AI English Translation, Pages 76-100
- p.2 -
Algebra. 4th Year. 18/5/67
⟦line⟧
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 ⟦line⟧ (1)
x³-5x-2=0 ⟦line⟧ (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).
⟦line⟧
Solutions to 4th Quarter Exam. May 1967.
4th year scientific
log 0.4007 = 1.6029 | 3 log 0.4007 = 2.8087
log tan 37° 19' = 1.8821 | 2 log tan 37° 19' = 1.7642
log 50.72 = 1.7052 | log Num. = 2.5729
log cos 14° 34' = 1.9858 | log Den. = 8.4834
⟦line⟧
5 log 50.72 = 8.5260 | 9 log x = 10.0895
3 log cos 14° 34' = 1.9574 | log x = 2.8988
log Den. = 8.4834 | x = 0.07921
| = 7.921 x 10^-2 Ans.
x = 9√((0.4007)^3 * tan^2 37° 19' / (50.72^5 * cos^3 14° 34'))
2. (i) simplify:
(x^(1 + p/q))^(p/(p+q)) ÷ p√(x^(2p) / (x^-1)^-p) = x^(p/(p+q) + q/(p+q)) ÷ p√(x^(2p) * x^-p)
= x^((p+q)/(p+q)) ÷ x^(p/p) = x ÷ x = 1 Ans.
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.
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.
Shamash Secondary School
4th Quarter Exam. May 5th, 1967
Subject: Mathematics
Date: 7/5/1967
Class: 4th Year Secondary
Time: 8:00 - 9:30
⟦line⟧
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)
Shamash Secondary School
4th Quarter Exam. May 5th, 1967
Subject: Mathematics
Date: 7/5/1967
Class: 4th Year Secondary
Time: 8:00 - 9:30
⟦line⟧
1. Compute by logarithm the expression:
9 / (0.4007)³ x tan² 37° 19'
/ ⟦line⟧
\/ 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) { ⟦line⟧ - ⟦line⟧ } ÷ { ⟦line⟧ - ⟦line⟧ }
{ 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)
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
⟦line⟧ | x²/³ + 4 x¹/³ + 4 x⁰ | ⟦line⟧
-2 x¹/³ - 4 - 16 x⁻²/³ - 32 x⁻¹
-2 x¹/³ - 8 - 8 x⁻¹/³
⟦line⟧
4 + 8 x⁻¹/³ - 16 x⁻²/³ - 32 x⁻¹
4 + 16 x⁻¹/³ + 16 x⁻²/³
⟦line⟧
-8 x⁻¹/³ - 32 x⁻²/³ - 32 x⁻¹
-8 x⁻¹/³ - 32 x⁻²/³ - 32 x⁻¹
⟦line⟧
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.
Shamash Secondary School
3rd Quarter Examination, March 1967.
Subject:: Algebra
Date:: 20/3/1967
Class:: 4th Secondary
Time:: 8:30 - 10:00 a.m.
⟦line⟧
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)
⟦line⟧
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.
⟦line⟧
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)...
-p.2-
Algebra. 4th Secondary. 6/2/1967
⟦line⟧
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)
⟦line⟧
x | y₁ | y₂
- 2 | - 11 | 9
- 1 | | - 7
0 | - 3 | - 15
½ | | - 16
1 | | - 15
2 | 5 | - 7
3 | | 9
4 | | 33
Solution to question :-
Mid-Year Exam.
21/12/1947
⟦graph showing a parabola and a straight line intersecting at points C and D⟧
y₁ = 4x - 3
y₂ = 4x² - 4x - 15
x | y₁ | y₂
-3 | | 33
-2 | -11 | 9
-1 | | -7
0 | -3 | -15
½ | | -16
1 | | -15
2 | 5 | -7
3 | | 9
4 | | 33
(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 <del>⟦illegible⟧</del> 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
Solution to Final Exam. Questions 1 to 4 4th year in Algebra
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
⟦line⟧
a + 1
= (a² + b² + c²)(a + 1) + (2ab - 2ac)(a + 1) - 2bc(a + 1)
⟦line⟧
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.
