AI Transcription, Pages 76-100
Page 76
Solution to Algebra paper cont. Final Exam 18/5/1967
page 3
⟦4(i)⟧ is 4, 12, 20, 28, ...
In this progression a=4 , d=8
∴ S = n/2 {2x4 + (n-1)x8} or S = n/2 {8 + 8n - 8} or
S = n/2 (8n) ∴ S = 4n² or S = (2n)² Ans.
∴ Whatever the value of n is, the sum S is always a perfect square
Q.E.D.
(10 marks)
⟦7(ii)⟧ 12, x, y, 4 from the statement of the question:
12, x, y are in A.P. ∴ x-12 = y-x ..... ①
also x, y, 4 are in G.P. ∴ y/x = 4/y ..... ②
from eq. ① : y = 2x-12 ..... ③ ∴ y² = (2x-12)² = 4(x-6)²
from eq. ② : y² = 4x ..... ④ or y² = 4(x²-12x+36)
∴ 4(x²-12x+36) = 4x ∴ x²-13x+36 = 0
∴ (x-4)(x-9) = 0 ∴ x=4 or x=9
when x=4 , y = 2(x-6) = 2(4-6) = -4
when x=9 , y = 2(x-6) = 2(9-6) = 6
∴ x=4 } Ans. 1 x=9 } Ans. 2
y=-4 y=6
(10 marks)
∴ the terms are either 12, 4, -4, 4
or 12, 9, 6, 4
Page 77
4th year Final Exam. 18/5/1961 Algebra.
| x | y₁ | y₂ | y₃ |
| -2 | -8 | -4 | -8 |
| -1 | -1 | -1 | -3 |
| 0 | 0 | 2 | 2 |
| 1 | 1 | ⟦illegible⟧ | ⟦illegible⟧ |
| 2 | 8 | 8 | 12 |
| 3 | 27 | | |
(6 marks) (i) The curve y = x³ is plotted as shown below
(7 marks) (ii) From the equation x³ = 3x + 2 we get the same
roots for x as the roots obtained from solving
the two simultaneous equations:
y₁ = x³ and y₂ = 3x + 2
the graph of y₂ = 3x + 2 is the st. line shown in the diagram.
This line touches the curve y₁ = x³ at B (-1, -1) and
intersects it at C (2, 8)
∴ the roots of the original equation
are x = -1 and
x = 2 Ans.
y
30
25
20
15
10
5
-3 -2 -1 0 1 2 3 x
B(-1,-1)
E(-0.4, -0.05)
A(-2,-8)
C(2,8)
D(2.4, 14)
y₁ = x³
y₃ = 5x + 2
y₂ = 3x + 2
Again the roots of the equation x³ - 5x - 2 = 0
are the same as the roots of the two simultaneous
equations y₁ = x³ and y₃ = 5x + 2
Since x³ = 5x + 2
Now the graph of y₃ = 5x + 2 is another
st. line which intersects the curve y₁ = x³
at A (-2, -8) and at D (2.4, 14) and at
E (-0.4, -0.05) ∴ the roots of x³ - 5x - 2 = 0 are
x = -2 and
x = 2.4
x = -0.4 Ans.
(7 marks)
(iii) From the diagram,
since the st. line
y₂ = 3x + 2 lies above the
curve y₁ = x³ between x = -1 and
x = 2 and also between x = -∞ and
x = -1 ∴ 3x + 2 > x³ when x < -1
and also when -1 < x < 2 } Ans.
Page 78
Shamash Secondary School
Final Examination, May, 1967
Subject: Algebra
Date: 18/5/1967
Class: 4th Year Secondary
Time: 8:00 - 11:00 a.m.
---
Answer all questions:
1. (i) If x = a + b/t and y = b + at, find an expression for y in terms
of a, b, x.
(6 marks)
(ii) If 2^(x-y) = 8 and 3^(x-2y) = 9 find the values of x and y.
(7 marks)
(iii) If m/n = 5/4 and p/q = 3/4, find the value of (3m+5p)/(n+q)
(7 marks)
2. A manufacturer calculated that, allowing for the initial outlay on
equipment, the cost of production of n tables was £(300+8n). Write
down the cost of production of 1 table when
(a) n tables are made,
(b) (50+n) tables are made.
If the cost of production of one table decreases by £1 when the extra
50 tables are made, find the value of n and the cost of production of
a table in each case.
(20 marks)
3. (i) Compute by logarithms the following expression, arranging your
work neatly:
7√ (Cos³ 42° 17' x tan⁵ 27° 34') / (1.009³ x 90.04⁵)
(10 marks)
(ii) If N = 2^4.136 and 10^0.301 = 2, find, without the use of tables,
the logarithm of N to the base 10 correct to five significant
figures.
