Format 210mm X 297mm Extent= 192 pgs (10.14 mm) (70gsm paper) Status: CRC Date: 14/3
Cambridge IGCSE TM redue70% DA1301
ACE YOUR
MATHEMATICS
Workbook
Cambridge IGCSE Ace Your Mathematics is written to improve students’
TM
approach to mathematics by providing them with high-quality educational
materials that uphold top academic standards, while allowing them to acquire
valuable techniques to excel in their examinations.
This workbook is written distinctively based on the Cambridge IGCSE 2020– Cambridge IGCSE TM ACE YOUR MATHEMATICS
2022 and 2023–2024 syllabuses for Mathematics (0580/0980) course at
Extended Level. The scope, sequence and level of the workbook has been
constructed to match the Cambridge IGCSE Year 10 and Year 11 syllabuses.
The workbook provides ample practice exercises for each topic. Answers
and fully-worked solutions are provided for all questions to enhance students’
understanding in how to master the approach to certain questions. It is
highly recommended as it serves as a good aid in evaluating students’
proficiency in mathematical skills, concepts and processes, helping them
achieve excellent results in examinations. Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
About the Author
Kung Girly received her Bachelor of Laws (Honours) from the University of London and practiced law
for a few months. She then decided to venture into her true passion in education and teaching. TM
Throughout her teaching pathway, she has educated many students who sat for the IGCSE exams Cambridge IGCSE
with excellent results. She has also prepared and conducted seminars for exam-going students,
providing them the foundation and tips to excel in the examinations. She also provides coaching
for students participating in international mathematics competitions. ACE YOUR
Ms Kung has seven years of experience in teaching IGCSE Additional Mathematics, and currently
MATHEMATICS
serves as a Director at Teras Murni Education Group. She is actively involved in educating students
to ensure they meet their academic potential.
Workbook
www.dickenspublishing.co.uk
DA1301
ISBN: 978-1-78187-259-8
Suite G7-G8, Davina House, 137-149 Goswell Road,
London, EC1V 7ET, United Kingdom.
E-mail: [email protected] Kung Girly
Contents
Preface iii
Chapter 4 Geometry 71
Assessment Overview iv ● Similar Shapes 71
● Circle Theorem 75
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List of Important Formulae v ● Geometric Construction 80
Chapter 1A Numbers 1
● Number Fact 1 Chapter 5 Mensuration 83
● Inequalities 2
● Rounding off 3 ● Mensuration 83
● Standard Form 5
● Recurring Decimals 7
● Direct and Inverse Proportion 8
● Fractions 10
● Multiples and Factors 13 Chapter 6 Trigonometry 95
● Upper Bound and Lower Bound 15 ● Trigonometry 95
● Powers and Roots 17 ● Trigonometry Graph 105
● Sets 19
Chapter 1B Numbers 20 Chapter 7 Vectors and Transformations 108
● Vectors 108
● Numbers 20 ● Transformations 118
Chapter 2 Algebra and Graphs 32 Chapter 8 Probability 123
● Algebra 32 ● Probability 123
● Functions 38
● Inequalities 43
● Sequences 46
● Speed, Distance and Time 52
● Graphs 57 Chapter 9 Statistics 135
● Statistics 135
Chapter 3 Coordinate Geometry 63 Answers 153
● Coordinate Geometry 63
viii Contents
Contents.indd 8 15/03/2022 10:32 AM
List of Important Formulae
Number
Simple interest
PRT
l =
100
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P = Principal
R = Rate of interest
T = Period of time
Compound interest
( R ) n
A = P 1 + 100
P = Principal
R = Rate of interest
n = Period of time
Speed, distance and time
Distance
Speed =
Time
Total distance
Average speed =
Total time
Proportion
Directly proportional
y = kx
Indirectly proportional
k
y =
x
Algebra
The law of indices
x × x = x a + b
a
b
x ÷ x = x a – b
a
b
(x ) = x
ab
a b
x = 1
–a
x a
a
b
x = √x
a
b
x = 1
0
Quadratic formula
–b + √b – 4ac
2
x =
2a
Regular polygons
Interior angles Exterior angles
Total angle (n – 2) × 180° 360°
Each angle (n – 2) × 180° 360°
n n
List of Important Formulae v
List Of Important Formulae.indd 5 15/03/2022 10:33 AM
Graphs
Equations of a straight line y = mx + c
y – y = m(x – x )
1
1
(where m is the gradient and (x , y ) is the known
1
1
point on the line)
Finding the distance between two points √(x – x ) + (y – y ) 2
2
1 2 1 2
( x + x 2 y + y 2 )
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1
1
Finding the midpoint of two points 2 , 2
Parallel lines have equal gradients. m = m
1 2
m × m = –1
Perpendicular lines have a gradient that is a 1 2
negative reciprocal
Mensuration
Rectangle Triangle Parallelogram Trapezium Circle
a
a
b r
a c
b
a
Area = a × b b
Area = a × b b (a + b)c Area = πr 2
a × b Area = 2
Area =
2
Diagram Surface area Volume
2
Cylinder Curved surface area = 2πr πr ℎ
Total surface area = 2πrℎ + 2πr 2
h
r
Cone Curved surface area = πrl 1
2
2
Total surface area = πr + πrl 3 πr ℎ
I
h
r
vi List of Important Formulae
List Of Important Formulae.indd 6 15/03/2022 10:33 AM
Diagram Surface area Volume
Sphere Surface area = 4πr 2 4 3
3 πr
r
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Pyramid Base area + Area of the shapes 1
Slant on each side 3 × base area × perpendicular height
height Height
Trigonometry
For a right-angled triangle, a = b + c 2
2
2
Hypotenuse
Opposite side
θ
Adjacent side
opposite side
sin θ =
hypotenuse
adjacent side
cos θ =
hypotenuse
opposite side
tan θ =
adjacent side
For other triangles
Sine rule:
a = b = c or sin A = sin B = sin C
sin A sin B sin C a b c
Cosine rule:
a = b + c – 2bc cos A or
2
2
2
b = a + c – 2ac cos B or
2
2
2
c = a + b – 2ab cos C
2
2
2
1
Area of triangle = ab sin C
2
List of Important Formulae vii
List Of Important Formulae.indd 7 15/03/2022 10:33 AM
1A Numbers
Answer all questions.
If working is needed for any question, write it clearly in the spaces provided below that question.
Number fact
1
22 23 24 25 26 27 28 29
From the list of numbers, write down
(a) the factor of 78
[1]
(b) the prime numbers
[1]
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(c) the cube number
[1]
(d) the square number
[1]
2
1
–3 100 0.73 √7 –28 6.1
6
From the list of numbers, write down
(a) a square number
[1]
(b) the smallest number
[1]
(c) a natural number
[1]
Chapter 1A Numbers 1
Chapter 1A.indd 1 15/03/2022 10:26 AM
(d) an irrational number
[1]
3 Considering the following numbers.
π
3
0 2.15 64 3 √4 0.5
2
3
List down all the
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(a) irrational number(s)
[1]
(b) whole number(s)
[1]
(c) recurring number(s)
[1]
(d) cube number(s)
[1]
4
18 15 59 31 4 63 7
Work out the difference between the biggest and the smallest prime number in the list above.
[2]
5 Write down the prime factors between 10 and 20.
[1]
Inequalities
1 Write the following in order of size, start with the smallest first.
tan 50° cos 50° 42 1.05 × 10 1
50
< < <
[2]
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2 Ace Your Mathematics
Chapter 1A.indd 2 15/03/2022 10:26 AM
2 Write the following in order of size, start with the smallest first.
3 1
√0.3 0.3 10 0.3
2
0.3
< < <
[2]
3 Write the following in order of size, start with the smallest first.
1 22 π 3.14
π 7
< < <
[2]
4 Write the following in descending order.
0.879 0.879 0.879 68
78
> > >
[2]
Write the following in descending order.
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2.84 × 10 1.089 × 10 9.72 × 10 2.35 × 10 3
–3
0
–2
> > >
[2]
Rounding off
1 Use your calculator to find the value of 3.52 – √6.08 .
3
(a) Write down all the numbers displayed on your calculator.
[1]
(b) Write your answer in part (a) correct to the nearest whole number.
[2]
Chapter 1A Numbers 3
Chapter 1A.indd 3 15/03/2022 10:26 AM
5
2
2 Use your calculator to work out the value √0.7 + (10 × 3.1) and write your answer correct to one
decimal place.
[2]
3 By correcting each number to 1 significant figure, estimate the value of √9.2 × 41.8 . Show all your
working. 3 √1345
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[2]
4 Calculate (0.78305 – 5.02 ) , giving your answer correct to 3 significant figures.
2 6
[2]
3.502
5 Evaluate . Give your answer correct to
0.897 – 25 × 10 –2
(a) 3 decimal places,
[1]
(b) 3 significant figures.
[1]
6 By writing each number correct to the nearest whole number, estimate the value of 24 × 0.573 .
Show your working clearly. 3.29
[2]
7 Calculate 5 1 (tan 38° + 3), give your answer correct to 1 decimal place.
3
[2]
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8 Calculate 3.54 × 10 × 0.0354 and correct your answer to the nearest thousand.
5
[2]
3 √569
9 Calculate . Give your answer correct to 2 significant figures.
4.1 (9.03 + 3.2 )
3
2
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[2]
10 It is given that 1 = 1 – 1 .
a b c
(a) Find the value of c when a = 3.5 × 10 and b = 3.145 × 10 . Give your answer in 3 decimal places.
3
–2
[2]
(b) Write c in terms of a and b.
[2]
11 Sebastian is cycling along Parkview Garden. The diameter of each wheel of his bicycle is 50 cm and each
wheel makes 100 revolutions a minute. Calculate the distance, in km to 3 decimal places, travelled by
Sebastian in 0.6 hours.
