AI English Translation, Pages 151-175
Shamash Secondary School
Conditional Examination, Sept. 1964
⟦line⟧
Subject: Algebra
Class: 4th Year Secondary, Scientific Section.
Date: 2/9/1964
Time: 8:00 - 11:00 a.m.
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All questions are to be attempted.
I. (a) The expression x^2+px+q reduces to zero when x equals 2 or 4.
Find the values of x for which this expression equals 48.
(10 marks)
(b) The expression x^4+ax^3+bx^2+cx+12 is factorable into (x-1), (x+1)
and (x+3). Find the values of a, b, c and the other factor.
(10 marks)
II. (a) At what time between eleven and twelve o'clock will the two hands
of a watch be at right angle for the second time ?
(10 marks)
(b) A car covers the distance between Mosul and Baghdad in five hours,
travelling on the route which lies on the right bank of the river
Tigris. Another car travelling at an average speed which is less
by 20 kilometres than the first, covers the distance between the
two cities which lies on the left bank of the Tigris and which is
50 kilometres longer, in 7 1/2 hours. Find the length of each course.
(10 marks)
III. (a) If log_2 (4x-4) = 2, find the value of log_4 x.
(10 marks)
(b) Compute by logarithms, arranging your work neatly:
7√((0.1062)^2 x (0.0071)^3 / (1.005) x (3.007)^5)
(10 marks)
IV. (a) If a body falls from rest, (neglecting the friction of the air ) it
will fall 16 ft during the first second, 48 ft during the second
second, 80 ft during the third second, 112 ft during the fourth
second and so on.
Find the number of seconds it will take a stone to reach the bottom
of a well 1936 ft. deep, if it is dropped from the top of the well.
(10 marks)
(b) In a Geometric progression, 1023 times the sum of the first five
terms is equal to 31 times the sum of the first ten terms. Find
the common ratio.
(10 marks)
( cont'd.p.2)..
-p.2-
Conditional Exam.in Algebra for the 4th Year, Sept.1964. (cont'd.)
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V. (a) Sketch the curve of the function (x⁴-8x²) for values of x from -3 to 3
choosing ½ inch as one unit on the x-axis and one tenth of an inch as
one unit on the y-axis.
(8 marks)
(b) From this curve find the values of x at which the function (x⁴-8x²)
has minimum or maximum values. Find also these minimum and maximum
values.
(6 marks)
(c) Plot on the same diagram the graph of y= 3x-10 and from these two
graphs find the solution of the equation x⁴-8x²+10 = 3x.
(6 marks).
⟦line⟧
Solutions to Conditional Exam in Algebra
4th Year, Sept., 1964.
①
1. (a) Substitute x = 2 in the expression x² + px + q + you get 4 + 2p + q = 0 ... (1)
substitute x = 4 " " " " " " " " 16 + 4p + q = 0 ... (2)
subtracting equation (1) from (2), we get 12 + 2p = 0 or p = -6
∴ from (1) 4 - 12 + q = 0 ∴ q = 8 ∴ q = 8
∴ when x² - 6x + 8 = 48, we get x² - 6x - 40 = 0 or (x - 10)(x + 4) = 0
x = 10 or x = -4 Ans.
(b) x⁴ + ax³ + bx² + cx + 12 will be equal to zero when x = 1, -1, or -3
since (x - 1) and (x + 1) and (x + 3) are factors. Substituting x = 1,
we get 1 + a + b + c + 12 = 0 ... (1). Substituting x = -1, we get:
1 - a + b - c + 12 = 0 ... (2). " " x = -3, we get:
81 - 27a + 9b - 3c + 12 = 0 ... (3). Adding equations (1) + (2), we get:
2 + 2b + 24 = 0 or 2b = -26 or b = -13
dividing eq (3) by 3, we get 27 - 9a + 3b - c + 4 = 0 or 27 - 9a + 3(-13) - c + 4 = 0 or
9a + c + 8 = 0 ... (4). But from (1) we get
⟦a + c⟧ = 0 ... (5). Subtracting (5) from (4) we get
8a + 8 = 0 or a = -1. substituting in (1)
we get 1 - 1 - 13 + c + 12 = 0 or c = 1 ∴ a = -1, b = -13, c = 1 Ans
Hence the expression is x⁴ - x³ - 13x² + x + 12
But (x - 1)(x + 1)(x + 3) = (x² - 1)(x + 3) = x³ + 3x² - x - 3 ∴ the fourth factor =
(x⁴ - x³ - 13x² + x + 12) ÷ (x³ + 3x² - x - 3) = x - 4 Ans.
x⁴ - x³ - 13x² + x + 12 | x³ + 3x² - x - 3
x⁴ + 3x³ - x² - 3x | x - 4
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-4x³ - 12x² + 4x + 12
-4x³ - 12x² + 4x + 12
⟦line⟧
2. (a) Let the time be x minutes after 11, then
the hour hand will be pointing x/12 divisions
after 11. Now arc BC = 5 - x/12
∴ x + arc BC = 45 or x + 5 - x/12 = 45
∴ 11x/12 = 40 or 11x = 480 or x = 480/11
or x = 43 7/11 min. = 43.63 min = 43 min 38 2/11 seconds <del>⟦illegible⟧</del>
∴ the time is 43 min 38 2/11 sec. after eleven Ans.