(10 marks)
4. (i) Prove that the sum of any number of terms of the progression
4, 12, 20, 28, ... is a perfect square.
(ii) The first three of the four terms, 12, x, y, 4, are in arithmetical
progression and the last three are in geometrical progression.
Find x and y.
(p.2)..
Page 79
- p.2 -
Algebra. 4th Year. 18/5/67
---
5. (i) Draw the graph of y = x³ from x = -3 to x = +3, using 1 inch for
1 unit on the x-axis and 1 inch for 10 units on the y-axis.
(6 marks).
(ii) By drawing two straight-line graphs on the same diagram find the
roots of each of the following equations correct to one decimal
place.
x³ = 3x+2 .......(1)
x³-5x-2=0 .......(2)
(7 marks)
(iii) Find from the resulting diagram³ the range of values of x for
which (3x+2) is greater than x³.
(7 marks).
-----
Page 80
Solution to 4th Quarter Exam. May 1967.
4th year scientific
1.
log 0.4007 = ̅1.6028 | 3 log 0.4007 = ̅2.8084
log tan 37° 19' = ̅1.8821 | 2 log tan 37° 19' = ̅1.7642
log 50.72 = 1.7052 | log Num. = ̅2.5726
log cos 14° 34' = ̅1.9858 | log Den. = 8.4834
---------------------------------- | log x = ̅10.0892
5 log 50.72 = 8.5260 | log x = ̅2.0892
3 log cos 14° 34' = ̅1.9574 | x = 0.01228
log Den. = 8.4834 | = 1.228 x 10⁻² Ans.
x = ∜ [ (0.4007)³ × tan² 37° 19' / 50.72⁵ × cos³ 14° 34' ]
2 (i) simplify:
(x¹⁺ᵖ/ᵑ)ᵖ⁺ᵑ ÷ ᵖ√[x²ᵖ / (x⁻¹)⁻ᵖ] = xᵖ/ᵖ⁺ᵑ + ᵑ/ᵖ⁺ᵑ ÷ ᵖ√x²ᵖ ÷ xᵖ
= xᵖ⁺ᵑ/ᵖ⁺ᵑ ÷ xᵖ/ᵖ = x ÷ x = 1 Ans.
Page 81
2
1 (ii) Simplify:
{ (y^½ + y^-½) / (y² - y + 1) - (y^½ - y^-½) / (y² + y + 1) } ÷ { (y^½ + 2y^-½) / (y³ - 1) - (y^½ - 2y^-½) / (y³ + 1) }
{ (y + 1) / (y^½ (y² - y + 1)) - (<del>⟦illegible⟧</del> (y - 1)) / (y^½ (y² + y + 1)) } ÷ { (y + 2) / (y^½ (y³ - 1)) - (y - 2) / (y^½ (y³ + 1)) }
((y + 1)(y² + y + 1) - (y - 1)(y² - y + 1)) / (y^½ (y² - y + 1)(y² + y + 1)) ÷ ((y + 2)(y³ + 1) - (y - 2)(y³ - 1)) / (y^½ (y³ - 1)(y³ + 1))
= (y³ + y² + y + y² + y + 1 - (y³ - y² + y - y² + y - 1)) / (y^½ (y² - y + 1)(y² + y + 1)) × (y^½ (y³ - 1)(y³ + 1)) / (y⁴ + y + 2y³ + 2 - (y⁴ - y - 2y³ + 2))
= (4 y² + 2) / (y^½ (y² - y + 1)(y² + y + 1)) × (y^½ (y³ - 1)(y³ + 1)) / (4 y³ + 2y) =
= (<del>2</del> (2y² + 1)) / (<del>y^½</del> (y² - y + 1)(<del>y² + y + 1</del>)) × (<del>y^½</del> (<del>y - 1</del>)(<del>y² + y + 1</del>) (y + 1)(y² - y + 1)) / (<del>2</del> y (2y² + 1))
= ((y - 1)(y + 1)) / y = (y² - 1) / y = y - 1/y Ans.
Page 82
3
⟦illegible⟧ 2 x 3^{2x} = 4^{x-1}
log 2 + 2x log 3 = (x-1) log 4
2x log 3 - x log 4 = - log 4 - log 2
2x log 3 - 2x log 2 = - 2 log 2 - log 2
2x (log 3 - log 2) = - 3 log 2
x = - 3 log 2 / 2 (log 3 - log 2) = - 3 x 0.3010 / 2 (0.4771 - 0.3010)
∴ x = - 3 x 0.3010 / 2 x 0.1761 = - 0.9030 / 0.3522 = - 2.5638...
∴ x = - 2.564 Correct to four significant figures.
