AI English Translation, Pages 201-225
Shamash Secondary School
Final Examination, June 1960.
Subject: Algebra
Class: 4th Year Secondary
Date: 17/6/1960
Time: 8:00-11:00
⟦line⟧
All questions are to be attempted.
1. (a) If X = (2Y-1)/(3Y-4) , and Y = (Z + 1)/(Z - 1) , find Z in terms of X. (8 marks)
(b) Prove that, if a + b = C, and none of these quantities is zero, the
expression
1 / (a² + b² - C²) + 1 / (b² + C² - a²) + 1 / (C² + a² - b²)
is equal to zero. (8 marks)
2. (i) Find the value of b for which the expression X³ - 2 - b (X-1) is equal
to Zero when X = 2. (5 marks)
(ii) Factorize the expression for this value of b, and find the other
values of X for which the expression is Zero.
( 12 marks)
3. A train left station P at 10 a.m. on a non-stop run of 300 miles to station
Q where it was due to arrive at 4:15 p.m. At a station B some miles from Q
it was 3¾ minutes behind the Scheduled time. But by travelling from B to Q
at 60 miles per hour the train arrived at its destination on time.
How far is it from B to Q ? (17 marks)
4. (a) Compute by logarithms the following expression:
⁷√( (1.004² X 0.000491³) / (516.2 X 2.003²) ) (8 marks )
(b) Use logarithms to solve the equation 4²ˣ - 8 x 4ˣ + 12 = 0
(8 marks)
5. (a) The 21st term of an arithmetical progression is 2½ times the 8th term,
and the arithmetic mean of the 5th and 13th terms is 29. Find the sum
of the first 15 terms. (8 marks)
(b) Prove that in any Geometric series the sum of the 4th, 5th, and 6th
terms is the Geometric mean of the sum of the 1st, 2nd, and 3rd terms
and the sum of the 7th, 8th, and 9th terms. (8 marks)
6. (a) Draw the graph of Y = X² - 3X + 2 for values of X between -2 and 5.
(6 marks).
(b) Use your graph to solve the equations:
( i) X² - 3X + 2 = 0 (6 marks)
(ii) X² - 3X - 4 = 0 (6 marks)
⟦line⟧
⟦Final Examination, June 1960⟧
Subject: Algebra Date: 17/6/1960
Class: 4th year Secondary Time: 8:00 - 11:00
All questions are to be attempted.
1. (a) If x = (2y-1)/(3y-4), and y = (z+1)/(z-1), find z in terms of x.
(8 marks)
(b) Prove that, if a + b = c, and none of these quantities is zero, the
expression
1/(a² + b² - c²) + 1/(b² + c² - a²) + 1/(c² + a² - b²)
is equal to zero. (8 marks)
2. (i) Find the value of b for which the expression x³ - 2 - b(x - 1) is
to zero when x = 2. (5 marks)
(ii) Factorize the expression for this value of b, and find the
values of x for which the expression is zero. (12 marks)
3. A train left station P at 10 a.m. on a non-stop run
300 miles to station Q where it was due to arrive at 4:15 p.m.
At a station B some miles from Q it was 3 3/4 minutes behind the
scheduled time. But by travelling from B to Q at 60 miles
per hour the train arrived at its destination on time.
How far is it from B to Q? (17 marks)
4. (a) Compute by logarithms the following expression:
⁷√((1.004² × 0.000401³)/(516.2 × 2.003⁵)) (8 marks)
(b) Use logarithms to solve the equation 4²ˣ - 8 × 4ˣ + 12 = 0
⟦(8 marks)⟧
⟦(a)⟧ ⟦If⟧ ⟦the⟧ 2nd term of an arithmetic progression is 2 1/2
times the 5th term, and the arithmetic mean of the 5th and
12th terms is 27. Find the sum of the first 15 terms.