OR x = 11x5 + x/12 - 15
x = 55 + x/12 - 15
x = 40 + x/12
2
II. (b) Let x kilometres = the distance between the two cities on the right Bank
∴ x/5 k.p.hr. is the average speed of the first car
∴ x+50/7.5 k.p.h. " " " " " second car } ∴ x/5 - x+50/7.5 = 20
Multiplying the terms of the equation by 15, we get: 3x - 2(x+50) = 300 or
3x - 2x - 100 = 300 or x = 400 km. Ans. 1 = the length of the first course
x+50 = 400+50=450 km. Ans. 2 = " " " 2nd "
III (a) If log₂(4x-4) = 2 , then log₄x = ?
From the equation we get 4x-4 = 2² or 4x-4 = 4 ∴ x = 2
Now log₄x = log₄2 = y ∴ 2 = 4ʸ or 2 = 2²ʸ ∴ 2y = 1 ∴ y = 1/2
∴ log₄x = log₄2 = 1/2 Ans.
(b) Let x = ⁷√[(0.1062)²(0.0071)³ / (1.005)(3.007)⁵]
log 0.1062 = 1̄.0261 | 2 log 0.1062 = 2̄.0522 | log 1.005 = 0.0021
log 0.0071 = 3̄.8513 | 3 log 0.0071 = 7̄.5539 | 5 log 3.007 = 2.3905
log 1.005 = 0.0021 | log Num. = 9̄.6061 | log Den. = 2.3926
log 3.007 = 0.4781 | log Den. = 2.3926 |
| 7 log x = 11̄.2135 |
log x = 2̄.45907 = 2̄.4591 Correct to 4 dec. pl.
∴ x = 0.02878 or = 2.878 x 10⁻² Ans.
IV (a) The distances fallen during the consecutive seconds beginning with
the first are : 16 + 48 + 80 + 112 + ... which is the sum of an A.P. in which
a = 16 , d = 32 , s = 1936 , n = ?
From the formula s = n/2 {2a + (n-1)d} we get: 1936 = n/2 {2x16 + (n-1)x32} or
2 x 1936 = n {32 + 32n - 32} or 32n² = 3872 or n² = 121 or n = 11
∴ the number of seconds which will take the stone to reach the bottom = 11 seconds
Ans.
3
Ⅳ (b) Suppose that the G.P. is a, ar, ar², ... Then S₅ = a (r⁵-1) / r-1 and
the sixth term = ar⁵. The sum of the terms from the sixth to the tenth inclusive = ar⁵ (r⁵-1) / r-1
∴ S₁₀ = S₅ + ar⁵ (r⁵-1) / r-1 or S₁₀ = a (r⁵-1) / r-1 + ar⁵ (r⁵-1) / r-1
But 1023 S₅ = 31 S₁₀ or 1023 [a (r⁵-1) / r-1] = 31 [a (r⁵-1) / r-1 + ar⁵ (r⁵-1) / r-1] or
1023 [a (r⁵-1) / r-1] = 31 (a (r⁵-1) / r-1) [1 + r⁵] or 1023 = 31 (1 + r⁵) or
1 + r⁵ = 1023 / 31 or 1 + r⁵ = 33 or r⁵ = 32 or r⁵ = 2⁵ ∴ r = 2 Ans.
x | y = x^4 - 8x^2 + 3x - 10
-3 | 9
-2 |
| -16 | -4
| 3
Graph of the function f(x) = x^4 - 8x^2 is the curve shown below
Ans. 1
In the graph it is clear that:
when x = -2 , x^4 - 8x^2 = -16 minimum value
when x = 0 , x^4 - 8x^2 = 0 maximum value
when x = 2 , x^4 - 8x^2 = -16 a second minimum value
2) The solution of the equation x^4 - 8x^2 + 10 = 3x is the
same as the solution of equation x^4 - 8x^2 = 3x - 10
But the L.H.S. of the last equation is f(x) and the R.H.S. is y
The two graphs intersect at x = -2 (they touch)
x = 1
and x = 2.8 approximately
Ans. 2
⟦graph with axes and curves⟧
f(x) = x^4 - 8x^2
y = 3x - 10
(0,0)
(-2, -16)
(2, -16)
(1, -7)
(2.8, -1.3) approximately
Shamash Secondary School
Conditional Examination, Sept. 1964
Subject: Algebra Date: 2/9/1964
Class: 4th Year Secondary, Time: 8:00 - 11:00 a.m.