Page 83
Shamash Secondary School
4th Quarter Exam. May 5th, 1967
Subject: Mathematics
Date: 7/5/1967
Class: 4th Year Secondary
Time: 8:00 - 9:30
-----
1. Compute by logarithm the expression:
9 / (0.4007)³ x tan² 37° 19'
/ 50.72⁵ x Cos³ 14° 34'
(30 marks)
2. Simplify: (i) (x^(1+q/p))^(p/(p+q)) ÷ p√(x^(2p) / (x⁻¹)^(-p))
(20 marks)
(ii) { (y^(1/2) + y^(-1/2)) / (y² - y + 1) - (y^(1/2) - y^(-1/2)) / (y² + y + 1) } ÷ { (y^(1/2) + 2y^(-1/2)) / (y³ - 1) - (y^(1/2) - 2y^(-1/2)) / (y³ + 1) }
(20 marks)
3. Solve the equation: 2x3^(2x) = 4^(x-1) finding the answer correct to
four significant figures.
(30 marks)
Page 84
Shamash Secondary School
4th Quarter Exam. May 5th, 1967
Subject: Mathematics
Date: 7/5/1967
Class: 4th Year Secondary
Time: 8:00 - 9:30
-----
1. Compute by logarithm the expression:
9 / (0.4007)³ x tan² 37° 19'
/ -----------------------
\/ 50.72⁵ x Cos³ 14° 34' (30 marks)
2. Simplify: (i) (x^(1+q/p))^(p/(p+q)) ÷ p\/ (x^2p / (x^-1)^-p)
(20 marks)
{ y^1/2 + y^-1/2 y^1/2 - y^-1/2 } { y^1/2 + 2y^-1/2 y^1/2 - 2y^-1/2 }
(ii) { --------------- - --------------- } ÷ { ---------------- - ---------------- }
{ y^2 - y + 1 y^2 + y + 1 } { y^3 - 1 y^3 + 1 }
(20 marks)
3. Solve the equation: 2x3^2x = 4^x-1 finding the answer correct to
four significant figures.
(30 marks)
Page 85
Solutions to 3rd Quarter Exam. in Algebra
4th Year Secondary
20/3/1967
2. (i) b / √a . ∛ac . ⁴√c³ / √b . √b⁻¹ / a⁻¹/⁶ = b / a¹/² . a¹/³ . c¹/³ . c³/⁴ / b¹/² . b⁻¹/² / a⁻¹/⁶
= a⁻¹/² + ¹/³ + ¹/⁶ . b¹⁻¹/²⁻⁻¹/² . c¹/³ + ³/⁴ = a⁻³⁺²⁺¹/₆ . b²⁻¹⁻¹/₂ . c⁴⁺⁹/₁₂
= a⁰ . b⁰ c¹³/¹² = ¹²√c¹³ = c ¹²√c Ans.
(ii) [(9ⁿ⁺¹/⁴)(√3.3ⁿ) / 3 √3ⁿ⁻²]¹/ⁿ = [(3²)(ⁿ⁺¹/⁴) (3ⁿ⁺¹/²)¹/² / 3.3ⁿ⁻²/²]¹/ⁿ = [3²ⁿ⁺¹/² . 3ⁿ⁺¹/⁴ . 3⁻ⁿ/²⁻¹]¹/ⁿ
= [3 ⁴ⁿ⁺¹⁺ⁿ⁺¹⁺ⁿ⁻² / 2]¹/ⁿ = [3⁶ⁿ/²]¹/ⁿ = 3³ⁿ/ⁿ = 3³ = 27 Ans.
3. (∛x² + 2 x¹/³ - 16 x⁻²/³ - 32/x) ÷ (x¹/⁶ + 4 x⁻¹/⁶ + 4/√x)
(x²/³ + 2 x¹/³ - 16 x⁻²/³ - 32 x⁻¹) ÷ (x¹/⁶ + 4 x⁻¹/⁶ + 4 x⁻¹/²)
x¹/⁶ + 4 x⁻¹/⁶ + 4 x⁻¹/² | x²/³ + 2 x¹/³ - 16 x⁻²/³ - 32 x⁻¹ | x¹/² - 2 x¹/⁶ + 4 x⁻¹/⁶ - 8 x⁻¹/²
____________________| x²/³ + 4 x¹/³ + 4 x⁰ |________________________
-2 x¹/³ - 4 - 16 x⁻²/³ - 32 x⁻¹
-2 x¹/³ - 8 - 8 x⁻¹/³
____________________
4 + 8 x⁻¹/³ - 16 x⁻²/³ - 32 x⁻¹
4 + 16 x⁻¹/³ + 16 x⁻²/³
____________________
-8 x⁻¹/³ - 32 x⁻²/³ - 32 x⁻¹
-8 x⁻¹/³ - 32 x⁻²/³ - 32 x⁻¹
___________________________
Page 86
y₁ = x³
y₂ = 3x² - 4
| x | y₁ = x³ | y₂ = 3x² - 4 |
| -3 | -27 | 23 |
| -2 | -8 | 8 |
| -1 | -1 | -1 |
| 0 | 0 | -4 |
| 1 | 1 | -1 |
| 2 | 8 | 8 |
| 3 | 27 | 23 |
y₁ = x³
y₂ = 3x² - 4
y₂ = 3x² - 4
30
25
20
15
10
5
-1
A(-1, -1)
B(2, 8)
5
10
15
20
25
30
(i) The two curves are plotted as they appear in the figure.