[2]
Standard form
1 Given that A = 3.51 × 10 , B = 2.6 × 10 , C = 8.01 ×10 , evaluate the following expression and write your
–2
4
–1
answer to the nearest whole number.
(a) 2A – B + C
[2]
Chapter 1A Numbers 5
Chapter 1A.indd 5 15/03/2022 10:26 AM
(b) AC – B
[2]
(c) A – BC
2
[2]
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(d) (ABC) 2
[2]
(e) A
BC
[2]
A
(f)
(B – C) 2
[2]
2 The world population was 6.2 × 10 in 2015. It is predicted that in 2030, the world population will
10
increase to 8.97 × 10 . Find the predicted number of increase in the population, write your answer in
12
standard form.
[2]
3 Work out, give your answer in standard form.
4.05 × 10 + 4.05 × 10 40
38
[2]
4 Calculate 1.3 × 10 ÷ (2 ÷ 10 ). Give your answer in standard form.
5
–2
[2]
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Chapter 1A.indd 6 15/03/2022 10:26 AM
5 The price of a theme park ticket is $85 for adult and $55 for children below 12 years old.
(a) Calculate the amount received when 11809 of adult tickets and 8532 of children tickets were sold.
[2]
(b) Write your answer in part (a) in standard form.
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[2]
6 A clock is always behind by 4.5 minutes every hour. Calculate the number of seconds the clock becomes
slower in a year, and write your answer in standard form.
[2]
7 A sheet of paper is 1.085 × 10 cm thick. Find the total thickness of 80000 of these papers. Give your
–3
answer in standard form.
[2]
Recurring decimals
Write the recurring decimal as a fraction in its simplest form.
1 0.26
[2]
2 0.34
[2]
Chapter 1A Numbers 7
Chapter 1A.indd 7 15/03/2022 10:26 AM
3 0.807
[2]
4 9.16
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[2]
5 2.534
[2]
Direct and inverse proportion
1 A is directly proportional to the square of B.
When A = 300, B = 10.
(a) Find a formula for A in terms of B.
[2]
1
(b) Calculate the value of A when B = .
2
[2]
2 r is directly proportional to square root of s + 2.
It is given that when s = 3, r = 5√5.
When s = 7, find the value of r.
[2]
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3 A is proportional to B + 3B – 4.
2
A = 7 when B = 3.
Calculate the values of B when A= –2.
[2]
4 The time, t minutes, for a water heater to boil some water is directly proportional to the mass of water,
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m kg.
When m = 1200 grams, t = 0.96 minutes.
Find the value of m when t = 5.
[2]
5 The time, t minutes, for a water heater to boil a fixed mass of water is inversely proportional to the
power supply, P watts, of the water heater.
When P = 220 watts, t = 0.25 minutes.
Find the value of t when P = 200.
[2]
6 The weight of a book is directly proportional to its number of pages.
Book A has 50 pages and a weight of 0.08 kg.
Book B has 720 pages, calculate its weight.
[2]
7 The amount of money Boey earns, m, is directly proportional to the number of hours she works, h.
If she works for 12.5 hours, she earns $131.25.
Find the amount of money she earns if she works 25 days in a month, 11 hours a day.
[2]
Chapter 1A Numbers 9
Chapter 1A.indd 9 15/03/2022 10:27 AM
8 If 10 workers can build 8 houses in 5 months. How long would it take for 20 workers to build the same
number of houses?
[2]
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9 Sean takes 150 days to reduce 12 kilograms of his weight by doing 30 minutes exercise every day. If he
does exercise 1 hour 15 minutes every day, how many days would he take to reduce the same weight?
[2]
Fractions
1 Without using a calculator, show that
(a) 3 2 + 1 1 = 4 13
7 3 21
[2]
(b) 8 3 – 6 2 = 1 19
10 3 30
[2]
2 Without using a calculator, work out 2 3 – 1 ÷ 2.
4 6
Write down all the steps of your working and give your answer as a fraction in its simplest form.
[2]
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Chapter 1A.indd 10 15/03/2022 10:27 AM
3 Without using your calculator, show that 1 5 ÷ 3 = 2 4 .
6 4 9
[2]
22 5
4 Without using your calculator, work out + 3 × .
Write your answer as a mixed number. 7 9
[2]
5 Without using a calculator, work out 4 ÷ 2 2 .
9 3
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Write down all the steps of your working and give your answer as a fraction in its simplest form.
[2]
6 Without using a calculator, work out 5 1 × 1 .
3 8
Show all your working and give your answer as a fraction in its lowest terms.
[2]
Chapter 1A Numbers 11
Chapter 1A.indd 11 15/03/2022 10:27 AM
1
1
7 Work out + 2 – , giving your answer as a fraction in its lowest terms.
5 3 6
Do not use a calculator and show all steps of your working.
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[2]
5 5
8 Find the fraction that is exactly halfway between and .
7 9
[2]
9 On the school concert day, 5 of the audience were parents, 4 of the remainder were students, the
12 7
rest of them were teachers. What fraction were teachers over parents?
[2]
10 Express 625 g as a fraction of 1 kg, and write your answer in its simplest form.
[2]
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Chapter 1A.indd 12 15/03/2022 10:27 AM
Multiples and factors
1 Three numbers P, Q and R, are written as the products of their prime factors.
P = 2 × 3 × 7 2
3
Q = 2 × 7 2
4
R = 2 × 3 × 7 2
5
Give your answers in index form, find
(a) the lowest common multiple of P and Q
[2]
(b) the highest common multiple of P, Q and R
[2]
(c) the square root of Q
[2]
2 Express 45 and 60 as the product of their prime factors.
(a) Find the highest common factor of 45 and 60.
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[2]
(b) Write down the smallest positive integer, a, such that 45a is a perfect square.
[2]
3 Lucas wants to pack 72 mint sweets, 60 orange sweets and 96 chocolate sweets into small containers as
many as possible. He wants to ensure that no sweets are leftover. The number of each flavoured sweet
must be the same in each container.
Find
(a) the greatest number of containers that Lucas would need to use
[2]
(b) the number of orange sweets in each container
[2]
Chapter 1A Numbers 13
Chapter 1A.indd 13 15/03/2022 10:27 AM
4 Express 160 as a product of its prime factors, give your answer in index notation.
Hence, find the smallest possible value of x such that 160x is a perfect cube.
[2]
5 (a) It is given that 400 = 2 × 5 . Express 9000 as a product of its prime factors, give your answer in index
4
2
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notation.
[2]
(b) Find the smallest positive integer m for which 400m is a multiple of 9000.
[2]
4
(c) Find the smallest positive integer n for which √9000n is a whole number.
[2]
6 Wei Wei bought a cardboard measuring 95 cm and 285 cm. She wants to cut out square cards of identical
size from the cardboard such that there is no wastage.
(a) Calculate the largest possible perimeter of each square that she can cut out.
[2]
(b) Find the total number of square cards she can cut out.
[2]
Three traffic lights at different locations are changing from green to red every 45, 48 and 60 seconds
respectively. If all three lights changed concurrently at 11:59 p.m., when would be the next time that
all three lights change at the same time again?
[3]
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Chapter 1A.indd 14 15/03/2022 10:27 AM
Upper bound and lower bound
1 It is given that
P = 3.29, correct to 2 decimal places
Q = 5, correct to the nearest whole number
R = 0.046, correct to 2 significant figures
(a) Find the lower bound of the following expression, correct your answer to 2 decimal places.
(i) (PQ) 2
[3]
R – Q
(ii)
P
[3]
Q
(iii)
P – R
[3]
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(b) Find the upper bound of the following expression, correct your answer to 2 decimal places:
P + Q
(i)
R
[3]
(ii) (R – Q) 2
[3]
(iii) Q
PR
[3]
2 The side of a square is 3.8 cm, correct to 1 decimal place.
Calculate the lower bound for the area and perimeter of the square.
[4]
Chapter 1A Numbers 15
Chapter 1A.indd 15 15/03/2022 10:27 AM
3 David has a dog, its weight was 1.7 kg six months ago, correct to 1 decimal place.
Today its weight is 4.8 kg, correct to 2 significant figures.
Work out the upper bound on the percentage for the increase in his dog’s weight.
[3]
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4 A jug holds 2 litres of orange juice, correct to the nearest whole number.
A cup holds 280 millilitres of water, correct to the nearest 10 millilitres.
Calculate the lower bound for the number of cups of orange juice which can be filled from the jug.
And how much of orange juice would left?
[3]
5 The sides of a triangle are 4.1 cm, 10.8 cm and 2.6 cm, each correct to the nearest millimetre. Calculate
the upper bound of the perimeter of the triangle.
[3]
6 Ashley runs for 2 hours, at a speed of 8 km/h.
Both values are correct to the nearest whole number.
Work out the lower bound of the distance that Ashley has run.
[3]
7 Aymil starts his journey with his car filled with 20 litres of fuel, correct to the nearest 5 litres. When
the car has travelled 125 km, correct to the nearest kilometres, the fuel tank is one-third full.
Find the upper bound of the number of kilometres travelled per litre of fuel.
