, find how many articles he must sell in order to obtain the same total profit as before ?

2.a) Compare the volume of the Moon with that of the Earth, if the diameter of the former be to the diameter of the latter as 27 to 100. Give your answer in the form of a ratio whose denomina- tor is unity.

b) In order to increase the weight of a block of steel by 1 oz. a cylindrical hole ¼ in. in diameter, is drilled in the block, and the hole is then filled with lead. To what depth must the hole be drilled if steel weighs .29 lb. per cub. in. and lead weighs .41 lb. per cub. in. ? (To nearest 1 in.) 100

3. The rate in a certain town is 13s. 10d. in the £. If the rateable value is increased by 5% and the rate reduced by 6d. in the £, will the income from rates be increased or decreased, and by how much per cent ?

4. Brass is made up of copper and zinc in the proportion 5:4 by volume. If 1cc of copper weighs 8.8 gm and 1 cc of zinc weighs 7.1 gm cal- culate:

(a) the weight of copper required to form brass with 50 cc of zinc, ⟦550 gm⟧

(b) the weight of copper in 1 kilogram of brass correct to the nearest gm, ⟦608 gm⟧

-----

**Traduction anglaise —**

Shamash Secondary Mid-Year Exam. 1965-1966 Subject:: Arithmetic Date:: 1.2.1966 Class :: 4th Year, Scientific Time:: 8:30 - 10:30 ⟦line⟧ Answer all questions: 1. A dealer sells 2640 articles for £ 341 his profit being 24% of his outlay. Find the cost price of each article. If the cost price to the dealer increased by 8% and he does not change his selling price, find how many articles he must sell in order to obtain the same total profit as before ? 2.a) Compare the volume of the Moon with that of the Earth, if the diameter of the former be to the diameter of the latter as 27 to 100. Give your answer in the form of a ratio whose denomina- tor is unity. b) In order to increase the weight of a block of steel by 1 oz. a cylindrical hole ¼ in. in diameter, is drilled in the block, and the hole is then filled with lead. To what depth must the hole be drilled if steel weighs .29 lb. per cub. in. and lead weighs .41 lb. per cub. in. ? (To nearest 1 in.) 100 3. The rate in a certain town is 13s. 10d. in the £. If the rateable value is increased by 5% and the rate reduced by 6d. in the £, will the income from rates be increased or decreased, and by how much per cent ? 4. Brass is made up of copper and zinc in the proportion 5:4 by volume. If 1cc of copper weighs 8.8 gm and 1 cc of zinc weighs 7.1 gm cal- culate: (a) the weight of copper required to form brass with 50 cc of zinc, ⟦550 gm⟧ (b) the weight of copper in 1 kilogram of brass correct to the nearest gm, ⟦608 gm⟧ (c) the volume of 1 kilogram of brass correct to the nearest cc. ⟦124 cc⟧ ⟦line⟧

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### csp_defca7b932445df4b289c1bddea06895

B01 · paragraph / latin for ⟦most⟧ children. The weather is usually good; so that one can spend most of one's time playing in the garden or, if one lives in the country, out in the woods and fields. Even if one lives in a big town, one can usually go to a park to play. The best place for a summer holiday, however, is the seaside. Some children are lucky enough to live near the sea, but for the others who <del>no</del> do not, a week or two at one of the big seaside towns is something which they will talk about for the whole of the following year.

**Traduction anglaise —**

for ⟦most⟧ children. The weather is usually good; so that one can spend most of one's time playing in the garden or, if one lives in the country, out in the woods and fields. Even if one lives in a big town, one can usually go to a park to play. The best place for a summer holiday, however, is the seaside. Some children are lucky enough to live near the sea, but for the others who <del>no</del> do not, a week or two at one of the big seaside towns is something which they will talk about for the whole of the following year.

---

### csp_df914ebe5f9b5b0eba55c4bb01f5fc8e

B01 · header / latin Solution :

B02 · paragraph / latin 3) (i) Yearly rates = (£ 13 1s 4d) x 2 = £ 26 2s 8d <del>Rateable</del> Rateable value of the House = £ 26 2s 8d ⟦line⟧ 16s 4d = 522 2/3 ⟦line⟧ 16 1/3 = 1568 = <del>313</del> £ 32 ⟦line⟧ 49

B03 · paragraph / latin (ii) (£ 13 1s 4d) 5s 7.3d ⟦line⟧ 16s 4d = 3136 x 67.3 = 16 x 67.3 ⟦line⟧ ⟦line⟧ 196 = 1076.8 d/year 1076.8 = 20.7 d ⟦line⟧ 52 or 1s 8.7d per ⟦line⟧ ⟦line⟧ week per child

B04 · marginalia / latin 5

**Traduction anglaise —**

Solution : 3) (i) Yearly rates = (£ 13 1s 4d) x 2 = £ 26 2s 8d <del>Rateable</del> Rateable value of the House = £ 26 2s 8d ⟦line⟧ 16s 4d = 522 2/3 ⟦line⟧ 16 1/3 = 1568 = <del>313</del> £ 32 ⟦line⟧ 49 (ii) (£ 13 1s 4d) 5s 7.3d ⟦line⟧ 16s 4d = 3136 x 67.3 = 16 x 67.3 ⟦line⟧ ⟦line⟧ 196 = 1076.8 d/year 1076.8 = 20.7 d ⟦line⟧ 52 or 1s 8.7d per ⟦line⟧ ⟦line⟧ week per child 5

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### csp_e046de4ba0425a50bdce0a3c670e62cd

B01 · paragraph / latin • Construct ⟦illegible⟧ ⟦illegible⟧

B02 · table / latin x | y₁ | y₂ -1 1/2 | -3 3/8 -1 | -1 -1/2 | -1/8 0 | 0 | 1 1/2 1/2 | 1/8 1 | 1 1 1/2 | 3 3/8 2 | 8 | 3 2.2 | 10.65

B03 · paragraph / latin (ii) Let y₂ = 3/4 x + 3/2 ⟦illegible⟧ straight line. Give some ⟦illegible⟧ for x Compute the corresponding values of y₂ & plot. The points of intersection satisfy both equations. ⟦illegible⟧ values of x at A, B ⟦illegible⟧ roots of the equation x³ = 3/4 x + 3/2 = 0 These roots are: x = -1 1/2 x = -1/2 ⟦illegible⟧ Ans. x = 2

B04 · other / latin y = x³ y = 3/4 x + 3/2 C (2, 8) B (-1/2, 1 1/8) A (-1 1/2, -3 3/8)

B05 · paragraph / latin (iii) The expression x³ - (3/4 x + 3/2) is positive when y₁ > y₂ This is the case when -1 1/2 < x < -1/2 and when x > 2 because in both these cases the curve of y₁ lies above the curve of y₂ Q.E.D.

**Traduction anglaise —**

• Construct ⟦illegible⟧ ⟦illegible⟧ x | y₁ | y₂ -1 1/2 | -3 3/8 | -1 | -1 | -1/2 | -1/8 | 0 | 0 | 1 1/2 1/2 | 1/8 | 1 | 1 | 1 1/2 | 3 3/8 | 2 | 8 | 3 2.2 | 10.65 | (ii) Let y₂ = 3/4 x + 3/2 ⟦illegible⟧ straight line. Give some ⟦values⟧ for x Compute the corresponding values of y₂ & plot. The points of intersection satisfy both equations. ⟦illegible⟧ values of x at A, B ⟦illegible⟧ roots of the equation x³ = 3/4 x + 3/2 = 0 These roots are: x = -1 1/2 x = -1/2 ⟦illegible⟧ Ans. x = 2 y = x³ y = 3/4 x + 3/2 C (2, 8) B (-1/2, 1 1/8) A (-1 1/2, -3 3/8) (iii) The expression x³ - (3/4 x + 3/2) is positive when y₁ > y₂ This is the case when -1 1/2 < x < -1/2 and when x > 2 because in both these cases the curve of y₁ lies above the curve of y₂ Q.E.D.

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### csp_e1203f985ef0526da0a1a0f5949ad98d

[Marginalia] Audrey [Marginalia] ⟦Samir 29)⟧

Shamash Secondary School Final Examination, May, 1966. Subject: Algebra Date: 29/5/1966 Class : 4th Year, Scientific. Time: 8:00 - 11:00 a.m.

--- Answer all f i v e questions:

3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: Log₃ (2x²-7x) = 2 (10 marks) (ii) Without using tables evaluate: (log₂ 9)(log₉ 32) (10 marks) (iii) Compute the value of y by logarithms, arranging your work neatly: y= ⁷√((tan 19°45')² . (cos 77°16')³ / (3.004)⁵ . (50.06)³) (10 marks)

(cont'd.p.2)..

**Traduction anglaise —**

Audrey ⟦Samir 29)⟧ Shamash Secondary School Final Examination, May, 1966. Subject: Algebra Date: 29/5/1966 Class : 4th Year, Scientific. Time: 8:00 - 11:00 a.m. ⟦line⟧ Answer all f i v e questions: 1. (a) Factor the following: (i) my⁴ + 16mx⁴ - 12mx²y² (4 marks) (ii) a²b²x² - a²b² - 2abx² + 2ab + x² - 1 (4 marks) (iii) 27x⁶y⁹ + 64y³ (4 marks) (b) Find the value of p and q which will make the expression 2x³ + px² + qx + 1 divisible by (x-1) and (x+1), and find the third factor. (8 marks) 2. (a) Use the method of completing the square to show that the sum of the roots of the equation ax²+bx+c=o is equal to (- b/a) and that their product is equal to ( c/a ). (10 marks) (b) Find the value of x from the following equation: 3.10²ˣ - 13.10ˣ + 4 = o (10 marks) 3. Solve only two sections from the following three sections: (i) Find the value of x from the following equation: Log₃ (2x²-7x) = 2 (10 marks) (ii) Without using tables evaluate: (log₂ 9)(log₉ 32) (10 marks) (iii) Compute the value of y by logarithms, arranging your work neatly: y= ⁷√((tan 19°45')² . (cos 77°16')³ / (3.004)⁵ . (50.06)³) (10 marks) (cont'd.p.2)..

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### csp_e1584e75c6bd52449f2e7fda64eb6f54

[Marginalia] تنزيل

(a) (i) Area of surface of sphere = 4 π r² (ii) Volume of sphere = 4/3 π r³ (iii) Area of curved surface of a cone = π r l (iv) Volume of a cone = 1/3 π r² h (v) Area of trapezium = (a+b) h / 2

(b) Volume of bar = 15" x 9" x 8" cubic inches Net volume (after allowing for 10% loss = 15 x 9 x 8 x 9 / 10 cubic inches

Volume of a sphere = 4/3 x 3.142 x 27/8 = 1.571 x 9 cubic inches. No. of spheres = 15 x 9 x 8 x 9 / 10 x 9 x 1.571 = 68 spheres Wt. of 68 spheres = 320/1728 x 68 x 1.571 x 9 = 5474.935 / 12 = 456 lb to nearest lb.

205 17 --- 1435 205 --- 3485 1.571 --- 3485 24395 17425 3485 --- 5474.935 456.2 12 ) 5474.935 67 74 29

**Traduction anglaise —**

Download (a) (i) Area of surface of sphere = 4 π r² (ii) Volume of sphere = 4/3 π r³ (iii) Area of curved surface of a cone = π r l (iv) Volume of a cone = 1/3 π r² h (v) Area of trapezium = (a+b) h / 2 (b) Volume of bar = 15" x 9" x 8" cubic inches Net volume (after allowing for 10% loss = 15 x 9 x 8 x 9 / 10 cubic inches Volume of a sphere = 4/3 x 3.142 x 27/8 = 1.571 x 9 cubic inches. No. of spheres = 15 x 9 x 8 x 9 / 10 x 9 x 1.571 = 68 spheres Wt. of 68 spheres = 320/1728 x 68 x 1.571 x 9 = 5474.935 / 12 = 456 lb to nearest lb. 205 17 ⟦line⟧ 1435 205 ⟦line⟧ 3485 1.571 ⟦line⟧ 3485 24395 17425 3485 ⟦line⟧ 5474.935 456.2 12 ) 5474.935 67 74 29

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### csp_e28a2a5706c75287b2ce0c31238401e8

B01 · header / latin Solution to ⟦illegible⟧ Paper ⟦illegible⟧, May 19th, 66.