(ii) The roots of the equation x³ - 3x² + 4 = 0 are the same as the
roots of the equation x³ = 3x² - 4 . These roots are the same as
the abscissas of the points of intersections of the two curves
y₁ = x³ and y₂ = 3x² - 4 , which are x = -1 and x = 2 .
Since the two curves have points A(-1, -1) & B(2, 8) as common
points between them.
(iii) ⟦as⟧ The expression x³ - 3x² + 4 is always negative when x³ < 3x² - 4
or when y₁ < y₂ . But y₁ < y₂ for all values of x < -1
since the curve y₁ = x³ lies below the curve y₂ = 3x² - 4 .
(b) Also the expression x³ - 3x² + 4 is always positive when y₁ > y₂
and this is the case for all values of x > -1 ⟦except x=2⟧, since for all these
values of x, the curve y₁ lies above the curve y₂ . Q.E.D.
Page 87
Shamash Secondary School
3rd Quarter Examination, March 1967.
Subject:: Algebra
Date:: 20/3/1967
Class:: 4th Secondary
Time:: 8:30 - 10:00 a.m.
---
1- (i) Draw the graphs of y = x³ and y = 3x²-4 on the same axes,
for values of x from x = -3 to x = 3
(20 marks)
(ii) From your graphs find the roots of the equation x³-3x²+4=0
(15 marks).
(iii) For what values of x is the expression x³-3x²+4 always negative?
always positive ? Use your graphs to explain why.
(15 marks)
2- Simplify (i) / √a . ∛/ac . √/c³ / √b . √b⁻¹ / a⁻¹/6
(15 marks)
(ii) [ (9ⁿ⁺¼) . √((3)(3ⁿ)) / 3√3⁻ⁿ ]¹/ⁿ
(15 marks)
3- Divide (∛x² + 2x⅓ - 16x⁻⅔ - 32/x) by (xℙ + 4x⁻ℙ + 4/√x)
(20 marks)
------
Page 88
Shamash Secondary School
Mid-Year Examination, Feb.1967
Subject: Algebra
Date: 6/2/1967
Class: 4th Year,Secondary
Time: 8:30 - 11:30 a.m.
-----
Attempt all questions:
1. Revolve into factors:
(i) 3x²-(4a+2b)x+a²+2ab (6 marks)
(ii) 8x³-27y³+z³+18xyz (6 " )
(iii) Divide (a²+b²+c²)(a+1)+(2ab-2ac)(a+1)-2abc-2bc by (a+1)
and express the quotient as a perfect square. (8 marks)
2. (i) If x+y = 2a and x-y=2b, find in the shortest possible way,
the value of x⁴+x²y²+y⁴. (10 marks)
(ii) Find the value of p which will make the expression 2x³+px²-5x+2
divisible by (x+2) and find the other two factors.
(10 marks)
3. (i) A man can row upstream at 'a' miles an hour and downstream
at 'b' miles an hour. He rows up to a certain point and then
returns to his starting point, and finds that his average
speed is 's' miles an hour for the double journey. Express
each of the letters in terms of the other two. Find the value
of 'b' if a=2 and s=3.
(10 marks)
(ii) Solve the two simultaneous equations:
x²+4y²+80 = 15x+30y ........(1)
xy = 6 . ........(2)
(10 marks)
4. Two men started at the same time to meet each other from points
which were 26 miles apart. If one took 4½ minutes longer than
the other to walk a mile, and they met 2 hours after starting,
find the speed of each in miles per hour.
(20 marks)
(cont'd.p.2)...
Page 89
-p.2-
Algebra. 4th Secondary. 6/2/1967
-- --
5. (i) Draw on the same diagram the graphs of the function 4x-3,
and of the function 4x²-4x-15, taking ½ inch as one unit
on the x-axis and one tenth of an inch as one unit on the
y-axis. (8 marks)
(ii) From your diagram, find the roots of the two simultaneous
equations y₁ = 4x-3 ........(1)
y₂ = 4x²-4x-15 ....(2) (7 marks)
(iii) From the graph of the function 4x²-4x-15, find the
roots of the equation 4x²-4x-15=0.