(8 marks)
(b) Prove that in any Geometric series the sum of the 4th,
5th, and 6th terms is the Geometric mean
⟦sum⟧ of the 1st, 2nd, and 3rd terms and the
the 7th, 8th, and 9th terms. (8 marks)
6. (a) Draw the graph of y = x² - 3x + 2 for values
x between -2 and 5. (6 marks)
(b) Use your graph to solve the equations:
(i) x² - 3x + 2 = 0 (6 marks)
(ii) x² - 3x - 4 = 0 (6 marks)
All questions are to be attempted
1. (a) Resolve into factors : (i) 3 (2x-5)² + 4(2x-5) - 32
⟦illegible⟧ (ii) 5 (2y-4)² + 12(2y-4) - 9
⟦illegible⟧ (iii) 4(x²-3y²)² + (x²-3y²)x - 3x²
a = 5bk / 4(5b-k)
b = 4ak / 5(4a-k)
4a / 5b
⟦illegible⟧ k = 20ab / 4a+5b , find: (i) "a" in terms of "b" and "k" (2 marks)
(ii) "b" in terms of "a" and "k" (2 marks)
(iii) The value of √((k-4a)/(k-5b)) in terms of a/b
(6 marks)
2. (i) Using tables, compute by logarithms the value :
³√((0.002037² x 2.005) / 40.03²) (10 marks)
(ii) Solve for x the equation 3²ˣ - 28 x 3ˣ⁻¹ + 3 = 0 (10 marks)
3. A man in a speed-boat sees the flash of a gun fired a ship
directly towards him and hears the report 20 seconds later. The boat is
travelling to meet the ship at 30 miles per hour. Find the distance
between the two when the shot was fired, taking the velocity of sound
to be 1100 ft. per second. (20 marks)
4. Draw the graph of y = 6 + 3x - x² for values of x from -2 to 5 taking
1 inch as unit on the x-axis and 1/2 inch as unit on the y-axis. From
your graph find:
(i) the maximum value of y
(ii) the values of x between which the function 6 + 3x - x² is positive.
5. (i) What kind of series is 1/4, 3/10, 7/20, 2/5, ...? Find its nth term
and the sum of the first ten terms. (10 marks)
(ii) ⟦illegible⟧ 36
SHAMASH SECONDARY SCHOOL
Final Examination, 1958-1959.
Subject: Algebra
Date: 28/5/1959
Class: 4th Year Secondary
Time: 8:00-10:30 a.m.
All questions are to be attempted.
1. (a) Resolve into factors: (i) (7x + 8)² - 2(7x + 8) - 15. (3 marks)
(ii) 2(x - y)² - 3x + 3y - 5. (4 marks)
(iii) a(a - 4) - b(b - 4). (3 marks)
(b) If 15(2x² - y²) = 7xy, and if x and y are both positive, find
the ratio of x to y. Use the shortest possible way. (10 marks)
2(i) Using tables, compute by logarithms the value of :
⁵/ (0.004678)² x 1.002
/ ⟦line⟧ (10 marks).
\/ (30.04)³
(ii) Given : log70 = 1.8451, log110 = 2.0414, log34.62 = 1.5394,
compute, without using tables, the value of : ³√41503 ,
correct to four sugnificant figures. (10 marks).
3. A certain alloy contains 6 parts by weight of a metal A and 5 parts
by weight of a metal B; another alloy contains ⟦7⟧ parts by weight
of A and ⟦3⟧ parts by weight of B. If these alloys are melted and
mixed together, how many pounds of the second alloy must be mixed
with 11 pounds of the first alloy to make a mixture which contains
40 per cent. of A ? (20 marks).
4. (a) The expression 2 - 2ⁿ⁺¹ / 3ⁿ is a formula for the sum of 'n'
terms of a certain geometric series, n being any positive
integer. Find the first term of the series, the common ratio,
and the formula for the n-th term. (10 marks).
(b) The first and second terms of a series are 'a' and 'b'
respectively. Find the n th term (i) if the series is an
arithmetic series; (ii) if it is a geometric one. (10 marks).
5. (i) Taking ½ in. as one unit on the x-axis and on the y-axis, plot
the curve y = 3/4 x² for values of x between x = -4 and x = 4.
(7 marks).
(ii) On the same axes of coordinates draw the graph of the equation
3x + 2y = 12. (6 marks).
(iii) From the above graphs find two roots for the simultaneous
equations y = 3/4 x² and 3x + 2y = 12. Verify your graphical
answers by solving algebraically. (7 marks).
S = 2 - ⟦2^{n+1}/3^{n}⟧
Sum of 1st & 2nd terms = 2 - ⟦2^3/3^2 = 2 - 8/9 = 10/9⟧
∴ Common ratio r = ⟦4/9 ÷ 2/3 = 4/9 × 3/2 = 2/3 Ans. 2⟧
∴ the nth term l = ar^{n-1} = 2/3 (2/3)^{n-1} = (2/3)^n Ans.
