CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Answers
The questions and example answers that appear in this resource were written by the author. In examination, the way marks would
be awarded to answers like these may be different.
Chapter 1
Getting started
Exercise 1.1
1
1
a
b
2
3
a
Twenty-one thousand, eight hundred and
thirty-seven
a
b
c
93
122
75
2
1
__
(3)
0
4
__
(3)
e
f
g
h
i
2
a
b
c
d
e
3
a
b
c
d
e
f
4
a
_1
92
14 000 010 019
a
Any real-world measurement problems
involve a level of approximation, as do
problems where you have to work out if
you have enough money, or have catered
enough food, estimated times of arrivals,
estimates for building materials and costs
of doing different jobs.
Encourage students to share ideas and
discuss their own methods of deciding.
Answers will vary, but could include that
estimating allows you find errors and
judge the size an answer should be, avoid
mistakes due to button push or place
value errors.
c
a
b
c
d
e
f
f
g
b
1
There are many possible answers for each
value. For example, (a) could be 92 or
9(2 + 7) or 8 × 10 + 1. Let students use
calculators to check that each other’s clues
work.
b
d
4
Student answers will vary based on what
they already know and feel confident
doing.
Some students will select the things they
are less confident in, but other may select
things they enjoy doing or are good at.
Encourage them to say why they have
made each selection.
b
c
{3, 4, 6, 11, 16, 19, 25}
{4, 6, 16}
{3, 11, 19, 25}
{−4, −1, 0, 3, 4, 6, 11, 16, 19, 25}
{−4, −1}
1
__
{2}
{4, 16, 25}
{3, 11, 19}
1 , 0.75, 6}
{−4, −1, 0, __
2
{109, 111, 113, 115}
Various, e.g. {2010, 2012, 2014, 2016} or
{2020, 2022, 2024, 2026} etc.
{995, 997, 999, 1001, 1003, 1005}
{1, 4, 9, 16, 25}
Various, e.g. {0.49, 048, 0.47, 0.46, 0.45}
or {0.4, 0.3, 0.2, 0.1}
13 7
3 ___
7 2 ___
1 , __
Various, e.g. __
, , __
, 11 , ___, ___
3 5 12 3 20 20 10
Even
Even
Odd
Odd
Even
Even
A perfect number is one where the sum of
its factors, including 1, but excluding the
number itself, is that number. 6 is perfect
number because 1 + 2 + 3 = 6.
A palindromic number is a ‘symmetrical’
number like 16461 that remains the same
when its digits are reversed.
A narcissistic number is one that is the
sum of its own digits each raised to the
power of the number of digits,
e.g. 371 = 33 + 73 + 13.
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 1.2
1
2
3
c
a
b
19 , 45
12 + 18 = 30
1
c 0.5 = __
2
d 0.8 ≠ 8.0
e −34 , 2 × −16
___
f ∴ x = √ 72
g x < − 45
h p is approximately equal to 3.14
i
5.1 . 5.01
j
3+4≠3×4
k 12 − (−12) .12
l
(−12) + (−24) , 0
m 12x is approximately equal to −40
a
b
c
d
e
f
g
h
i
j
k
l
m
n
False
True
True
True
True
True
False
True
True
True
False
False
True
False
f
g
h
3
a
b
c
4
576, 396, 792, 1164
5
816 and 1116
32, 36, 40, 44, 48, 52
50, 100, 150, 200, 250, 300, 350
4100, 4200, 4300, 4400, 4500, 4600, 4700,
4800, 4900
1
a
b
c
d
e
f
2
No – the common multiples are infinite.
10
40
12
9
385
66
Exercise 1.5
Exercise 1.3
2
e
Exercise 1.4
Students’ own discussions.
1
a
b
c
d
e
f
g
h
2, 4, 6, 8, 10
3, 6, 9, 12, 15
5, 10, 15, 20, 25
8, 16, 24, 32, 40
9, 18, 27, 36, 45
10, 20, 30, 40, 50
12, 24, 36, 48, 60
100, 200, 300, 400, 500
2
a
29, 58, 87, 116, 145, 174, 203, 232, 261,
290
44, 88, 132, 176, 220, 264, 308, 352, 396,
440
b
d
75, 150, 225, 300, 375, 450, 525, 600, 675,
750
114, 228, 342, 456, 570, 684, 798, 912,
1026, 1140
299, 598, 897, 1196, 1495, 1794, 2093,
2392, 2691, 2990
350, 700, 1050, 1400, 1750, 2100, 2450,
2800, 3150, 3500
1012, 2024, 3036, 4048, 5060, 6072, 7084,
8096, 9108, 10 120
9123, 18 246, 27 369, 36 492, 45 615,
54 738, 63 861, 72 984, 82 107, 91 230
1
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
F4 = 1, 2, 4
F5 = 1, 5
F8 = 1, 2, 4, 8
F11 = 1, 11
F18 = 1, 2, 3, 6, 9, 18
F12 = 1, 2, 3, 4, 6, 12
F35 = 1, 5, 7, 35
F40 = 1, 2, 4, 5, 8 , 10, 20, 40
F57 = 1, 3, 19, 57
F90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
F100 = 1, 2, 4, 5, 10, 20, 25, 50, 100
F132 = 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132
F160 = 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, 160
F153 = 1, 3, 9, 17, 51, 153
F360 = 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20,
24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
3
4
a
b
c
d
e
4
45
14
22
8
a
b
c
d
e
f
g
h
false
true
true
true
true
true
true
false
conjecture is much more difficult to prove
and that the method used to prove the
weak conjecture won’t work for the strong
one.
2
b
3
The smallest factor is 1 and the largest factor
is the number itself.
a
b
c
d
e
f
g
h
3
8
5
14
4
2
22
6
2
a
b
Any two from: 4, 6, 10, 14
12 and 18 are the only possible two, less
than 20
3
1 because each prime number has only 1 and
itself as factors.
4
18 m
5
20 students
6
150 bracelets
If you write prime backwards you get emirp.
An emirp is a prime number that when you
write it backwards gives you a different prime.
For example, 17 and 71. The first few emirps
are: 13, 17, 31, 37, 71, 73, 79, 97, 107, 113,
149, 157.
1
2
2
14
3
a
b
Why do mathematicians find prime numbers
exciting?
1 a Every even integer greater than 2 can be
written as the sum of two prime numbers.
b The weak conjecture is that every odd
integer greater than 5 can be written as
the sum of three odd prime numbers.
Harald Helfgott’s proof uses complicated
mathematics to prove that this is correct.
His proof is largely accepted by the
mathematics community but they also
acknowledge (as does he) that the strong
3
The prime number theorem shows that
prime numbers become less common as
they get bigger using the rate at which
prime numbers occur.
Yes. Euclid (325–265BCE) proved there
are infinitely many prime numbers. This
proof is known as Euclid’s theorem.