(5 marks)
-----
| x | y₁ | y₂ |
| - 2 | - 11 | 9 |
| - 1 | | - 7 |
| 0 | - 3 | - 15 |
| ½ | | - 16 |
| 1 | | - 15 |
| 2 | 5 | - 7 |
| 3 | | 9 |
| 4 | | 33 |
Page 90
Ans to question 2
Mid Year Exam.
12/1967
y
30
25
20
15
10
D(3, 9)
5
(-1.5, 0) A
B (2.5, 0)
-3 -2 -1 1 2 3 4 x
-5
C(-1, -7)
-10
-15
-20
| x | y₁ | y₂ |
| -3 | | 33 |
| -2 | -11 | 9 |
| -1 | | -7 |
| 0 | -3 | -15 |
| 1/2 | | -16 |
| 1 | | -15 |
| 2 | 5 | -7 |
| 3 | | 9 |
| 4 | | 33 |
y₁ = 4x - 3
⟦y₂⟧ = 4x² - 4x - 15
(i) The two graphs are drawn as shown above.
(ii) The roots of the two simultaneous equations
are the coordinates of the two points of intersection
C(-1, -7) and D(3, 9). (i.e.)
x = -1 } Ans. 1 x = 3 } Ans. 2
y = -7 } y = 9 }
(iii) The roots of the equation 4x² - 4x - 15 = 0
are the abscissas of the points of intersection
of the curve with the x-axis, i.e.:
x = -1.5 Ans. 1
x = 2.5 Ans. 2
Page 91
Solution to Final Exam. Questions to 4th Year in Algebra
Cont.
February 6th 1967
⟦1. (i)⟧ 3x² - (4a+2b)x + a² + 2ab = 3x² - (4a+2b)x + a(a+2b)
= [3x - (a+2b)][x - a]
= (3x - a - 2b)(x - a) Ans.
(ii) 8x³ - 27y³ + z³ + 18xyz = (2x)³ + (-3y)³ + z³ - 3(2x)(-3y)(z)
= (2x - 3y + z)(4x² + 9y² + z² + 6xy - 2xz + 3yz)
Ans.
(iii) (a²+b²+c²)(a+1) + (2ab-2ac)(a+1) - 2abc - 2bc
------------------------------------------------
a+1
= (a²+b²+c²)(a+1) + (2ab-2ac)(a+1) - 2bc(a+1)
------------------------------------------------
a+1
= a² + b² + c² + 2ab - 2ac - 2bc = <del>⟦illegible⟧</del>
= a² + b² + (-c)² + 2ab + 2a(-c) + 2b(-c)
= (a + b - c)² Ans.
2. (i) x + y = 2a | or x = a + b | Now x⁴ + x²y² + y⁴ = x⁴ + 2x²y² + y⁴ - x²y²
x - y = 2b | y = a - b | = (x² + y²)² - (xy)²
2x = 2(a+b) | | = (x² + 2xy + y² - 2xy)² - (xy)²
2y = 2(a-b) | ∴ xy = (a+b)(a-b) | = [(x+y)² - 2xy]² - (xy)²
| or xy = a² - b² |
= [(x+y)² - 2xy + xy][(x+y)² - 2xy - xy] = [(x+y)² - xy][(x+y)² - 3xy]
= [(2a)² - (a² - b²)][(2a)² - 3(a² - b²)] = (4a² - a² + b²)(4a² - 3a² + 3b²)
= (3a² + b²)(a² + 3b²) Ans.
Page 92
Solutions to Mid-Year Exam in Algebra Cont. page 2.
6/2/1967
(2 x³ + px² - 5x + 2) ÷ (x + 2) = 2x² + (p-4)x - 2p + 3 + (4p-4)/(x+2)
∴ 4p - 4 = 0 ∴ p = 1
2x³ + px² - 5x + 2 | 2x² + (p-4)x - 2p + 3
2x³ + 4x²
(p-4)x² - 5x + 2
(p-4)x² + 2(p-4)x
(-2p + 3)x + 2
(-2p + 3)x - 4p + 6
4p - 4 = Remainder = 0
∴ p = 1 and the other two factors are (2x-1) and (x-1) Ans.