(b) (i) In the Arithmetic Series, the common difference: d = b - a
∴ the nth term l = a + (n-1) (b - a) Ans. 1
(ii) In the Geometric Series, the common ratio: r = b/a
∴ the nth term: l = a (b/a)^{n-1} = b^{n-1}/a^{n-2} Ans. 2
5. (i) y = 3/4 x² ...... ①
x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4
y | 12 | 6 3/4 | 3 | 3/4 | 0 | 3/4 | 3 | 6 3/4 | 12
(ii) 3x + 2y = 12 ...... ②
the st. line is determined by the two points: (0, 6) and (4, 0)
x | 0 | 4
y | 6 | 0
⟦Graph showing parabola and straight line intersection⟧
(-4, 12)
3x + 2y = 12
(2, 3)
y = 3/4 x²
(iii) The points of intersections of the two graphs
are: (-4, 12) and (2, 3)
∴ the solutions are: x = -4, y = 12 } Ans. 1
and x = 2, y = 3 } Ans. 2
To verify algebraically, from equation ② y = (12 - 3x)/2, substituting
12 - 3x / 2 = 3/4 x² ∴ 24 - 6x = 3x²
∴ 3x² + 6x - 24 = 0
⟦x² + 2x - 8 = 0⟧
⟦(x + 4)(x - 2) = 0⟧
⟦x = -4 or x = 2⟧
(i) (7x+8) - 2 (7x+8) - 15 = (7x+8) - 5 (7x+8) + 3 (7x+8)
(ii) 2(x-y)^2 - 3x + 3y - 2 = 2(x-y)^2 - 3(x-y) - 2 = (2x-2y+1)(x-y-2)
(iii) a(a-4) - b(b-4) = a^2 - 4a - b^2 + 4b = (a^2 - b^2) - 4(a-b)
= (a-b)(a+b) - 4(a-b) = (a-b)(a+b-4)
15(x^2-y^2) = 7xy ∴ 15x^2 - 7xy - 15y^2 = 0 ∴ (3x-5y)(5x+3y) = 0
∴ 3x-5y = 0 ∴ x/y = 5/3 Ans. or 5x+3y = 0 or x/y = -3/5 to be discarded
or divide 15(x^2-y^2) = 7xy by y^2 and we get 15((x/y)^2 - 1) = 7(x/y)
∴ 15(x/y)^2 - 7(x/y) - 15 = 0 ∴ (3x/y - 5)(5x/y + 3) = 0 ∴ x/y = 5/3
or x/y = -3/5 to be discarded
5√((0.004678)^2 × 1.002) / (30.04)^3 = x
log 0.004678 = ̅3.6700 | 2 log 0.004678 = ̅5.3400
log 1.002 = 0.0008 | log 1.002 = 0.0008
log 30.04 = 1.4777 | log Num. = ̅5.3408
| log Den. = 4.4331
| 5 log x = ̅10.9077
| log x = ̅2.18154
| x = 0.01519 Ans.
(ii) log 70 = 1.8451 | ∛√41503 = ∛√7^3 × 11^2 = x
log 110 = 2.0414 | log 7 = 0.8451 ∴ 3 log 7 = 2.5353
log 34.62 = 1.5394 | log 11 = 1.0414 ∴ 2 log 11 = 2.0828
| ∴ 3 log x = 4.6181
| ∴ log x = 1.5394
| ∴ x = 34.62 Ans.
Since log 34.62 = 1.5394
3. In the 11 lbs of the first alloy there are 6 lbs of metal A + 5 lbs of metal B
In x lbs of the second alloy there are (7/(7+13) x) lbs of metal A + (13/20 x) lbs of metal B
∴ (6 + 7/20 x) lbs = 40/100 (11+x) ∴ 6 + 7x/20 = 2/5 (11+x) or 120 + 7x = 8(11+x)
or x = 32 lbs. Ans.
SHAMASH SECONDARY SCHOOL
Final Examination, 1958-1959.
Subject: Algebra
Date: 28/5/1959
Class: 4th Year Secondary
Time: 8:00-10:30 a.m.
All questions are to be attempted.