Scientific Section.
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All questions are to be attempted.
I. (a) The expression x² +px+q reduces to zero when x equals 2 or 4.
Find the values of x for which this expression equals 48.
(10 marks)
(b) The expression x⁴ +ax³ +bx² +cx+12 is factorable into (x-1), (x+1)
and (x+3). Find the values of a, b, c and the other factor.
(10 marks)
II. (a) At what time between eleven and twelve o'clock will the two hands
of a watch be at right angle for the second time ?
(10 marks)
(b) A car covers the distance between Mosul and Baghdad in five hours,
travelling on the route which lies on the right bank of the river
Tigris. Another car travelling at an average speed which is less
by 20 kilometres than the first, covers the distance between the
two cities which lies on the left bank of the Tigris and which is
50 kilometres longer, in 7½ hours. Find the length of each course.
(10 marks)
III. (a) If log₂ (4x-4) = 2, find the value of log₄ x.
(10 marks)
(b) Compute by logarithms, arranging your work neatly:
⟦line⟧
7 / (0.1062)² x (0.0071)³
\/ ⟦line⟧
(1.005) x (3.007)⁵
(10 marks)
IV. (a) If a body falls from rest, (neglecting the friction of the air ) it
will fall 16 ft during the first second, 48 ft during the second
second, 80 ft during the third second, 112 ft during the fourth
second and so on.
Find the number of seconds it will take a stone to reach the bottom
of a well 1936 ft. deep, if it is dropped from the top of the well.
(10 marks)
(b) In a Geometric progression, 1023 times the sum of the first five
terms is equal to 31 times the sum of the first ten terms. Find
the common ratio.
(10 marks)
( cont'd.p.2)..
-p.2-
Conditional Exam. in Algebra for the 4th Year, Sept. 1964. (cont'd.)
⟦line⟧
V. (a) Sketch the curve of the function (x⁴-8x²) for values of x from -3 to 3
choosing ½ inch as one unit on the x-axis and one tenth of an inch as
one unit on the y-axis.
(8 marks)
(b) From this curve find the values of x at which the function (x⁴-8x²)
has minimum or maximum values. Find also these minimum and maximum
values.
(6 marks)
(c) Plot on the same diagram the graph of y= 3x-10 and from these two
graphs find the solution of the equation x⁴-8x²+10 = 3x.
(6 marks).
⟦line⟧
⟦(8 marks)⟧
⟦(6 marks)⟧
⟦(6 marks)⟧
⟦(6 marks)⟧
⟦(6 marks)⟧
⟦(6 marks)⟧
⟦...⟧ taken total from G.C.E. past papers
not considered as an exam.
Shamash Secondary School
Final Exams. June, 1 9 6 4.
Paper II
Subject: Algebra
Date: 16/6/1964
Class: 4th Year (Scientific Section)
Time: 8:00-10:00 a.m.
All questions are to be attempted.
I. A and B are two towns 44 miles apart. A cyclist and a motorist
travel from A to B. The motorist leaves A 1 hr. 36 min. later
than the cyclist but they reach B at exactly the same time.
If the average speed of the motorist is 18 m.p.h. greater than
that of the cyclist, find the average speed of each.
(20 marks)
II. (i) Find the values of x and y if
x y x y
- + - = 4 = - + - . (10 marks)
2 4 4 2
(ii) Find the values of a and b so that 3x²-4x+1 can be expressed
in the form a(x-1)²+b(x-1). (10 marks)
III. (i) The first, second and last terms of an arithmetic
progression are x, y and z respectively.
a- Express the number of terms of the progression in terms
of x, y and z. (5 marks)
b- Show that the sum of the progression is
(x+z)(y+z-2x)
⟦line⟧ (5 marks)
2(y - x)
(ii) Four positive numbers are in geometric progression. The
product of the first and third is 36 and the product of
the second and fourth is 324. Find the numbers.
(10 marks)
IV. (i) If S denotes the sum 1+2+3+....+n, and
T denotes the sum 1+2+3+....+(n-1),
find in terms of n the value of S²-T². Give
your answer in its lowest terms. (10 marks)
(ii) What number must be added to each of the numbers 3, 6, 10½
to form the first three terms of a geometric progression?
Find the sum of the first six terms of this progression.
(10 marks)
V.
(cont'd.p.2)
- p.2 -
Algebra. 4th Year 16/6/1964
(cont'd.)