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 }
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 ⟦line⟧ 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 ⟦line⟧ 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. }
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.
⟦line⟧
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)...
-p.2-
Algebra. 4th Secondary, 6/2/1967
⟦line⟧
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)
⟦line⟧
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)
⟦line⟧
Solutions to 2nd Quarter Exam in Algebra Cont.
27/12/1966 page 1.
3 (a). ⟦line⟧5⟦line⟧ = ⟦line⟧5⟦line⟧ = ⟦line⟧5⟦line⟧
6 - ⟦line⟧5⟦line⟧ 6 - ⟦line⟧5⟦line⟧ 6 - ⟦line⟧5(6-x)⟦line⟧
6 - ⟦line⟧5⟦line⟧ 36-6x-5 31-6x
6-x 6-x
∴ ⟦line⟧5⟦line⟧ = ⟦line⟧5 (31-6x)⟦line⟧ = ⟦line⟧155 - 30x⟦line⟧
186 - 36x - 30 + 5x 156 - 31x 156 - 31x
⟦line⟧
31 - 6x
∴ ⟦line⟧155 - 30x⟦line⟧ = 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.
(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.
(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
⟦line⟧ -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
⟦line⟧ ⟦line⟧ ⟦line⟧
∴ 2x = 12 2x = 4 2x = 8
x = 6 } Ans.1 x = 2 } Ans.2 x = 4 } Ans.3
y = 2 } y = 6 } y = 4 }
(10 marks)
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.
(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)
(20 marks)
Shamash Secondary School
2nd Quarter Examination, December, 1966
Subject: Algebra Date: 27/12/1966
Class: 4th Scientific Year Time: 11:00 - 12:30 morning.
⟦line⟧
All questions are to be attempted.
1. (a) Given that x = 3ay - 5bz ; make a, b respectively the subject of
⟦line⟧
3ay + 5bz
the formula. (8 marks)
(b) If a/b = k, express 4a - 5b in terms of k.
⟦line⟧
√ 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
⟦line⟧
2
(8 marks)
(b) Solve the equation 1-1.4x = 0.7(x-1)
⟦line⟧ ⟦line⟧
0.2+x 0.1-0.5x (8 marks)
3. (a) Solve 5 = x (8 marks)
⟦line⟧
6 - 5
⟦line⟧
6 - 5
⟦line⟧
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)
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)
5. (i) Divide 3x³ - 2x² + Bx - 26 by x - 2 .
Hence find the value of 'B' that makes the expression
3x³ - 2x² + Bx - 26 factorable into (x - 2), and find
the other factor. (9 marks)
(ii) A man can swim at x m.p.h. in still water. His <del>speed</del> rate
increases y m.p.h. when he swims with the current, and decreases
y m.p.h. when swims against the current. The difference
in his time to swim 2 miles with the current and 2 miles
against the current is z hours. Find a formula for z in
terms of x and y. (7 marks)
6. Give the English equivalent to the following:
1. Even numbers and odd numbers
2. The square root of a fractional number