1. (a) Resolve into factors: (i) (7x + 8)² - 2(7x + 8) - 15. (3 marks)
(ii) 2(x - y)² - 3x + 3y - 5. (4 marks)
(iii) a(a - 4) - b(b - 4). (3 marks)
(b) If 15(2x² - y²) = 7xy, and if x and y are both positive, find
the ratio of x to y. Use the shortest possible way. (10 marks)
2(i) Using tables, compute by logarithms the value of :
⁵√ (0.004678)² x 1.002 (10 marks).
⟦line⟧
(30.04)³
(ii) Given : log70 = 1.8451, log110 = 2.0414, log34.62 = 1.5394,
compute, without using tables, the value of : ³√ 41503 ,
correct to four sugnificant figures. (10 marks).
3. A certain alloy contains 6 parts by weight of a metal A and 5 parts
by weight of a metal B; another alloy contains 7 parts by weight
of A and 13 parts by weight of B. If these alloys are melted and
mixed together, how many pounds of the second alloy must be mixed
with 11 pounds of the first alloy to make a mixture which contains
40 per cent. of A ? (20 marks).
4. (a) The expression 2 - (2ⁿ⁺¹ / 3ⁿ) is a formula for the sum of 'n'
terms of a certain geometric series, n being any positive
integer. Find the first term of the series, the common ratio,
and the formula for the n-th term. (10 marks).
(b) The first and second terms of a series are 'a' and 'b'
respectively. Find the n th term (i) if the series is an
arithmetic series; (ii) if it is a geometric one. (10 marks).
5. (i) Taking ½ in. as one unit on the x-axis and on the y-axis, plot
the curve y = ¾x² for values of x between x = -4 and x = 4.
(7 marks).
(ii) On the same axes of coordinates draw the graph of the equation
3x + 2y = 12. (6 marks).
(iii) From the above graphs find two roots for the simultaneous
equations y = ¾x² and 3x + 2y = 12. Verify your graphical
answers by solving algebraically. (7 marks).
Shamash Secondary School
Final Examination, 1955-1956
Subject: Algebra
Date: 30/5/56
Class: 4th Year
Time: 8:00 - 10:00 a.m.
All questions are to be attempted:
1. In 1955 a housewife could buy 7 more eggs for 10s. 6d. than
she can buy for 14s. in 1956, when the price per egg has
increased by one penny. Find the price of eggs, per dozen,
in 1955. (Ans. 2s.)
2. A motorist travels a certain distance x at a certain
uniform speed. If his speed had been 4 miles per
hour greater, he would have saved 10 minutes on the
journey; and if his speed had been 9 miles per hour
greater he would have saved 20 minutes. Find the
distance x. (Ans. 60 miles)
3. Using one pair of axes draw graphs of x² and of ½x + 2 between
the values of x = -3 and x = +3, choosing your own
scales. From your graphs read off the solutions of
x² = ½x + 2. By drawing a further graph find the
solutions of x² = ½x + 1. (Ans. 1.7 or -1.2,
1.3 or -0.8).
5. (i) Compute by logarithms ⁷√((0.00092)² x (4.006)³ / (0.006204)⁵)
(ii) Find the value of x from the equation 4^(2-x) x 3^x = 243
4. (i) The sum of the first 29 terms of an Arithmetical progression,
whose common difference is (-0.8), is zero. Find the
first term. (Ans. 11.2)
(ii) Find the tenth term + the sum of the first ten terms
of the Arithmetical progression whose nth term is 3 - ½n
(Ans. -2, 2.5).
⟦illegible arrow pointing to question 5⟧
⟦illegible arrow pointing to question 4⟧
⟦Delinquent⟧ Examination, September 1954
Algebra Date: 21st Sept., 1954
4th (Scientific) Time: 8:30 — 10:30 a.m.
All questions are to be attempted.
1. Find by how much (x+a)² + (x-a)² exceeds twice x². Hence find the
difference between (4.3)² + (4.1)² and 2(4.2)².
2. Solve the equations: (i) 3x² + 1.7x - 2.6 = 0 (correct to two decimal places)
(ii) 4x² - 8xy + 4y² - 3x + 3y - 1 = 0 ; x + 2y = 7.
3. Factorise: (i) 6x² + 7x - 20
(ii) (p² + pq + q²)² - (p² - pq + q²)²
(iii) x³ - x² + 2x - 1
4. (i) What kind of series is 1/4, 3/10, 7/20, 2/5, ...? Find its nth term
and the sum of the first ten terms.
