AI en Translation, Pages 201-225
Page 201
Shamash Secondary School
Final Examinations, June 1961.
Subject: Algebra
Date: 12/6/1961
Class: 4th Year Secondary
Time: 8:00-11:00 a.m.
Attempt all questions :
1. How much are pencils a gross when 120 more for a sovereign lowers
the price 2d. a score ? ( 20 marks).
2. (i) Solve the equation √a - x + √b + x = √a + √b . (10 marks).
(ii) Simplify and express with positive indices :
⟦{ (a^(p-q) / q√(a^(q^2-pq))) * a^(x(p-q)) }^n⟧ (10 marks).
3. (i) Compute by logarithms the following :
7√{ (0.001021)^2 . (4.003) / (16.02)^5 . (3.001)^4 } (10 marks).
(ii) Derive a formula for finding the logarithms of a number to the
base 'y' having given tables of logarithms to the base 'x'.
Hence use your common logarithmic tables to compute the value
of log70.56. (10 marks).
17
4. (i) A man drives a distance of 14 17/20 miles in 18 minutes. He
drives the longest stretch in the first minute; and in every
subsequent minute the distance he drives is 1/20 miles shorter
than the distance driven in the previous minute. What are the
distances driven in the first and last minutes ? Use arithmetic
progressions to solve the question. (10 marks).
(ii) The sum of n terms of a geometric progression is 4(5^n -1). Find
the fifth term and the number of terms that must be taken for
their sum to be 312496. (10 marks).
5. In the equation y = 16x(4 - x), y represents the height in feet
risen by a stone thrown vertically upward from a point A on the
roof of a building 20 ft. above the level of the ground; and x
represents the time in seconds taken by the stone to reach the height
y from that point.
Draw a graph between x = 0 and x = 5 showing the relationship
between y and x. (Take 1 in. = 1 second and 20 ft. respectively).
From the graph find :
(a) the maximum height above the level of the ground reached by the
stone,
(b) how long the stone remains at least 68 ft. above the ground,
(c) how many seconds elapse from the time the stone was thrown
to the time the stone strikes the ground. (20 marks).
Page 202
Shamash Secondary School
3rd Quarter Examination
Subject: Algebra
Date: 17/3/1961.
Class: 4th Secondary
Time: 8:30-10:00 a.m.
⟦line⟧
Attempt all Questions:
1. A boy can swim at v m.p.h. in still water. When he swims downstream
the current increases his speed by u m.p.h., and when he swims
upstream the current decreases his speed by u m.p.h. The difference
in his time to swim one mile downstream and one mile upstream is
t hours. Find a formula for t in terms of v and u, and find v if
u = 2, t = 4/5.
2. A dealer bought a horse, expecting to sell it again at a price
that would have given him 10 per cent. profit on his purchase; but
he had to sell it for £50 less than he expected, and he then found
that he had lost 15 per cent. on what it cost him. What did he
pay for the horse ?
3. At what time between 12 o'clock and 1 o'clock are the two hands of
a watch at right angles for the second time?
4. By lowering the price of eggs and selling them one penny per egg
cheaper, a man finds that he can sell 5 more than he used to do for
5s. At what price per egg did he sell them at first?
5. A cistern can be filled by two pipes running together in 24 minutes.
The larger pipe would fill the cistern in 20 minutes less than the
smaller one. Find the time taken by each.
⟦line⟧
Page 203
Shamash Secondary School
3rd Quarter Examination
Subject: Algebra
Class: 4th Secondary
Date: 17/5/1961.
Time: 8:30-10:00 a.m.
⟦line⟧
Attempt all Questions:
1. A boy can swim at v m.p.h. in still water. When he swims downstream
the current increases his speed by u m.p.h., and when he swims
upstream the current decreases his speed by u m.p.h. The difference
in his time to swim one mile downstream and one mile upstream is
t hours. Find a formula for t in terms of v and u, and find v if
u = 2, t = 4/5.
2. A dealer bought a horse, expecting to sell it again at a price
that would have given him 10 per cent. profit on his purchase; but
he had to sell it for £50 less than he expected, and he then found
that he had lost 15 per cent. on what it cost him. What did he
pay for the horse ?
3. At what time between 12 o'clock and 1 o'clock are the two hands of
a watch at right angles for the second time?
4. By lowering the price of eggs and selling them one penny per egg
cheaper, a man finds that he can sell 5 more than he used to do for
5s. At what price per egg did he sell them at first?
5. A cistern can be filled by two pipes running together in 24 minutes.
The larger pipe would fill the cistern in 20 minutes less than the
smaller one. Find the time taken by each.
⟦line⟧
Page 204
Shamash Secondary School
Final Examination, June 1960.
Subject: Algebra
Date: 17/6/1960
Class: 4th Year Secondary
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 0.000401³) / (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⟧
Page 205
⟦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)⟧
Page 206
⟦(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)
Page 208
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).
Page 210
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⟧
Page 212
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).
Page 213
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⟧
Page 214
⟦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.
Page 215
⟦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⟧
Page 216
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⟧
Page 218
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.
Page 219
2/1
2.(a) The total rateable value of a town is £30000.
How much does the town receive in a year if the
rate levied is 16 s 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 leave
the rate in the £ at 16 s 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
⟦line⟧
(b) 3750 / 30,000 = £ 0.125 = 2 s 6 d Increase
in rate
New Rate = 16 s + 2 s 6 d = 18 s 6 d
⟦line⟧
(c) New Rateable Value / Old Rateable Value = New tax (Rates) / Old tax (Rates)
(provided the rate
in the £ 1 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
Page 220
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>⟦illegible⟧</del> £ 32
⟦line⟧
49
(ii) (£ 13 1s 4d) 5s 7.3d
⟦line⟧
16s 4d
= 3136 x 67.3 = 16 x 67.3
⟦line⟧
196 = 1076.8 d/year
1076.8 = 20.7 d
⟦line⟧
52 or 1s 8.7 d per
⟦line⟧
week per child
Page 221
3
New R. V = 32 X 27,750
⟦line⟧
24,000
= £ 37 Ans.
Page 222
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
Page 223
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 Exchanges questions for the Fourth Preparatory Grade
For the second session which took place on 10/17/1967
Arith. + ⟦Ex⟧, 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.
There are 100 minutes between 8.30 a.m. (true time)
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.
Page 224
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⟧
Page 225
(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 %
3