3. Reciprocal of the number
4. Literal coefficient
5. The base of the power and the exponent of the power
6. Absolute error and relative error
7. Moving terms from one side to the other side of the equation
8. Find the value 4.419 correct to the nearest three significant figures
9. Perimeter of the polygon
10. Chord of an arc in a circle
(20 marks)
⟦illegible⟧ ⟦illegible⟧ ⟦illegible⟧ in Algebra
July 8th 1966
The Scientific Year
⟦illegible⟧ (xy + 8z) / (2x - yz) = 5/4 ∴ 16x + 16z = 50x - 25y
x(16y + 16z - 50) x = -25y / (16y + 16z - 50) Ans. I (2 marks)
y(16x + 25) = 50x - 16z x y = (50x - 16z) / (16x + 25) Ans. II (2 marks)
z(16x + 5) = 50x - 16xy ∴ z = (50x - 16xy) / (16x + 25) Ans. III (2 marks)
(ii) Simplify by removing brackets
35 [ (3x - 4y) / 5 - 1/10 { 2x - 5/7 (7x - 4y) } ] + 8(y - x) =
= 35 [ (3x - 4y) / 5 - 1/10 { 2x - 5x + 20/7 y } ] + 8y - 16x
= 7(3x - 4y) - 7/2 (3x - 5x + 20/7 y) + 8y - 16x
= 21x - 28y - 21/2 x + 35/2 x - 10y + 8y - 16x =
= x(21 - 21/2 + 35/2 - 16) + y(-28 - 10 + 8) =
= 12x - 30y Ans. (8 marks)
2 (i) ① a^7 - b^7 = (a - b)(a^6 + a^5b + a^4b^2 + a^3b^3 + a^2b^4 + ab^5 + b^6) Ans. I 3
② a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4) Ans. II
③ a^6 - b^6 = (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5) Ans. III
④ a^4 + b^4 = not factorable
(ii) (2a)^5 - (3b)^5 / (2a - 3b) = (2a)^4 + (2a)^3(3b) + (2a)^2(3b)^2 + (2a)(3b)^3 + (3b)^4
= 16a^4 + 24a^3b + 36a^2b^2 + 54ab^3 + 81b^4
3 (i) 2.4 = 0.24 / 0.6 - (0.16x - 7.6) / 0.8 ∴ 2.4 = 24/60 - (16x - 760) / 80
2.4 = 0.4 - (x/5 - 7.6) / 0.8 ∴ 2.0 = - (x/5 - 7.6) / 0.8
1.6 = - (x/5 - 7.6) ∴ 8 = -x + 38 ∴ x = 30
⟦illegible⟧ 48 = -5x + 228 ∴ 5x = 180
x = 36 Ans.
at 5 1/2 m/h 3 1/2 m/h
A ⟦line⟧ B ⟦line⟧ C
(2 1/2) (5 1/2)
AB = 5/2 x 7/2 = 35/4 miles
Let t hours = ⟦time taken A to⟧ reach C
∴ 5 1/2 t - 3 1/2 t = 35/4 ∴ t(5 1/2 - 3 1/2) = 35/4 ∴ 2t = 35/4 ∴ t = 35/8 hrs.
∴ t = 4 3/8 hours or 4 hours 22.5 minutes. Ans. (8 marks)
4. (i) boys men acres days 1 man = u boys
{ b y d
{ m a ?
b boys = b/u men (9 marks)
men acres days
{ b/u y d } ∴ No. of days = d.a.u / y.m
{ m a ?
= adb / myu days Ans.
(ii) √16x⁴ + 8x² + 16/3 x²y + 4/3 y + 4/9 y² + 1 | 4x² + 1 + 2/3 y Ans.
16x⁴
8x² + 1 | 8x² + 16/3 x²y + 4/3 y + 4/9 y² + 1
| 8x² + 1
8x² + 2 + 2/3 y | 16/3 x²y + 4/3 y + 4/9 y²
+ 2/3 y | 16/3 x²y + 4/3 y + 4/9 y² (8 marks)
5 (i) x - 2 | 3x³ - 2x² + Bx - 26 | 3x² + 4x + 8 + B
3x³ - 6x²
4x² + Bx - 26
4x² - 8x
(B + 8)x - 26
(B + 8)x - 16 - 2B
2B - 10
Hence 2B - 10 = 0 ∴ B = 5
Hence the other factor is the quotient 3x² + 4x + 8 + B or
3x² + 4x + 8 + 5 or (9 marks)
3x² + 4x + 13 Ans.
(ii) 3 / (x - y) - 2 / (x + y) = z or z = 3x + 3y - 2x + 2y / (x - y)(x + y) or z = x + 5y / x² - y² Ans. (7 marks)