(ii) Three numbers in A.P. add up to 36. When they are increased
by 1, 4, 43 respectively they form a G.P. What are the numbers?
5. A manufacturer produces a motor car at a cost of £ 440. He sells it
to a dealer at a loss, and the dealer sells it to a customer for £ 480.
Given that the dealer's percentage profit is double the manufacturer's
percentage loss, find at what price the manufacturer sold the car
to the dealer.
6. Draw the graph of y = 6 + 3x - x² for values of x from -2 to 5, taking
1 in. as unit on the x-axis and 1/2 in. as unit on the y-axis.
(correct to one dec. pl.)
From your graph find (i) the maximum value of y, and (ii) between
what values of x the function is positive.
⟦Solutions to the Final Exam 1953-1954 Questions⟧
Algebra
Class : 4th Secondary
1. (i) { a^(p-q) / ⟦illegible⟧ }^(p+q) . { a^(q-r) / ⟦illegible⟧ }^(q+r) . { a^(r-p) / ⟦illegible⟧ }^(r+p) = { a^(p^2-q^2) / a^(p^2-q^2) } . { a^(q^2-r^2) / a^(q^2-r^2) } . { a^(r^2-p^2) / a^(r^2-p^2) } = a^(p^2-q^2+q^2-r^2+r^2-p^2) = a^0 = 1 Ans.
(ii) (6√x - 7) / (√x - 1) - 5 = (√x - 26) / (7√x - 21) ∴ (6√x - 7)(7√x - 21) - 5(√x - 1)(7√x - 21) = (√x - 26)(√x - 1)
∴ 42x - 175√x + 147 - (35x - 140√x + 105) = 7x - 33√x + 26
2√x = 16 ∴ √x = 8 ∴ x = 64 Ans.
2. (i) ∛1.002 ⁷√(0.005001)² / ⁵√(0.03)³ x (4.003)⁵ = x
log 1.002 = 0.0003
log 0.005001 = 3.6991
log 0.03 = 2.4771
log 4.003 = 0.6024
1/3 log 1.002 = 0.0001
2/7 log 0.005001 = 1.3426
log Num. = 1.3427
log Den. = 1.8584
log x = 1.4843
∴ x = 0.3051 Ans.
(ii) 2^x = 8^(y+1) ; 9^y = 3^(x-9) ∴ 2^x = (2^3)^(y+1) ; (3^2)^y = 3^(x-9)
∴ 2^x = 2^(3y+3) and 3^(2y) = 3^(x-9) ∴ x = 3y+3 and 2y = x-9
or x - 3y = 3 } y = 6 } Ans.
x - 2y = 9 } x = 21 }
3. (i) l_3 = 15 | l_100 = ? | 15 = a + 2d } ∴ 75 = 15d | ∴ a = 15 - 10 = 5
l_18 = 90 | S_100 = ? | 90 = a + 17d } ∴ d = 5 | ∴ a = 5 , d = 5
l_100 = a + 99d = 5 + 99x5 = 500 Ans. 1
S_100 = n/2(a+l) = 100/2(5+500) = 505 x 50 = 25250 Ans. 2
(ii) (1 + x + x^2 + ... + x^(n-1))(1 - x + x^2 - x^3 + ... + x^(n-1)) = 1 + x^2 + x^4 + ... + x^(2n-2)
S_1 = 1(x^n - 1) / (x - 1) , S_2 = 1[1 - (-x)^n] / (1 + x) ; S_3 = 1[(x^2)^n - 1] / (x^2 - 1) = (x^(2n) - 1) / (x^2 - 1)
S_1 x S_2 = (x^n - 1) / (x - 1) x (x^n + 1) / (x + 1) = (x^(2n) - 1) / (x^2 - 1) = S_3 Q.E.D.