∴ Original expression =
2x³ + x² - 5x + 2 =
= (x+2) [2x² + (p-4)x - 2p + 3]
= (x+2) (2x² - 3x + 1) =
= (x+2)(2x-1)(x-1)
(i) Rate of rowing upstream = a m.p.h. } let the distance rowed each way = x miles
" " " downstream = b m.p.h. } ∴ x/a + x/b = 2x/s or
1/a + 1/b = 2/s ∴ bs + as = 2ab or 2ab - as = bs or a(2b-s) = bs
or a = bs / (2b-s) Ans. 1
also 2ba - bs = as ∴ b(2a-s) = as ∴ b = as / (2a-s) Ans. 2
also bs + as = 2ab ∴ s(a+b) = 2ab ∴ s = 2ab / (a+b) Ans. 3.
but b = as / (2a-s) = (2x3) / (2x2-3) = 6/1 = 6 Ans. 4
(ii) { x² + 4y² + 80 = 15x + 30y ... ① } from ② : 4xy = 24 ... ③
{ xy = 6 ... ② } add eq. ① + ③ and you get:
x² + 4xy + 4y² + 80 = 15x + 30y + 24 or (x+2y)² + 80 = 15(x+2y) + 24 or
(x+2y)² - 15(x+2y) + 56 = 0 ∴ [(x+2y)-7][(x+2y)-8] = 0 or
x + 2y - 7 = 0 or x = 7 - 2y ... ④ also
x + 2y - 8 = 0 or x = 8 - 2y ... ⑤
from eq. ② + ④ (7-2y)y = 6 or 7y - 2y² = 6 ∴ 2y² - 7y + 6 = 0 ∴ (2y-3)(y-2) = 0
∴ y = 2 and y = 3/2 ∴ from ④ : x = 7 - 4 = 3 and x = 7 - 3 = 4
x = 3 } Ans. x = 4 } Ans. || again from ② + ⑤ : (8-2y)y = 6 ∴ 8y - 2y² = 6
y = 2 } y = 3/2 } || ∴ y² - 4y + 3 = 0 ∴ (y-1)(y-3) = 0 ∴ y = 1, y = 3
from ⑤ : x = 8 - 2 = 6 and x = 8 - 6 = 2 ∴ x = 6 } Ans. x = 2 } Ans.
y = 1 } y = 3 }
x = 3 } Ans. 1 x = 4 } Ans. 2 x = 6 } Ans. 3 x = 2 } Ans. 4
y = 2 } y = 3/2 } y = 1 } y = 3 }
Page 93
Solution to Final Exam. Questions in Algebra to 4th year Comt. page 3.
February 6th 1967.
← 26 miles →
A * * B
→ ←
Let the speed of A be x m.p.h.
" " " B " y m.p.h.
∴ 2x + 2y = 26 or x + y = 13
or y = 13 - x ---- ①
also A walks one mile in 1/x hrs or in 60/x minutes
B " " " " 1/y hrs. " " 60/y minutes
∴ 60/x - 60/y = 4 1/2 or 20/x - 20/y = 3/2 or 40y - 40x = 3xy or
40(y - x) = 3xy ---- ②
Substitute from ① in ② : 40(13 - 2x) = 3x(13 - x) or
520 - 80x = 39x - 3x² or 3x² - 119x + 520 = 0 or
(3x - 104)(x - 5) = 0 or x = 5 and x = 104/3 = 34 2/3
∴ y = 13 - x or y = 13 - 5 = 8 or y = 13 - 34 2/3 = -21 2/3 inadmissible
∴ x = 5 m.p.h. } Ans.
y = 8 m.p.h. }
Page 94
Shamash Secondary School
Mid-Year Examination, Feb. 1967
Subject: Algebra Date: 6/2/1967
Class: 4th Year, Secondary Time: 8:30 - 11:30 a.m.
-----
Attempt all questions:
1. Revolve into factors:
(i) 3x²-(4a+2b)x+a²+2ab (6 marks)
(ii) 8x³-27y³+z³+18xyz (6 " )
(iii) Divide (a²+b²+c²)(a+1)+(2ab-2ac)(a+1)-2abc-2bc by (a+1)
and express the quotient as a perfect square. (8 marks)
2. (i) If x+y = 2a and x-y=2b, find in the shortest possible way,
the value of x⁴+x²y²+y⁴ . (10 marks)
(ii) Find the value of p which will make the expression 2x³+px²-5x+2
divisible by (x+2) and find the other two factors.
(10 marks)
3. (i) A man can row upstream at 'a' miles an hour and downstream
at 'b' miles an hour. He rows up to a certain point and then
returns to his starting point, and finds that his average
speed is 's' miles an hour for the double journey. Express
each of the letters in terms of the other two. Find the value
of 'b' if a=2 and s=3.
(10 marks)
(ii) Solve the two simultaneous equations:
x²+4y²+80 = 15x+30y ........(1)
xy = 6 . .........(2) (10 marks)
4. Two men started at the same time to meet each other from points
which were 26 miles apart. If one took 4½ minutes longer than
the other to walk a mile, and they met 2 hours after starting,
find the speed of each in miles per hour.
(20 marks)
(cont'd.p.2)...