[3]
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Chapter 1A.indd 16 15/03/2022 10:27 AM
Powers and roots
1 Evaluate each of the following and write your answer in index form.
(a) 2 × 2 –5
3
[2]
1
2
(b) 4 × 8 ÷ 2 5
[2]
(c) 3 ÷ 9 × 27 2
4
–2
[2]
(d) 5 × 5 7
(5 × 5 )
0
–3 –3
[2]
2
( ) –2 ( ) –1 0 ( ) –2
1
1
1
A = 5 B = 5 C = 25 D = 5 E = 125
–2
From the list of numbers, write down
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(a) the letter with the smallest number
[2]
(b) the letter with the biggest number
[2]
(c) the two letters with the equal numbers
[2]
3 Evaluate without using calculator, [7 + (–1) ] − [2(–2) ÷ 4] , show all your working clearly.
121
9
[3]
Chapter 1A Numbers 17
Chapter 1A.indd 17 15/03/2022 10:27 AM
4 Given that 8 × 4 = 1, find the value of a.
–3
a
[3]
2 1
5 Show all your working, simplify fully 8 3 + 27 3 .
1
196 2
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[4]
3 1 1
6 Evaluate √625 – 4913 .
2
3
[3]
3
2
7 Calculate 13 – √5 – 5 .
2
[3]
–1
8 Write your answer in index form ( √256 ) .
[3]
9 Write your answer in fraction.
4 – 3 2 × 2 3
2
3 3 √64
[3]
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Chapter 1A.indd 18 15/03/2022 10:27 AM
Sets
Shade the required region in each of the Venn diagrams.
1 (A ∪ B)’ 4 (A ∪ B ∩ C)’
A B
A B
[2] C [2]
2 A’ ∩ B 5 A ∪ B ∪ C’
A B
A B
[2] C [2]
3 A ∪ B’ 6 A ∩ B’ ∩ C
A B
A B
C
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[2]
Write down the set of the Venn diagrams.
7
A B
C
[3]
8
A B
C
[3]
9
A B
C
[3]
Chapter 1A Numbers 19
Chapter 1A.indd 19 15/03/2022 10:27 AM
3 Coordinate Geometry
Answer all questions.
If working is needed for any question, write it clearly in the spaces provided below that question.
Coordinate geometry
1 Find the gradient of the line that passes through points (–7, –1) and (–2, –7).
[2]
2 Find the distance between point A (0, 1) and point B (6, 9).
[3]
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The midpoint of the line segment joining (4, –7) and (10, y) is (7, 1). What is the value of y?
[2]
4 The diagram shows the straight line l, which passes through the points (–7, 10) and (–1, 0).
(a) Find the equation of line l, in the form of ay + bx = c. y
(–7, 10)
x
–1
[3]
Chapter 3 Coordinate Geometry 63
Chapter 3.indd 63 15/03/2022 10:28 AM
(b) Another line is perpendicular to line l. Write down the gradient of line l.
[2]
5 Find the equation of the line that passes through the x-axis at 5 and y-axis at –3.
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[3]
5
6 The line ay + bx = c is parallel to line y = x – 5. The distance PQ is 9 units. Find the values of a, b and
c. 3
y
ay + bx = c
5
y = x – 5
9 units 3
P Q x
[4]
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Chapter 3.indd 64 15/03/2022 10:28 AM
7 The points P(–1, 3), Q(4, 5) and R(9, k) all lie on a straight line.
(a) Find the value of k.
[3]
(b) Find the equation of the line.
[3]
(c) Show that point Q is the midpoint of line PR.
[2]
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y
P(–6, 3)
x
Q(0, –5)
(a) Find the equation of line PQ.
[3]
(b) Find the length of line PQ.
[3]
Chapter 3 Coordinate Geometry 65
Chapter 3.indd 65 15/03/2022 10:28 AM
(c) Write down the coordinates of point R such that PQR is an isosceles triangle.
[1]
(d) Find the area of the triangle PQR.
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[2]
9 The points A and B have coordinates (–8, 12) and (6, –2).
(a) Calculate the length of line AB.
[3]
(b) Find the coordinates of the midpoint of the line joining points A and B.
[2]
(c) Find the equation of line AB.
[2]
(d) Find the equation of the perpendicular bisector of AB.
[2]
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10 It is given that the area of triangle ABC is 12 square units.
(a) Find the value of p.
y
A(6, 5)
B(6, 2)
x
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2
C(p, q)
[2]
(b) (i) Write the equation of line BC.
[3]
(ii) Hence, proof that q = –2.
[2]
(c) Find the equation of the line passing through A and parallel to BC.
[3]
(d) Write down the coordinates of point D such that ABCD is a parallelogram.
[2]
Chapter 3 Coordinate Geometry 67
Chapter 3.indd 67 15/03/2022 10:28 AM
11 A line joins the points P(–10, 4) and Q(2, –5).
(a) Find the coordinates of the midpoint of PQ.
[2]
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(b) Find the equation of the line passing through P and Q.
[3]
(c) Another line RS is parallel to PQ passes through the point (1, 3). Write down the equation of this
line RS.
[3]
(d) A perpendicular line to RS passes through (1, 3) and point T on line PQ. Find the coordinates of
point T.
[4]
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68 Ace Your Mathematics
Chapter 3.indd 68 15/03/2022 10:28 AM
( )
2
12 Find the equation of the line that is perpendicular to 3x – 2y + 5 = 0 and passing through 2, 3 .
Write your answer in the form of ay + bc + c = 0.
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[6]
13 Line AB joins points (1, 5) and (–3, –3). Find the equation of line CD, which is the perpendicular
bisector of line AB.
[6]
Chapter 3 Coordinate Geometry 69
Chapter 3.indd 69 15/03/2022 10:28 AM
14 Line X joins the points (10, 3) and (–8, –5) and line Y joins the points (8, –5) and (2, –15). Determine
whether X is perpendicular to Y.
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[3]
15 Line X joins the points (–10, –3) and (–2, 9) and line Y joins the points (0, –9) and (2, –6).
Determine whether X and Y are parallel lines.
[3]
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70 Ace Your Mathematics
Chapter 3.indd 70 15/03/2022 10:28 AM
Answers
1A
1A Numbers
Number fact = 1 − 1
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3145 0.03
1 (a) 26 [1] c = −0.030 [2]
(b) 23 and 29 [1] 1 a – b
(c) 27 [1] (b) c = ab
(d) 25 [1] ab
2 (a) 100 [1] c = a – b [2]
(b) –28 [1] 11 0.6 hours = 36 minutes
(c) 100 [1] Distance travelled = 50π × 100 × 36
(d) √7 [1] 100 × 1000
3
3 (a) π and √4 [1] = 5.655 km [2]
3
(b) 0, 64 and 3 2 [1] Standard form
(c) 0.5 [1]
(d) 64 [1] 1 (a) 70200.05 70200 [2]
∼
4 59 – 7 = 52 [2] (b) 2811.484 2811 [2]
∼
5 11, 13, 17, 19 [1] (c) 1232010000 [2]
∼
(d) 5343.5 5344 [2]
Inequalities (e) 16853932.58 16853933 [2]
∼
∼
(f) 11992578.95 11992579 [2]
1 cos 50° < 42 < tan 50° < 1.05 × 10 1 [2] 2 (8.97 × 10 ) – (6.2 × 10 )
10
12
50 = 10 (8.97 × 10 – 6.2)
10
2
1
12
3
2 0.3 < √0.3 < 10 < 0.3 [2] = 8.908 × 10 2 [2]
0.3
2
38
3 1 < 3.14 < π < 22 [2] 3 10 (4.05 + 4.05 × 10 ) [2]
= 4.0905 × 10
π
40
7
4 0.879 > 0.879 > 0.879 > 68 [2] 4 6.5 × 10 2 [2]
78 5 (a) 11809 × 85 + 8532 × 55
5 2.35 × 10 > 2.84 × 10 > 9.72 × 10 > 1.089 × 10 –3 = 1473025 [2]
0
–2
3
[2] (b) 1.473025 × 10 6 [2]
6 (4.5 × 60) × 24 × 365
Rounding off = 2365200
= 2.3652 ×10 6 [2]
–3
1 (a) 41.1484424 [1] 7 1.085 × 10 × 80000
(b) 41 [2] = 8.68 × 10 1 [2]
2 3.95 4.0 [2]
∼
√9 × 40 3 × 40 Recurring decimals
3 = =12 [2]
3
√1000 10 1 Let x = 0.26
4 211929381.5 212000000 [2] 10x = 2.6
∼
5 (a) 5.413 [1] 100x = 26.6
(b) 5.41 [1] 90x = 24
6 24 × 1 = 8 [2] x = 24 = 4
3 90 15
∼
7 1.047 1.0 [2] 2 Let x = 0.34 [2]
∼
8 12531.6 13000 [2] 100x = 34.34
9 0.006239 0.0062 [2] 99x = 34
∼
1 1 1
10 (a) = − a x = 34
c
b
99 [2]
Answers 153
Answers.indd 153 15/03/2022 11:08 AM
3 Let x = 0.807 k = 55
1000x = 807.807 t = 55
999x = 807 200
x = 807 = 269 = 0.275 min [2]
999 333 [2] 6 Weight of book = W
4 9.16 = 9 + 0.16 Number of pages = N
Let x = 0.16 W = kN
10x = 1.6 0.08 = k(50)
100x = 16.6 k = 0.0016
90x = 15 w = 0.0016(720)
x = 15 =1.152 kg [2]
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90 7 m = kh
9.16 = 9 + 15 131.25 = k(12.5)
= 9 1 90 k = 10.5
6 [2] m = 10.5(11 × 25)
5 2.534 = 2 + 0.534 = $2887.5 [2]
Let x = 0.534 8 Number of months = M
10x = 5.34 Number of workers = W
1000x = 534.34 k
990x = 529 M = w
x = 529 5 = k
990 10
2.534 = 2 + 529 = 2 529 k = 50
50
990 990 [2] M = 20
= 2.5 months [2]
Direct and inverse proportion
9 Number of days =
1 A = kB 2 Amount of time (in hours) = T
300 = k(10) D = k
2
k = 3 T
(a) A = 3B 2 [2] 150 = k
( ) 2 3 0.5
1
(b) A = 3 2 = 4 [2] k = 75
D = 75
2 r = k√s + 2 1.25
5√5 = k√3 + 2 = 60 days [2]
k = 5
r = 5√7 + 2 Fractions
=15 [2]
1 (a) 3 6 + 1 7 = 4 13 (shown) [2]
3 A = k(B + 3B – 4) 21 21 21
2
7 = k(3 + 3(3) – 4) (b) 249 − 200 = 49 = 1 19 (shown) [2]
2
k = 1 30 30 30 30
2 2 11 − 1 × 1
–2 = 1 (B + 3B – 4) 4 6 2
2
2 11 1
B(B + 3) = 0 = 4 − 12
B = 0 or B = –3 [2] 33 – 1
=
4 = km 12
0.96 = k(1.2) = 32
k = 0.8 12
5 = 0.8m 8 2
m = 6.25 kg [2] = 3 or 2 3 [2]
k
5 t = p 3 11 × 4
0.25 = k 6 22 3
220 =
9
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2
= 2 4 (shown) 2 45 = 3 × 5
60 = 2 × 3 × 5
2
9 [2] (a) HCF = 3 × 5 = 15 [2]
4 22 + 5 (b) a = 5 [2]
7 3 3 72 = 2 × 3 2
3
= 66 + 35 60 = 2 × 3 × 5
2
21 96 = 2 × 3
5
= 101 (a) HCF = 2 × 3 = 12 [2]
2
21 (b) 5 [2]
= 4 17 4 160 = 2 × 5
5
21 [2] x = 2 × 5 2 [2]
5 4 ÷ 8 5 (a) 9000 = 2 × 3 × 5 3 [2]
3
2
9 3 (b) LCM = 2 × 3 × 5 = 18000
3
4
2
= 4 × 3 400m = 18000
9 8 m = 45 [2]
= 1 (c) n = 2 × 3 × 5 = 90 [2]
2
6 [2] 6 95 = 5 × 19
6 16 × 1 285 = 3 × 5 × 19
3 8 (a) HCF = 5 × 19 = 95
= 2 95 × 4 = 380 cm [2]
3 [2] (b) 285 ÷ 95 = 3 cards [2]
2
7 1 + 2 − 1 7 45 = 3 × 5
5 3 6 48 = 2 × 3
4
2
= 6 + 20 – 5 60 = 2 × 3 × 5
30 LCM = 2 × 3 × 5
4
2
= 7 = 720 seconds
10 [2] = 12 minutes
8 ( 5 + 5 ) ÷ 2 11:59 p.m. + 12 mins [3]
=12.11 a.m.