4. x^2 - 3x + 2 = y ; y = x^2 - 3x + 1 ; y_2 = 1 } x = 2 } Ans. 2
x = 1 }
If we subtract 1 from y_1, we get y_2
∴ the solution of x^2 - 3x + 1 amounts to solving
y_1 and y_2 simultaneously where
we get: x = 2, 1
⟦Graph showing parabola y = x^2 - 3x + 2 and line y = 1 intersecting at x=1 and x=2⟧
Frank Iny School
Intermediate and Primary
Baghdad
Telephone Number 91693
FRANK INY SCHOOL
INTERMEDIATE & PRIMARY
Baghdad
Telephone No. 91693
No.: | الرقم:
Date: Final Exam. in Algebra Cont. | التاريخ:
5. (i) y = 3 cos 60°, ∴ AC = 4 + y = 4 + 3 cos 60°
z = 3 sin 60°, ⟦thus⟧ tan α = z / AC = 3 sin 60° / (4 + 3 cos 60°)
∴ tan α = (3√3 / 2) / (4 + 3/2) = 3√3 / (8 + 3) = 3√3 / 11 = (3 × 1.732) / 11 = 5.196 / 11 = 0.47236
tan α = (3 × 0.8660) / (4 + 1.500) = 2.5980 / 5.5 = 0.47236
∴ α = 25° 17' ∴ θ = 90 - α = 64° 43' Bearing of B from A ⟦Ans.⟧
x / z = sin α ∴ x = z / sin α = 3 sin 60° / sin α = (3 × 0.8660) / 0.4271 = 2.5980 / 0.4271 = 6.0828
Ans.
(ii) Let each side of the equilateral triangle = 2x,
then its height AD = x√3
y = x√3 sin 60° = x√3 . √3 / 2 = 3x / 2
sin θ = y / 2x = (3x / 2) / 2x = 3x / 4x = 3 / 4 = 0.7500
<del>θ = 36° 54'</del> Ans.
θ = 48° 36' or 48° 35' Ans.
⟦illegible mathematical sketches and faded text in blue ink⟧
⟦illegible⟧ SCHOOL
⟦illegible⟧ PRIMARY
Baghdad
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧ = ⟦illegible⟧ + ⟦illegible⟧
⟦illegible⟧ = ⟦illegible⟧
tan ⟦illegible⟧ = ⟦illegible⟧
⟦illegible⟧ = ⟦illegible⟧
⟦illegible⟧ = ⟦illegible⟧
⟦illegible⟧ = ⟦illegible⟧
⟦illegible⟧ = ⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
Illustrative Examples:
In the financial year 1954-55 a man had an
earned income of £ 1350 and a further unearned
income of £ 200 from investments. In that
year income-tax was levied according to the
following rules:
The first £ 210 of the man's income was free
of tax and there was a further tax free allowance
of 2/9 th. of the earned income. The remainder
of the earned income, called "taxable income"
was taxed as follows: The first £ 100
of the taxable income was taxed at 2s, 6d.
in the £ 1, the next £ 150 at 5s, the
next £ 150 at 7s. and the remainder at
9 s. in the £. How much tax did the man
pay?
solution:
Total income = £ 1350 + £ 200 = £ 1550.
Taxable income = £ 1550 - { £ 210 + £ 1350 x 2 / 9 = £ 1040
£ 100 @ 2 s 6 d £ 100 x 0.125 = £ 12.500
£ 150 @ 5 s £ 150 x 0.250 = £ 37.500
£ 150 @ 7 s £ 150 x 0.350 = £ 52.500
Remainder (£ 640) £ 640 x 0.450 = £ 288.000
Total tax = £ 390.500
or £ 390 10 s
Ans.
P.T.O.
SHAMASH SECONDARY SCHOOL
FINAL EXAM. MAY 1964.
Subject :: Arithmetic & Trigonometry
Date ::
Class :: 4th Year Secondary
Time :: 8:00 - 10:30 a.m.
⟦line⟧
Answer All Questions
1. Two lighthouses A and B are 5 miles apart, B being due east of A. A ship
at P is due north of A, and on a bearing 322° (N.38°W.) from B. The
ship then sails in a direction 056° (N.56° E.) to a position Q which is
due north of B. Calculate PQ, BQ, and the bearing of Q from A.
PQ = 6.031 Mi. BQ = 9.772 Mi
27° 6' E
2. The elevation of a spire from a point A due N. of it is 28°, and from a
point B due E. of it 18°. Find the height of the spire if AB is 100
yards.
27.7 yds
3. A city council requires an annual rate of 16s. 4d. in the £ of rateable
value. Out of each 16s. 4d. received the sum of 5s. 7.3d. is spent on
education.
(i) What is the rateable value of a house, the holder of which pays
£ 13 1s. 4d. half-yearly in rates ?
(ii) If this householder has two children, both attending the council's
school, what sum, to the nearest 1/10 d., is he contributing per week
towards the education of each child ? (Take a year to be 52 weeks.)