Exercise 1.7
Exercise 1.6
1
a
6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22,
24, 25, 26, 27, 28
6 = 3 + 3, 8 = 3 + 5,
9 = 2 + 7, 10 = 5 + 5,
12 = 5 + 7, 14 = 3 + 11,
15 = 2 + 13, 16 = 5 + 11,
18 = 5 + 13, 20 = 3 + 17,
21 = 2 + 19, 22 = 5 + 17,
24 = 5 + 19 or 17 + 7, 25 = 2 + 23,
26 = 3 + 23 or 13 + 13, 27 = not possible,
28 = 5 + 23
4
3 and 5, 5 and 7, 11 and 13, 17 and 19, 29 and
31, 41 and 43, 59 and 61, 71 and 73
5
149 is prime. Determined
_by trial division by
all integers from 2 to √ 149
Exercise 1.8
1
a
b
c
d
e
f
g
h
i
j
30 = 2 × 3 × 5
24 = 2 × 2 × 2 × 3
100 = 2 × 2 × 5 × 5
225 = 3 × 3 × 5 × 5
360 = 2 × 2 × 2 × 3 × 3 × 5
504 = 2 × 2 × 2 × 3 × 3 × 7
650 = 2 × 5 × 5 × 13
1125 = 3 × 3 × 5 × 5 × 5
756 = 2 × 2 × 3 × 3 × 3 × 7
9240 = 2 × 2 × 2 × 3 × 5 × 7 × 11
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 1.9
Exercise 1.11
1
a
b
c
d
e
f
g
h
12
24
18
26
25
22
78
5
1
2
a
b
c
d
e
f
g
h
540
216
360
240
360
2850
270
360
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
2<8
4<9
12 > 3
6 > −4
−7 < 4
−2 < 4
−2 > −11
−12 > −20
−8 < 0
−2 < 2
−12 < −4
−32 < −3
0 > −3
−3 < 11
12 > −89
−3 < 0
3
a
HCF = 36
LCM = 216
HCF = 25
LCM = 200
HCF = 5
LCM = 2280
HCF = 12
LCM = 420
2
a
b
c
d
−12, −8, −1, 7, 10
−10, −8, −4, −3, 4, 9
−12, −11, −7, −5, 0, 7
−94, −90, −83, −50, 0
3
a
b
c
d
e
f
g
h
i
j
k
l
a
b
c
d
e
−4
10
−14
−3
−2.7
5
−6
−6
−27
−4
−4
−5
1 °C
1 °C
−3 °C
12 °C
−3 °C
b
c
d
4
120 listeners
5
36 minutes
6
a
b
c
8
16
2n
Exercise 1.10
1
4
a
b
c
d
e
f
g
h
i
j
k
l
+$100
−25 km
−10 marks
+2 kg
−1.5 kg
8000 m
−10 °C
−24 m
−$2000
+$250
−2 h
+400 m
4
5
$28.50
6
a
b
c
7
−11 m
−$420
$920
−$220
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
8
−8°C
9
a
b
c
d
10 a
b
3
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
x=5
x=2
x = 11
x=9
x = 18
x = 20
x = 20
x = 15
x=1
x = 81
x=1
x = 6561
x=8
x=1
x=4
4
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
3
8
1
2
10
0
9
20
36
42
2
1
−3
4
10
−6
8
9
−12
18
8 p.m.
12 p.m.
10 p.m.
1 a.m.
17.1 litres per day
578 litres
Exercise 1.12
1
2
5
a
b
c
d
e
f
g
h
i
j
9
49
121
144
10 000
196
1
27
64
1000
a
b
c
d
e
f
g
h
i
j
441
361
1024
4624
216
729
1 000 000
5832
27 000
8 000 000
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
a
324
____
√ 324
_
√ 324
2×2 × 3×3 × 3×3
⏟
⏟
⏟
= 2
× 3
× 3
= 18
=
b
225
=
____
√ 225 =
_
√ 225 =
c
784
_
√ 784
_
√ 784
d
2025
_
_
3×3 × 3×3 × 5×5
⏟
⏟
⏟
= 3
× 3
× 5
= 45
=
2×2 × 2×2 × 5×5 × 7×7
⏟
⏟
⏟
= 2
× 2
× 5
× 7
= 140
19 600
=
_
√ 19 600
_
√ 19 600
250 000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5 × 5
⏟ ⏟ ⏟ ⏟
_______ ⏟
√ 250 000 = 2 × 2 × 5 × 5 × 5
_______
√ 250 000
6
a
5×5
5
2×2 × 2×2 × 7×7
⏟
⏟
⏟
= 2
× 2
× 7
= 28
√ 2025
f
×
×
=
√ 2025
e
3×3
3
15
= 500
27 = 3 × 3 × 3
_ ⏟
√ 27 = 3
3
b
3×3×3 × 3×3×3
⏟
⏟
3
×
3
729 =
3
_
√ 729 =
_
√ 729 =
3
c
9
2197 = 
13 × 13 × 13
3
_
√ 2179 = 13
d
1000
3
_
√ 1000
_
√ 1000
3
e
2×2×2 × 5×5×5
⏟
⏟
=
2
×
5
=
=
5×5×5 × 5×5×5
⏟
⏟
√ 15 625 =
5
×
5
15 625 =
3
_
_
√ 15 625 =
3
f
10
25
32 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
⏟ ⏟ ⏟ ⏟ ⏟
_
3
√ 32 768 =
2
×
2
×
2
×
2
×
2
_
√ 32 768 = 32
3
6
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
a
b
c
d
e
5
6
7
1
__
2
3
1
__
4
1
__
2
1
___
16
1
___
16
a
4−1
b
c
d
e
f
g
h
5−1
7−1
9−1
10 000−1
256−1
49−1
18−1
a
b
c
d
e
f
g
h
5.0625
1000
2.25
0.015 625
36
8
13
17
a
b
c
d
e
f
g
h
31
32
36
3−3
3−1
30
3−5
−(32)
3
4
52
b
83
c
13 3
d
11 4
e
93
f
63
g
32 4
h
2(12 5)
a
b
c
d
e
f
g
h
i
j
k
l
5
a
c = 70 × (√ m )
b
251.40 calories
c
41 622.25 calories
1
a
b
c
8
___
b
√ 25
__
3
√3
c
√ 40
d
e
√6
_
8
√3
f
(√ 2 )
g
(√ 12 )
h
(√ 5 )
a
d
___
_
4
3
9
e
_ 3
_ 2
_ 2
_1
_1
_1
_2
_4
_3
_7
3
4
8
36
0.5
6.78
0.0016
0.5
16
36
64
4
_ 3
Exercise 1.16
Exercise 1.15
1
_1
a
f
(4 + 7) × 3
= 11 × 3
= 33
(20 − 4) ÷ 4
= 16 ÷ 4
=4
50 ÷ (20 + 5)
= 50 ÷ 25
=2
6 × (2 + 9)
= 6 × 11
= 66
(4 + 7) × 4
= 11 × 4
= 44
(100 − 40) × 3
= 60 × 3
= 180
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
g
h
i
j
k
l
2
3
a
b
c
d
e
f
g
h
i
108
72
3
10
32
9
5
1
140
a
5 × 10 + 3
= 50 + 3
= 53
5 × (10 + 3)
= 5 × 13
= 65
2 + 10 × 3
= 2 + 30
= 32
(2 + 10) × 3
= 12 × 3
= 36
23 + 7 × 2
= 23 + 14
= 37
b
c
d
e
9
16 + (25 ÷ 5)
= 16 + 5
= 21
19 − (12 + 2)
= 19 − 14
=5
40 ÷ (12 − 4)
= 40 ÷ 8
=5
100 ÷ (4 + 16)
= 100 ÷ 20
=5
121 ÷ (33 ÷ 3)
= 121 ÷ 11
= 11
15 × (15 − 15)
= 15 × 0
=0
f
6 × 2 ÷ (3 + 3)
= 12 ÷ 6
=2
g
15 − 5
______
2×5
10
= ___
10
=1
h
(17 + 1) ÷ 9 + 2
= 18 ÷ 9 + 2
=2+2
=4
i
16 − 4
______
j
4−1
12
= ___
3
=4
17 + 3 × 21
= 63 + 17
= 80
k
48 − (2 + 3) × 2
= 48 − 5 × 2
= 48 − 10
= 38
l
12 × 4 − 4 × 8
= 48 − 32
= 16
m 15 + 30 ÷ 3 + 6
= 15 + 10 + 6
= 31
n 20 − 6 ÷ 3 + 3
= 20 − 2 + 3
= 21
o 10 − 4 × 2 ÷ 2
= 10 − 4 ÷ 1
= 10 − 4
=6
4
a
b
c
d
e
f
7
7
3
0
3
10
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
a
b
c
d
e
f
g
h
13
8
58
192
12 000
1660
260
868
6
a
b
c
d
e
f
18
3
3
8
4
9
7
a
b
c
d
a
b
c
d
e
f
g
h
i
j
k
l
8
9
10
Exercise 1.17
1
False
True
False
True
3 × (4 + 6) = 30
(25 − 15) × 9 = 90
(40 − 10) × 3 = 90
(14 − 9) × 2 = 10
(12 + 3) ÷ 5 = 3
(19 − 9) × 15 = 150
(10 + 10) ÷ (6 − 2) = 5
(3 + 8) × (15 − 9) = 66
(9 − 4) × (7 + 2) = 45
(10 − 4) × 5 = 30
6 ÷ (3 + 3) × 5 = 5
BODMAS means that brackets are not
needed.