9
7
1
80
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=
×
63 2 Upper bound and lower bound
= 40 1 (a) (i) (PQ) = (3.285 × 4.5) = 218.52 [3]
2
2
63 [2] R – Q 0.0455 – 5.5
[(
9 1 − 5 − 1 − 5 ) × 4 ] (ii) P = 3.295 = –1.66 [3]
12 12 7 Q 4.5
= 1 − 5 − 1 (iii) P – R = 3.295 – 0.0455 = 1.38 [3]
12 3 P + Q 3.295 + 5.5
= 1 (b) (i) R = 0.0455 = 193.30 [3]
4 (ii) (R − Q) = (0.0465 − 4.5) = 19.83 [3]
2
2
(iii) Q = 5.5 = 36.80 [3]
Fraction of teachers over parents PR 3.285 × 0.0455
1 2 Area = 3.75 × 3.75 = 14.0625 cm 2
= 4 Perimeter = 3.75 × 4 = 15 cm [4]
5 3 Percentage increase
12 = 4.85 – 1.65 × 100%
1.65
= 3 = 193.94% [3]
5 [2] 4 Number of cups
10 625 g = 58 [2] = 1.5 l
1 kg 285 ml
= 5.26
Multiples and factors ≈ 5 cups
1.5 l − 0.285 l × 5 = 0.075 l or 75 ml [3]
1 (a) 2 × 3 × 7 = 21168 [2] 5 4.15 + 10.85 + 2.65
4
2
3
2
(b) 2 × 7 = 98 [2] = 17.65 cm [3]
2
4
(c) √2 × 7 = 2 × 7 = 28 [2] 6 Distance = 7.5 × 1.5
2
= 11.25 km [3]
Answers 155
Answers.indd 155 15/03/2022 11:08 AM
125.5 3
7 = 12.55 km/l [3] ξ
15 × 2 A B
3
Powers and roots
1 (a) 2 –2 [2] [2]
(b) 2 × 2 ÷ 2 = 2 2 [2] 4
5
4
3
(c) 3 ÷ 3 × 3 = 3 14 [2] ξ A B
–4
6
4
(d) 5 8 = 5 –1 [2]
5 9
2 (a) A [2]
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(b) E [2] C [2]
(c) B and C [2]
3 [7 + (–1)] – [(–4) ÷ 4] 121 5 ξ
= 6 – (–1) A B
121
= 6 – (–1)
= 7 [3]
4 2 × 2 = 2 0 C
–9
2a
–9 + 2a = 0 [2]
a = 4.5 [3] 6
2 1 ξ A B
2 3 + 3 3
3 ×
3 ×
5 1
2
2
(2 × 7 ) 2
2
= 2 + 3 C
2 × 7 [2]
= 1 [4] 7 (A ∪ B)’ ∩ C [3]
2
1 1 1 8 A’ ∩ B ∪ C [3]
6 (5 2 – 17 1 3) 3
3 ×
4 ×
= (25 – 17) 3 9 A ∪ B ∪ C’ [3]
= 2 [3]
7 169 – √100
1A
= 159 1 1 –1 [3] 1B Numbers
8 ( ( (2 ) ) 2 ) 54
8 2
= 1 [3] 1 (a) Time = 72 = 0.75 h = 45 min [1]
4 (b) Time = 54 = 3 h = 36 min
9 16 – 9 × 8 90 5
27 8 45 – 36 × 100% = 20% [2]
= 7 [3] 45
27 2 (a) Amount received
= (2000 × 1.85) × 0.99
Sets
= $3663 [2]
1 ξ (b) (i) Amount received
A B = (2000 × 0.005 × 6) + 2000 [2]
= $2060
(ii) Amount received
= (2060 × 1.8) − $30
= $3678 [2]
[2] (c) Danny, by $15 more [1]
(d) $500 ÷ $1.94
2 ξ = £257.73
A B = £258 [2]
3 (a) (i) $10000 × 0.3
= $3000
A : B : C
2.1 : 1.5 : 3.6
[2] 7 : 5 : 12
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156 Ace Your Mathematics
Answers.indd 156 15/03/2022 11:08 AM
7 7 : 6 : 3
A = 3000 × 24 7
= 875 36000 × 6 = 42000 [3]
5 16
B = 3000 × 24 (ii) 36000 × = 96000 [2]
= 625 100 6
12 (b) 36000 × = 28800 [2]
C = 3000 × 24 125
10
= 1500 [3] (c) 36000 = 25000(1 + R)
36000
5 ( 25000 ) 0.1
(ii) $1000 ÷ = $4800 R = – 1
24
30 = 0.0371
$4800 ÷ = $16000 [2] = 3.71% [2]
100
35000 – 28500 (d) Number of children = 36000 × 3 =18000
(b) × 100% 6
35000 Number of secondary students = 18000 × 0.52
= 18.57% [1] = 9360
4 Price of the car = $20000 × 0.4 (i) Number of boys = 9360 × 0.4 = 3744
= $8000 Number of girls = 9360 × 0.6 = 5616 [2]
Money she received from selling off her car 3744
= $8000 × 0.75 (ii) 96000 × 100%
= $6000 = 3.9% [1]
Amount invested in the bank 7 (a) (i) 1.35 p.m. + 12.5 h – 8 h = 6.05 p.m. (Sunday)
= $20000 × 0.6 [2]
= $12000 12000 km 12000 × 1000 m
Interest (ii) 12.5 h = 12.5 × 60 × 60 s
= $12000 × 0.045 × 2 800
= $1080 = 3 m/s or 266.67 m/s [2]
Total amount she withdrew from the bank 3
= $1080 + $12000 (b) (i) £18.50 × 5 = £11.10 [1]
= $13080 (ii) Business class ticket
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Percentage of the money left
8
13080 + 6000 = £18.50 × 5 × 55%
= × 100% = 95.4% [6] = £16.28
20000
5 (a) Number of tickets sold Difference in price
= 400 × 0.76 = £16.28 − £11.10
= 304 = £5.18 [2]
Money received (c) Time = 1.5 km + 280 m
= (80 × $150) + (304 – 80)($300) 100 km/h
= $12000 + $67200 = 0.0178 h
= $79200 = 64.08 s [2]
= 7.92 × 10 4 [3] (d) Let time = t
(b) (i) 117000 = 150y + (400 – y)300 [1] 14 = (3.50t) × 0.8
(ii) 117000 = 150y + 120000 – 300y t = 5 h [2]
150y = 3000 8 (a) $6.90 × 3729 = $25730.10
y = 2 [2] = $25700 [1]
(c) (i) Wages (b) 3729 × 1 =3300 [2]
3 13 1.13
= 1.03 × 10 × 4 × 25 5
6
= $401700 [2] (c) 3510 × 18 = 975 [2]
(ii) Profit 1
1 (d) (3729 + 3300 + 3510) × = 1171 [2]
= 1.03 × 10 × 4 9
6
= $257500 (e) Total percentage of fashion content
Interest received 3
