AI en Translation, Pages 126-150
Page 126
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
Page 127
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)⟧
Page 128
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)
Page 129
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 pt. of intersections give
the roots x = -0.8 } Ans.
and x = 2.47 }
(8 marks)
A
B
y = 1/2
x = -0.8
x = 2.47
Solution to Conditional Exam in Algebra
Sept., 1965
Page 130
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..)
Page 131
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⟧
Page 132
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 -
Page 133
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⟧
Page 134
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 -
Page 135
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⟧
Page 136
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⟧
Page 137
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)
Page 138
⟦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⟧
Page 139
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.
Page 140
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
Page 141
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.
Page 143
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⟧
Page 144
x | y
0 | -9
1 | 0
1 1/2 | 1 1/8
2 | 1
2 1/2 | 3/8
3 | 0
3 1/2 | 5/8
4 | 3
4 1/2 | 6 1/8
5 | 16
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
Answer:
(i) The graph is plotted as shown.
(ii) When x = 1 1/2 y = 1 1/8 a ⟦approximate⟧ maximum
When x = 3 y = 0 an exact minimum
(iii) The roots of the equation (x-1)(x-3)² = 3x-7
are the abscissae of the points A, B & C
at which:
x = 0
x = 2
x = 5
(iv) The function (x-1)(x-3)² is greater
than 3x-7 when 0 < x < 2
and x > 5
since for all these values of x, the curve of
the function (x-1)(x-3)² lies above the curve y = 3x-7
Graph
Maximum point
Minimum point
Page 145
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.
Page 146
SHAMASH SECONDARY SCHOOL
FINAL EXAMINATION JUNE, 1958
Date: ⟦illegible⟧
Time: 8:00-11:00 a.m.
4th year Secondary sections A & B
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)
⟦illegible⟧
P.T.O. Look for question 2 at the back of this sheet.
⟦The rest of the page contains faint, mirrored text from the reverse side of the paper, which is not part of the primary document content on this face.⟧
Page 147
SHAMASH SECONDARY SCHOOL, BAGHDAD.
MID-YEAR EXAMINATION, JANUARY 1958.
4th Year Secondary - Sections A, B, C, D & E.
Date: 13/1/1958.
Time: 8:00-11:00 a.m.
2. 'A' 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)
Page 148
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)
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).
⟦Look for questions at the back of this sheet.⟧
Page 149
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⟧
Page 150
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⟧