£ 32
1s 8.7d
4. A man invests £ 30,155, partly in 3 1/2% stock at 86 and the rest in 4 1/2%
stock at 99. He divides the money so as to obtain the same income
from each stock. Find the total income.
1295
5. Find, correct to three significant figures, the weight in pounds of a
cylindrical iron pipe 10 ft. long, whose outer diameter is 1 ft. 6 in.
and inner diameter 1ft. 4 in., given that 1 cu. ft. of iron weights
494 lb. (Take π as 3.142.)
If the inner diameter is increased to 1 ft. 5 in., the outer diameter
and the length remaining unaltered, find the ratio of the new weight
to the old.
1830
35 / 66
⟦line⟧
PQ = 5 cosec 56°
BQ = 5 cot 58° + ⟦illegible⟧ cot 38°
tan = ⟦illegible⟧
27
2(a) The total rateable value of a town is £30,000.
How much does the town receive in a year if the
rate levied is 16s in the £,
(b) If the town estimates that it will need
an extra £3750 next year, by how
much will the rate in the £ have
to be increased if the total rateable value
remains unchanged?
(c) If the town decided to have
the rate in the £ at 16s but to get
the extra £3750 by increasing the
rateable value of all property, what
ought to be the new rateable
value of a house which was
formerly rated at £32 -
solution:-
(a) Total rates = £30,000 X 0.8 = £24,000
(b) 3750 / 30,000 = £0.125 = 2s 6d Increase in rate
New Rate = 16s + 2s 6d = 18s 6d
(c) New Rateable Value / Old Rateable Value = New tax (Rate) / Old tax (Rate)
(provided the rate
in the £ remains
the same)
New Rateable Value of the House / Old " " " " = 27,750 / 24,000
New R.V. = Old R.V. X 27,750 / 24,000
Solution :
3)
(i)
Yearly rates = (£ 13 1s 4d) x 2
= £ 26 2s 8d
<del>Rateable</del>
Rateable value
of the House = £ 26 2s 8d
⟦line⟧
16s 4d
= 522 2/3
⟦line⟧
16 1/3
= 1568 = <del>313</del> £ 32
⟦line⟧
49
(ii) (£ 13 1s 4d) 5s 7.3d
⟦line⟧
16s 4d
= 3136 x 67.3 = 16 x 67.3
⟦line⟧ ⟦line⟧
196 = 1076.8 d/year
1076.8 = 20.7 d
⟦line⟧
52 or 1s 8.7d per
⟦line⟧ ⟦line⟧
week per child
5
3
New R. V = 32 X 27,750
⟦line⟧
24,000
= £ 37 Ans.
stencil
Taxation
(a) Income tax Income tax
(b) Rates = tax on <del>rateable</del> real property Property tax
(a) Income tax
Income tax is levied in accordance with
ascending scale (progressive rate) i.e. the
rate of tax increases with higher income.
Earned income = Income <del>as a</del> resulting
from one's toil (person's labor)
e.g. salaries.
Unearned income = Income from deposits,
shares etc.
In the U.K income tax is <del>colle</del> levied
at the rate of so many shillings in
the £ e.g 10s 6d of £1 of income.
Total income = Total earnings
Taxable income = Total income - tax-
free allowance
⟦line⟧
(b) Rates
Rates - Property tax
Rent - Rent
Rateable value - assessed value
Estimated amount
The rates are collected on
P T O
the rateable value which may or
may not be equal to the rent.
Rates are <del>⟦illegible⟧</del> levied on the basis
of so many shillings in the £ 1 of
rateable value
e.g the rate of a house
is 5s in the £
If the rateable value
is £ 100, the rates are
£ 25.
The rates may sometimes exceed
the rateable value; Reason: the
assessment was made many years
ago when rents were low. Instead
of making reassessment, it is
sometimes decided to raise the
rate or the tax per £ 1. We thus
find in certain boroughs the rate is 25s
or more in the £.
Penny rate = a rate of 1d in the
£ 1 of rateable value.