⟦line⟧
V. A ball is thrown vertically upwards into the air from a point
which is 20 feet above the sea. After x seconds the height y
feet of the ball above the point from which it is thrown is
given by y = 16x (4-x).
Draw a graph between x=0 and x=5 showing the relationship
between y and x. (Take 1 in. = 1 second and 20 feet respectively).
(5 marks)
From the graph, find:
(a) the maximum height above the sea reached by the ball. (5 marks)
(b) how long the ball remains at least 30 feet above (5 marks)
the sea,
(c) how many seconds elapse from the time the ball was (5 marks)
thrown to the time the ball strikes the sea.
⟦line⟧
Exam, June 1964
4th Year Secondary
Let x m.p.h. = average speed of cyclist
∴ (x + 18) m.p.h. = " " " Motorist
∴ <del>44</del> / x = 44 / (x + 18) + 1 36/60 or 44 / x = 44 / (x + 18) + 1 3/5 or 44 / x = 44 / (x + 18) + 8/5
∴ 5 × 44(x + 18) = 5 × 44x + 8x(x + 18) or 220x + 3960 = 220x + 8x² + 144x
or 8x² + 144x - 3960 = 0 or x² + 18x - 495 = 0 or
(x + 33)(x - 15) = 0 ∴ x = -33 (to be discarded) + x = 15 m.p.h. Ans.
= av. sp. of cyclist
∴ x + 18 = 15 + 18 = 33 m.p.h. Ans.
= av. sp. of motorist
II. (i) x/4 + y/4 = 4 = x/4 + y/2 ∴ x/2 + y/4 = 4 .... ①
x/4 + y/2 = 4 .... ②
∴ 2x + y = 16 .... ①a ∴ { 4x + 2y = 32 .... ③
x + 2y = 16 .... ②a x + 2y = 16 .... ②
∴ 3x = 16 ∴ x = 16/3 from ①a, y = 16 - 2x = 16 - 32/3 = 16/3
∴ x = 16/3 } Ans.
y = 16/3
(ii) In order that the expression 3x² - 4x + 1 be expressed in the form
a(x - 1)² + b(x - 1), the two forms should be identical. Hence the
equation: 3x² - 4x + 1 = a(x - 1)² + b(x - 1) is an identity which is true for all
values of x, Now let x = 0, then 1 = a(-1)² + b(-1) or
a - b = 1 .... ① } Also let x = 2, then 5 = a + b or
a + b = 5 .... ②
∴ 2a = 6 ∴ a = 3 and b = 2 ∴ { a = 3 } Ans.
b = 2
III. (i) (a) Let the No. of terms be n. Now the 1st term = x and the
common difference d = y - x ∴ the <del>⟦illegible⟧</del> n th term z = x + (n - 1)(y - x)
∴ (n - 1)(y - x) = z - x ∴ n - 1 = (z - x) / (y - x) ∴ n = (z - x) / (y - x) + 1
or n = (z + y - 2x) / (y - x) Ans.
page 1
⟦page 2⟧
⟦IV⟧ (i) b) Sum of the Progression S_n = n/2 (a+l) = (z+y-2x)/2(y-x) [x+z]
or S_n = (x+z)(y+z-2x)/2(y-x) Ans.
(ii) Let the four terms be a, ar, ar², ar³ . Then :
(a) (ar²) = 36 or a² r² = 36 ⟦line⟧ (1) } dividing (2) by (1),
and (ar)(ar³) = 324 or a² r⁴ = 324 ⟦line⟧ (2) } we get
r² = 324/36 = 9 ∴ r² = 9 ∴ r = 3 (the negative root is to be excluded)
from (1) a² r² = 36 ∴ 9 a² = 36 ∴ a² = 4 ∴ a = 2 . Therefore the
four positive numbers are : 2 , 6 , 18 , 54 Ans.
V (i) S = 1 + 2 + 3 + ... + n ∴ S = n/2 (1+n) = n(n+1)/2
T = 1 + 2 + 3 + ... + (n-1) ∴ T = (n-1)/2 [1+(n-1)] = n(n-1)/2
but S² - T² = (S+T)(S-T) = [n(n+1)/2 + n(n-1)/2] [n(n+1)/2 - n(n-1)/2]
∴ S² - T² = (n²+n+n²-n)/2 x (n²+n-n²+n)/2 = n² * n = n³ Ans.
(ii) 3 , 6 , 10 1/2 Let x be the number to be added to each of
the three numbers to make a G. P. out of them . Then the three terms
are (3+x) , (6+x) , (21/2 + x) . Hence r = (21/2 + x)/(6+x) = (6+x)/(3+x)
multiplying across, we get : (6+x)² = (3+x)(21/2 + x) or
36 + 12x + x² = (3+x)(21+2x)/2 or 36 + 12x + x² = (63 + 27x + 2x²)/2 or
72 + 24x + 2x² = 63 + 27x + 2x² or 3x = 9 ∴ x = 3 Ans.