m (1 + 4) × (20 ÷ 5) = 20
n (8 + 5 − 3) × 2 = 20
o 36 ÷ (3 × 3 − 3) = 6
p 3 × (4 − 2) ÷ 6 = 1
q (40 ÷ 4) + 1 = 11
r BODMAS means that brackets are not
needed.
a 2 − 10 ÷ 5 = 0
b 13 − 18 ÷ 9 = 11
c 8 ÷ (16 − 14) − 3 = 1
d (9 + 5) − (6 − 4) = 12
or (9 + 5) − (12 − 4) = 6
2
3
4
5
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
a
b
c
d
e
f
a
b
c
d
e
f
−10
8.86
13
29
−22
8.75
20
0
4
70
12
20
8
15
20
Correct
Incorrect = 608
Correct
Correct
Incorrect = 368
Incorrect = 10
12 ÷ (28 − 24) = 3
84 − 10 × 8 = 4
3 + 7(0.7 + 1.3) = 17
23 × 11 − 22 × 11 = 11
40 ÷ 5 ÷ (7 − 5) = 4
9 + 15 ÷ (3 + 2) = 12
a
b
c
d
0.5
2
0.183
0.5
e
1 is approximately equal to 0.333 (3 s.f.)
__
f
g
3
1
2
h
2 is approximately equal to 0.667 (3 s.f.)
__
3
Correct to 3 significant figures
a 0.0112
b 0.0950
c −0.317
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
6
7
Correct to 3 significant figures
a 89.4
b 20.8
c 7.52
d 19.6
e 2.94
f 1.45
g 0.25
h 2.16
c
a
b
c
d
e
f
g
h
i
f
d
e
1
0.5
−26.94
0.28
14.5
6.54
1728.69
−1999
0.339
g
h
You may find that your calculator gives an exact
answer rather than a decimal. This may include a
root or a fraction. Check your calculator manual
to find out how to change this to a decimal.
i
j
Exercise 1.18
1
2
3
4
a
b
c
d
e
3.19
0.06
38.35
2.15
1.00
a
b
c
d
e
500
53 400
3000
0
10 100
a
b
c
d
e
630 000
100 000
10 000
10 000
160 000
a
i
ii
iii
i
ii
iii
b
11
k
l
5
a
b
c
d
e
f
i
ii
iii
i
ii
iii
i
ii
iii
i
ii
iii
i
ii
iii
i
ii
iii
i
ii
iii
i
ii
iii
i
ii
iii
i
ii
iii
65 240
65 200
70 000
320.6
321
300
25.72
25.7
30
0.0007650
0.000765
0.0008
1.009
1.01
1
7.349
7.35
7
0.009980
0.00998
0.01
0.02814
0.0281
0.03
31.01
31.0
30
0.006474
0.00647
0.006
2.556
2.56
2.6
2.56
2.6
3
Exercise 1.19
4512
4510
5000
12 310
12 300
10 000
1
a
49
___
= 4.9, which is close to 5, so not sensible
b
10
4 × 3 × 9 = 108, so not sensible
c
5 × 8 = 40, so not sensible
d
50 × 8 = 400, so sensible
e
3 × 300 = 900, so not sensible
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
_
f
6 × √ 20 =_
6 × 4.5 (approximate
root
_
between √ 16 and √ 25 ) = 27, so sensible
a
23.6
24 = 4
_____
is approximately equal to ___
b
6.3
6
4
4 is
________
is approximately equal to _____
0.09 × 4
0.36
approximately equal to 11
c
7 × 0.5
3.5
_______
is approximately equal to ____ is
d
9
approximately equal to 0.39
2.5 + 1
approximately equal to 8.6
3.5
12
√ 49
7
_______
is approximately equal to ____ is
2.5 + 4
approximately equal to 1
a
b
5×6
30
_______
is approximately equal to ____ is
3
6.5
f
(0.5 + 2)(6.5 − 2) is approximately equal
to (2.5)(4.5) is approximately equal to 11.3
g
24 + 20
44 = 4
_______
is approximately equal to ___
Zaf changed decimals to fractions to
easily divide by 2. Marwan cancelled
before rounding to have fewer numbers to
deal with.
Once you have rounded, you are
calculating exact values, so even if 2 and 3
are rounded values, 2 + 3 is equal to 5, not
approximately equal to 5.
Possible examples:
a Overestimate the cost of buying several
items to make sure you definitely have
enough money
b Underestimate the size of a doorway to
make sure you have enough room to move
furniture though it.
5+6
11
110 − 45
65
________
is approximately equal to ___ = 13
5
19 − 14
Practice questions
i
3 2 × √ 49 is approximately equal to
9 × 7 = 63
1
49 − 30 = 19
2
9 and −4 or −9 and 4
j
√_
224 × 45 is approximately equal to
√ 10 080 is approximately equal to 100
3
15
4
216 216
k
√ 9 × √ 100 is approximately equal to 3 ×
10 = 30
5
l
43 × 24 is approximately equal to
64 × 16 = 1024
735
736
737
738
739
741
742
743
744
6
1080 = 23 × 33 × 5
1080 is not a cube number. Not all the factors
are powers with indices that are multiples of 3.
7
a
b
33 and 61
26 and 45
8
a
b
c
32
340
25
h
3
2
9
_
e
Making decisions about accuracy
1 a Whole numbers
b 2 d.p.
c Millions
d 4 d.p.
e 3 s.f.