B02 · paragraph / latin ⟦illegible⟧ = m(16x⁴ - 16x²y² + y⁴) = m(16x⁴ - 8x²y² + y⁴ - 8x²y²) = = m[(4x² - y²)² - (⟦illegible⟧)²] = m(4x² - y² + 2xy)(4x² - y² - 2xy) = m(4x² + 2xy - y²)(4x² - 2xy - y²) Ans. 1

B03 · paragraph / latin (ii) a²b²x² - a²b² - 2abx² + 2ab + x² - 1 = a²b²(x² - 1) - 2ab(x² - 1) + (x² - 1) = (x² - 1)(a²b² - 2ab + 1) = (x - 1)(x + 1)(ab - 1)²

B04 · paragraph / latin (iii) 27x⁶y³ + 64y⁹ = y³(27x⁶y⁶ + 64) = y³[(3x²y²)³ + 4³] = y³(3x²y² + 4)(9x⁴y⁴ - 12x²y² + 16) = y³(3x²y² + 4)(9x⁴y⁴ + 24x²y² + 16 - 36x²y²) = y³(3x²y² + 4)[(3x²y² + 4)² - (6xy)²] = y³(3x²y² + 4)(3x²y² + 6xy + 4)(3x²y² - 6xy + 4) Ans. 3

B05 · paragraph / latin (b) 2x³ + px² + qx + 1 will be divisible by (x - 1) if the expression = 0 when x = 1 ∴ 2 + p + q + 1 = 0 ... ① also when (x + 1) = 0 then x = -1 when the expression ∴ -2 + p - q + 1 = 0 ... ② ∴ 2p + 2 = 0 ∴ p + 1 = 0 ∴ p = -1 from equation ① -1 + p + q + 1 = 0 ∴ 2 - 1 + q + 1 = 0 ∴ q = -2 ∴ the expression is 2x³ - x² - 2x + 1 = 0 2(x³ - x) - (x² - 1) = 0 or 2x(x² - 1) - (x² - 1) = 0 ∴ (x² - 1)(2x - 1) = 0 (x - 1)(x + 1)(2x - 1) = 0 ∴ the third factor is (2x - 1) Ans. 3 Another method would be to divide (2x³ + px² + qx + 1) ⟦illegible⟧ (x² - 1) getting a remainder of (p + p + q + 1) in the first case and a remainder of (-2 + p - q + 1) second case. Equating each remainder to zero, we obtain the same values.

B06 · paragraph / latin 2. (a) ax² + bx + c = 0 ∴ a(x² + b/a x + c/a) = 0 ∴ a[x² + b/a x + b²/4a² + c/a - b²/4a²] = 0 ∴ a[(x + b/2a)² - (b² - 4ac)/4a²] = 0 ∴ a[(x + b/2a)² - (√(b² - 4ac)/2a)²] = 0 ∴ a(x + b/2a + √(b² - 4ac)/2a)(x + b/2a - √(b² - 4ac)/2a) = 0 x + b/2a + √(b² - 4ac)/2a = 0 ∴ x₁ = (-b - √(b² - 4ac))/2a x + b/2a - √(b² - 4ac)/2a = 0 ∴ x₂ = (-b + √(b² - 4ac))/2a ∴ the sum of the two roots = x₁ + x₂ = (-b - √(b² - 4ac))/2a + (-b + √(b² - 4ac))/2a = -2b/2a = (-b/a) Ans. 1 the product = x₁x₂ = ((-b - √(b² - 4ac))/2a) ((-b + √(b² - 4ac))/2a) = ((-b)² - (√(b² - 4ac))²) / 4a² = (b² - b² + 4ac) / 4a² = 4ac / 4a² = (c/a) Ans. 2

**Traduction anglaise —**

Solution to ⟦illegible⟧ Paper ⟦illegible⟧, May 19th, 66. ⟦illegible⟧ = m(16x⁴ - 16x²y² + y⁴) = m(16x⁴ - 8x²y² + y⁴ - 8x²y²) = = m[(4x² - y²)² - (⟦illegible⟧)²] = m(4x² - y² + 2xy)(4x² - y² - 2xy) = m(4x² + 2xy - y²)(4x² - 2xy - y²) Ans. 1 (ii) a²b²x² - a²b² - 2abx² + 2ab + x² - 1 = a²b²(x² - 1) - 2ab(x² - 1) + (x² - 1) = (x² - 1)(a²b² - 2ab + 1) = (x - 1)(x + 1)(ab - 1)² (iii) 27x⁶y³ + 64y⁹ = y³(27x⁶y⁶ + 64) = y³[(3x²y²)³ + 4³] = y³(3x²y² + 4)(9x⁴y⁴ - 12x²y² + 16) = y³(3x²y² + 4)(9x⁴y⁴ + 24x²y² + 16 - 36x²y²) = y³(3x²y² + 4)[(3x²y² + 4)² - (6xy)²] = y³(3x²y² + 4)(3x²y² + 6xy + 4)(3x²y² - 6xy + 4) Ans. 3 (b) 2x³ + px² + qx + 1 will be divisible by (x - 1) if the expression = 0 when x = 1 ∴ 2 + p + q + 1 = 0 ... ① also when (x + 1) = 0 then x = -1 when the expression ∴ -2 + p - q + 1 = 0 ... ② ∴ 2p + 2 = 0 ∴ p + 1 = 0 ∴ p = -1 from equation ① -1 + p + q + 1 = 0 ∴ 2 - 1 + q + 1 = 0 ∴ q = -2 ∴ the expression is 2x³ - x² - 2x + 1 = 0 2(x³ - x) - (x² - 1) = 0 or 2x(x² - 1) - (x² - 1) = 0 ∴ (x² - 1)(2x - 1) = 0 (x - 1)(x + 1)(2x - 1) = 0 ∴ the third factor is (2x - 1) Ans. 3 Another method would be to divide (2x³ + px² + qx + 1) ⟦illegible⟧ (x² - 1) getting a remainder of (p + p + q + 1) in the first case and a remainder of (-2 + p - q + 1) second case. Equating each remainder to zero, we obtain the same values. 2. (a) ax² + bx + c = 0 ∴ a(x² + b/a x + c/a) = 0 ∴ a[x² + b/a x + b²/4a² + c/a - b²/4a²] = 0 ∴ a[(x + b/2a)² - (b² - 4ac)/4a²] = 0 ∴ a[(x + b/2a)² - (√(b² - 4ac)/2a)²] = 0 ∴ a(x + b/2a + √(b² - 4ac)/2a)(x + b/2a - √(b² - 4ac)/2a) = 0 x + b/2a + √(b² - 4ac)/2a = 0 ∴ x₁ = (-b - √(b² - 4ac))/2a x + b/2a - √(b² - 4ac)/2a = 0 ∴ x₂ = (-b + √(b² - 4ac))/2a ∴ the sum of the two roots = x₁ + x₂ = (-b - √(b² - 4ac))/2a + (-b + √(b² - 4ac))/2a = -2b/2a = (-b/a) Ans. 1 the product = x₁x₂ = ((-b - √(b² - 4ac))/2a) ((-b + √(b² - 4ac))/2a) = ((-b)² - (√(b² - 4ac))²) / 4a² = (b² - b² + 4ac) / 4a² = 4ac / 4a² = (c/a) Ans. 2

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### csp_e5bea5a7a77a52c9bc3ab09f92dc5480

B01 · header / latin Shamash School Monthly Quiz

B02 · form / latin Subject: Trigonometry Date: 25/1/50 Class: 4th Time: 45 minutes, 1 hour.

B03 · paragraph / latin All questions are to be attempted.

B04 · paragraph / latin 1. (a) Find θ, φ and ψ when: cot θ = 2.4363 , sec φ = 3.1241 , csc ψ = 4.0213 22° 19' 71° 20' 14° 24'

B05 · paragraph / latin (b) Solve the triangle ABC, given that: C = 90°, B = 71° 31', b = 76 in. A = 18° 29' a = 25.4068 c = 81.19

B06 · paragraph / latin 2. (a) Evaluate as shortly as possible 1/sin 19° , 1/cos 29° , 1/tan 42° , 3.0713 1.1430 1.1106 cos 17°/sin 17° , sin 42°/sin 48° , cos 63/cos 27 3.2709 0.9004 0.5095

B07 · paragraph / latin (b) If tan 19° + tan 31° = tan θ , Find θ using tables 0.3443 + 0.6009 = 0.9452 ∴ θ = 43° 23'

B08 · paragraph / latin 3. The sun is due ⟦line⟧ W. at an elevation of 25°. (a) Find the length of the shadow thrown on the ground by a vertical pole 30 ft high. 30 x 2.1445 = 30/.4663 = 64.335 ft. (b) Find the height of this shadow on a vertical wall running N. + S. and 10 yds away from the pole. 16.08 ft. = 34.335 x 0.4663

**Traduction anglaise —**

Shamash School Monthly Quiz Subject: Trigonometry Date: 25/1/50 Class: 4th Time: 45 minutes, 1 hour. All questions are to be attempted. 1. (a) Find θ, φ and ψ when: cot θ = 2.4363 , sec φ = 3.1241 , csc ψ = 4.0213 22° 19' 71° 20' 14° 24' (b) Solve the triangle ABC, given that: C = 90°, B = 71° 31', b = 76 in. A = 18° 29' a = 25.4068 c = 81.19 2. (a) Evaluate as shortly as possible 1/sin 19° , 1/cos 29° , 1/tan 42° , 3.0713 1.1430 1.1106 cos 17°/sin 17° , sin 42°/sin 48° , cos 63/cos 27 3.2709 0.9004 0.5095 (b) If tan 19° + tan 31° = tan θ , Find θ using tables 0.3443 + 0.6009 = 0.9452 ∴ θ = 43° 23' 3. The sun is due ⟦line⟧ W. at an elevation of 25°. (a) Find the length of the shadow thrown on the ground by a vertical pole 30 ft high. 30 x 2.1445 = 30/.4663 = 64.335 ft. (b) Find the height of this shadow on a vertical wall running N. + S. and 10 yds away from the pole. 16.08 ft. = 34.335 x 0.4663

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### csp_e5f7090430445824afd5f7526d95ace2

Solution to Algebra 4th Year Final Exam, 14/5/1969 page 3

3 (ii) x = ⁷√((0.5002)² sin³ 14° 25' / (4.003)³ cos² 15° 27')

(i) If (b+c)⁻¹ , (c+a)⁻¹ , (a+b)⁻¹ are in A.P., prove that a², b², c² are also in A.P. By hypothesis: (a+b)⁻¹ - (c+a)⁻¹ = (c+a)⁻¹ - (b+c)⁻¹ or 1/(a+b) - 1/(c+a) = 1/(c+a) - 1/(b+c) or (c+a-a-b)/((a+b)(c+a)) = (b+c-c-a)/((c+a)(b+c)) or (c-b)/((a+b)(c+a)) = (b-a)/((c+a)(b+c)) ∴ (c-b)/(a+b) = (b-a)/(b+c) or (c-b)(c+b) = (b-a)(b+a) or c² - b² = b² - a² which makes a², b², c² in A.P. by definition of A.P. Q.E.D.

(ii) ⟦diagram of a bouncing ball with vertical arrows and labels 10, 5, 2.5⟧ + soon.