∴ the numbers are 6 , 9 , 27/2 ...
<del>S_6 = 6/2 {2x6 + (6-1) 3/2} = 3 (12 + 15/2) = 3x39/2 = 117/2 = 58 1/2 Ans.</del>
S_6 = a(rⁿ - 1)/(r-1) = 6[(3/2)⁶ - 1]/(3/2 - 1) = 6[729/64 - 1]/(1/2) = 12 [665/64] = 1995/16 =
⟦124⟧ 11/16
x | y
0 | 0
1 | 48
2 | 64
3 | 48
4 | 0
5 | -80
y = 10
x
4.3 sec.
5
-20
sea level
⟦illegible⟧ height above the sea of the ball = 64 + 20 = 84 ft. Ans.
b) The ball remains at least 30 ft above the sea when it
is always above the line y = 10.
Therefore, the ball remains in this height between
t = 0.1 seconds + t = 3.9 seconds or
time = 3.9 - 0.1 = 3.8 seconds approximately Ans.
c) The ball strikes the sea after 4.3 seconds Approximate Ans.
⟦illegible⟧ [ 8 + x ] ⟦illegible⟧ = ( ⟦illegible⟧ + 8 ) ⟦illegible⟧ = A
⟦illegible⟧
(ii) Let the four nos. be ⟦illegible⟧. Then
⟦illegible⟧ = 6 ⟦illegible⟧ (1)
and (⟦illegible⟧) (⟦illegible⟧) = ⟦illegible⟧ ⟦illegible⟧ (2)
⟦illegible⟧
From (1) ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
⟦illegible⟧
Thus ⟦illegible⟧
⟦line⟧
VII (i) ⟦illegible⟧ = (x+1) ⟦illegible⟧ = ⟦illegible⟧
⟦illegible⟧ = [ (1-x) + 1 ] ⟦illegible⟧ = ⟦illegible⟧
⟦illegible⟧
[ ⟦illegible⟧ ] [ ⟦illegible⟧ ] = (T - t) (T + t) = T² - t²
⟦illegible⟧ = T² - t²
⟦line⟧
(ii) ⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧
Let x be the number to be added to each of
the three numbers ⟦illegible⟧. Then the three
nos. (x + ⟦illegible⟧), (x + 6), (x + ⟦illegible⟧) are
in ⟦illegible⟧. Therefore (x + 6)² = (x + ⟦illegible⟧) (x + ⟦illegible⟧)
Simplifying ⟦illegible⟧
x² + 12x + 36 = x² + ⟦illegible⟧ x + ⟦illegible⟧
12x + 36 = ⟦illegible⟧ x + ⟦illegible⟧
3x = 9 or x = 3
The numbers are ⟦illegible⟧, 9, ⟦illegible⟧
⟦line⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧ simultaneously, we have:
⟦illegible⟧ = 3x(x-1) ∴ 4x-2 = 3x²-3x
∴ 3x²-7x+2 = 0 or (3x-1)(x-2) = 0 ∴ x=1/3 or x=2
the points of intersection A(2,2) + B(1/3, 1/3) the two
functions 2(2x-1)/3(x-1) and x are equal ∴ x=1/3 } Ans.
x=2 }
x | y
-2 | 1 1/9
-1 | 1 1/2
0 | 2/3
1/2 | 0
2/3 | - 2/3
1 1/3 | 3 1/3
2 | 2
4 | 2/3 | 1 1/2
y
4
3
x=1
y=x
2
A(2,2)
1
y₂ = 2(2x-1)/3(x-1)
B(1/3, 1/3)
-2
-1
0
1
2
3
4
x
-1
x=1 is a vertical asymptote
-2
Algebra Paper, Final Examination, June 3rd 1964
⟦line⟧
x³ + y³ = 9 ...... ① multiplying equation ② by 3 + adding to equation ①,
x²y + xy² = 6 ...... ② we get: 3x²y + 3xy² = 18 ...... ②
x³ + y³ = 9 ...... ①
∴ x³ + 3x²y + 3xy² + y³ = 27 ∴ (x + y)³ = 27 ∴ x + y = 3 ...... ③
from ② xy(x + y) = 6 ∴ 3xy = 6 ∴ xy = 2 ...... ④. From ③,
y = 3 - x, substituting in ④, we get x(3 - x) = 2 ∴ 3x - x² = 2
∴ x² - 3x + 2 = 0 or (x - 2)(x - 1) = 0 ∴ x = 2 and x = 1
y = 1 Ans. 1 y = 2 Ans. 2
(b) x⁻⁴ + 4 = 5x⁻² or x⁻⁴ - 5x⁻² + 4 = 0 or (x⁻¹)² - 5(x⁻¹)² + 4 = 0
∴ [(x⁻¹)² - 4] [(x⁻¹)² - 1] = 0 ∴ (x⁻¹)² = 4 or (x⁻¹)² = 1
∴ x⁻¹ = ± 2 or 1/x = ± 2 ∴ x = ± 1/2 also (x⁻¹)² = 1 ∴ x⁻¹ = ± 1
∴ <del>⟦illegible⟧</del> 1/x = ± 1 ∴ x = ± 1 ∴ x = +1, -1, +1/2, -1/2 Ans.