_
_
_
_
Answers given to 1 d.p.
a 3.7
b 12.7
c 0.4
d 8.0
e 1.0
f 10.8
g 4.2
h 11.7
i
44.4
j
100.5
k 30.4
l
898.2
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
9
d
e
f
33
9
−48
a
b
5 × 7 − 3 × 8 = 11
(5 − 32) × 6 + 8 ÷ (−2) = −28
2
Product = −36, which is negative ⇒ one
number is positive and the other is negative.
Factor pairs of 36: 1 × 36
2 × 18
3 × 12
4×9
6×6
You can make a difference of 13 with either
9 and −4 or −9 and 4.
3
The number if one fifteenth of its own square.
⇒ The number must be multiplied by 15 to
square it.
⇒ The number is 15.
4
2 × 2 × 2 × 3 × 3 × 3 × 7 × 11 × 13
= 216 216
5
Look at 154.4574 on a number line and you
will see that the number 154.45ABC must lie
between 154.45735 and 154.45744. (154.45745
rounds up to 154.45735.)
10 (7 + 14) ÷ (4 − 1) × 2 = 14
11 1.16
12 a
b
−4
0.276 to 3 s.f.
13 D, C, B, A
14 a
b
___
√ 338
17
___
c
2
5
d
216
____
e
3
__
15 a
b
c
125
2
154.4573
60 = 22 × 3 × 5
36 = 22 × 32
LCM = 22 × 32 × 5
= 180
28 August 2023
154.45735
Practice questions worked
solutions
The prime numbers smaller than 20 are:
2, 3, 5, 7, 11, 13, 17, 19
Sum of the three largest prime numbers
smaller than 20
= 13 + 17 + 19
= 49
Product of the three smallest prime numbers
=2×3×5
= 30
Difference = 49 − 30
= 19
154.4575
154.45745
So, A must be 7. The possibilities are:
16 BAD
1
154.4574
A
B
C
7
3
5
7
3
6
7
3
7
7
3
8
7
3
9
7
4
1
7
4
2
7
4
3
7
4
4
6
1080
540
270
135
27
9
3
2
2
2
5
3
3
1080 = 23 × 33 × 5
The power of 5 is not a multiple of 3 so 1080
is not a cube number.
13
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7
8
a
b
1170 = 2 × 3 × 3 × 5 × 13
= (3 × 3 × 5) × (2 × 13)
= 45 × 26
45 − 26 = 19 so the numbers are 45 and 26.
a
b
c
12 + 20 = 32
4 × 85 = 340
11 × 2 + (15 − 6) − 6
= 22 + 9 − 6
= 25
−15 − (−48)
= −15 + 48
= 33
−3 × (−11) + (−24)
= 33 − 24
=9
(−4)3 + 16
= −64 + 16
= −48
d
e
f
9
a
b
b
2013 = 3 × 11 × 61
= 33 × 61
33 + 61 = 94 so the numbers are 33 and 61.
(3 −2 + 2 −3) × 216 _23
2
1 + __
1 × (3√_
)
= __
216
(32 23)
1 + __
1 × 62
= (__
9 8)
17
17
= ___ × 36 = ___
72
2
((√ 2 ) + 23)
_ 2
c
= √ (2 + 23)
_
= √ 25
=5
d
√
e
2
1
____
( 81 ) = 4 ___
16
___
√ 81
1
= ____
2
__
(3)
3
= __
2
15 a
is approximately equal to
Calculator answer = −4.276 348 739 …
Difference = 0.276 348 739 …
= 0.276 (to 3 s.f.)
b
c
60 = 2 × 2 × 3 × 5
36 = 2 × 2 × 3 × 3
2 × 2 × 3 × 3 × 5 = 180
180 days after 1 March 2023 is 28 August
2023.
16
7500
1500
13 A = 4 × (4 + 16) = 4 × 20 = 80
64
B = ___ + 4 = 4 + 4 = 8
16
16 − 4
C = ______ = 3
4
D = 16 − 16 × 4 + 1
= 16 − 64 + 1
= −47
The order is D, C, B, A.
14
_
√ 98
_
_
3
− _14
16
___
5+7−3−8=1
(5 − 32) × 6 + 8 ÷ (−2)
= −4 × 6 + (−4)
= −24 − 4
= −28
20
5 −__5 2 ______
5 − 25
______
= − ___ = − 4
=
5
5
√ 25
14 a
___ 3
36
___
( 25 ) = ( 25 )
_3
36
___
6
= (__)
5
216
= ____
125
11 1.16 (to 3 s.f.)
b
2
_
10 (7 + 14) ÷ (4 − 1) × 2 = 14
12 a
_1
300
60
12
4
2
5
5
5
5
3
2
7500 = 22 × 31 × 54
=BAD
_
+ √ 72 = √ 49
× 2 +_ √ 36 × 2
_
√
= 7 2_+ 6√ 2
= 13√ 2
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 2
Getting started
1
2
a
b
c
a
___
m
c
an
am × an
d
(am)n
Example 1:
a Sign error
b 3x − x + 2
Example 2:
a Multiplied both numerator and
denominator by 3 instead of just the
3
3
numerator. (3 = __ and not __)
3
1
3x + 12
b _______
5
Example 3:
a Cancelled part of a term, but both x and
2 need to be divided by 2.
x+2
b _____ cannot be simplified further,
2
x
but can be written as __ + 1.
2
4
a
b
There are different options, But in
general, if you let one number be x, the
consecutive number is x + 1. The sum of
the numbers is x + x + 1 = 2x + 1. Any
multiple of 2 is even, so if you add 1, it
will be odd.
Using the same argument
x + x + 1 + x + 2 = 3x + 3.
3x can be odd or even (depending on the
value of x) so the answer can be either
odd or even.
Exercise 2.1
1
15
a
b
c
d
e
6xy
7ab
xyz
2y2
4ab
12xy
5ab
yz2
i
6
__
p
3(x + 1)
________
x
2x
4x ___
___
= y
j
2y
x+3
k _____
4
3
m
___ = m
l
m2
m 4x + 5y
n 7a − 2b
o 2x(x − 4)
Possible answers are:
a a0
b
3
C
A
A
f
g
h
f
a
b
c
d
e
f
2x
2(x + 4)
________
3
4x
2
___ = __
6x 3
m + 13
m+5
25 − m
m3
m
__
+3
3
4m − 2m = 2m
x+3
x−6
10x
−8 + x
x + x2
x + 2x = 3x
g
2x
_____
q
r
2
a
b
c
d
e
3
4
c
a
b
x+4
$(x − 10)
x
$ __
4
$15
m + 10 years
m − 10 years
c
m
__
years
a
b
5
2
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
6
a
b
p
$ __
3
p __
p
3p
__
$ , $ and $ ___
5
5 5
Exercise 2.2
1
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
2
a
b
c
d
e
f
g
h
i
j
k
l
9
30
10
27
18
7
16
36
4
6
6
30
5
2
1
__
2
c
d
e
30
45
16
5
13
16
31
450
24
8
24
5
f
g
26
m ___
3
n 10
o 4
p 3
q 6
r 225
s 12
t −10
u 129 600
3
16
a
i
b
y=0
h
i
ii
iii
iv
v
i
ii
iii
iv
v
i
ii
iii
iv
v
i
ii
iii
iv
v
i
ii
iii
iv
v
i
ii
iii
iv
v
i
ii
iii
iv
v
i
ii
iii
iv
v
i
ii
iii
iv
v
y = 12
y = 16
y = 40
y = 200
y=1
y = 10
y = 13
y = 31
y = 151
y = 100
y = 97
y = 96
y = 90
y = 50
y=0
y = 1.5
y=2
y=5
y = 25
y=0
y=9
y = 16
y = 100
y = 2500
undefined
y = 33.3 (3 s.f.)
y = 25
y = 10
y=2
y=4
y = 10
y = 12
y = 24
y = 104
y = −6
y=0
y=2
y = 14
y = 94
y=0
y = 81
y = 192
y = 3000
y = 375 000
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4
a
b
$(3x + 2y)
i
$18
ii $100
iii $350
5
a
b
c
d
P = 42 cm
P = 8m
P = 60 cm
P = 20 cm
a
i
43
ii 53
iii 71
iv 151
They’re all prime numbers.