② total distance it travels before coming to rest = = S_{n→∞} = 10 + (10 + 5 + 5/2 + ... to infinity) = 10 + (10 / (1 - 1/2)) = 10 + 10 / (1/2) = 10 + 20 = 30 ft. Ans. 2

① when the ball has struck the ground for the first time it has travelled 10 ft. When it strikes the ground for the 2nd time it will have gone upward 5 ft & downward 5 ft also ∴ it will have travelled another 10 ft, the next distance will be 5 ft + so on ∴ the total distance travelled by the end of the 10th time it strikes the ground will be 10 + 10 + 5 + 5/2 + 5/4 + ... to ten terms = 10 + (10 + 5 + 5/2 + 5/4 + ... to 9 terms) ∴ total distance = 10 + (10(1 - (1/2)⁹)) / (1 - 1/2) = 10 + (10(1 - 1/512)) / (1/2) = 10 + 20 [1 - 1/512] = 10 + 20 (511/512) = 10 + 10220 / 512 = 10 + 19 123/128 = 29 123/128 ft Ans. 1 [See above →]

**Traduction anglaise —**

Solution to Algebra 4th Year Final Exam, 14/5/1969 page 3 3 (ii) x = ⁷√((0.5002)² sin³ 14° 25' / (4.003)³ cos² 15° 27') log 0.5002 = 1̄.6992 | 2 log 0.5002 = 1̄.3984 | 3 log 4.003 = 1.8072 log sin 14° 25' = 1̄.3962 | 3 log sin 14° 25' = 2̄.1886 | 2 log cos 15° 27' = 1̄.9680 log 4.003 = 0.6024 | log Num = 3̄.5870 | log Den. = 1.7752 log cos 15° 27' = 1̄.9840 | log Den = 1.7752 | 7 log x = 5̄.8118 log x = 1̄.4017 x = 0.2522 or 2.522 x 10⁻¹ Ans. (i) If (b+c)⁻¹ , (c+a)⁻¹ , (a+b)⁻¹ are in A.P., prove that a², b², c² are also in A.P. By hypothesis: (a+b)⁻¹ - (c+a)⁻¹ = (c+a)⁻¹ - (b+c)⁻¹ or 1/(a+b) - 1/(c+a) = 1/(c+a) - 1/(b+c) or (c+a-a-b)/((a+b)(c+a)) = (b+c-c-a)/((c+a)(b+c)) or (c-b)/((a+b)(c+a)) = (b-a)/((c+a)(b+c)) ∴ (c-b)/(a+b) = (b-a)/(b+c) or (c-b)(c+b) = (b-a)(b+a) or c² - b² = b² - a² which makes a², b², c² in A.P. by definition of A.P. Q.E.D. (ii) ⟦diagram of a bouncing ball with vertical arrows and labels 10, 5, 2.5⟧ + soon. ② total distance it travels before coming to rest = = S_{n→∞} = 10 + (10 + 5 + 5/2 + ... to infinity) = 10 + (10 / (1 - 1/2)) = 10 + 10 / (1/2) = 10 + 20 = 30 ft. Ans. 2 ① when the ball has struck the ground for the first time it has travelled 10 ft. When it strikes the ground for the 2nd time it will have gone upward 5 ft & downward 5 ft also ∴ it will have travelled another 10 ft, the next distance will be 5 ft + so on ∴ the total distance travelled by the end of the 10th time it strikes the ground will be 10 + 10 + 5 + 5/2 + 5/4 + ... to ten terms = 10 + (10 + 5 + 5/2 + 5/4 + ... to 9 terms) ∴ total distance = 10 + (10(1 - (1/2)⁹)) / (1 - 1/2) = 10 + (10(1 - 1/512)) / (1/2) = 10 + 20 [1 - 1/512] = 10 + 20 (511/512) = 10 + 10220 / 512 = 10 + 19 123/128 = 29 123/128 ft Ans. 1 [See above →]

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### csp_e75a3539237d5898a62676f0effc6ed5

B01 · paragraph / latin 3. (a) 4(x²-1) + 2(x+3) = 2 + 2x(1+2x) (b) x(6x+1) = 2x+1 (c) x(x+2) = 2(x-2)

B02 · paragraph / latin 6 Simplifying (a), we get: 4x²-4+2x+6 = 2+2x+4x² or 4x²+2x+2 = 4x²+2x+2 (always true, being an identity)

B03 · paragraph / latin 8 from (b): 6x²+x = 2x+1 ∴ 6x²-x-1 = 0 or (3x+1)(2x-1) = 0 ∴ x = -1/3 and x = 1/2 (conditional equation) and the values of x are x = -1/3 } Ans. x = 1/2

B04 · paragraph / latin 6 from (c): x²+2x = 2x-4 or x² = -4 (never true, x being real)

B05 · paragraph / latin 4 (i) 1/x + 1/y + 1/z = 2 1/2 .... ① Dividing eq. ② by 2 & subtracting with eq. ①, we get: 2/x + 2/y + 2/z = 5 .... ② { 1/x + 1/y + 1/z = 2 1/2 .... ① 3/x - 5/y + 7/z = 2 5/6 .... ③ { 1/x + 3/y - 3/z = 1 .... ⟦illegible⟧ ∴ 2/x + 3/y = 3 1/2 .... ④ from ① multiplying by 7: 7/x + 7/y + 7/z = 17 1/2 from ③ (⟦illegible⟧) 3/x - 5/y + 7/z = 8 1/2 25/y = 12 1/2 ∴ y = 25 / 12 1/2 = 2 -2/x + 22/y = 9 .... ⑦ y = 2 from ④ 2/x + 3/2 = 3 1/2 ∴ 2/x = 2 ∴ x = 1 from ① 1/1 + 1/2 + 1/z = 2 1/2 or 1/z = 1 ∴ z = 1 ∴ x = 1 y = 2 } Ans. z = 1

B06 · marginalia / latin (10 marks)

B07 · paragraph / latin (ii) (x/y² + y/x² - 1) / (x²/y² + y²/x² + 1) . (1 + y/x) / (x - y) ÷ (1 + y/x) / (x² - y²) = (x³+y³-xy²) / (x²+x²y+xy²) . (x+y) / (x²-xy) ÷ (x³+y³) / (x⁵-x²y³) = (y(y²+x²-xy)) / (x(y²+x²+xy)) . (x+y) / (x(x-y)) . (x²(x³-y³)) / (y(x³+y³)) = (y(x+y)(x²+xy+y²)(x/y)(x²+xy+y²)) / (x²y(x-y)(x²+xy+y²)(x+y)(x²-xy+y²)) = 1 Ans.

B08 · marginalia / latin (10 marks)

**Traduction anglaise —**

3. (a) 4(x²-1) + 2(x+3) = 2 + 2x(1+2x) (b) x(6x+1) = 2x+1 (c) x(x+2) = 2(x-2) 6 Simplifying (a), we get: 4x²-4+2x+6 = 2+2x+4x² or 4x²+2x+2 = 4x²+2x+2 (always true, being an identity) 8 from (b): 6x²+x = 2x+1 ∴ 6x²-x-1 = 0 or (3x+1)(2x-1) = 0 ∴ x = -1/3 and x = 1/2 (conditional equation) and the values of x are x = -1/3 } Ans. x = 1/2 6 from (c): x²+2x = 2x-4 or x² = -4 (never true, x being real) 4 (i) 1/x + 1/y + 1/z = 2 1/2 .... ① Dividing eq. ② by 2 & subtracting with eq. ①, we get: 2/x + 2/y + 2/z = 5 .... ② { 1/x + 1/y + 1/z = 2 1/2 .... ① 3/x - 5/y + 7/z = 2 5/6 .... ③ { 1/x + 3/y - 3/z = 1 .... ⟦illegible⟧ ∴ 2/x + 3/y = 3 1/2 .... ④ from ① multiplying by 7: 7/x + 7/y + 7/z = 17 1/2 from ③ (⟦illegible⟧) 3/x - 5/y + 7/z = 8 1/2 25/y = 12 1/2 ∴ y = 25 / 12 1/2 = 2 -2/x + 22/y = 9 .... ⑦ y = 2 from ④ 2/x + 3/2 = 3 1/2 ∴ 2/x = 2 ∴ x = 1 from ① 1/1 + 1/2 + 1/z = 2 1/2 or 1/z = 1 ∴ z = 1 ∴ x = 1 y = 2 } Ans. z = 1 (10 marks) (ii) (x/y² + y/x² - 1) / (x²/y² + y²/x² + 1) . (1 + y/x) / (x - y) ÷ (1 + y/x) / (x² - y²) = (x³+y³-xy²) / (x²+x²y+xy²) . (x+y) / (x²-xy) ÷ (x³+y³) / (x⁵-x²y³) = (y(y²+x²-xy)) / (x(y²+x²+xy)) . (x+y) / (x(x-y)) . (x²(x³-y³)) / (y(x³+y³)) = (y(x+y)(x²+xy+y²)(x/y)(x²+xy+y²)) / (x²y(x-y)(x²+xy+y²)(x+y)(x²-xy+y²)) = 1 Ans. (10 marks)

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### csp_e85411d7833a5641a04c3db0d29f6af5

B01 · paragraph / latin (i) 3x(x-1) + 2 = 5(x²-4) + 13 - 11x ∴ 3x² - 3x + 2 = 5x² - 20 + 13 - 11x or 2x² - 8x - 9 = 0 or x² - 4x - 4.5 = 0 ⟦illegible⟧ (ii) x²(x-5) + 3(x-1) = 2x³ - x(5x-3) - 3 ∴ x³ - 5x² + 3x - 3 = 2x³ - 5x² + 3x - 3 which is an identity and ⟦illegible⟧ true for all values of x (iii) x(x²-1) + 2(1+x)(1-x) = 0 or x(x+1)(x-1) + 2(x+1)(1-x) = 0 ∴ (x+1) [x(x-1) + 2(1-x)] = 0 or (x+1) [x(x-1) - 2(x-1)] = 0 ∴ (x+1)(x-1) [x-2] = 0 ∴ x = 1, -1, 2 This is a conditional equation which is true for these values only. In other words it is not true. 3. (i) 2x³ - 14x² + 22x = 0 By trial x = 2 satisfies the equation since 2(2)³ - 14(2)² + 22(2) = 16 - 56 + 44 = 4 ∴ x = 2 is a root and hence (x-2) is a factor ∴ 2x³ - 14x² + 22x = (x-2)(2x² - 10x - 2) = 0 2(x-2)(x² - 5x - 1) = 0 ∴ x = 2 x = ⟦illegible⟧ x + y + 2xy + x² + y² = 0 ... ① ∴ (x+y) + (x+y)² = 0 ... (1a) x - y + 2xy + x² + y² = 6 ... ② ∴ (x-y) + (x-y)² = 6 ... (2a) from (1a) (x+y)[1+x+y] = 0 ∴ x+y = 0 or y = -x ... ③ also 1+x+y = 0 or y = -x-1 ... ④ substitute from ③ into (2a), since (x+x) + (x+x)² = 6 or 2x + 4x² - 6 = 0 ∴ 2x² + x - 3 = 0 ∴ (2x+3)(x-1) = 0 ∴ x = 1 from ③ ∴ y = -1 x = -3/2 y = 3/2 Ans. 1 x = 1, y = -1 Ans. 2 x = -3/2, y = 3/2 substitute from ④ into (2a), since (x+x+1) + (x+x+1)² = 6 ∴ 2x+1 + (2x+1)² = 6 ∴ 4x² + 6x - 4 = 0 or 2x² + 3x - 2 = 0 ∴ (2x-1)(x+2) = 0 or x = 1/2 or x = -2 x = 1/2 y = -3/2 Ans. 3 x = -2, y = 1 Ans. 4 x = -2, y = 1