3 (a)(i) x³ + y³ + 5x²y + 5xy² = (x + y)(x² - xy + y²) + 5xy(x + y) = (x + y) [x² - xy + y² + 5xy]
= (x + y)(x² + 4xy + y²) Ans.
(ii) (x + y + z)² + x² - y² - z² = x² + y² + z² + 2xy + 2xz + 2yz + x² - y² - z²
= 2x² + 2xy + 2xz + 2yz = 2 [x² + xy + xz + yz]
= 2 [x(x + y) + z(x + y)] = 2(x + y)(x + z) Ans.
(iii) (a² - b²)(x² - y²) + 4abxy = a²x² - a²y² - b²x² + b²y² + 4abxy
= a²x² + 2abxy + b²y² - (b²x² - 2abxy + a²y²)
= (ax + by)² - (bx - ay)² = (ax + by + bx - ay)(ax + by - bx + ay)
= [(a + b)x + (b - a)y] [(a - b)x + (b + a)y] Ans.
⟦illegible⟧ graphs
2
3 (b) x + 1/x = √3 Prove that x³ + 1/x³ = 0
x³ + 1/x³ = (x + 1/x)(x² - 1 + 1/x²) = (x + 1/x)(x² + 2 + 1/x² - 3) = (x + 1/x)[(x + 1/x)² - 3]
= √3 [(√3)² - 3] = √3 (3 - 3) = zero Q.E.D.
4 (a) Let the length of the course be = x yards
∴ No. of seconds taken by the first boat = x/4 seconds = t₁
also " " " " " 2nd " = x/3½ + x/4½ = x/7/2 + x/9/2 = 16x/63 seconds = t₂
∴ t₁ = x/4 = 63x/4x63 seconds } ∴ the first boat wins the race by
∴ t₂ = 16x/63 = 64x/4x63 " } 64x - 63x / 4 x 63 = x / 4x63 seconds Ans.
(b) Let the distance of riding be = x miles
x/b hours = time taken in riding } ∴ x/b + x/c = a
x/c hours = time " " walking } ∴ cx + bx / bc = a
or x(b + c) = a.b.c ∴ x = a.b.c / b + c mile Ans.
5 (i) It is required to find the sum of 2 + 4 + 6 + ... to (n+1) terms
∴ Sum: S = n/2 [2a + (n-1)d] or S = n+1/2 [2(2) + (n+1-1)(2)] = <del>⟦illegible⟧</del>
S = n+1/2 {4 + 2n} = (n+1)(n+2) = n² + 3n + 2 Ans.
(ii) S' = 22 + 33 + 44 + ... to ten terms
S' = 10/2 {2 x 22 + (10-1) x 11} = 5(44 + 99) = 5 x 143 = 715 Ans.
(iii) The distance fallen in the different seconds are: 16, 48, 80, ...
l₁₂ = a + (n-1)d = 16 + 11 x 32 = 16 + 352 = 368 ft. Ans. 1
S = n/2 {a + l} = 12/2 (16 + 368) = 6 x 384 = 2304 ft. Ans. 2
(iv) Total distance covered = 10 + 10 + 5 + 2½ + 1¼ + ... to infinity
<del>⟦illegible⟧</del> = 10 + (10 / 1 - ½) = 10 + 20 = 30 Ans.
3
6. (a) X = ⁵√[ (1.005)³ × (0.0004007)⁵ / (0.06109)² × (10.71)⅓ ]
log 1.005 = 0.0021 / 3 log 1.005 = 0.0063 | 2 log 0.06109 = 3̅.5718
log 0.0004007 = 4̅.6029 / 5 log 0.0004007 = 1̅7̅.0145 | ⅓ log 10.71 = 0.3433
log 0.06109 = 2̅.7859 / | +
log 10.71 = 1.0298 / log Num. = 1̅7̅.0208 | log Den. = 3̅.9151
log Den. = 3̅.9151 | ③
5 log x = 1̅5̅.1057
log x = 3̅.02114 ②
x = 1.050 X 10⁻³ = 0.001050
Ans.