When n = p, n2 + n + p becomes n(n + 2);
in other words, it has factors n and n + 2,
so is not prime.
6
b
c
l
m
n
o
p
q
r
13xyz
2x2
5y2
−y2
12ab2
5x2y
2xy2
3
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
5x + y
4x + 2y
7x
4 + 4x
6xy − 2y
−x2 + 2x
−x + 4y
3x + 3y
8x + 6y
8x − 2y
14x2 − 4x
10x2
12xy − 2x
8xy − 2xz
−x2 − 2y2
8x2 y − 2xy
6xy − x
6xy − 2
4
a
b
c
d
e
f
g
h
i
a
b
c
d
e
f
g
h
2y − 8
4x2 − 5x
7x + 4y
y2 + 5y − 7
x2 − 5x + 3
x2 + 5x − 7
3xyz − 3xy + 2xz
8xy − 10
−3x2 + 6x − 4
P = 8x
P = 4x + 14
P = 6x + 3
P = 5x + 4
P = 12y − 6
P = 8y2 + 2y + 14
P = 12y − 4
P = 18x − 1
What is the point of algebra?
Students’ own work. There are many accessible
examples that students could use, for example,
conversion formulae, using algebra to find break
even points (finance and economics), calculating
doses of medicine based on person’s mass or other
factors, working out trajectories in sports (such
as basketball or snooker), using BMI and other
factors to work out health and fitness and so on.
Exercise 2.3
1
2
17
a
6x, 4x, x
b
c
d
e
f
3
−3y, __ y, −5y
4
ab, −4ba
−2x, 3x
5a, 6a and 5ab, ab
−1xy, −yx
a
b
c
d
e
f
g
h
i
j
k
8y
7x
13x
22x
5x
0
−x
−3y
4x
7xy
4pq
5
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Magic squares
1 No, the sum of numbers in a 3 × 3 magic
square is always 3 times the centre number.
In this case, that would be 12a + 15y for each
row, column and diagonal and they do not all
sum to that.
2
For example:
a−c
a+b+c
a−b
a−b+c
a
a+b−c
a+b
3
a−b−c
x
x−1
x+1
x+1
x
x−1
x−1
x+1
x
x+1
x−1
x
x−1
x
x+1
x
x+1
x−1
a+c
a
x−1
x+1
x
x+1
x
x−1
b
Yes. Student’s own magic squares or
explanation.
Exercise 2.4
18
x
x−1
x+1
x
x+1
x−1
x−1
x
x+1
x+1
x−1
x
1
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
12x
8y
12m
6xy
8xy
27xy
24yz
12xy
8x2y2
8x2y
27xy2
24xy2
8a2b
12ab2c
12a2bc
16a2b2c
24abc
72x2y2
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
3
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
24x
30x2y
12x2y2
x3yz
48x
24x3y
4x2y2
12a2bc
60xy
8xy
9x3y
8x3y3
42x2y2z2
56x3y2
36x2y2z
18x4y4
54x4y
6x3y3
a
b
c
d
e
f
5r
4r
3r
6s
7r
2s
s
__
4
1
___
4s
t
__
2
6s
1
__
4
1
__
9
g
h
i
j
k
l
4
19
ab
___
k
5
g
6
2
a
___
12
5 a2 b
______
6
10a
____
3b
3ab
____
8
25
a2
_____
4
2
h
a
___
i
2ab
j
8a
___
k
1
__
b
c
d
e
f
1
c
4x
___
d
8
e
7x
____
f
3x
x
__
3
1
___
4y
h
a
j
2
3
3
4
a2
Exercise 2.5
4x
6y
g
l
7y
9y
___
4
4xy
4y
___
x
l
a
b
y
2
y2
i
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
2x + 12
3x + 6
8x + 12
10x − 60
4x − 8
6x − 9
5a + 20
24 + 6a
9a + 18
14c − 14d
6c − 4d
4c + 16d
10x − 10y
18x − 12y
12y − 6x
4s − 16t2
9t2 − 9s
28t + 7t2
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
3
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
2x2 + 2xy
3xy − 3y2
2x2 + 4xy
12x2 − 8xy
x2y − xy2
12xy + 6y
18ab − 8ab2
6a2 − 4a2b
12a2 − 12a3
36a − 8ab
10b − 5ab
12a − 3ab
2x2y2 − 4x3y
12xy2 − 8x2y2
3x2y2 + 3xy3
2x3y + x2y2
81x2 − 18x3
12xy2 − 4x2y2
a
b
c
A = x2 + 7x
A = 2x3 − 2x
A = 4x2 − 4x
Maths jokes
Student’s own discussions, but could include puns,
play on words, misinterpretation of concepts.
2
20
10 + 5x
7y − 6
4x − 8
6x − 6
2t2 + 8t − 5
4x + 1
3x
8x + 6
6x + 9
3h + 2
8d + 6
3y + xy − 4
2x2 + 8x − 4
−4y2 + 4xy + 8y
10s − 12s2
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
6x + 154
4x + 2
7x + 26
92
2x2 + 16
6p2 + 10px
24pq + 4p
2xy + 4x
−3x − 18xy
21x − 12y − 2xy
22x2 − 7x3
x2 − xy + 6x − 3y
16s − 3st − 8
2x2
4x2 + 8xy
2x2 − 3x + 15
9k − 17
7xy + 9x
1
a
b
c
d
e
f
g
h
i
j
k
l
−30p − 60
−15x − 21
−20y − 1
−3q + 36
−24t + 84
−12z + 6
−6x − 15y
−24p − 30q
−27h + 54k
−10h − 10k + 16j
−8a + 12b + 24c − 16d
−6x2 − 36y2 + 12y3
2
a
b
c
d
e
f
g
h
i
−5x − 8
−5x + 12
10x − 38
−13f
−36g + 37
12y − 20
−26x2 − 76x
−x2 + 77x
−9x2 + 30x
Exercise 2.6
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
6x2 + 12x − 9
−y2 + 6y
6x − 6
Exercise 2.7
You could extend this by asking students to
develop their own funny maths memes to share
with the class.