**Traduction anglaise —**

(i) 3x(x-1) + 2 = 5(x²-4) + 13 - 11x ∴ 3x² - 3x + 2 = 5x² - 20 + 13 - 11x or 2x² - 8x - 9 = 0 or x² - 4x - 4.5 = 0 ⟦illegible⟧ (ii) x²(x-5) + 3(x-1) = 2x³ - x(5x-3) - 3 ∴ x³ - 5x² + 3x - 3 = 2x³ - 5x² + 3x - 3 which is an identity and ⟦illegible⟧ true for all values of x (iii) x(x²-1) + 2(1+x)(1-x) = 0 or x(x+1)(x-1) + 2(x+1)(1-x) = 0 ∴ (x+1) [x(x-1) + 2(1-x)] = 0 or (x+1) [x(x-1) - 2(x-1)] = 0 ∴ (x+1)(x-1) [x-2] = 0 ∴ x = 1, -1, 2 This is a conditional equation which is true for these values only. In other words it is not true. 3. (i) 2x³ - 14x² + 22x = 0 By trial x = 2 satisfies the equation since 2(2)³ - 14(2)² + 22(2) = 16 - 56 + 44 = 4 ∴ x = 2 is a root and hence (x-2) is a factor ∴ 2x³ - 14x² + 22x = (x-2)(2x² - 10x - 2) = 0 2(x-2)(x² - 5x - 1) = 0 ∴ x = 2 x = ⟦illegible⟧ x + y + 2xy + x² + y² = 0 ... ① ∴ (x+y) + (x+y)² = 0 ... (1a) x - y + 2xy + x² + y² = 6 ... ② ∴ (x-y) + (x-y)² = 6 ... (2a) from (1a) (x+y)[1+x+y] = 0 ∴ x+y = 0 or y = -x ... ③ also 1+x+y = 0 or y = -x-1 ... ④ substitute from ③ into (2a), since (x+x) + (x+x)² = 6 or 2x + 4x² - 6 = 0 ∴ 2x² + x - 3 = 0 ∴ (2x+3)(x-1) = 0 ∴ x = 1 from ③ ∴ y = -1 x = -3/2 y = 3/2 Ans. 1 x = 1, y = -1 Ans. 2 x = -3/2, y = 3/2 substitute from ④ into (2a), since (x+x+1) + (x+x+1)² = 6 ∴ 2x+1 + (2x+1)² = 6 ∴ 4x² + 6x - 4 = 0 or 2x² + 3x - 2 = 0 ∴ (2x-1)(x+2) = 0 or x = 1/2 or x = -2 x = 1/2 y = -3/2 Ans. 3 x = -2, y = 1 Ans. 4 x = -2, y = 1

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### csp_e912c6261ee95f46a1733ac4444072f5

SHAMASH SECONDARY SCHOOL Final Examination, 1958-1959.

Subject: Algebra Date: 28/5/1959 Class: 4th Year Secondary Time: 8:00-10:30 a.m.

All questions are to be attempted.

2(i) Using tables, compute by logarithms the value of : ⁵/ (0.004678)² x 1.002 / ------------------- (10 marks). \/ (30.04)³ (ii) Given : log70 = 1.8451, log110 = 2.0414, log34.62 = 1.5394, compute, without using tables, the value of : ³√41503 , correct to four sugnificant figures. (10 marks).

3. A certain alloy contains 6 parts by weight of a metal A and 5 parts by weight of a metal B; another alloy contains ⟦7⟧ parts by weight of A and ⟦3⟧ parts by weight of B. If these alloys are melted and mixed together, how many pounds of the second alloy must be mixed with 11 pounds of the first alloy to make a mixture which contains 40 per cent. of A ? (20 marks).

4. (a) The expression 2 - 2ⁿ⁺¹ / 3ⁿ is a formula for the sum of 'n' terms of a certain geometric series, n being any positive integer. Find the first term of the series, the common ratio, and the formula for the n-th term. (10 marks). (b) The first and second terms of a series are 'a' and 'b' respectively. Find the n th term (i) if the series is an arithmetic series; (ii) if it is a geometric one. (10 marks).

5. (i) Taking ½ in. as one unit on the x-axis and on the y-axis, plot the curve y = 3/4 x² for values of x between x = -4 and x = 4. (7 marks). (ii) On the same axes of coordinates draw the graph of the equation 3x + 2y = 12. (6 marks). (iii) From the above graphs find two roots for the simultaneous equations y = 3/4 x² and 3x + 2y = 12. Verify your graphical answers by solving algebraically. (7 marks).

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Final Examination, 1958-1959. Subject: Algebra Date: 28/5/1959 Class: 4th Year Secondary Time: 8:00-10:30 a.m. All questions are to be attempted. 1. (a) Resolve into factors: (i) (7x + 8)² - 2(7x + 8) - 15. (3 marks) (ii) 2(x - y)² - 3x + 3y - 5. (4 marks) (iii) a(a - 4) - b(b - 4). (3 marks) (b) If 15(2x² - y²) = 7xy, and if x and y are both positive, find the ratio of x to y. Use the shortest possible way. (10 marks) 2(i) Using tables, compute by logarithms the value of : ⁵/ (0.004678)² x 1.002 / ⟦line⟧ (10 marks). \/ (30.04)³ (ii) Given : log70 = 1.8451, log110 = 2.0414, log34.62 = 1.5394, compute, without using tables, the value of : ³√41503 , correct to four sugnificant figures. (10 marks). 3. A certain alloy contains 6 parts by weight of a metal A and 5 parts by weight of a metal B; another alloy contains ⟦7⟧ parts by weight of A and ⟦3⟧ parts by weight of B. If these alloys are melted and mixed together, how many pounds of the second alloy must be mixed with 11 pounds of the first alloy to make a mixture which contains 40 per cent. of A ? (20 marks). 4. (a) The expression 2 - 2ⁿ⁺¹ / 3ⁿ is a formula for the sum of 'n' terms of a certain geometric series, n being any positive integer. Find the first term of the series, the common ratio, and the formula for the n-th term. (10 marks). (b) The first and second terms of a series are 'a' and 'b' respectively. Find the n th term (i) if the series is an arithmetic series; (ii) if it is a geometric one. (10 marks). 5. (i) Taking ½ in. as one unit on the x-axis and on the y-axis, plot the curve y = 3/4 x² for values of x between x = -4 and x = 4. (7 marks). (ii) On the same axes of coordinates draw the graph of the equation 3x + 2y = 12. (6 marks). (iii) From the above graphs find two roots for the simultaneous equations y = 3/4 x² and 3x + 2y = 12. Verify your graphical answers by solving algebraically. (7 marks).

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### csp_eb76a50f640f5ef7849451fcc8695722

Solution to Mid-year exam. in Arithmetic ⟦illegible⟧ 4th Year, ⟦February⟧ Jan., 29th 1967

(a) +32,000 / 540,000 = £ 0.8 or 16 s (b) £ 63 x 0.8 = £ 50.4 or £ 50 8 s (c) 540,000 / 240 = £ 2250

4 Let radius of whole roll = r inches π r² - 9 π = 4800 x 12 x 1/120 = 480 π r² = 480 + 9 π r² = 480/π + 9 = 480/3.1416 + 9 = 152.79 + 9 = 161.79 r = 12.72 inches.

5) 990 x 20 / 27 1/2 = 990 x 20 x 2 / 55 = 720 (£ 1) shares 720 x 32 s = £ 720 x 1.6 = £ 1,152 he realized from the sale 1152 x 20 / 9 = 2560 (10 s) shares he bought Change of income: Income from (£ 1) shares = 720 x 10 / 100 = £ 72 " " (10 s) = 2560/2 x 5/100 = £ 64 Income decreased by £ 72 - £ 64 = £ 8

**Traduction anglaise —**

Solution to Mid-year exam. in Arithmetic ⟦illegible⟧ 4th Year, ⟦February⟧ Jan., 29th 1967 (a) +32,000 / 540,000 = £ 0.8 or 16 s (b) £ 63 x 0.8 = £ 50.4 or £ 50 8 s (c) 540,000 / 240 = £ 2250 4 Let radius of whole roll = r inches π r² - 9 π = 4800 x 12 x 1/120 = 480 π r² = 480 + 9 π r² = 480/π + 9 = 480/3.1416 + 9 = 152.79 + 9 = 161.79 r = 12.72 inches. 5) 990 x 20 / 27 1/2 = 990 x 20 x 2 / 55 = 720 (£ 1) shares 720 x 32 s = £ 720 x 1.6 = £ 1,152 he realized from the sale 1152 x 20 / 9 = 2560 (10 s) shares he bought Change of income: Income from (£ 1) shares = 720 x 10 / 100 = £ 72 " " (10 s) = 2560/2 x 5/100 = £ 64 Income decreased by £ 72 - £ 64 = £ 8

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### csp_edb2a65ba44255559ec98bb613c45ab1

B01 · header / latin 7 marks

B02 · paragraph / latin The graph of ⟦illegible⟧ ⟦illegible⟧ the equation x³ - x² + 1 = 0 is ⟦illegible⟧ ⟦illegible⟧ equation ⟦illegible⟧ = -1 x³ - x² - x = x(x² - x - 1) ⟦illegible⟧ the graph of y = -1 ⟦illegible⟧ the graph x³ - x² - x

B03 · table / latin x | y | y -1.5 | -2 5/8 | -2.625 -1 | 0 | 0 Max -0.5 | 0.6 | 0.625 -0.5 | 5/8 | 0.625 0 | 0 | 0 0.5 | -1 1/8 | -1.125 1 | -1 | -1 Min 1.2 | -1.2 | -1.2 1.5 | -1 1/8 | -1.125 2 | 0 | 0 2.5 | 4 3/8 | 4.375

B04 · marginalia / latin ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ x = -1.2 ans 6 marks

B05 · other / latin ⟦Graph showing a cubic function with points plotted and axes labeled⟧ x = -1.2 x = 0.4 x = 1.8 y = -1

**Traduction anglaise —**

7 marks The graph of ⟦illegible⟧ ⟦illegible⟧ the equation x³ - x² + 1 = 0 is ⟦illegible⟧ ⟦illegible⟧ equation ⟦illegible⟧ = -1 x³ - x² - x = x(x² - x - 1) ⟦illegible⟧ the graph of y = -1 ⟦illegible⟧ the graph x³ - x² - x x | y | y -1.5 | -2 5/8 | -2.625 -1 | 0 | 0 Max -0.5 | 0.6 | 0.625 -0.5 | 5/8 | 0.625 0 | 0 | 0 0.5 | -1 1/8 | -1.125 1 | -1 | -1 Min 1.2 | -1.2 | -1.2 1.5 | -1 1/8 | -1.125 2 | 0 | 0 2.5 | 4 3/8 | 4.375 ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ x = -1.2 ans 6 marks ⟦Graph showing a cubic function with points plotted and axes labeled⟧ x = -1.2 x = 0.4 x = 1.8 y = -1

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### csp_ee4d58f4389d577e9c318d923a375c08

3 (b) x + 1/x = √3 Prove that x³ + 1/x³ = 0 x³ + 1/x³ = (x + 1/x)(x² - 1 + 1/x²) = (x + 1/x)(x² + 2 + 1/x² - 3) = (x + 1/x)[(x + 1/x)² - 3] = √3 [(√3)² - 3] = √3 (3 - 3) = zero Q.E.D.

4 (a) Let the length of the course be = x yards ∴ No. of seconds taken by the first boat = x/4 seconds = t₁ also " " " " " 2nd " = x/3½ + x/4½ = x/7/2 + x/9/2 = 16x/63 seconds = t₂ ∴ t₁ = x/4 = 63x/4x63 seconds } ∴ the first boat wins the race by ∴ t₂ = 16x/63 = 64x/4x63 " } 64x - 63x / 4 x 63 = x / 4x63 seconds Ans.

(b) Let the distance of riding be = x miles x/b hours = time taken in riding } ∴ x/b + x/c = a x/c hours = time " " walking } ∴ cx + bx / bc = a or x(b + c) = a.b.c ∴ x = a.b.c / b + c mile Ans.

5 (i) It is required to find the sum of 2 + 4 + 6 + ... to (n+1) terms ∴ Sum: S = n/2 [2a + (n-1)d] or S = n+1/2 [2(2) + (n+1-1)(2)] = <del>⟦illegible⟧</del> S = n+1/2 {4 + 2n} = (n+1)(n+2) = n² + 3n + 2 Ans.