Page 95
-p.2-
Algebra. 4th Secondary, 6/2/1967
--
5. (i) Draw on the same diagram the graphs of the function 4x-3,
and of the function 4x²-4x-15, taking ½ inch as one unit
on the x-axis and one tenth of an inch as one unit on the
y-axis. (8 marks)
(ii) From your diagram, find the roots of the two simultaneous
equations y = 4x-3 ........(1)
y = 4x²-4x-15 ....(2) (7 marks)
(iii) From the graph of the function 4x²-4x-15, find the
roots of the equation 4x²-4x-15=0.
(5 marks)
------
Page 96
Solutions to 2nd Quarter Exam. algebra, 4th year. page 2.
27/12/1966
1. (a) x = 3ay - 5bz / 3ay + 5bz ∴ 3axy + 5bxz = 3ay - 5bz ∴ 3ay - 3axy = 5bxz + 5bz
∴ 3ay(1-x) = 5bz(x+1) ∴ a = 5bz / 3y . 1+x / 1-x Ans. 1
also 5bxz + 5bz = 3ay - 3axy ∴ 5bz(x+1) = 3ay(1-x) (8 marks)
∴ b = 3ay / 5z . 1-x / 1+x Ans. 2
(b) a/b = k ∴ 4a - 5b / √18a² - 4b² = 4 a/b - 5 / b/b √18a² - 4b² = 4 a/b - 5 / √18 a²/b² - 4 = 4k - 5 / √18k² - 4 Ans.
(8 marks)
2. (a) x = y / y+1 and y = a-2 / 2 . Prove that x(y+2) + x/y + 1/x = a
x = (a-2)/2 / (a-2)/2 + 1 = a-2 / a-2+2 = a-2 / a . Now substitute x = a-2 / a , y = a-2 / 2
∴ the expression x(y+2) + x/y + 1/x = a-2 / a ( a-2 / 2 + 2 ) + a-2 / a / a-2 / 2 + 1 / a-2 / a
∴ the expression = a-2 / a . a+2 / 2 + 2a-4 / a(a-2) + a²-2a / 2(a-2) = a²-4 / 2a + 2(a-2) / a(a-2) + a(a-2) / 2(a-2)
= a²-4 / 2a + 2/a + a/2 = a²-4+4+a² / 2a = 2a² / 2a = a Q.E.D.
(8 marks)
(b) 1 - 1.4x / 0.2 + x = 0.7(x-1) / 0.1 - 0.5x multiply the first fraction (both numerat. + Den.)
by 5 + the second fraction by 10. , we get
5 - 7x / 1 + 5x = 7(x-1) / 1 - 5x ∴ (5-7x)(1-5x) = 7(1+5x)(x-1)
∴ 5 - 32x + 35x² = 7(5x² - 4x - 1) or 35x² - 32x + 5 = 35x² - 28x - 7
∴ 4x = 12 ∴ x = 3 Ans. (8 marks)
________________________________________________________________________________
Page 97
Solutions to 2nd Quarter Exam in Algebra Cont.
27/12/1966 page 1.
3 (a). _________________5_________________ = _______________5_______________ = _______________5_______________
6 - _________5_________ 6 - _________5_________ 6 - ____5(6-x)____
6 - ____5____ 36-6x-5 31-6x
6-x 6-x
∴ _______________5_______________ = ____5 (31-6x)____ = ____155 - 30x____
186 - 36x - 30 + 5x 156 - 31x 156 - 31x
___________________
31 - 6x
∴ ____155 - 30x____ = x & 155 - 30x = 156x - 31x²
156 - 31x
∴ 31x² - 186x + 155 = 0 ∴ x² - 6x + 5 = 0
∴ (x-1)(x-5) = 0 ∴ x = 1 & x = 5 Ans.
[Marginalia] (8 marks)
(b) 3x³ + x² + 4 = 8x ∴ 3x³ + x² - 8x + 4 = 0
when x = 1, then the LHS = 3 + 1 - 8 + 4 = 0 ∴ (x-1) is a factor
3x³ + x² - 8x + 4 = <del>⟦illegible⟧</del> 3x²(x-1) + 4x(x-1) - 4(x-1)
3x³ + x² - 8x + 4 = (x-1)(3x² + 4x - 4) = (x-1)(3x-2)(x+2) = 0
∴ x = 1 , x = -2 , x = 2/3 Ans.
[Marginalia] (10 marks)
(c) x²y² + 192 = 28xy ----- ①
x + y = 8 ----- ②
x²y² - 28xy + 192 = 0 ∴ (xy - 16)(xy - 12) = 0
∴ xy = 16 or xy = 12.
squaring eq. ②, x² + 2xy + y² = 64 --- ③
-4xy = -48 or x² + 2xy + y² = 64
__________________ -4xy = -64
∴ x² - 2xy + y² = 16 or x² - 2xy + y² = 0
(x-y)² = 16 (x-y)² = 0
x - y = ±4 x - y = 0
Now x + y = 8 x + y = 8 x + y = 8
x - y = 4 x - y = -4 x - y = 0
_________ __________ _________
∴ 2x = 12 2x = 4 2x = 8
x = 6 } Ans.1 x = 2 } Ans.2 x = 4 } Ans.3
y = 2 } y = 6 } y = 4 }
[Marginalia] (10 marks)
Page 98
Solutions to 2nd Quarter Exam in Algebra Cont.