(b) log₁₀ 2 = 0.30103 , log₁₀ 3 = 0.47712 , log₃ 648 = ?
log₃ 648 = log₃ 2³ × 3⁴ = log₃ 2³ + log₃ 3⁴ = 3 log₃ 2 + 4 log₃ 3 =
= 4 + 3 ( log₁₀ 2 / log₁₀ 3 ) = 4 + 3 × 0.30103 / 0.47712 = 4 + .90309 / .47712
= 4 + 1.892794 = 5.892794 = 5.89279 Correct to five decimal places
Ans.
② ⑥
3⁴ = (648)
3 log 2 : 3 × 0.30103 = .90309
4 log 3 : 4 × 0.47712 = 1.90848
2.81157
log₃ 648 = log₁₀ 648 / log₁₀ 3 = 2.81157 / 0.47712 = 5.892794 = 5.89279 Correct to 5 dec. pl.
Ans
Shamash Secondary School
Final Exams. June, 1964.
Subject: Algebra
Date: 3/6/1964
Class: 4th Secondary (Scientific Section)
Time: 8:00-10:45 a.m.
⟦line⟧
1. (a) Draw the graph of Y = 2(2X-1) / 3(X-1) for values of X from X=-2 to X =4,
choosing one inch to represent one unit on the X-axis and 9 tenths
of an inch to represent one unit on the Y-axis.
(8 marks)
(b) Plot another graph on the same axes to find a value of X for
which 2(2X-1) / 3(X-1) = X and verify the result by solving this equation
algebraically.
( 8 marks)
2. (a) Solve simultaneously:
X³ + Y³ = 9
X²Y + XY² = 6
( 8 marks)
(b) Find all values of X which satisfy the equation:
X⁴ + 4 = 5X²
( 8 marks)
3. (a) Find the factors of:
i- X³ + Y³ + 5X²Y + 5XY² ( 4 marks)
ii- (X+Y+Z)² + X² - Y² - Z² ( 4 marks)
iii- (a² - b²)(x² - y²) + 4abxy ( 4 marks)
(b) If X + 1/X = √3, prove that X³ + 1/X³ = 0 ( 8 marks)
4. (a) One boat in a race was rowed over the course at an average pace
of 4 yards a second; the other moved over the first half of the
course at the rate of 3½ yards a second, and over the last half
at the rate of 4½ yards a second. Which of them won and by how
many seconds ?
( 8 marks)
(b) A person has 'a' hours free. How far can he ride at 'b' miles an
hour so that walking back at 'c' miles an hour he may reach home
in time ?
( 8 marks)
(cont'd.p.2)
- p. 2-
Algebra. 4th Second. 3/6/1964
(cont'd.)
⟦line⟧
5. (i) What is the sum of the first (n+1) even numbers ?
( 4 marks)
(ii) What is the sum of the first ten numbers beginning with 22 that
are divisible by 11 ?
( 4 marks)
(iii)A body falling freely falls approximately 16 ft., in the first
second, and in each succeeding second 32 ft. more than in the
second immediately preceding. If a stone dropped from a
stationary balloon reaches the ground in 12 seconds, how far
does it fall in the last second ? How high is the balloon ?
( 4 marks)
(iv) If it were possible for a rubber ball to fall 10 ft., and bound
back 5 ft., then to fall 5 ft., and bound back 2½ ft., and to
continue this forever, what is the limit of the total distance
through which the ball would pass ?
( 4 marks)
6. (a) Compute by logarithms the value of the following, arranging your
work neatly:
⟦line⟧
5 / (1.005)³ x (0.0004007)⁵
/ ⟦line⟧
V (0.06109)² x (10.71)⟦⅕⟧
( 8 marks)
(b) Find, correct to five decimal places, the value of log 648,
3
having given:
log 2 = 0.30103 and log 3 = 0.47712
10 10
( 8 marks)
⟦line⟧
Shamash Secondary School
3rd Quarter Examination, March 1964
Subject: Algebra
Date: 26/3/1964
Class: 4th Year Secondary
Time: 10:15-11.45
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Attempt all questions:
1. An education Committee spent £P in one year awarding n scholarships
at Secondary schools. The school fees of the scholars, amounting
to £F, were paid in each case; and in addition some of the scholars
received a grant of £a and the remainder a grant of £b. How many
received the grant of £a ?
(20 marks)
2. Solve each of the following equations:-
(i) (X²+2)² + 198 = 29(X²+2) (10 marks)
(ii) X³+7X²+ 7X-15 = 0 (10 marks)
3. Solve the two simultaneous equations:-
X²+4Y²+80 = 15X+30Y (20 marks)
XY = 6
4. At what time between ten and eleven O'clock are the hands of a
watch at right angles for the second time.