(ii) S' = 22 + 33 + 44 + ... to ten terms S' = 10/2 {2 x 22 + (10-1) x 11} = 5(44 + 99) = 5 x 143 = 715 Ans.

(iii) The distance fallen in the different seconds are: 16, 48, 80, ... l₁₂ = a + (n-1)d = 16 + 11 x 32 = 16 + 352 = 368 ft. Ans. 1 S = n/2 {a + l} = 12/2 (16 + 368) = 6 x 384 = 2304 ft. Ans. 2

(iv) Total distance covered = 10 + 10 + 5 + 2½ + 1¼ + ... to infinity <del>⟦illegible⟧</del> = 10 + (10 / 1 - ½) = 10 + 20 = 30 Ans.

**Traduction anglaise —**

2 3 (b) x + 1/x = √3 Prove that x³ + 1/x³ = 0 x³ + 1/x³ = (x + 1/x)(x² - 1 + 1/x²) = (x + 1/x)(x² + 2 + 1/x² - 3) = (x + 1/x)[(x + 1/x)² - 3] = √3 [(√3)² - 3] = √3 (3 - 3) = zero Q.E.D. 4 (a) Let the length of the course be = x yards ∴ No. of seconds taken by the first boat = x/4 seconds = t₁ also " " " " " 2nd " = x/3½ + x/4½ = x/7/2 + x/9/2 = 16x/63 seconds = t₂ ∴ t₁ = x/4 = 63x/4x63 seconds } ∴ the first boat wins the race by ∴ t₂ = 16x/63 = 64x/4x63 " } 64x - 63x / 4 x 63 = x / 4x63 seconds Ans. (b) Let the distance of riding be = x miles x/b hours = time taken in riding } ∴ x/b + x/c = a x/c hours = time " " walking } ∴ cx + bx / bc = a or x(b + c) = a.b.c ∴ x = a.b.c / b + c mile Ans. 5 (i) It is required to find the sum of 2 + 4 + 6 + ... to (n+1) terms ∴ Sum: S = n/2 [2a + (n-1)d] or S = n+1/2 [2(2) + (n+1-1)(2)] = <del>⟦illegible⟧</del> S = n+1/2 {4 + 2n} = (n+1)(n+2) = n² + 3n + 2 Ans. (ii) S' = 22 + 33 + 44 + ... to ten terms S' = 10/2 {2 x 22 + (10-1) x 11} = 5(44 + 99) = 5 x 143 = 715 Ans. (iii) The distance fallen in the different seconds are: 16, 48, 80, ... l₁₂ = a + (n-1)d = 16 + 11 x 32 = 16 + 352 = 368 ft. Ans. 1 S = n/2 {a + l} = 12/2 (16 + 368) = 6 x 384 = 2304 ft. Ans. 2 (iv) Total distance covered = 10 + 10 + 5 + 2½ + 1¼ + ... to infinity <del>⟦illegible⟧</del> = 10 + (10 / 1 - ½) = 10 + 20 = 30 Ans.

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### csp_ef0c0c33c9d9560e9cc22bea68d99568

B01 · header / latin SHAMASH SECONDARY SCHOOL FINAL EXAM. MAY 1964.

B02 · form / latin Subject :: Arithmetic & Trigonometry Date :: Class :: 4th Year Secondary Time :: 8:00 - 10:30 a.m.

B03 · paragraph / latin ⟦line⟧ Answer All Questions

B04 · paragraph / latin 1. Two lighthouses A and B are 5 miles apart, B being due east of A. A ship at P is due north of A, and on a bearing 322° (N.38°W.) from B. The ship then sails in a direction 056° (N.56° E.) to a position Q which is due north of B. Calculate PQ, BQ, and the bearing of Q from A.

B05 · marginalia / mixed PQ = 6.031 Mi. BQ = 9.772 Mi 27° 6' E

B06 · paragraph / latin 2. The elevation of a spire from a point A due N. of it is 28°, and from a point B due E. of it 18°. Find the height of the spire if AB is 100 yards.

B07 · marginalia / latin 27.7 yds

B08 · paragraph / latin 3. A city council requires an annual rate of 16s. 4d. in the £ of rateable value. Out of each 16s. 4d. received the sum of 5s. 7.3d. is spent on education. (i) What is the rateable value of a house, the holder of which pays £ 13 1s. 4d. half-yearly in rates ? (ii) If this householder has two children, both attending the council's school, what sum, to the nearest 1/10 d., is he contributing per week towards the education of each child ? (Take a year to be 52 weeks.)

B09 · marginalia / latin £ 32 1s 8.7d

B10 · paragraph / latin 4. A man invests £ 30,155, partly in 3 1/2% stock at 86 and the rest in 4 1/2% stock at 99. He divides the money so as to obtain the same income from each stock. Find the total income.

B11 · marginalia / latin 1295

B12 · paragraph / latin 5. Find, correct to three significant figures, the weight in pounds of a cylindrical iron pipe 10 ft. long, whose outer diameter is 1 ft. 6 in. and inner diameter 1ft. 4 in., given that 1 cu. ft. of iron weights 494 lb. (Take π as 3.142.) If the inner diameter is increased to 1 ft. 5 in., the outer diameter and the length remaining unaltered, find the ratio of the new weight to the old.

B13 · marginalia / latin 1830 35 / 66

B14 · other / mixed ⟦line⟧ PQ = 5 cosec 56° BQ = 5 cot 58° + ⟦illegible⟧ cot 38° tan = ⟦illegible⟧

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL FINAL EXAM. MAY 1964. Subject :: Arithmetic & Trigonometry Date :: Class :: 4th Year Secondary Time :: 8:00 - 10:30 a.m. ⟦line⟧ Answer All Questions 1. Two lighthouses A and B are 5 miles apart, B being due east of A. A ship at P is due north of A, and on a bearing 322° (N.38°W.) from B. The ship then sails in a direction 056° (N.56° E.) to a position Q which is due north of B. Calculate PQ, BQ, and the bearing of Q from A. PQ = 6.031 Mi. BQ = 9.772 Mi 27° 6' E 2. The elevation of a spire from a point A due N. of it is 28°, and from a point B due E. of it 18°. Find the height of the spire if AB is 100 yards. 27.7 yds 3. A city council requires an annual rate of 16s. 4d. in the £ of rateable value. Out of each 16s. 4d. received the sum of 5s. 7.3d. is spent on education. (i) What is the rateable value of a house, the holder of which pays £ 13 1s. 4d. half-yearly in rates ? (ii) If this householder has two children, both attending the council's school, what sum, to the nearest 1/10 d., is he contributing per week towards the education of each child ? (Take a year to be 52 weeks.) £ 32 1s 8.7d 4. A man invests £ 30,155, partly in 3 1/2% stock at 86 and the rest in 4 1/2% stock at 99. He divides the money so as to obtain the same income from each stock. Find the total income. 1295 5. Find, correct to three significant figures, the weight in pounds of a cylindrical iron pipe 10 ft. long, whose outer diameter is 1 ft. 6 in. and inner diameter 1ft. 4 in., given that 1 cu. ft. of iron weights 494 lb. (Take π as 3.142.) If the inner diameter is increased to 1 ft. 5 in., the outer diameter and the length remaining unaltered, find the ratio of the new weight to the old. 1830 35 / 66 ⟦line⟧ PQ = 5 cosec 56° BQ = 5 cot 58° + ⟦illegible⟧ cot 38° tan = ⟦illegible⟧

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### csp_f1c5e09a0e1e593cac615b717fa294d9

⟦line⟧ (ii) \frac{\frac{x}{y} + \frac{y}{x} - 1}{\frac{x^2}{y^2} + \frac{y^2}{x^2} + 1} \cdot \frac{x + y}{x - y} \div \frac{x^3 + y^3}{x^4 - y^4} = \frac{x^2 + y^2 - xy}{x^2 y^2} \cdot \frac{x^2 y^2}{x^4 + x^2 y^2 + y^4} \cdot \frac{x + y}{x - y} \div \frac{x^3 + y^3}{x^4 - y^4} = y(y^2+x^2-xy) / x(y^2+x^2+xy) " (x+y) / x(x-y) " x^2(x^2-y^2) / y(x^3+y^3) = x^2y(x+y)(x^2-xy+y^2)(x-y)(x+y) / xy^2(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2) = 1 Ans.

**Traduction anglaise —**

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### csp_f33b77c1227e514d8995e79f10d4c405

SHAMASH SECONDARY SCHOOL 2nd Quarter Examination

Subject: Algebra Date: 31/12/1967. Class: 4th Year, Scientific Section. Time: 10:15-11:45 a.m.

------------------------------------------------------------ Attempt all questions: 12 12 1. (i) Resove into five factors : X - Y (10 marks) (ii) Find the value of 'A' which will make the expression 3 2 6x + Ax + x - 6 divisible by (x + 3) and find the other factors. (10 marks)

2. Find the square root of : 9 6 5 74 4 61 3 62 2 2 1 - x - 2x + --x - --x + --x - -x + - . (20 marks) 4 45 30 75 5 4

a - b a² - b² 1 + ----- 1 + ------- a + b a² + b² 3. (i). Reduce to simplest form : --------- ÷ ----------- (10 marks) a - b a² - b² 1 - ----- 1 - ------- a + b a² + b² 5x - 8 6x - 44 10x - 8 x - 8 (ii). Solve the equation : ------ + ------- - ------- = ----- (10 marks) x - 2 x - 7 x - 7 x - 6

4. Find the values of x, y, and z from the following equations : 3x - 2y + 4z = 3y - 2x + 7 = 7x + 2z - 2 = 11 . (20 marks)

5. A basket of eggs is emptied by one person taking half of them and one more, a second person taking half of the remainder and one more, and a third person taking half of the remainder and six more. How many did the basket contain at first? (20 marks)

------------------------------------------------------------

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL 2nd Quarter Examination Subject: Algebra Date: 31/12/1967. Class: 4th Year, Scientific Section. Time: 10:15-11:45 a.m. ⟦line⟧ Attempt all questions: 12 12 1. (i) Resove into five factors : X - Y (10 marks) (ii) Find the value of 'A' which will make the expression 3 2 6x + Ax + x - 6 divisible by (x + 3) and find the other factors. (10 marks) 2. Find the square root of : 9 6 5 74 4 61 3 62 2 2 1 - x - 2x + --x - --x + --x - -x + - . (20 marks) 4 45 30 75 5 4 a - b a² - b² 1 + ----- 1 + ------- a + b a² + b² 3. (i). Reduce to simplest form : --------- ÷ ----------- (10 marks) a - b a² - b² 1 - ----- 1 - ------- a + b a² + b² 5x - 8 6x - 44 10x - 8 x - 8 (ii). Solve the equation : ------ + ------- - ------- = ----- (10 marks) x - 2 x - 7 x - 7 x - 6 4. Find the values of x, y, and z from the following equations : 3x - 2y + 4z = 3y - 2x + 7 = 7x + 2z - 2 = 11 . (20 marks) 5. A basket of eggs is emptied by one person taking half of them and one more, a second person taking half of the remainder and one more, and a third person taking half of the remainder and six more. How many did the basket contain at first? (20 marks) ⟦line⟧

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### csp_f3b9455bbc17544198ac606a6cd00a0d

⟦illegible⟧ Trigonometry questions 4th year Conditional Exam Sept., 1966.