= 257500 (1 + 0.05) – 257500 (100% – 15%) × 41 17 %
3
= $40588 [2] =
3 100%
6 (a) (i) 1.75 : 150% : 4 = 35%
7 3 3 Advertisement = 15%
: : Fashion = 35%
4 2 4
Answers 157
Answers.indd 157 15/03/2022 11:08 AM
Business + Education = 100% – 15% – 35% (d) Original airfare
= 50% = $250 × 100
Business content = 60 83
50% 90 = $301.20 [1]
Business content = 33 1 % [4] 12 ξ
3 A B
9 Let Y = each other art pieces that are sold 46 12 16
($150 × 0.68 × 50) + ($150 × 0.55 × 25) + ($150 ×
0.48 × Y) = 8000 35 5 2
Y = 11.63
8
Total amount of artwork C
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= 50 + 25 + 12 [5]
= 87 [4] 13 (a) 6174 = 2 × 3 ×7 3 [2]
2
3
3
2
2
174 km (b) √2 × 3 × 7 × (2 × 3)
3
h 1740 km = √2 × 3 × 7
3
3
3
10 (a) (i) = [2]
6 h h 2 = 42
60 k = 2 × 3
2
( 46 – 4 + 52 ) (174) (c) 180 = 2 × 3 × 5 [2]
= 12
2
2
2
2
(ii) Distance = 60 60 2 × 3 × 5 × m = multiple of 2 × 3 × 7 3
2
2 m = 7 = 343
3
2
= 136.3 km [2] 180m = 2 × 3 × 5 × 7 3
2
(iii) Average speed = 61740 [2]
Total distance (d) 61740 = 2 × 3 × 5 × 7 3
2
2
= b = 2, c = 2, d = 5, e = 1
Total time a = 2 × 3 × 5
2
2
= 136.3 km = 180 [2]
52 h (e) 180 × f = 2 × 3 × 5 × f
2
2
60 = 2 × 3 × 5 × (2 × 3 × 5 )
2
4
4
2
5
= 157.269 km/h = 2 × 3 × 5 6
6
6
= 43.69 m/s [2] f = 4050000 [2]
(b) Time 14 (a) Time taken for the first part of the journey
1.1 km + 250 m
= = 30.9 s [2] 3 × 420
43.69 m/s 4
(c) Speed = 375 km = 60
156.5 min = 3 hours
= 143.77 km/h [2] Time taken for the second part of the journey
11 (a) (i) 4, 3, 7, 1 [2] 4
(ii) 12 [2] 7 × 420
979 =
(b) (i) Time = = 2.2 h 80
445 = 3 hours
10.10 a.m. + 2.2 h Total distance
= 10.10 a.m. + 2 h 12 min Average speed = Total time
= 12.22 p.m. [2] 420
(ii) 2.2 h × 7 = 3 + 3 + 0.25
11
= 132 min × 7 = 67.2 km/h [4]
= 84 min 11 (b) Speed of car B = 67.2 × 100
= 1 h 24 min [2] 105
(c) Fuel used = 64 km/h
= 979 km Time car B arrives at City Q
55 m/l = 21.30 + 420
= 979000 m 64
55 m/l = 21.30 + 6.5625
= 17800 l = 21.30 + 6 h 34 min
= 1.78 × 10 l [2] = 4.04 a.m. [3]
4
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158 Ace Your Mathematics
Answers.indd 158 15/03/2022 11:08 AM
(c) Time car C arrives at City Q x = 2.54 or x = –0.29 [3]
420 (x + 4)(x – 4)
= 21.30 + (c)
67.5 3(x + 4)
= 21.30 + 56 = x – 4 [2]
9 3
= 21.30 + 6 h 13 min (d) 1 – 4x + 4x – xy + 4x
2
= 3.43 a.m. [3] = 4x – xy + 1 [2]
2
(d) Time car A arrives at City Q 2 (a) 3g(1 – 4g + 2g ) [2]
2
= 21.30 + 470 (b) a(c + d) + b(c + d)
67.2 = (a + b)(c + d) [3]
= 21.30 + 6.25 (c) 5(x – 2) + (x – 2)
2
= 21.30 + 6 h 15 min = (x – 2)[5(x – 2) + 1]
= 3.45 a.m. = (x – 2)(5x – 9) [2]
4.04 a.m. – 3.45 a.m. 3 (a) x = 3 – 6y
2
= 19 min [3] x = ±√3 – 6y [2]
15 1 cm : 280000 cm
1 cm : 2800 m (b) 3y – 2 = √x – 1
1 cm : 2.8 km x – 1 = (3y –2)
2
2
(a) 17 = 6.071 cm x = 9y – 12y + 4 + 1
2
2.8 = 9y – 12y + 5 [3]
∼ 6.1 cm [2] (c) −2xy – 6y = 1 + x
(b) 1 cm : 7.84 km x + 2xy = –6y – 1
2
2
3.2 = 0.4082 cm x(1 + 2y) = –6y – 1
2
7.84 x = – 6y + 1 [3]
∼ 0.41 cm 2 [2] 1 + 2y
(c) 6.25 × 7.84 = 49 km ( ) 2
1
2
= 49 × 1000 × 100 cm 4 (a) –3 = 4 2 – p
2
2
2
= 4.9 × 10 cm 2 [2] p = 1+ 3
11
(d) Distance between Cyberjaya and Putrajaya = 4 [2]
= 12 × 0.75 (b) p = ± – r
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= 9 km
q
Speed on the return journey 1
9 = ± – 2
=
0.75 – 0.15 –3
= 15 km/h 1
Percentage increased = ± 6 [2]
15 – 12
= × 100% p + 2qr
12 (c) p =
= 25% [4] 2q
2pq = p + 2qr
p(2q – 1) = 2qr
2 2 Algebra and Graphs p = 2qr
2q – 1
1
Algebra 2(–3) ( )
2
1 (a) x + 4x + 4 – 15 + 6x = 0 = 2(–3) – 1
2
2
7x + 4x – 11 = 0
2
(7x + 11)(x – 1) = 0 = 3 [2]
7
x = – 11 or x = 1 [2]
7 x × 1
81
3x – 15 – 8x – 12 3
(b) = 2 5 (a)
2x – 7x – 15 1 1
2
9
3
–5x – 27 = 4x – 14x – 30 3 × 3 x × 3
2
4x – 9x – 3 = 0 = x 27
2
2
–(–9) + √(–9) – 4(4)(–3) 3x 3
x = x 24
2(4) = [3]
3
Answers 159
Answers.indd 159 15/03/2022 11:08 AM
(b) 2x 2 (c) PQ = 3 (3) cm
13 1 4
(x 3 ) 2 = 2.25 cm [1]
= 2x 13 2 8 (a) Volume
x 6 = (x + 6)(2x)(2x – 1)
–
= 2x 6 1 [3] = (2x + 12x)(2x – 1)
2
3
(c) 2 × 2 = 2 = 4x + 22x – 12x cm 3 [2]
–
–2a
2
3
7
2
( )
–2a + – 3 = 7 (b) (i) Length = 2x – 4 cm
Width = 2x – 1 – 4
2
−2a = 17 = 2x – 5 cm
2 Height = x + 6 – 2
(ii) (x – 9)(3x + 32) = 0Reserved.