(20 marks)
5. Find the value of X⁴-47X²Y²+Y⁴ in terms of p and q
when X+Y = p and X-Y = q.
(20 marks)
⟦line⟧
Shamash Secondary School
Conditional Examination, Sept. 63
Subject: Algebra
Date: 12/9/1963
Class: 4th Year Secondary
Time: 8.30-11.00 a.m.
⟦line⟧
Attempt all questions:
1. (i) The equation 7X + 2 / X² - 4 = A / X-2 + B / X+2 is true for all values of X.
Find the values of A & B. (8 marks)
(ii) The expression X³ + pX² + qX + 6 is factorable into (X-1)
and (X+2). Find the values of p and q and find the third
factor. (8 marks)
2. Solve for X the following equations, rejecting all extraneous
roots:
(i) 1 / 1 - X + 1 / √X +1 + 1 / √X -1 = 0 (6 marks)
(ii) (X-7)^(1/2) = √X - 7 (6 marks)
(iii) 3X^(-2/3) - 10X^(-1/3) + 3 = 0 (6 marks)
3. (i) Solve for X, using tables if necessary:
20^x = 2^(x+2)
(ii) Compute by logarithms:
⁵√((0.0012)² X (1.003)³) / (7515 X 2.004) (8 marks)
4. A man travels 108 miles, and finds that he could have made the
journey in 4 1/2 hours less, had he travelled 2 miles an hour
faster. At what rate did he travel ? (16 marks)
5. (i) Find by series the value of the recurring fraction 0.3205 (8 marks)
(ii) The three digits of a number are in arithmetical progression.
The number itself divided by the sum of the digits is 48.
The number formed by the same digits in reverse order is
396 less than the original number. What is the number ?
(8 marks)
- p.2 -
Shamash Secondary School
Cond.Exam. September, 1963.
Algebra 12/9/63
4th Year Secondary.
⟦line⟧
6. Draw the graph of Y = ½X³ for values of X between -3 and 4
taking ½ inch to represent one unit on the X-axis and two
tenths of an inch to represent one unit on the Y-axis.
By drawing other graphs on the same figure, solve the
equations:
(i) ½X³ - ⁷/₂X -3 = 0 (6 marks)
(ii) ½X³ + ³/₂X-2 = 0 (6 marks)
(iii) From the graph find the range of values of X for
which ½X³ is greater than ⁷/₂X + 3.
(6 marks).
⟦line⟧
Shamash Secondary School
Conditional Examination, Sept. 63
Subject: Algebra
Date: 12/9/1963
Class: 4th Year Secondary
Time: 8.30-11.00 a.m.
Attempt all questions:
1. (i) The equation 7X + 2 / X² - 4 = A / X-2 + B / X+2 is true for all values of X.
Find the values of A & B. (8 marks)
(ii) The expression X³ + pX² + qX + 6 is factorable into (X-1)
and (X+2). Find the values of p and q and find the third
factor.
(8 marks)
2. Solve for X the following equations, rejecting all extraneous
roots:
(i) 1 / 1 - X + 1 / √X + 1 + 1 / √X - 1 = 0 (6 marks)
(ii) (X-7)⅓ = √X - 7 (6 marks)
(iii) 3X⁻⅔ - 10X⁻⅓ + 3 = 0 (6 marks)
3. (i) Solve for X, using tables if necessary:
20ˣ = 2ˣ⁺²
(ii) Compute by logarithms:
⁵√ ( (0.0012)² X (1.003)³ ) / ( 7515 X 2.004 ) (8 marks)
4. A man travels 108 miles, and finds that he could have made the
journey in 4½ hours less, had he travelled 2 miles an hour
faster. At what rate did he travel ?
(16 marks)
5. (i) Find by series the value of the recurring fraction 0.3205
(8 marks)
(ii) The three digits of a number are in arithmetical progression.
The number itself divided by the sum of the digits is 48.
The number formed by the same digits in reverse order is
396 less than the original number. What is the number ?
(8 marks)
- p.2 -
Shamash Secondary School
Cond. Exam. September, 1963.
Algebra 12/9/63
4th Year Secondary.
⟦line⟧
6. Draw the graph of Y = ½X³ for values of X between -3 and 4
taking ½ inch to represent one unit on the X-axis and two
tenths of an inch to represent one unit on the Y-axis.
By drawing other graphs on the same figure, solve the
equations:
(i) ½X³ - 7/2 X - 3 = 0 (6 marks)
(ii) ½X³ + 3/2 X - 2 = 0 (6 marks)
(iii) From the graph find the range of values of X for
which ½X³ is greater than 7/2 X + 3.
(6 marks).
⟦line⟧