Q.4. PQ = ? , ∠θ = ? In Δ PDQ , PQ/5 = cosec 56° ∴ PQ = 5 cosec 56° DP/5 = tan 34° ∴ DP = 5 tan 34° AP/5 = tan 52° ∴ AP = 5 tan 52° AD = BQ = DP + AP = 5 tan 34° + 5 tan 52° = 5 (tan 34° + tan 52°) from Δ AQD , tan θ = 5/AD ∴ tan θ = 5 / (5 (tan 34° + tan 52°)) ∴ tan θ = 1 / (tan 34° + tan 52°)

⟦line⟧ PQ = 5 cosec 56° = 5 x 1.2062 = 6.0310 miles Ans. 1 BQ = 5 (tan 34° + tan 52°) = 5 (0.6745 + 1.2799) = 5 x 1.9544 = 9.7720 miles Ans. 2 tan θ = 1 / (tan 34° + tan 52°) = 1 / (0.6745 + 1.2799) = 1 / 1.9544 = 0.51165 = 0.5117 Correct to 4 dec. pl. ∴ θ = 27° 6' Ans. 3

Q.5. ∠xoy = 143° 28' ∴ ∠xoT = (143° 28') / 2 = 71° 44' In Δ oxT 3.2 / oT = cos 71° 44' ∴ oT = 3.2 / cos 71° 44' = 3.2 sec 71° 44' oT = 3.2 x 3.1903 = 10.20896 in = 10.21 in Correct to 2 dec. pl. Ans. 1 In Δ oxB , xB / 3.2 = sin 71° 44' ∴ xB = 3.2 sin 71° 44' ∴ xy = 2 xB = 2 x 3.2 sin 71° 44' = 6.4 sin 71° 44' = 6.4 x 0.9496 = 6.07744 in = 6.08 in Correct to 2 dec. pl. Ans. 2

⟦Diagram of a rectangle with diagonals and angles labeled A, B, Q, D, P, θ⟧ ⟦Diagram of a circle with a tangent triangle labeled T, x, y, o, B⟧

**Traduction anglaise —**

⟦illegible⟧ Trigonometry questions 4th year Conditional Exam Sept., 1966. Q.4. PQ = ? , ∠θ = ? In Δ PDQ , PQ/5 = cosec 56° ∴ PQ = 5 cosec 56° DP/5 = tan 34° ∴ DP = 5 tan 34° AP/5 = tan 52° ∴ AP = 5 tan 52° AD = BQ = DP + AP = 5 tan 34° + 5 tan 52° = 5 (tan 34° + tan 52°) from Δ AQD , tan θ = 5/AD ∴ tan θ = 5 / (5 (tan 34° + tan 52°)) ∴ tan θ = 1 / (tan 34° + tan 52°) ⟦line⟧ PQ = 5 cosec 56° = 5 x 1.2062 = 6.0310 miles Ans. 1 BQ = 5 (tan 34° + tan 52°) = 5 (0.6745 + 1.2799) = 5 x 1.9544 = 9.7720 miles Ans. 2 tan θ = 1 / (tan 34° + tan 52°) = 1 / (0.6745 + 1.2799) = 1 / 1.9544 = 0.51165 = 0.5117 Correct to 4 dec. pl. ∴ θ = 27° 6' Ans. 3 Q.5. ∠xoy = 143° 28' ∴ ∠xoT = (143° 28') / 2 = 71° 44' In Δ oxT 3.2 / oT = cos 71° 44' ∴ oT = 3.2 / cos 71° 44' = 3.2 sec 71° 44' oT = 3.2 x 3.1903 = 10.20896 in = 10.21 in Correct to 2 dec. pl. Ans. 1 In Δ oxB , xB / 3.2 = sin 71° 44' ∴ xB = 3.2 sin 71° 44' ∴ xy = 2 xB = 2 x 3.2 sin 71° 44' = 6.4 sin 71° 44' = 6.4 x 0.9496 = 6.07744 in = 6.08 in Correct to 2 dec. pl. Ans. 2 ⟦Diagram of a rectangle with diagonals and angles labeled A, B, Q, D, P, θ⟧ ⟦Diagram of a circle with a tangent triangle labeled T, x, y, o, B⟧

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### csp_f3fdbacb84aa5ae392d51716834eb02c

SHAMASH SECONDARY SCHOOL

Subject: Algebra Class: 4th Year Secondary Date: 7/1/1969 Time: 10:15-11:45

Attempt all questions:

1. The expression 2 x³ + Ax² + Bx - 4 is exactly divisible by x²-4. Find the values of A and B and find the remaining factor. (20 marks)

2. Find the square root of: 4x⁴ - 3x⁵ - 3x³ + 9/4 x⁶ + 5/3 x² - 2/3 x + 1/9 (20 marks)

3. Which of the following equations is always true, which is sometimes true and which is never true? Find the values of x which satisfy the equatioh which is sometimes true. (a) 4(x²-1) + 2(x + 3) = 2 + 2x(1 + 2x) (b) x(6x + 1) = 2x + 1 (c) x(x + 2) = 2(x - 2) (20 marks)

4. (i) Solve simultaneously the following equations: 1/x + 1/y + 3/z = 2½ ....................(1) 2/x + 4/y - 6/z = 2 ....................(2) 3/x + 5/y + 7/z = 2 5/6 ....................(3) (ii) Simplify the following expression to simplest form: x/y + y/x - 1 / (x²/y² + x/y + 1) . (1 + y/x) / (x - y) ÷ (1 + y³/x³) / (x²/y - y²/x) (20 marks)

5. A man bought "A" lbs of coffee for a certain sum of money. He kept "B" lbs to himself and sold the remainder at "C" shillings a pound more than he paid for it. He found that he received for this portion an amount equal to the original sum of money which he paid for the whole. Find the original sum of money which he paid for the whole. (20 marks)

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Subject: Algebra Class: 4th Year Secondary Date: 7/1/1969 Time: 10:15-11:45 Attempt all questions: 1. The expression 2 x³ + Ax² + Bx - 4 is exactly divisible by x²-4. Find the values of A and B and find the remaining factor. (20 marks) 2. Find the square root of: 4x⁴ - 3x⁵ - 3x³ + 9/4 x⁶ + 5/3 x² - 2/3 x + 1/9 (20 marks) 3. Which of the following equations is always true, which is sometimes true and which is never true? Find the values of x which satisfy the equatioh which is sometimes true. (a) 4(x²-1) + 2(x + 3) = 2 + 2x(1 + 2x) (b) x(6x + 1) = 2x + 1 (c) x(x + 2) = 2(x - 2) (20 marks) 4. (i) Solve simultaneously the following equations: 1/x + 1/y + 3/z = 2½ ⟦line⟧ (1) 2/x + 4/y - 6/z = 2 ⟦line⟧ (2) 3/x + 5/y + 7/z = 2 5/6 ⟦line⟧ (3) (ii) Simplify the following expression to simplest form: x/y + y/x - 1 / (x²/y² + x/y + 1) . (1 + y/x) / (x - y) ÷ (1 + y³/x³) / (x²/y - y²/x) (20 marks) 5. A man bought "A" lbs of coffee for a certain sum of money. He kept "B" lbs to himself and sold the remainder at "C" shillings a pound more than he paid for it. He found that he received for this portion an amount equal to the original sum of money which he paid for the whole. Find the original sum of money which he paid for the whole. (20 marks)

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### csp_f4d20ccd011d5a098827d23b8e396b66

Make-up Exam. 2nd Quarter ⟦الصف الرابع⟧ 17/1/1969

I. (i) Find the value of x² - 1/y when x = -1/2 & y = -3 (6 marks) (ii) Solve the equation x/3 + (x-1)/2 = 7 (7 marks) (iii) Find x if 2x + y = 4 and 3y + 4 = 6x (7 marks)

II (i) If t = ∛((x² + 4y) / 2yz) , find z in terms of x, y and t. (10 marks) (ii) If F = av - b/v² and if F = 4 when v = 5 and F = 36 when v = 10, find the values of "a" and "b" and the value of F when v = 20 (10 marks)

III. a man can cycle at x m.p.h. in still air. His speed increases y m.p.h. when he cycles with the wind, and decreases y m.p.h. when he cycles against the wind. The difference in his time to cycle one mile with the wind and one mile against the wind is z hours. Find a formula for z in terms of x and y, and find x if y = 2, z = 1/3. (20 marks)

IV. In how many days will "a" horses eat 1/n th of the corn of a field the whole of which can be eaten by "b" horses in "c" days. (20 marks)

V. Find the square root of: 16x⁴ + 16/3 x²y + 8x² + 4/9 y² + 4/3 y + 1 showing your steps neatly. (20 marks)

**Traduction anglaise —**

Make-up Exam. 2nd Quarter ⟦Fourth Grade⟧ 17/1/1969 I. (i) Find the value of x² - 1/y when x = -1/2 & y = -3 (6 marks) (ii) Solve the equation x/3 + (x-1)/2 = 7 (7 marks) (iii) Find x if 2x + y = 4 and 3y + 4 = 6x (7 marks) II (i) If t = ∛((x² + 4y) / 2yz) , find z in terms of x, y and t. (10 marks) (ii) If F = av - b/v² and if F = 4 when v = 5 and F = 36 when v = 10, find the values of "a" and "b" and the value of F when v = 20 (10 marks) III. a man can cycle at x m.p.h. in still air. His speed increases y m.p.h. when he cycles with the wind, and decreases y m.p.h. when he cycles against the wind. The difference in his time to cycle one mile with the wind and one mile against the wind is z hours. Find a formula for z in terms of x and y, and find x if y = 2, z = 1/3. (20 marks) IV. In how many days will "a" horses eat 1/n th of the corn of a field the whole of which can be eaten by "b" horses in "c" days. (20 marks) V. Find the square root of: 16x⁴ + 16/3 x²y + 8x² + 4/9 y² + 4/3 y + 1 showing your steps neatly. (20 marks)

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### csp_f78f562c983f5b08ad0490c44956f5cb

B01 · header / latin - p.2 - Shamash Secondary School Cond.Exam. September, 1963. Algebra 12/9/63 4th Year Secondary.

B02 · other / latin ⟦line⟧

B03 · paragraph / latin 6. Draw the graph of Y = ½X³ for values of X between -3 and 4 taking ½ inch to represent one unit on the X-axis and two tenths of an inch to represent one unit on the Y-axis. By drawing other graphs on the same figure, solve the equations:

B04 · paragraph / latin (i) ½X³ - ⁷/₂X -3 = 0 (6 marks) (ii) ½X³ + ³/₂X-2 = 0 (6 marks)

B05 · paragraph / latin (iii) From the graph find the range of values of X for which ½X³ is greater than ⁷/₂X + 3. (6 marks).

B06 · other / latin ⟦line⟧

**Traduction anglaise —**

- p.2 - Shamash Secondary School Cond.Exam. September, 1963. Algebra 12/9/63 4th Year Secondary. ⟦line⟧ 6. Draw the graph of Y = ½X³ for values of X between -3 and 4 taking ½ inch to represent one unit on the X-axis and two tenths of an inch to represent one unit on the Y-axis. By drawing other graphs on the same figure, solve the equations: (i) ½X³ - ⁷/₂X -3 = 0 (6 marks) (ii) ½X³ + ³/₂X-2 = 0 (6 marks) (iii) From the graph find the range of values of X for which ½X³ is greater than ⁷/₂X + 3. (6 marks). ⟦line⟧

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### csp_fa36ba61675f55f0a2df55d84ef30185

Solutions to 3rd Quarter Exam. in Algebra 4th Year Secondary 20/3/1967

2. (i) b / √a . ∛ac . ⁴√c³ / √b . √b⁻¹ / a⁻¹/⁶ = b / a¹/² . a¹/³ . c¹/³ . c³/⁴ / b¹/² . b⁻¹/² / a⁻¹/⁶ = a⁻¹/² + ¹/³ + ¹/⁶ . b¹⁻¹/²⁻⁻¹/² . c¹/³ + ³/⁴ = a⁻³⁺²⁺¹/₆ . b²⁻¹⁻¹/₂ . c⁴⁺⁹/₁₂ = a⁰ . b⁰ c¹³/¹² = ¹²√c¹³ = c ¹²√c Ans.