a = − 17 [3] = x + 4 cm [3]
4 (ii) Volume = (2x – 4)(2x – 5)(x + 4)
6 (a) 2x – 25y = 135 = (4x – 18x + 20)(x + 4)
2
2
y = 4x – 2x – 52x + 80 [2]
3
2x = 9 – 5 (c) (i) (4x + 22x – 12x) – (4x – 2x – 52x + 80)
3
2
2
3
y = 2224
9 – = 135 + 25y
5 24x + 40x – 80 – 2224 = 0
2
45 – y = 675 + 125y 3x + 5x – 288 = 0 (shown) [2]
2
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126y = –630
y = –5 x = 9 or x = – 32 (invalid)
2x = 9 – (–5) 3
x = 5 5 [4] Length = 2(9) = 18 cm
(b) y = 3 – x Width = 2(9) – 1 = 17 cm
x + 3x = 3 – x + 2 Height = 9 + 6 = 15 cm
2
x + 4x – 5 = 0 2x cm is the longest side. [3]
2
(x + 5)(x – 1) = 0 9 (a) 50 kg [1]
x = –5 or x = 1 x
y = 8 or y = 2 [4] (b) 50 – 50 = 1.25
x x + 2
(c) y = 21 – x – 4 50(x + 2) – 50x
7 x(x + 2) = 1.25
y = 17 – x 1.25x + 2.5x – 100 = 0
2
7 x + 2x – 80 = 0 (shown) [3]
2
y + 4 = 9 + 3 – 6x (c) x + 2x – 80 = 0
2
5 (x – 8)(x + 10) = 0
y = 5 + 3 – 6x x = 8 or x = –10 [2]
5
28 – 6x (d) Increased price = $8 + $2
y =
5 = $10 [1]
17 – x 28 – 6x 10 (a) Kaylee’s age on April 2018 = x – 2
=
7 5 Lily’s age on April 2018 = 4(x – 2)
85 – 5x = 196 – 42x = 4x – 8 [2]
37x = 111 (b) (x + 2)(4x – 6) = 9(x – 2) 2
2
2
x = 3 9x – 4x – 36x – 2x + 36 + 12 = 0
2
y = 17 – 3 5x – 38x + 48 = 0 (shown) [2]
7 (c) (x – 6)(5x – 8) = 0
y = 2 [4] x = 6 or x = 8
( ) 5
3y
(y – 3 + y) Lily’s age on April 2018 = 4(6) – 8 = 16
7 (a) Area = 4 Lily’s age on April 2010 = 16 – 8 = 8 [2]
2
6y – 9y
2
= [2] Functions
8
(b) 6y – 9y = 27 ( 3 ) 2
2
2y – 3y − 9 = 0 1 (a) hg(x) = x + 1 – 2
2
(y – 3)(2y + 3) = 0 9 6
3 = 2 + – 1
y = 3 or y = – [2] x x
2
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160 Ace Your Mathematics
Answers.indd 160 15/03/2022 11:08 AM
2
9 + 6x – x 2 (b) –9x – 3x
[
= [2] 1 1 2 1 2
x 2 = –9 x + x + ( ) ( )]
–
2
(b) gf(100) 3 2 3 3
[(
1
= 3 + 1 = –9 x + 1 ) ]
–
√100 – 4 9 9
3 ( 1 ) 2
= [2] = –9 x + + 1
2 9
x
(c) √ – 4 = 5 a = –9, b = – 1 , c = 1
x
(
√ = 9 9 1
x = 81 [2] Maximum point − 9 , 1 ) [4]
(d) Let g (x) = y (c) 3x + 1 = 6 + 4
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–1
3 x
+ 1 = x
2
y 3x + x = 6x + 4
2
g (x) = 3 3x – 5x – 4 = 0
–1
x – 1 –(–5) + √(–5) – 4(3)(–4)
2
3 = 1 x = 2(3)
x – 1 x – 2
2
x = 2.26 or x = −0.59
3x – 6 = x – 1 4 (a) f(–2) = 3(–2) – (–2) – 2 [3]
2
2
3x – x – 5 = 0 = 12 [1]
2
–(–1) + √(–1) – 4(3)(–5) (b) 3x – x – 2 = 2
2
2
x =
2(3) 3x – x – 4 = 0
2
x = 1.47 or x = –1.14 [4] (3x – 4)(x + 1) = 0
4
2 (a) fg(x) = 6 ( x – 3 ) – 2 x = 4 or x = –1 [2]
3
–1
= 30 – 2x [2] (c) Let g (x) = y
x – 3 4 – 5y = x
(b) fg(x) = x y = x – 4
30 – 2x = x –5
x – 3 g (x) = 4 – x [2]
–1
30 – 2x = x – 3x 5
2
2
x – x – 30 = 0 (shown) [2] (d) 3(4 – 5x) – x – 2 = 0
2
2
(c) y = x – x – 30 5x – 7x + 2 = 0
2
= (x – 6)(x + 5) (5x – 2)(x – 1) = 0
y x = 2 or x = 1 [2]
5
(e) (i) B [1]
(ii) F [1]
2
x 5 (a) gf(x) = (x – 2) – 25
–5 6 = x – 4x – 21 [2]
2
(b) gf(x) = 0
x – 4x – 21 = 0
2
(x – 7)(x + 3) = 0
–30 (c) x = 7 or x = –3 [2]
[3] gf(x)
(d) (i) x – x – 30 2
2
(
= x – 1 ) – 121 –3 7 x
2
4
a = 1 , b = – 121 [3]
2 4
(ii) Minimum value ( 1 2 , – 121 ) [1]
4
3 (a) fh(x) = (3x + 1) – (3x + 1) 2 –21 [3]
= 3x + 1 – (9x + 6x + 1) (d) gf(x) = x – 4x – 21
2
2
= –9x – 3x (shown) [2]
2
= (x – 2) – 25
2
a = –2, b = –25 [3]
Answers 161
Answers.indd 161 15/03/2022 11:08 AM
(e) Minimum point = (2, –25) [1] (d) $60 × 4 + $70 × 14 = $1220 [3]
(f) x = 2 [1] 5 (a) x > 2y
x > 120
Inequalities y > 80
x + y ≤ 500 [4]
1 (a) 4 ≥ 3x – 4x (b)
2
3x – 4x – 4 ≤ 0
2
(3x + 2)(x − 2) ≤ 0 y
2
– ≤ x ≤ 2 [2]
3 500
(b) 10x + 6 > –6 + 15 x
5x < 12 400
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12
x < [2]
5
(c) –9 ≤ 2x + 1 < 15 300
–10 ≤ 2x < 14
–5 ≤ x < 7 [3]
2. 200
y
25 100
20
x
15 0 100 200 300 400 500
10
[4]
5 (c) 160 × $15 + 80 × $20 = $4000 [2]
x
0 5 10 15 20 25 30 35 40 45 50
Sequences
[4]
3 L , y ≤ x 1 (a) 29 + 5 = 34 [1]
1
L , 10y – 2 ≥ 3 (b) T = a + (n – 1)d
n
2
L , y < 4 [3] = 14 + (n – 1)(5)
3
4 (a) (i) x ≥ 4 [1] = 9 + 5n [2]
(ii) y ≥ 6 [1] (c) 9 + 5n = 324
(iii) x + y ≤ 18 [1] n = 63 [2]
(b) 60x + 70y ≥ 840 2 (a) 20a – 6a = 14a [1]
6x + 7y ≥ 84 [1] (b) T = a + (n – 1)d
n
(c) = 38a + (n – 1)(–6a)
y = 38a – 6an + 6a
= 44a − 6an [2]
20 (c) (i) T = 44(2) – 6(2)n
n
T = 88 – 12n
n
88 – 12n < 0
15 –12n < –88
n > 7 1
3
n = 8 [2]
10 (ii) 88 – 12n = –1172
n = 105 [2]
3 (a) T = a + (n – 1)d
n
5 = 5 + (n – 1)(2)
= 3 + 2n
n
T = [2]
x n 3 + 2n
0 5 10 15 20 118
(b) T =
118
[4] 3 + 2(118)
118
= [1]
239
Cambridge IGCSE
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162 Ace Your Mathematics
Answers.indd 162 15/03/2022 11:08 AM
4 (a) T = n 2 [2] = 8 + (n – 1)(6)
n
(b) T = n + 1 [2] = 6n + 2
2
n
(c) T = 2(n + 1) Total perimeter
2
n
= 2n + 2 [2] = [6(1) + 2] + [6(2) + 2] + [6(3) + 2]… +
2
5 (a) 2a = 2 [6(49) + 2] + [6(50) + 2]
3a + b = 6 = 6(1) + 6(2) + 6(3)… + 6(49) + 6(50) +
a + b + c = 3 (2 × 50)
a = 1, b = 3, c = –1 = 6(1 + 2 + 3 + … + 49 + 50) + 100
T = n + 3n – 1 [3] = 6(1275) + 100 = 7750 [4]
2
n
(b) n + 3n – 1 = 269 9 (a)
2
n + 3n – 270 = 0 Diagram 5 6
2
(n – 15)(n + 18) = 0 Number of dots 21 28
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n = 15 or n = –18 (invalid) [2] Number of lines 45 63
6 (a) 2a = 4 Number of triangles 25 36
3a + b = 3
a + b + c = –1 [3]
a = 2, b = –3, c = 0 (b) (i) 2a = 1
T = 2n – 3n 3a + b = 3
2
n a + b + c = 3
2, 5 ,8, 11 1
T = a + (n – 1)d a = 2
n
= 2 + (n – 1)(3) 3
= 3n – 1 b = 2
2
T = 2n – 3n [3] c = 1
n
3n – 1
2
2
(b) T = 2(10) – 3(10) T = 1 2 n + 3 n + 1 [3]
2
n
10
3(10) – 1
2
= 170 [1] (ii) 1 n + 3 n + 1 = 136
2
2
29 n + 3n – 270 = 0
2
2
(c) 2n – 3n > 10 (n – 15)(n + 18) = 0
3n – 1
n = 15 or n = –18 (invalid)
2n – 3n > 30n – 10 (c) (i) 2a = 3 [2]
2
2n – 33n + 10 > 0 3a + b = 6
2
n < 0.309 (invalid) or n > 16.19 a + b + c = 3
n = 17 [2]
7 (a) T = 4 – 16 = 240 [1] a = 3
4
5
(b) 4, 7, 10, 13 2
3
a + (n – 1)d b =
= 4 + (n – 1)(3) c = 0 2
= 3n + 1
T = 4 n – 1 – (3n + 1) T = 3 n + 3 n [3]
2
n
T = 4 n – 1 – 3n – 1 [2] n 2 2
n
(c) T = 4 10 – 1 – 3(10) – 1 (ii) T = 3 (10) + 3 (10)
2
10
= 4 – 31 10 2 2
9
= 262113 [2] = 165 [1]
8 (a) a = 70 (d) T = n
2
n
b = 96 [1] T = 20 2
20
(b) 2a = 4 = 400 [2]
