AI English Translation, Pages 176-200
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)
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
⟦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.
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)⟧
⟦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)⟧
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.
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
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.
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⟧
⟦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
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
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.
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
(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⟧
⟦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.
⟦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⟧
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⟧
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.
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)
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
⟦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)
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)
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).
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⟧
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⟧