1
p
q
r
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
j
k
l
24q
−42pq + 84p
−48m + 48n
a
b
c
d
e
f
g
h
i
j
k
l
12x − 6
13x − 6
−2x + 17
x + 13
23 − 7x
10x − 8
7x − 5
x2 − 5x + 8
3x2 − 7x + 2
2x2 + 3x + 6
2x − 18
6x2 + 6x − 6
k
l
1
m ___
3x
n 4ab
o 1
3
Exercise 2.8
1
2
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
x8
a10
y2
x13
y9
y7
y6
t5
6x7
9y6
2m4
6s7
15x3
8x7
8z7
4x7
a
b
c
d
e
f
g
h
i
x2
j
21
g9
y
k2
s4
x2
3x2
3p3
4y
x
__
2
3
3b
4
a
a4
b
c
d
e
f
g
h
i
j
k
l
m
n
o
v6
f 12
y6
32x10
9c4d 4
1
125x6
a6b6
x10y20
x3y12
16g2h4
81x8
x4y24
1
a
b
c
12x6
24x3y
4k4
d
x
___
2
j
4
44x3a4b2
4x3 + 28x
4x3 − x5
x2
7
___
x4
2x2
k
a
___
o
2xy3
e
f
g
h
i
12
b6
x4 y8
l _____
16
m 1
n 8x5
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 2.9
1
2
a
b
c
d
true
false
false
false
a
1
___
b
c
d
e
f
g
h
3
b
b
6
___
d
e
f
x
h
1
___
i
j
k
l
4
22
1
x5
a7
___
b5
x 10
___
y 12
2y
____
11
x4
1
12n 15
m _____
m3
a x=4
b x=5
c x=2
d x = −3
e x=3
f x=3
_2
a
x3
b
x6
c
1
___
_7
d
y3
x3
___
y
e
a7
f
7b
____
g
h
4
__
2
4
2
___
x2
3
___
4k
_7
i
3x 4
j
1
____
o
x 12
p
x6
___
_3
4x 2
s
k __
4
_3
3x 4
____
l
2
x
__
m
8
1
n ____
_3
4x 2
x6
1
____
3s 4
1
___
h 11
1
____
8x 6
1
___
c6
g
x=2
x=4
Exercise 2.10
x2
1
___
y3
1
_____
x2 y2
2
___
x2
12
___
x3
7
___
y3
8x
___
y3
12
_____
x3 y4
a
c
g
h
2
a
b
c
d
e
f
g
h
i
j
k
l
13
_
_1
_7
y6
x=6
1
x = __
2
x = 16 807
x = 257
x=4
x=4
x=6
x=5
x=2
x = −4
1
x = __
6
3
x = __
4
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Practice questions
1
2
a
b
c
d
n + 12
2n − 4
(nx)2
(n2)3 or (n3)2 or n6
a
b
Any of 2n, 4n, 6n,…
2n is always even because it is a multiple
of 2. Every even number is ‘next to’ an
odd number and 2n + 1 is ‘next to’ 2n.
p = 2n + 3
2n + 1 + 2n + 3 = 4n + 4 = even + even =
even
c
d
3
a
b
c
4
a
b
15xy + x
5xy + 3y
5
a
b
c
d
e
f
g
a2b
2x6
6x4y2
1
4x5y3
15x2
x3
h
16
16x−10 or ___
x 10
27x 3
_____
64y 3
i
j
x
_____
k
7p
_____
6
0
7
a
b
c
8
7.35
9
a
b
10 a
b
23
p + 2q + r
8q, 3r + 4q
Top brick = 2h + 2j + 2k = even + even +
even = even
11 a
x3
b
4
___
c
1
1
________
= ___________________
(2x − 2) 3 8x 3 − 24x 2 + 24x − 8
12 a
b
c
14 a
b
c
15x
9y3
4x
15 a
−3
b
−3
c
1
__
d
1
− __
3
pq
pq
___
4
p3
b
c
17 a
b
4
15
___
8
1
a
b
c
d
n + 12
2n − 4
(nx)3
(n2)3
2
a
b
2n (or 2n + any even number)
2n is always even and 2n + 1 is one more
than an even number. Whole numbers
alternate
odd
even
odd
even …
so 2n + 1 must be odd.
p + 2 = 2n + 1 + 2
= 2n + 3
(2n + 1) + (2n + 3)
= 4n + 4
= 2(2n + 2), which is a multiple of 2 and
hence even.
3y 12
6
10
10
10
c
m2 − n2
0
3
Practice questions worked
solutions
19
8x − 4
x2 + 37xy
3
3
3
13 18
16 a
6q 20
x2
d
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
a
p+q+q+r
= p + 2q + r
b
3r + 4q
2h
j
a
9xy + 3x + 6xy − 2x
= 3x − 2x + 9xy + 6xy
= x + 15xy
b
6xy − xy + 3y
= 5xy + 3y
a
a b
______
= a 2b
b
c
2(x3)2 = 2x6
3x × 2x3y2
= (3 × 2) × x × x3 × y2
= 6x4y2
(4ax2)0 = 1
4x2y × x3y2
= 4x2 × x3 × y × y2
= 4x5y3
3x−4 × 5x6
= 3 × 5 × x−4 × x6
= 15 × x−4 + 6
= 15x2
g
32
2k
2 14 x −4
1 (u + v)t
s = __
2
9
1 __
2 + __
= __
×3
2(5 2)
3 4 + 45
= __(______) × 3
2
10
3 × 49
= ______
20
147
= ____
20
9
a
b
10 a
= x3
h
i
(4x−5)2 = 42(x−5)2
= 16x−10
(3x) 3
= _____3
(4y )
(4y)
3x
___
3
−4 q 5
8
= _____4 × _______
−6
17 x
26 x
= x 5 − 4 − 4 − (−6)
12 p
−7
x + 5 − (x − 5)
=x+5−x+5
= 10 for all values of x.
So, a 10
b 10
c 10
−6
13 x 5
1
7
6x
3x
____
÷ ______
14 x −4
4 q4
−4
7x3y2 × (2x)3 − (4x3y)2 − 4xy2 × 10x5
= 7x3y2 × 8x3 − 16x6y2 − 40x6y2
= 56x6y2 − 16x6y2 − 40x6y2
=0
41
5
630 p
5
6
1 a1b 3
7 x4
7
3
1 x 19 y −12
= __
3
x 19
_____
= 12
3y
7
= __ p 5 + 1 − 4 − (−4) q −4 − 4 − 7 − 5
6
7
= __ p 10 q −20
6
7p 10
= _____
6q 20
j + 2k
Top brick = 2h + 2j + 2k
= three even numbers added
together
= even number
f
14 p q ________
5 pq
_________
×
−3
12 x −7 y 9
− 3r
2h + 2j + 2k
2h + j
d
e
k
12
2q − 3r
2q
c
5
4x y
1 x 12 − −(7) y −3 − 9
________
= __
8q
3r + 6q
4
j
b
5(x − 2) + 3(x + 2)
= 5x − 10 + 3x + 6
= 8x − 4
5x(x + 7y) − 2x(2x − y)
= 5x2 + 35xy − 4x2 + 2xy
= x2 + 37xy
m(m − n) − n(n − m)
= m2 − mn − n2 + mn
= m2 − n2
x( y − z) + y(z − x) + z(x − y)
= xy−xz + yz−xy + xz − yz
=0
27x 3
= _____3
64y
24
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
11 a
x5 × x−2 = x5 − 2 = x3
48 x 21
b
4
_____ = ___
c
1
(2x − 2) −3 = _________
(2x − 2) 3
12 x
42
13
4x = 43 ⇒x = 3
b
3x − 5 = 22
3x = 27
= 33
x=3
c
4 × 6 p = 864
6 p = 216
= 63
p=3
14 a
b
ca
c
125x = 5
(5 3) x = 5
5 3x = 5 1
3x = 1
1
So, x = __
3
1
125 x = __
5
(5 3) x = 5−1
16 a
_1
) = 81_( )
_1
81y 6 2
_1
2
_1
y6 2
√ 81 y 3
=
= 9y 3
c
1 = 3 −3
1 = __
3 x = ___
27 3 3
x = −3
5 3x = 5 −1
3x 2 × 5x 2
_1 _1
= 15x 2 + 2
= 15x
(
b
ba
_1
x
2 −x = 2 3
−x = 3
So, x = −3
d
− +
= 32 − (−1)3 + 23
= 9 − (−1) + 8
= 17 + 1
= 18
ab
(2) = 8
1
__
(2 −1) x = 2 3
x2
1
= _________
2 3(x − 1)
1
= ________
8(x − 1)
12 a
15 a
(64x 3) _13 = 64 _13 (x 3) _13
_
3
√ 64 x 1
=
= 4x
3x = − 1
1
x = − __
3
2x + y = 2x2y
= pq
b
pq
2 x 2 y = ___
2 x + y − 2 = _____
2
4
2
c
23x = (2x)3 = p3
17 a
b
n−1 = 2−2
n−1 = (22)−1
n = 22 = 4
_ 3
4 n = (√ 32 )
4
(2 2) n = [(2 5) _14 ]
3
_5
2 2n = 2 4×3
15
2n = ___
4
15
So n = ___
8
25
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 3
Diagrams provided as answers are NOT TO
SCALE and are to demonstrate construction lines
or principles only.