3a + b = 10 (e) Sequence of perimeter: 3, 6, 9, 12
a + b + c = 6 T = a + (n – 1)d
n
a = 2, b = 4, c = 0 T = 3 + (n – 1)(3)
n
t = 2n + 4n [3] T = 3n
2
n
(c) 390 = 2n + 4n 3n = 459
2
n + 2n – 195 = 0 n = 153 [2]
2
(n – 13)(n + 15) = 0 10 (a) (i) Plan A: $90
n = 3 or n = –15 (invalid) [2] Plan B: $80
(d) Perimeter: 8, 14, 20, 26… Plan C: $25 [1]
T = a + (n – 1)d
n
Answers 163
Answers.indd 163 15/03/2022 11:08 AM
(ii) Plan A: (b)
T = a + (n – 1)d Distance (km)
n
= 50 + (n – 1)(10)
= 40 + 10n
2000
Plan B:
T = ar n – 1
n
= 5(2) n – 1
Plan C: 1000
T = n 2 [2]
n
(b) (i) Plan A:
Total: 5(8) + 45(8) = $680
2
0
Plan B: 1300 1500 1700 1900 2100 2300 0100
Total: 5(2 – 1) = $1275 Time (hours) [2]
8
Plan C: 3 (a) 8.45 a.m. [1]
Total: 8(8 + 1)(2 × 8 + 1) = $204 [3] (b) 300 m [1]
6 (c) 11.15 a.m. [1]
(ii) 5(2 – 1) = 4500 (d) 11.15 a.m. – 9.45 a.m. = 1.5 h [2]
n
2 = 901 (e) 1200 m × 2 = 2400 m [1]
n
n = 9.815 (f) Speed = 1200 m
n ∼10th month [2] 1 1 h
= 1 m/s
3
Speed, distance and time 1200 m
Speed =
2
1 (a) (i) Total distance travelled 2 1.5 h
= 18 km + 5 km = 9 m/s
= 23 km 1
Difference = m/s [2]
Total time used v 30 9
18 5 4 (a) =
15
k
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=
+
7.5 4 v = 2k [2]
= 2.4 + 1.25 = 3.65 h 30
Complete = 7.38 a.m. + 3.65 h (b) Deceleration = Time
= 7.38 a.m. + 3 h + (0.65 × 60) min Time = 30
= 11.17 a.m. [3] 1.5
(ii) Average speed = 23 = 20 s
3.65 Time it starts decelerating = 60 – 20
= 6.301 km/h [1] = 40 s [2]
(b) (i) x + x + 7 = 23 (c) Distance = (60 + 25)(30)
x = 8 [1] 2
(ii) Total distance travelled = 23 km = 1275 m
Total time used = 1.275 km [2]
= 8 + 15 = 3.5 h [2] 5 (a) Total distance = [(12 – 5) + 17]v
4 10 2
Average speed 96 = 24v
23 km v = 4 m/s [2]
= (T – 25)(10)
3.5 h (b) = 48
= 46 km/h 2 T – 25 = 9.6
7
(c) Time = 7.38 a.m. + 3.5 h T = 34.6 s [2]
= 7.38 a.m. + 3 h + 30 min (c) (i) Acceleration = 4 m/s 2 [1]
= 11.08 a.m. 5
Susan will complete the journey first. [2] (ii) Acceleration = 10
2 (a) Average speed = 2000 = 363.64 km/h [1] 34.6 – 25
5.5 = 1.04 m/s 2 [2]
Cambridge IGCSE
TM
164 Ace Your Mathematics
Answers.indd 164 15/03/2022 11:08 AM
(d) Distance (m) (e) x = –1.5 to –1.6
x = 0.2 to 0.3 [2]
40 3 (a) –0.75, –1.11, –1, 1.5 [4]
34
30 (b) y
24
20 7
10 6
Time (s)
0 5 12 17 20 5
[3]
4
Graphs 3
1 (a) 2.5, 5.2, –5.2, –2.5, –2.9 [4] 2
(b) and (d) 1
y
3 x
0 1 2 3 4 5 6
2 –1
1 –2
x
–5 –4 –3 –2 –1 0 1 2 3 4 5 [5]
–1 (c) x = 0.5
–2 x = 3.7 [4]
x + 10
–3 (d) (i) Draw the line of y = . [3]
5
[5] (ii) x = 0.4 to 0.6
(c) k =1 or k = –1 [2] x = 4.6 to 4.8 [2]
x + 10
x
2
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2
(d) See the graph in (b)
[2]
+
– 3 =
(iii)
(e) x = 4.8 to 5 [2] 4 x 5
3
2
(f) x + 4 = 2x – 7 x + 10 = x + 8 – 12x
2x 5 4x
x + 4 = 4x – 14x 5x – 4x – 100x + 40 = 0
2
2
2
3
3x – 14x – 4 = 0 5 : 100 : 40
2
a = 3, b = –14, c = –4 [3] 1 : 20 : 8 [3]
2 (a) 2.25, 1, –3.186, 1 [4] 1 1
(b) y 4 (a) 2, – 3 , – 2 , –0.7 [4]
(b)
y
3
9
2
8
1
7
x
–2 –1 0 1 6
–1
5
–2
4
–3 3
–4 2
[5] 1
(c) (i) x = –2, –1, 1 [3]
(ii) x = 1 to 1.1 0 1 2 3 4 5 6 7 8 9 x
x = –0.8 to –0.9 [2] –1
(d) –3 ≤ k ≤ 2 [2]
[5]
Answers 165
Answers.indd 165 15/03/2022 11:08 AM
(c) (1, 2) –7 + y
(3.5, –0.5) [4] 3 2 = 1
(d) –6 to –7 [3] −7 + y = 2
(e) x – 6x + 7 ≤ 2x y = 9 [2]
2
x – 6x + 7
2
≤ 2 10 – 0
x 4 (a) m = –7 – 1
y ≤ 0 10
1 ≤ x ≤ 7 [2] = –8
2
(f) 2x – 17x + 14 = 0 5
2 = – 4
x – 17 x + 7 = 0 5
2
5All Rights Reserved.
2 10 = – (–7) + c
x – 6x – 5 x + 7 = 0 4
2
2 c = 5
5 4
x – 6x + 7 = x
2
2 y = mx + c
x – 6x + 7 5 5 5
2
= y = – x +
x 2 4 4
5 4y + 5x = 5 [3]
y =
2 (b) m = –1 ÷ – 5
x = 0.9 to 1 4 4
x = 7.5 to 7.6 [5] = 5 [2]
5 (a) 25, 30, 30, 25 [4] 5 (5, 0)
(b) (0, –3)
y 0 – (–3) 3
m = =
5 – 0 5
35 3
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y = x – 3 [3]
30
6 0 = 3 x – 5
25 5
x = 3
20
Q(3, 0)
15 P(–6, 0)
5
10 0 = ( ) (–6) + c
3
5 c = 10
0 x y = 5 x + 10
1 2 3 4 5 6 7 8 3
3y – 5x = 30
[5] a = 3
(c) 34 metres [1] b = –5
(d) 1 metre [1] c = 30 [4]
(e) –1 ± 0.05 [3] 5 – 3 k – 5
7 (a) =
4 – (–1) 9 – 4
3 3 Coordinate Geometry 2 = k – 5
5
5
k = 10 + 5
Coordinate geometry 5
k = 7 [3]
–7 – (–1)
1 Gradient = (b) m = 2
–2 – (–7) 5
6
= – [2] 5 = 2 (4) + c
5 5
2 Distance between AB = √(0 – 6) + (1 – 9) 2 17
2
= √36 + 64 c = 5
= 10 units [3] 2 17
y = x + [3]
5 5
Cambridge IGCSE
TM
166 Ace Your Mathematics
Answers.indd 166 15/03/2022 11:08 AM
(c) Midpoint 11 (a) Midpoint = ( –10 + 2 , 4 – 5 )
( –1 + 9 3 + 7 ) 2 2
= 2 , 2 1
= (4, 5) (shown) [2] = (–4, – 2 ) [2]
3 – (–5) 8 4 4 – (–5) 9
8 (a) m = = = – (b) m = =
–6 – 0 –6 3 –10 – 2 –12
4 3
y = – x – 5 [3] = –
3 4
(b) Length = √(–6 – 0) + (3 – (–5)) 3
2
2
( )
4 = – (–10) + c
= √36 + 64 4
=10 units [3] c = – 7
(c) R(6, 3) [1] 2
12 × 8 3 7
(d) Area = y = – x – [3]
2 4 2
= 48 square units [2] (c) 3 = – 3 (1) + c
9 (a) Length = √(–8 – 6) + (12 – (–2)) 2 15 4
2
2
2
= √(–14) + (14) c = 4
= 14√2 [3] y = – 3 x + 15 [3]
(b) Midpoint = ( –8 + 6 , 12 + (–2) ) 4 3 4 4
2
2
= (–1, 5) [2] (d) m = –1 ÷ – 4 = 3
(c) Gradient, m = 12 – (–2) = –1 3 = 4 (1) + c
–8 – 6
1
12 = –1(–8) + c 5 3
c = 4 c = 3
y = –x + 4 [2] Equation for the perpendicular line to RS:
(d) m = –1 = 1 4 5
2 –1 y = 3 x + 3
5 = (1)(–1) + c
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7
3
5
4
c = 6 3 x + 3 = – 4 x – 2
y = x + 6 [2] 25x 31
Length × 3 = –
10 (a) = 12 12 6
2
Length of triangle = 8 units x = – 62
p = 6 – 8 units 25
p = –2 [2] y = – 41
2 – 0 1 25
(
(b) (i) m = = 62 41
6 – 2 2 T – 25 , – 25 ) [4]
1
0 = ( ) (2) + c 12 2y = 5 + 3x
2
c = –1 y = 3 x + 5
y = 1 x – 1 [3] 2 2
2 m × m = –1
1
2
1
(ii) q = (–2) – 1 2
2 m = –1 × 3
2
q = –2 [2] 2
1 = –
(c) 5 = (6) + c 3
2 2 2
c = 2 3 = – 3 (2) + c
1
y = x + 2 [3] 2 4
2 c = 3 + 3
1
(d) y = (–2) + 2 = 2
2 2
= 1 y = – 3 x + 2
D(–2, 1) [2] 3y = –2x + 6
3y + 2x – 6 = 0 [6]
Answers 167
Answers.indd 167 15/03/2022 11:08 AM
13 Midpoint of AB = ( 1 – 3 , 5 – 3 ) EC = 2.5 × 8
10
2
2
= (–1, 1) EC = 2 cm [2]
ED
– 3 – 5 Area of ΔEDC ( ) 2
Gradient of line AB = (c) =
–3 – 1 Area of ΔABC AB
2.5
–8 ( ) 2
= Area of ΔEDC = × 100
–4 10
= 2 = 6.25 cm 2 [2]
2 (a) ∠A = ∠E and ∠B = ∠D
1 (Alternate angles are equal.)
