AI English Translation, Pages 101-125
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)
⟦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⟧
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
⟦...⟧ 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
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
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)
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
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
⟦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>
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.
• 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.
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)..
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⟧
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)..
⟦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⟧
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)..
⟦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⟧
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)..
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⟧
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)..
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⟧
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
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)⟧
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)
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