Exercise 3.1
1
Getting started
Each student will produce a different spider
diagram depending on their prior knowledge.
The information shown might include facts about
angles in triangles and quadrilaterals, names and
properties of special triangles and quadrilaterals,
how to measure lines and angles, and so on.
You can ask students to complete their own and
then to compare with others and add any points
they have missed out (but know). You could also
use the student contributions to develop a class
spider diagram, which will give you some idea of
what students already know so that you can focus
the work on the new concepts in this chapter.
3
4
a
b
(Hexagonal) prism
Yes, if the base is a hexagon, the prism
has six rectangular faces.
It speeds up the process and allows
different members of the team to put
pieces together on one model to check for
overlaps/errors and gives a view of the
finished process. Building information
modelling (BIM) allows them to strip the
model down to beams and walls so that
they can decide where to install or place
infrastructural elements.
Computer models can be moved, changed
and rescaled as needed digitally. It is
easy to share and collaborate ideas and
different members of the team can work
on the design at the same time.
iii
a
acute
Answers will vary
40º
b
acute
70º
c
obtuse
130º
d
acute
30º
e
obtuse
170º
f
right
90º
g
acute
70º
h
acute
60º
i
obtuse
140º
290°
3
a
b
Answers will vary, but students should be
able to find rectangles, trapezia and general
quadrilaterals, pentagons, hexagons and some
triangles inside the building, formed by the
structures.
a
b
ii
2
Shapes and solids
Answer suggestions:
1 Students can show each other where they have
found the different elements. It is possible to
find them all.
2
i
c
This protractor is able to measure angles
from 0° to 360°.
Student’s own answer. Something like:
ensure that the 0°/360° marking of the
protractor is aligned with one of the arms of
the angle you are measuring, and the vertex
of the angle is aligned with the centre of the
protractor. Whether you use the inner or
outer scale will be determined by which arm
you aligned with 0. Use the scale that gives
an angle <180°.
You would use the scale that gives you an
angle >180°.
Exercise 3.2
1
a
A
80°
B
C
b
P
30°
Q
c
R
X
135°
Y
26
Z
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
d
E
90°
F
e
4
60° and 120°
5
53°, 127° and 53°.
Exercise 3.4
G
1
a
K
b
M
L
c
210°
f
355°
K
J
5°
L
Exercise 3.3
1
a
b
e
f
EBF and FBC; or ABD and DBE
ABE and EBC; or DBA and CBG; or
DBC and ABG
ABD and DBC; or ABE and EBC;
or ABF and FBC;
or ABG and CBG;
or DBE and EBG;
or DBF and FBG;
or DBC and CBG;
or DBA and ABG;
or ABG and GBC
DBE, EBF, FBC and CBG
or DBA and ABG
or DBF, FBC and CBG
or DBF and FBG
or DBC and CBG
(and combinations of these)
FBC
EBA
a
b
c
d
e
f
g
h
i
x = 68°
x = 40°
x = 65°; y = 115°
x = 59°; y = 57°
x = 16°; y = 82°; z = 16°
x = 47°; y = 43°; z = 133°
x = 57°
x = 71°
x = 38°
a
b
c
30°
15°
30°
c
d
2
3
27
d
e
f
g
h
i
2
a
b
c
a = 112° (alternate angles equal)
b = 112° (vertically opposite angles equal)
x = 105° (alternate angles equal)
y = 30° (sum of triangle)
z = 45° (alternate angles equal)
c = 40° (vertically opposite angles equal)
b = 72° (corresponding angles equal)
a = 68° (angles on a line)
d = 68° (vertically opposite angles equal)
e = 40° (alternate angles equal)
a = 39° (corresponding angles equal)
b = 102° (angle sum of triangle)
x = 70° (angle on a line)
y = 70° (corresponding angles equal)
z = 85° (corresponding angles equal
(180 − 95 = 85°, angles on a line, z is
corresponding angle equal to 85°)
x = 45° (alternate angles equal)
y = 60° (alternate angles equal)
x = 82° (co-interior angles
supplementary)
y = 60° (corresponding angles equal)
z = 82° (angles on a line)
x = 42 (alternate angles equal)
y = 138° (angles on a line)
z = 65° (alternate angles equal)
a = 40° (alternate angles equal)
b = 140° (angles on a line)
d = 75° (angles on a line)
c = 75° (corresponding angles equal)
e = 105° (corresponding angles equal)
AB ∥ DC (alternate angles equal)
AB ∦ DC (co-interior angles not
supplementary)
AB ∥ DC (co-interior angles
supplementary)
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
General results
1 a y
b 180 − x
c y is equal to z.
d x and z are equal.
2
a
b
c
Angle ABF is equal to angle CDF
(corresponding angles since lines AB and
CE are parallel). Angle CDF is equal to
angle CEG (corresponding angles since
lines BF and EG are parallel). So angle
ABC = angle CEG, so x = y.
Angle AFE is equal to angle HFB
(vertically opposite angles). Angle BFH
is equal to angle DGH (corresponding
angles since lines AB and CD are parallel).