(ii) [(9ⁿ⁺¹/⁴)(√3.3ⁿ) / 3 √3ⁿ⁻²]¹/ⁿ = [(3²)(ⁿ⁺¹/⁴) (3ⁿ⁺¹/²)¹/² / 3.3ⁿ⁻²/²]¹/ⁿ = [3²ⁿ⁺¹/² . 3ⁿ⁺¹/⁴ . 3⁻ⁿ/²⁻¹]¹/ⁿ = [3 ⁴ⁿ⁺¹⁺ⁿ⁺¹⁺ⁿ⁻² / 2]¹/ⁿ = [3⁶ⁿ/²]¹/ⁿ = 3³ⁿ/ⁿ = 3³ = 27 Ans.

3. (∛x² + 2 x¹/³ - 16 x⁻²/³ - 32/x) ÷ (x¹/⁶ + 4 x⁻¹/⁶ + 4/√x) (x²/³ + 2 x¹/³ - 16 x⁻²/³ - 32 x⁻¹) ÷ (x¹/⁶ + 4 x⁻¹/⁶ + 4 x⁻¹/²)

x¹/⁶ + 4 x⁻¹/⁶ + 4 x⁻¹/² | x²/³ + 2 x¹/³ - 16 x⁻²/³ - 32 x⁻¹ | x¹/² - 2 x¹/⁶ + 4 x⁻¹/⁶ - 8 x⁻¹/² ____________________| x²/³ + 4 x¹/³ + 4 x⁰ |________________________ -2 x¹/³ - 4 - 16 x⁻²/³ - 32 x⁻¹ -2 x¹/³ - 8 - 8 x⁻¹/³ ____________________ 4 + 8 x⁻¹/³ - 16 x⁻²/³ - 32 x⁻¹ 4 + 16 x⁻¹/³ + 16 x⁻²/³ ____________________ -8 x⁻¹/³ - 32 x⁻²/³ - 32 x⁻¹ -8 x⁻¹/³ - 32 x⁻²/³ - 32 x⁻¹ ___________________________

**Traduction anglaise —**

Solutions to 3rd Quarter Exam. in Algebra 4th Year Secondary 20/3/1967 2. (i) b / √a . ∛ac . ⁴√c³ / √b . √b⁻¹ / a⁻¹/⁶ = b / a¹/² . a¹/³ . c¹/³ . c³/⁴ / b¹/² . b⁻¹/² / a⁻¹/⁶ = a⁻¹/² + ¹/³ + ¹/⁶ . b¹⁻¹/²⁻⁻¹/² . c¹/³ + ³/⁴ = a⁻³⁺²⁺¹/₆ . b²⁻¹⁻¹/₂ . c⁴⁺⁹/₁₂ = a⁰ . b⁰ c¹³/¹² = ¹²√c¹³ = c ¹²√c Ans. (ii) [(9ⁿ⁺¹/⁴)(√3.3ⁿ) / 3 √3ⁿ⁻²]¹/ⁿ = [(3²)(ⁿ⁺¹/⁴) (3ⁿ⁺¹/²)¹/² / 3.3ⁿ⁻²/²]¹/ⁿ = [3²ⁿ⁺¹/² . 3ⁿ⁺¹/⁴ . 3⁻ⁿ/²⁻¹]¹/ⁿ = [3 ⁴ⁿ⁺¹⁺ⁿ⁺¹⁺ⁿ⁻² / 2]¹/ⁿ = [3⁶ⁿ/²]¹/ⁿ = 3³ⁿ/ⁿ = 3³ = 27 Ans. 3. (∛x² + 2 x¹/³ - 16 x⁻²/³ - 32/x) ÷ (x¹/⁶ + 4 x⁻¹/⁶ + 4/√x) (x²/³ + 2 x¹/³ - 16 x⁻²/³ - 32 x⁻¹) ÷ (x¹/⁶ + 4 x⁻¹/⁶ + 4 x⁻¹/²) x¹/⁶ + 4 x⁻¹/⁶ + 4 x⁻¹/² | x²/³ + 2 x¹/³ - 16 x⁻²/³ - 32 x⁻¹ | x¹/² - 2 x¹/⁶ + 4 x⁻¹/⁶ - 8 x⁻¹/² ⟦line⟧ | x²/³ + 4 x¹/³ + 4 x⁰ | ⟦line⟧ -2 x¹/³ - 4 - 16 x⁻²/³ - 32 x⁻¹ -2 x¹/³ - 8 - 8 x⁻¹/³ ⟦line⟧ 4 + 8 x⁻¹/³ - 16 x⁻²/³ - 32 x⁻¹ 4 + 16 x⁻¹/³ + 16 x⁻²/³ ⟦line⟧ -8 x⁻¹/³ - 32 x⁻²/³ - 32 x⁻¹ -8 x⁻¹/³ - 32 x⁻²/³ - 32 x⁻¹ ⟦line⟧

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### csp_fc02030361ed57cfb36995c7c14170c7

Shamash Secondary School Conditional Examination, September 1961

Subject: Algebra Date: 1/9/1961 Class: 4th Year Secondary Time: 8:00-10:30

⟦line⟧ Attempt all questions:

1. (i) Factor completely: 4a²(a-1) - 9a+9 (6 marks) (ii) Find the numerical value of (7A-3B)/(5A+6B), having given A/B = 11 (7 marks) (iii) A basket contains oranges, lemons and apples. The total number of of these fruits is X. There are 2X/5 oranges and X/6 lemons. What fraction of the total is the number of apples? (7 marks)

2. (i) If X =1, X=-1 and X=2/3 satisfy the equation Ax³+Bx²+Cx+2=0, find the values of A,B and C. (10 marks) (ii) Given that √(Ay-1)/(Ax+1) = y/x , obtain in its simplest form an expression for A in terms of x and y. (10 marks)

3. (i) Calculate the values of (729)^(5/6) , (32/3125)^(-3/5) and (343)^(5/6) ÷ (343)^(1/3) (7 marks) (ii) Express ¹¹√0.0001 as a power of 10. (6 marks) (iii) Use logarithms to calculate the value of ⁷√( (0.00561)² / 1.008 ) correct to four decimal places. (7 marks)

4. (i) The sum of seven numbers which are in arithmetic progression is 35 and the difference between the first and the seventh is 18. Find the seven numbers. (8 marks) (ii) The sum of the fourth term and six times the fifth term of a geometric progression is equal to the third term. Find the two possible values of the common ratio. If the second term is 16 and the common ratio is negative, find the sum of the first six terms. (12 marks)

5. Find the time between 7 and 8 o'clock when the hands of a watch are separated by 15 minutes for the first time. (20 marks)

⟦line⟧ P.T.O.

[Marginalia] ⟦illegible⟧ [Marginalia] ⟦illegible⟧ [Marginalia] ⟦illegible⟧

**Traduction anglaise —**

Shamash Secondary School Conditional Examination, September 1961 Subject: Algebra Date: 1/9/1961 Class: 4th Year Secondary Time: 8:00-10:30 ⟦line⟧ Attempt all questions: 1. (i) Factor completely: 4a²(a-1) - 9a+9 (6 marks) (ii) Find the numerical value of (7A-3B)/(5A+6B), having given A/B = 11 (7 marks) (iii) A basket contains oranges, lemons and apples. The total number of of these fruits is X. There are 2X/5 oranges and X/6 lemons. What fraction of the total is the number of apples? (7 marks) 2. (i) If X =1, X=-1 and X=2/3 satisfy the equation Ax³+Bx²+Cx+2=0, find the values of A,B and C. (10 marks) (ii) Given that √(Ay-1)/(Ax+1) = y/x , obtain in its simplest form an expression for A in terms of x and y. (10 marks) 3. (i) Calculate the values of (729)^(5/6) , (32/3125)^(-3/5) and (343)^(5/6) ÷ (343)^(1/3) (7 marks) (ii) Express ¹¹√0.0001 as a power of 10. (6 marks) (iii) Use logarithms to calculate the value of ⁷√( (0.00561)² / 1.008 ) correct to four decimal places. (7 marks) 4. (i) The sum of seven numbers which are in arithmetic progression is 35 and the difference between the first and the seventh is 18. Find the seven numbers. (8 marks) (ii) The sum of the fourth term and six times the fifth term of a geometric progression is equal to the third term. Find the two possible values of the common ratio. If the second term is 16 and the common ratio is negative, find the sum of the first six terms. (12 marks) 5. Find the time between 7 and 8 o'clock when the hands of a watch are separated by 15 minutes for the first time. (20 marks) ⟦line⟧ P.T.O. ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧

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### csp_fc451146863958a68c7e047d01604dd3

B01 · header / latin Shamash Secondary School Final Examination, May, 1967.

B02 · form / latin Subject:: Arithmetic & Trigonometry Date:: 17/5/1967 Class:: 4th Year Secondary. Time:: 8:00 - 10:30 a.m.

B03 · other / latin ---

B04 · marginalia / latin 2 3

B05 · paragraph / latin Answer five questions which must include questions 2 & 5.

B10 · footer / latin (cont'd.p.2)..

**Traduction anglaise —**

Shamash Secondary School Final Examination, May, 1967. Subject:: Arithmetic & Trigonometry Date:: 17/5/1967 Class:: 4th Year Secondary. Time:: 8:00 - 10:30 a.m. ⟦line⟧ 2 3 Answer five questions which must include questions 2 & 5. 1. (a) How much stock is obtained by investing £2,286 in a 4½ per cent stock at 95¼? After receiving the first annual dividend on this stock, it is immediately resold at 98. Calculate the total gain on the transaction. (b) I have a watch which gains six minutes in every true hour. I put the watch right at 8.30 a.m. What is the latest time indicated by the watch at which I must set out to catch a train which leaves at 10.25 a.m. if it takes me 15 minutes to walk to the station ? 2. Following a storm, water is pumped out of a flooded area through a pipe of 8 in. diameter at the rate of 1,000 gallons per minute. Taking 1 cu.ft as 6¼ gallons and π as 22/7, calculate: (a) the speed in ft. per sec. at which the water is passing through the pipe. (b) how many tons of sediment will be pumped out in two days if it is known that the flood water contains ½ oz. of sediment in every cu.ft of water. 3. A merchant bought 15 tons of potatoes from a farmer at £18 per ton. (a) He sold 4 ton 12 cwt of the potatoes in 1 cwt bags at £1 5s. per bag. The additional cost to the merchant in selling the potatoes in this way was 6s. 3d. per ton. (b) He sold 15 cwt retail at 4d. per lb for which he incurred additional labour costs of £5 10s. (c) He sold the remainder of the potatoes in bulk at £20 per ton. Calculate: (i) the merchant's total costs, including the initial cost of the potatoes, cost of selling the potatoes in bags, and additional labour cost for the retail sales. (ii) the total amount the merchant received from his sales. (iii) the merchant's profit calculated as a percentage, correct to 2 significant figures, of his total costs. 4. A borough is divided into two districts whose rateable values are respectively £52,320 and £127,460. The rate in the first district is 12s. 9d. in the £, and in the second district it is 19s. 3d. in the £. Find the average rate for the whole borough to the nearest farthing. (cont'd.p.2)..

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### csp_fcdbbcdaa65d5b89a07ffd3d9594547f

B01 · header / latin SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965.

B02 · form / latin Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m.

B03 · paragraph / latin ⟦line⟧ Attempt all questions :

B04 · paragraph / latin 1. (i) If m = (2x + y) / (x + 2y) , find an expression for y in terms of m and x. If also y = mx , find the values of m. (7 marks). (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks).

B05 · paragraph / latin 3. (i) Compute by logarithms, arranging your work neatly : ⁷√((cos² 18° 47')(sin³ 48° 21')) / ((10.09)³ (0.0002049)) (6 marks). (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks).

B06 · paragraph / latin 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ⟦line⟧ (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks).

