AI English Translation, Pages 126-150
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..)
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⟧
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 -
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⟧
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 -
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⟧
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⟧
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)
⟦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⟧
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.
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
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.
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⟧
⟦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⟧
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
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.
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⟧
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⟧
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.
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⟧
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⟧
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⟧
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⟧
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⟧
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.