Gradient of line CD = –
2 ∠ACB = ∠DCE
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1 = – 1 (–1) + c (Vertically opposite)
2 ΔABC ΔEDC [1]
∼
c = 1 (b) 6 = 7x + 4
2 7 11x – 1
Equation of line CD, 66x – 6 = 49x + 28
y = – 1 x + 1 [6] 17x – 34 = 0
2 2 x – 2 = 0
–5 – 3 x = 2 [2]
14 Gradient of line X =
–8 – 10 (c) CE = 11(2) – 1
4 = 21 [1]
=
7
9 (d) Area of ΔABC = ( ) 2
–15 – (–5) Area of ΔEDC 21
Gradient of line Y =
1
2 – 8 = ( ) 2
5 3
=
3 = 1
4 5 20 9
× = (≠ – 1)
9 3 27 Area of ΔABC : area of ΔEDC = 1:9 [2]
X is not perpendicular to Y. [3] 3 (a) (i) 3 = 4x – 5
x 4x – 5 + 2x
9 – (–3) 2
15 Gradient of line X = 18x – 15 = 4x – 5x
–2 – (–10) 4x – 23x + 15 = 0 (shown) [2]
2
3 (ii) (x – 5)(4x – 3) = 0
=
2 3
–6 – (–9) x = 5 or x = 4 (invalid) [1]
Gradient of line Y =
2 – 0 Area of ΔRQS ( ) 2
3
3 (b) (i) Area of ΔRPT = 5
=
2 9
X and Y are parallel lines. [3] = 25
Area of ΔRQS : Area of ΔRPT = 9:25 [2]
4 4 Geometry Area of ΔRQS
(ii)
Area of QPTS
Similar shapes = Area of ΔRQS
Area of ΔRPT – Area of ΔRQS
1 (a) ∠A = ∠E and ∠B = ∠D = 9
(Alternate angles are equal.) 25 – 9
∠ACB = ∠ECD = 9
(Vertically opposite) 16
ΔBAC ΔDEC [1] Area of ΔRQS : Area of QPTS = 9:16 [2]
∼
(b) BC = AB Volume of A 1 3 1
DC ED 4 Volume of A + B = ( 1 + 1 ) = 8
BC = 10 × 3 Volume of A 1 3 1
2.5 = ( ) =
BC = 12 cm Volume of A + B + C 1 + 1 + 2 64
EC ED Volume of B = 8 – 1 = 7 = 1
= Volume of C 64 – 8 56 8
AC AB
Cambridge IGCSE
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168 Ace Your Mathematics
Answers.indd 168 15/03/2022 11:08 AM
Volume of B : volume of C = 1:8 [6] Angle at the centre of a circle is twice the angle
5 (a) (i) 1 : 2 [1] at the circumference.
(ii) 1 : 4 [2] ∠AEC = 180° – ∠ADC
(iii) 1 : 8 [2] = 180° – 57°
3 = 123°
(b) 100 × = 75 cm 2 [2]
4 Cyclic quadrilaterals [2]
(c) Volume = 7 V [2] (c) ∠OAB = ∠OCB = 90°
8 The angle between the tangent and radius is
always 90°.
Circle theorem ∠OABC = ∠OAB + ∠OCB + ∠AOC + ∠ABC
360° = 90° + 90° + 114° + ∠ABC
1 ∠OTS = 90° ∠ABC = 66°
The angle between a tangent and a radius is 90°. Interior angles of quadrilateral is 360°. [2]
∠ORT = 90° (d) ∠DAE = 90°
The perpendicular from the centre to the chord Angle in semicircle is 90°.
bisects the chord. ∠ADE = 180° – 90° – 54°
∠ROT = 180° – 90° – ∠OTR = 36°
= 90° – (90° – ∠STQ) ∠ADE = ∠ACE
= ∠STQ (shown) [4] ∠ACE = 36°
2 ∠POQ = 2 × ∠PSQ Angles in the same segments are equal.
= 2 × 50° ∠OCA + ∠ACE + ∠ECB = 90°
= 100° The angle between a tangent and a radius is
180° – 100° 90°.
∠QPO =
2 ∠ECB = 90° – 33° – 36°
= 40° = 21° [3]
∠SPO = 180° – 40° – 77° 6 (a) (i) ∠OTP = 180° – 68°
= 63° [3] 2
3 (a) ∠BAD = 90° – ∠DAO = 56°
= 90° – 50° The base angles of an isosceles triangle are
equal.
[2]
= 40°
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(b) ∠AOD = 180° – 50° × 2 (ii) ∠TQS = ∠OPT
= 80° = 56°
∠ABC = (180°− 90°− 80°) × 2 Angles in the same segment are equal. [1]
= 20° [2] (iii) ∠QSP + ∠RSQ = ∠SPT
4 (a) ∠RTS = ∠RQS ∠QSP = 56° – 21°
= 180° – 78° – 65° = 35°
= 37° SR // TP, alternate angles are equal. [1]
∠TRS = ∠TQS (iv) ∠PRS = 90°
= 65° Angles in the semicircle is 90°.
∠RST = 180° – 37° – 65° ∠QRP = ∠QSP
= 78° [3] = 35°
(b) ∠RSQ = ∠RTQ Angles in the same segment are equal.
∠RTQ + ∠RTS = 78° + 20° ∠QRS = ∠QRP + ∠PRS
∠RTQ = 98° – 37° = 35° + 90°
= 61° = 125° [2]
∠RSQ = 61° [2] (b) ∠PST = 68°
(c) ∠QRT = 180° – ∠RQS – ∠RSQ – ∠SRT 2
= 180° – 37° – 61° – 65° = 34°
= 17° [2] ∠RST = ∠RSQ + ∠QSP + ∠PST
5 (a) Obtuse ∠AOC = 180° – 33° × 2 = 21° + 35° + 34°
= 114° = 90°
(Isosceles triangle base angles are equal.) RT is the diameter of the circle, thus extension
Reflex ∠AOC = 360° – 114° of TO will pass by R. [2]
= 246° [2]
∠AOC
(b) ∠ADC = Geometric construction
2
114° 1 (a) Correct triangle shape ABC with arc seen
=
2 AB = 12 cm [1]
= 57°
Answers 169
Answers.indd 169 15/03/2022 11:08 AM
(b) Bearing of C from B = 190° (b) Total surface area
BC = 12.2 cm [2] = 2π(12) + 2π(12)(27)
2
2 Accurate plan with 3 arcs seen. = 2940.91 cm 2 [2]
∠PTS = 120° (c) Volume of cylinder A + Volume of cylinder of
∠RST = 130° B
TS = 5 cm [7] = π(8) (18) + π(12) (27)
2
2
(6 – 2) × 180° = 15835.68 cm 3
3 (a) Interior angle = = 120° (shown)
6 [2] Total mass =15835.68 × 11.3 g
(b) Accurate plan with 4 arcs seen with each = 178943.18 g [3]
= 178.94 kg
interior angle of 120°. [3] 2 2 2
4 (a) Correct triangle shape ABC with 2 arcs seen 5 (a) OU = UV – OV
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
= 13 – 12
2
2
∠CAB = 64° OU = √25
∠CBA = 33° [3] = 5 cm [2]
(b) Correct triangle shape ABC with 2 arcs seen (b) (i) RU = OR – OU 2
2
2
∠CAB = 133° = 7 – 5 2
2
∠ACB = 17° [3]
RU = 4.899 cm
SR = 9.798 cm [2]
5 5 Mensuration (ii) Area = [ RQ × OU ] × 5
2
[ 9.798 × 5 ]
Mensuration = 2 × 5
1 (a) 82 [1] = 122.475 cm 2 [2]
(b) 164.9 [1] (c) Volume = 1 × Base area × Height
(c) 30300 [1] 3
(d) 8.76 [1] = 1 × 122.475 × 12
(e) 5290 [1] 3
(f) 6100 [1] = 489.9 cm 3
(g) 450 [1] Price = 489.9 × 19.28 × $55
(h) 38800 [1] = $519470 [3]
(i) 4010 [1] 6 (a) Perimeter of the butterfly
(j) 22.7; 22700 [1] = (AB + BXC + CD + DE + EF + AF) × 2
)
[
5
65
2 (a) ∠POR = 2 × tan –1 ( ) = 3 + 2π × 1.5 + 3 + ( 360 × 2π × 3 + 3 +
2
4.8
= 92.38° ( 1 )]
Reflex angle of ∠POR = 267.661 [2] 4 × 2π × 6 × 2
(b) (i) OP = 4.8 +5 2 13
2
2
OP = 6.931 = 6 + 3π + 6 + 6 π + 6 + 6π
Perimeter = 267.661 × 2π × 6.931 + 10 = 53.09 cm [4]
360 (b) Area of the butterfly
= 42.38 cm [3] = (Area of ACF − Area of BXC +
(ii) Area = 267.661 × π × 6.931 + 10 × 4.8 Area of FED) × 2
2
360 2 ( 1 2 1 2 65 2 )
= 136.21 cm 2 [2] = 4 × π × 6 – 2 π × 1.5 + 360 × π × 3 × 2
3 PQ = x – 8 2 = 59.69 cm 2 [4]
2
2
PQ = √x – 64 7 (a) Curved surface area
2
Area of trapezium = 25 × Area of ΔPQS = π × 5 × 13
(PQ + SR)PS = 25 × PQ × PS = 204.2 cm 2 [2]
2 2 (b) PP' = 2 × π × 5
2
2
(√x – 64 + SR)8 = 25 × √x – 64 × 8 = 31.416 cm
2 2 ∠POP' × 2 × π × 13 = 31.416
SR + √x – 64 = 25√x – 64 360
2
2
∠POP'= 138.462°
2
SR = 24 √x – 64 [5] 138.462 2
Height of cylinder B 12 Area = 360 × π × 13
4 (a) =
18 8 = 204.2 cm 2 [4]
Height of cylinder B = 27 cm [2]
Cambridge IGCSE
TM
170 Ace Your Mathematics
Answers.indd 170 15/03/2022 11:08 AM
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