B07 · paragraph / latin 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axis and 0.2 inch for your unit on the y-axis. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks). ⟦line⟧

B08 · marginalia / latin ⟦illegible⟧

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL FINAL EXAMINATION, JUNE, 1965. Subject: Algebra. Date: 1/6/1965. Class: 4th year, secondary, sections A & B. Time: 8:00-11:00 a.m. ⟦line⟧ Attempt all questions : 1. (i) If m = (2x + y) / (x + 2y) , find an expression for y in terms of m and x. If also y = mx , find the values of m. (7 marks). (ii) Resolve into two factors : c³ - 27b³ + a³ + 9abc (7 marks). (iii) Resolve the expression 5x² - 14x + 9 into two factors and show that the value of this expression is negative when x lies between 1 and 1.8. (6 marks). 3. (i) Compute by logarithms, arranging your work neatly : ⁷√((cos² 18° 47')(sin³ 48° 21')) / ((10.09)³ (0.0002049)) (6 marks). (ii) If 2 log a - 5 log b = 3 log c, find 'a' in terms of 'b' and 'c'. (4 marks). (iii) Given logₐ 4.41 = 2 , calculate the value of 'a'. (4 marks). (iv) Solve the equation 2³⁻ˣ = 3²ˣ⁺¹ giving your answer correct to three decimal places. (6 marks). 4. (i) Write down and simplify an expression for the nth term of the arithmetic progression 3 , 7 , 11 , ⟦line⟧ (4 marks). If the sum of n terms of this progression is bn + cn² find the values of b and c and the sum of the first thirty terms. (8 marks). (ii) The product of the first and seventh terms of a geometric progression is equal to the fourth term; and the sum of the first and fourth terms is 9. Find the sum of the first seven terms of the progression. (8 marks). 5. (i) Draw the graph of y = (x - 1)(x - 3)² for values of x from -½ to 5, choosing 0.5 inch for your unit on the x-axis and 0.2 inch for your unit on the y-axis. To get a good drawing of the curve, choose successive values of x at intervals of halves, beginning with -½. (5 marks). (ii) From this graph find an approximate maximum value and an exact minimum value for y and the corresponding values of x which make y a maximum or a minimum. (5 marks). (iii) By plotting another graph on the same diagram find the roots of the equation (x - 1)(x - 3)² = 5x - 9. (5 marks). (iv) From these two graphs find the values of x for which the function (x - 1)(x - 3)² is always greater than (5x - 9). (5 marks). ⟦line⟧ ⟦illegible⟧

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### csp_fe1d1bd1203852ea8d16534a7bb1da50

Algebra Paper, Final Examination, June 3rd 1964 ⟦line⟧

x³ + y³ = 9 ...... ① multiplying equation ② by 3 + adding to equation ①, x²y + xy² = 6 ...... ② we get: 3x²y + 3xy² = 18 ...... ② x³ + y³ = 9 ...... ① ∴ x³ + 3x²y + 3xy² + y³ = 27 ∴ (x + y)³ = 27 ∴ x + y = 3 ...... ③ from ② xy(x + y) = 6 ∴ 3xy = 6 ∴ xy = 2 ...... ④. From ③, y = 3 - x, substituting in ④, we get x(3 - x) = 2 ∴ 3x - x² = 2 ∴ x² - 3x + 2 = 0 or (x - 2)(x - 1) = 0 ∴ x = 2 and x = 1 y = 1 Ans. 1 y = 2 Ans. 2

(b) x⁻⁴ + 4 = 5x⁻² or x⁻⁴ - 5x⁻² + 4 = 0 or (x⁻¹)² - 5(x⁻¹)² + 4 = 0 ∴ [(x⁻¹)² - 4] [(x⁻¹)² - 1] = 0 ∴ (x⁻¹)² = 4 or (x⁻¹)² = 1 ∴ x⁻¹ = ± 2 or 1/x = ± 2 ∴ x = ± 1/2 also (x⁻¹)² = 1 ∴ x⁻¹ = ± 1 ∴ <del>⟦illegible⟧</del> 1/x = ± 1 ∴ x = ± 1 ∴ x = +1, -1, +1/2, -1/2 Ans.

3 (a)(i) x³ + y³ + 5x²y + 5xy² = (x + y)(x² - xy + y²) + 5xy(x + y) = (x + y) [x² - xy + y² + 5xy] = (x + y)(x² + 4xy + y²) Ans. (ii) (x + y + z)² + x² - y² - z² = x² + y² + z² + 2xy + 2xz + 2yz + x² - y² - z² = 2x² + 2xy + 2xz + 2yz = 2 [x² + xy + xz + yz] = 2 [x(x + y) + z(x + y)] = 2(x + y)(x + z) Ans. (iii) (a² - b²)(x² - y²) + 4abxy = a²x² - a²y² - b²x² + b²y² + 4abxy = a²x² + 2abxy + b²y² - (b²x² - 2abxy + a²y²) = (ax + by)² - (bx - ay)² = (ax + by + bx - ay)(ax + by - bx + ay) = [(a + b)x + (b - a)y] [(a - b)x + (b + a)y] Ans.

[Marginalia] ⟦illegible⟧ graphs

**Traduction anglaise —**

Algebra Paper, Final Examination, June 3rd 1964 ⟦line⟧ x³ + y³ = 9 ...... ① multiplying equation ② by 3 + adding to equation ①, x²y + xy² = 6 ...... ② we get: 3x²y + 3xy² = 18 ...... ② x³ + y³ = 9 ...... ① ∴ x³ + 3x²y + 3xy² + y³ = 27 ∴ (x + y)³ = 27 ∴ x + y = 3 ...... ③ from ② xy(x + y) = 6 ∴ 3xy = 6 ∴ xy = 2 ...... ④. From ③, y = 3 - x, substituting in ④, we get x(3 - x) = 2 ∴ 3x - x² = 2 ∴ x² - 3x + 2 = 0 or (x - 2)(x - 1) = 0 ∴ x = 2 and x = 1 y = 1 Ans. 1 y = 2 Ans. 2 (b) x⁻⁴ + 4 = 5x⁻² or x⁻⁴ - 5x⁻² + 4 = 0 or (x⁻¹)² - 5(x⁻¹)² + 4 = 0 ∴ [(x⁻¹)² - 4] [(x⁻¹)² - 1] = 0 ∴ (x⁻¹)² = 4 or (x⁻¹)² = 1 ∴ x⁻¹ = ± 2 or 1/x = ± 2 ∴ x = ± 1/2 also (x⁻¹)² = 1 ∴ x⁻¹ = ± 1 ∴ <del>⟦illegible⟧</del> 1/x = ± 1 ∴ x = ± 1 ∴ x = +1, -1, +1/2, -1/2 Ans. 3 (a)(i) x³ + y³ + 5x²y + 5xy² = (x + y)(x² - xy + y²) + 5xy(x + y) = (x + y) [x² - xy + y² + 5xy] = (x + y)(x² + 4xy + y²) Ans. (ii) (x + y + z)² + x² - y² - z² = x² + y² + z² + 2xy + 2xz + 2yz + x² - y² - z² = 2x² + 2xy + 2xz + 2yz = 2 [x² + xy + xz + yz] = 2 [x(x + y) + z(x + y)] = 2(x + y)(x + z) Ans. (iii) (a² - b²)(x² - y²) + 4abxy = a²x² - a²y² - b²x² + b²y² + 4abxy = a²x² + 2abxy + b²y² - (b²x² - 2abxy + a²y²) = (ax + by)² - (bx - ay)² = (ax + by + bx - ay)(ax + by - bx + ay) = [(a + b)x + (b - a)y] [(a - b)x + (b + a)y] Ans. ⟦illegible⟧ graphs

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### csp_fe31580312d458c3836060e28f446398

SHAMASH SECONDARY SCHOOL Mid-Year Examination, February, 1969.

Subject: Algebra Date: 17/2/1969. Class: 4th Year, Secondary Time: 8:30-11:30 a.m.

Five questions only are to be attempted.

1. The time now is x minutes after five and the two hands of the watch stand in a straight line on opposite sides of the centre of the dial of the watch. Find x and state, in words, the correct time. (20 marks)

2. (i) Find the value of x from the following equation : 6 x³ + 19 x² + x - 6 = 0. (10 marks) (ii) Solve the following equation, using the shortest possible method, by first reducing each fraction to a simpler form : 3 x - 7 2 x - 5 3 x + 7 2 x + 5 ------- + ------- = ------- + ------- . (10 marks) x - 2 x - 3 x + 2 x + 3

3. (i) In the following equation, A, B, and C are constants and the equation is true for all values of x. Find the values of A, B and C. A(x² - 2x) + B(x + 4) + C = 3x² + x + 25. (10 marks) (ii) Solve the following equations simultaneously : x² + xy + 2y² = 8 ...............(1) 2x² - 2xy - 3y² = 1 ...............(2) (10 marks)

4. (i) If x + 1/x = a and y + 1/y = b , find the value of the expression (x³ + y³ + 1/x³ + 1/y³) in terms of "a" and "b". Hence or otherwise find the value of (x³ + y³ + 1/x³ + 1/y³) if a = 1 and b = 2. (10 marks) (ii) Resolve the expression a³ + a - 8 b³ - 2b + c + 6abc + c³ into two factors one of which is (a - 2b + c). (10 marks)

5. A cask P is filled with 100 gallons of water, and a cask Q with 50 gallons of brandy; x gallons are drawn from each cask, mixed and replaced; and the same operation is repeated. Find x when there are 17 gallons of brandy in P after the second replacement. (20 marks)

6. Two trains A and B are travelling on two railway tracks which are parallel to each other. Train A is 240 ft long and it is travelling at 22.5 miles per hour. Train B is 200 ft long and is travelling at the rate 15 miles per hour. Find the length of time in seconds from the instant when the heads of the front cars of the two trains are together, to the instant when the

P. T. O.

**Traduction anglaise —**

SHAMASH SECONDARY SCHOOL Mid-Year Examination, February, 1969. Subject: Algebra Date: 17/2/1969. Class: 4th Year, Secondary Time: 8:30-11:30 a.m. Five questions only are to be attempted. 1. The time now is x minutes after five and the two hands of the watch stand in a straight line on opposite sides of the centre of the dial of the watch. Find x and state, in words, the correct time. (20 marks) 2. (i) Find the value of x from the following equation : 6 x³ + 19 x² + x - 6 = 0. (10 marks) (ii) Solve the following equation, using the shortest possible method, by first reducing each fraction to a simpler form : 3 x - 7 2 x - 5 3 x + 7 2 x + 5 ⟦line⟧ + ⟦line⟧ = ⟦line⟧ + ⟦line⟧ . (10 marks) x - 2 x - 3 x + 2 x + 3 3. (i) In the following equation, A, B, and C are constants and the equation is true for all values of x. Find the values of A, B and C. A(x² - 2x) + B(x + 4) + C = 3x² + x + 25. (10 marks) (ii) Solve the following equations simultaneously : x² + xy + 2y² = 8 ...............(1) 2x² - 2xy - 3y² = 1 ...............(2) (10 marks) 4. (i) If x + 1/x = a and y + 1/y = b , find the value of the expression (x³ + y³ + 1/x³ + 1/y³) in terms of "a" and "b". Hence or otherwise find the value of (x³ + y³ + 1/x³ + 1/y³) if a = 1 and b = 2. (10 marks) (ii) Resolve the expression a³ + a - 8 b³ - 2b + c + 6abc + c³ into two factors one of which is (a - 2b + c). (10 marks) 5. A cask P is filled with 100 gallons of water, and a cask Q with 50 gallons of brandy; x gallons are drawn from each cask, mixed and replaced; and the same operation is repeated. Find x when there are 17 gallons of brandy in P after the second replacement. (20 marks) 6. Two trains A and B are travelling on two railway tracks which are parallel to each other. Train A is 240 ft long and it is travelling at 22.5 miles per hour. Train B is 200 ft long and is travelling at the rate 15 miles per hour. Find the length of time in seconds from the instant when the heads of the front cars of the two trains are together, to the instant when the P. T. O.

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Exams, Shamash Secondary School

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