(a) Tax
Solutions to Arithmetic and Mathematics questions ⟦illegible⟧
The final exam that took place on 17/5/1967
Arith. + ⟦illegible⟧
⟦illegible⟧, 4th year, 1967
1 (a) Amount of stock obtained =
£ 2286 / 95 1/4 X 100 = £ 2,400
Dividend = 2400 / 100 X 4 1/2 = £ 108
Amount realized from sale of stock = £ 2400 / 100 X 98
= £ 2,352
Total gain = £ 2,352 + £ 108 - £ 2,286
= £ 174
(b) I must set out at 10.25 - 00.15 = 10.10 a.m
(true time)
There are 100 minutes between 8.30 a.m
and 10.10 a.m.
My watch gains in 100 minutes of true
time 6 / 60 X 100 = 10 minutes
Latest time indicated by watch at
which I must set out is 10.20 a.m
2 (a)
Speed in ft/sec
1000 / (6 1/4 x 60) ÷ (22/7 x 1/3^2)
= (1000 x 4 x 7 x 9) / (25 x 60 x 22) = 84/11 = 7 7/11 ft/sec
(b) Sediment
1000 x 4 / 25 x 60 x 48 x 1/2 x 1/16 x 1/2240 = 45/7
= 6 3/7 tons
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
⟦illegible⟧
3 (I)
Merchant's total Costs
15 X 18 + (6s 3d) 4.6 + £ 5 10s =
£ 270 + £ 1 8s 9d + £ 5 10s = £ 276 18s 9d
(II) Total amount merchant received from sale
92 X (£ 1 5s) + 15 X 112 X 4 / 240 + {15 - (4.6 + 0.75)} 20
= £ 115 + £ 28 + £ 193 = £ 336
(III) Profit 336 - 276 18s 9d = £ 59 1s 3d
Percentage profit = 59.0625 X 100 / 276.9375
= 59062500 / 2769375
Since to two significant figures
we need only take four in calculations
Percentage profit = 59060000 / 2769000
= 21 %
4) 52,320 X 0.6375 + 127,460 X 0.9625 =
£ 33,354 + £ 122,680 5s =
£ 156,034.25
Average rate =
£ 156,034.25
⟦line⟧
£ 52,320 + £ 127,460
= £ 156,034.25
⟦line⟧
£ 179,780
= £ 0.8679
= £ 1 17s 4 1/4 d to nearest
farthing
Shamash Secondary School
Final Examination, May, 1967.
Subject: Arithmetic & Trigonometry
Date: 17/5/1967
Class: 4th Year Secondary.
Time: 8:00 - 10:30 a.m.
⟦line⟧
Answer five questions which must include questions 2 & <del>5</del> 3.
1. (a) How much stock is obtained by investing £2,286 in a 4½ per cent stock
at 95¼? After receiving the first annual dividend on this stock, it
is immediately resold at 98. Calculate the total gain on the
transaction.
(b) I have a watch which gains six minutes in every true hour. I put the
watch right at 8.30 a.m. What is the latest time indicated by the
watch at which I must set out to catch a train which leaves at 10.25
a.m. if it takes me 15 minutes to walk to the station ?
2. Following a storm, water is pumped out of a flooded area through a pipe
of 8 in. diameter at the rate of 1,000 gallons per minute. Taking 1 cu.ft.
as 6¼ gallons and <del>⟦illegible⟧</del> π as 22/7 , calculate:
(a) the speed in ft. per sec. at which the water is passing through the
pipe.
(b) how many tons of sediment will be pumped out in two days if it is
known that the flood water contains ½ oz. of sediment in every cu.ft.
of water.
3. A merchant bought 15 tons of potatoes from a farmer at £18 per ton.
(a) He sold 4 ton 12 cwt of the potatoes in 1 cwt bags at £1 5s. per
bag. The additional cost to the merchant in selling the potatoes
in this way was 6s. 3d. per ton.
(b) He sold 15 cwt retail at 4d. per lb for which he incurred additional
labour costs of £5 10s.
(c) He sold the remainder of the potatoes in bulk at £20 per ton.
Calculate:
(i) the merchant's total costs, including the initial cost of
the potatoes, cost of selling the potatoes in bags, and
additional labour cost for the retail sales.
(ii) the total amount the merchant received from his sales.
(iii) the merchant's profit calculated as a percentage, correct
to 2 significant figures, of his total costs.
4. A borough is divided into two districts whose rateable values are
respectively £52,320 and £127,460. The rate in the first district
is 12s. 9d. in the £, and in the second district it is 19s. 3d. in
the £. Find the average rate for the whole borough to the nearest
farthing.
(cont'd.p.2)..