27/12/1966
page 3
4. y/x hrs. = No. of hrs. taken by fast train to cover the distance y miles.
Let v m.p.h. be the speed of the slower train.
∴ z/v hrs. = No of hrs taken by slower train to travel distance z
∴ z/v = y/x + t ∴ xz = vy + txv ∴ v(y + tx) = xz
∴ v = xz / (y + tx) & the difference between speeds = (x - v) m.p.h.
∴ diff. between the two speeds = x - xz / (y + tx) = (xy + tx² - xz) / (tx + y) = (tx² + x(y - z)) / (tx + y)
Ans.
[Marginalia] (20 Marks)
5. Let the time when he started be x minutes after 3
∴ x = 15 + x/12 ∴ 12x = 180 + x
∴ 11x = 180 ∴ x = 180/11 = 16 4/11 minutes
⟦Diagram of a clock face showing approx 3:16⟧ 1st case
Let the time when he finished = y minutes after five
∴ y = 25 + y/12 ∴ 12y = 300 + y
∴ 11y = 300 ∴ y = 300/11 = 27 3/11 minutes after five
⟦Diagram of a clock face showing approx 5:27⟧ 2nd case
he began at 16 4/11 min. after 3, and ended at 27 3/11 min past five.
he walked for a period of 2 hrs 10 10/11 minutes Ans.
= (5 hrs. 27 3/11 min - 3 hrs. 16 4/11 min)
[Marginalia] (20 marks)
Page 99
Shamash Secondary School
2nd Quarter Examination, December, 1966
Subject: Algebra Date: 27/12/1966
Class: 4th Scientific Year Time: 11:00 - 12:30 morning.
------
All questions are to be attempted.
1. (a) Given that x = 3ay - 5bz ; make a, b respectively the subject of
---------
3ay + 5bz
the formula. (8 marks)
(b) If a/b = k, express 4a - 5b in terms of k.
---------------
√ 18a² - 4b²
(8 marks)
2. (a) Prove that x(y+2) + x/y + y/x is equal to a, if x= y/(y+1) and y= a - 2
-----
2
(8 marks)
(b) Solve the equation 1-1.4x = 0.7(x-1)
------ --------
0.2+x 0.1-0.5x (8 marks)
3. (a) Solve 5 = x (8 marks)
-------------
6 - 5
---------
6 - 5
-----
6 - x
(b) Solve 3x³ + x² + 4 = 8x (10marks)
(c) Solve x²y² + 192 = 28xy .......(1) (10 marks)
x + y = 8 .......(2)
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 ⟦x, t,⟧ y and z.
(20 marks)
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
coincidents between five and six o'clock. What was the time when he
started, and how long did he walk ?
(20 marks)
Page 100
Shamash Secondary School
1st Quarter Examination, November, 1966.
Subject: Algebra
Date: 8/11/1966
Class: 4th Secondary, Scientific Section.
Time: 12:00 - 1:30 p.m.
all questions are to be attempted
1. (i) Given √xy + zx / √zx - yz = 3/4 solve each x, y, z in terms of the other
two (6 marks)
(ii) Simplify by removing brackets
35 [ 3x - 4y / 5 - 1/10 { 3x - 5/7 ( 7x - 4y ) } ] + 8 ( y - 2x ) ( 8 marks )
2. (i) Resolve into two factors (if possible) each of the following expressions
① a⁷ - b⁷ ② a⁶ + b⁶ ③ a⁶ - b⁶ ④ a⁹ + b⁹ (12 marks)
(ii) Write down by inspection the quotient of (2a)⁵ - (3b)⁵ / 2a - 3b (4 marks)
3. (i) Solve the equation 2.4 = 0.24 / 0.6 - 0.16x - 7.6 / 0.8 (9 marks)
(ii) Walking 5 1/2 miles an hour, I start 2 1/2 hours after a
friend whose pace is 3 1/2 miles an hour. How long shall
I be in overtaking him? (8 marks)
4. (i) How many days will "n" men take to mow "a" acres if "b"
boys can mow "y" acres in "d" days and each man's work
equals that of "u" boys? (9 marks)
(ii) Find the square root of :
16 x⁴ + 16/3 x²y + 8x² + 4/9 y² + 4/3 y + 1
showing your steps neatly. (8 marks)