So angle AFE = angle DGH, so x = y.
d
Exercise 3.5
1
2
a
b
c
a
b
3
a
b
28
x = 54° (angle sum of triangle)
x = 66° (base angle isosceles 4)
x = 115° (angle sum of triangle)
y = 65° (exterior angle of triangle equal
to sum of the opposite interior angles OR
angles on a line)
z = 25° (angle sum of triangle)
x = 60° (exterior angle of 4 equal to
sum of opposite interior angles),
so x + x = 120°, x = 60°
4x = 86° + (180° − 2x)
(exterior angle equals sum of opposite
interior angles, and angle on straight line)
6x = 266
x = 44.3°
angle BAC = 180° − 95°
(angles on a straight line) = 85°
angle ACB = 180° − 105°
(angles on a straight line) = 75°
180° = x + 75° + 85°
(angle sum of triangle)
x = 180° − 160°
x = 20°
angle CAB = 56°
(vertically opposite angles equal)
180° = 56° + 68° + x
(angle sum of triangle)
x = 180° − 124°
x = 56°
e
f
angle ACE = 53° (angles on straight line)
x = 53° (alt angles equal)
OR
angle CDE = 59° (alt angles equal)
180° = 68° + 59° + x (angle sum of ∆)
x = 180° − 127°
x = 53°
180° = 58° + angle ACB + angle CBA
(angle sum of triangle)
angle ACB = angle CBA (isosceles ∆)
⇒ 180° = 58° + 2y
2y = 122°
y = 61°
x = 180° − 61°
(exterior angles of a triangle equal to sum
of opposite interior angles)
x = 119°
angle AMN = 180° − (35° + 60°)
(angle sum of ∆)
angle AMN = 85°
x = 85°
(corresponding angles equal)
angle ACB = 360° − 295°
(angles around a point)
angle ACB = 65°
angle ABC = 65° (isosceles ∆)
x = 180° − (2 × 65°) (angle sum of ∆)
x = 50°
Exercise 3.6
1
a
b
Rhombus, kite or square
Square
2
a
angle QRS = 112° (vertically opposite
angles equal)
x = 112° (opposite angles in
parallelogram)
x = 62° (isosceles 4)
360° = 110° + 110° + 2x
(angle sum of quadrilateral)
140° = 2x
x = 70°
b
c
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
d
e
f
3
a
b
c
29
angle MLQ = 180° − 110°
(angles on a straight line)
angle LMN = 180° − 98°
(angles on a straight line)
360° = 70° + 82° + 92° + x
(angle sum of quadrilateral)
x = 116°
360° = 3x + 4x + 2x + x
(angle sum of quadrilateral)
360° = 10x
x = 36°
360° = (180° − x) + 50° + 110° + 90°
(angles on a straight line, and angle sum
of quadrilateral)
360° = (180° − x) + 250°
110° = 180° − x
x = 70°
180° = 70° + 2y (angle sum of a 4,
isosceles 4 to give 2y)
110° = 2y
y = 55°
∴ angle PRQ = 55°
angle MRS = 180° − (55° + 55°)
(angles on a straight line, and isosceles
triangle)
x = 70°
angle MNP = 98°
(opposite angles in parallelogram)
angle RNM = 180° − 98°
(angles on a straight line)
= 82°
180° = 2x + 82° (angle sum of a
triangle, and isosceles triangle)
2x = 98°
x = 49°
angle QRP = 55°
(angle sum of a triangle, and isosceles
triangle)
angle QRP = x
(alternate angles equal)
x = 55°
Exercise 3.7
1
Number
of sides
Angle
sum
5
6
7
8
540°
720°
900°
1080°
9
10
12
20
Number
of sides
Angle
sum
2
a
b
c
d
e
f
108°
120°
135°
144°
150°
165.6°
3
a
b
c
d
2340°
360°
156°
24°
4
24 sides
5
a
b
c
1260° 1440° 1800° 3240°
x = 135°
x = 110°
x = 72°
Exercise 3.8
1
a
b
c
d
e
f
2
a
Diameter
Major arc
Radius
Minor sector
Chord
Major segment
b
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
c
b
D
d
y
M
3
a
b
c
d
e
f
5 cm
5 cm
N
c
G
4 cm
I
Exercise 3.9
8 cm
25°
5 cm
NOT TO SCALE
1
H
a
A
6 cm
E
4 cm
F
Radius
Diameter
Minor arc
DO, FO or EO
Major arc
Sector
3
B
a
A
b
C
D
75 mm
7.2 cm
6.9 cm
c
E
2
F
5.5 cm
8.5 cm
B
C
b
a
Z
A
2.4 cm
B
1.7 cm
3.2 cm
86 mm
c
D
E
30
X
120 mm
Y
C
66 mm
6.5 cm
F
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
d
P
4
a
b
4 cm
Accurate drawing
i
5.5 cm
ii 4.2 cm
B
6.5 cm
a
b
c
d
e
f
a
b
31
a
b
c
360°
24° if a regular polygon
156°
8
a
Exterior angle of triangle is equal to
the sum of two opposite interior angles
x __
x
__
+ = x.
2 2
Opposite angles of parallelogram equal,
and vertically opposite angles equal.
b
125°
A
4
7
P
75°
3
Let angle MQN = x
Then angle PMQ = x (isosceles triangle)
So angle MPQ = 180° − 2x (angles in a
triangle add up to 180 degrees)
Therefore angle MPN = 180° − (180° − 2x) = 2x
(angles on a straight line)
So angle PMN = 2x (isosceles triangle)
and angle NMQ = x + 2x = 3x
NOT TO SCALE
C
2
6
R
Practice questions
1
720°
6.5 cm
6.5 cm
Q
5
x = 99° (co-interior angles supplementary)
x = 65° (corresponding angles equal)
x = 75° (angle sum of isosceles 4)
x = 112° (opposite angles of || gram)
x = 110°
If y = angle AEC
⇒ 360° = 90° + 110° + 90° + y
y = 70°
∴ angle AEC = 70°
angle ADE = 70° (isosceles triangle)
x = 180° − 70° (angles on a line)
x = 110°
x = 72.5°
Let y stand for base angles of isosceles 4.
2y + 35° = 180° (base angles isosceles 4
and angle sum of 4)
y = 72.5°
⇒ angle QRP = 72.5°
angle NRQ = 35° (alternate angles equal)
180° = x + 72.5° + 35°
x = 72.5°
Angles in a triangle add up to 180°.
x + y + 90° = 180°, so x + y = 90°
y = 53°
a = 70°, b = 110°, c = 100°
9
NOT TO SCALE
Q
5 cm
4.5 cm
P
10 b
d
7 cm
R
They all intersect at the same point.
The circle will always pass through all
three vertices if drawn correctly.
11 32
12 a
b
108°
36°
13 18
14 a
b
c
d
Angle UVP = x (alternate angles are equal)
Angle WVQ = c (alternate angles are
equal)
Angle UVP + angle PVQ + angle WVQ =
180° (angles on a straight line sum to 180°)
Therefore a + b + c = 180°, so the interior
angles of a triangle sum to 180°
Angle RQV = 180° − c
a + b = 180° − c = angle RQV
Therefore the exterior angle of a triangle
is equal to the sum of the two interior
opposite angles.
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Practice questions worked
solutions
1
f
a
A
b
180° − 35°
Angle RPQ = __________
2
145°
____
=
= 72.5°
2
x = angle RPQ = 72.5° (alternate angles
are equal)
B
C
D
3
75°
A
2
a
b
c
d
e
32
a
x + y + 90° = 180° (angles in a triangle
add up to 180°)
So x + y = 90°
b
37 + y = 90°
y = 90° − 37°
= 53°
125°
B
Angle CFG corresponds to angle AEF
so angle CFG = 81°
x + angle CFG = 180° because the angles
on a straight line add up to 180°.
x + 81° = 180° ⇒ x = 99°
x + 65° because angle NSR corresponds
to angle QTS.
Angle ABC = x because base angles of an
isosceles triangle are equal
x + x + 30° = 180° because the sum of the
interior angles of a triangle is 180°
2x = 150°
x = 75°
Angle PNM = 180° − 112°
= 78° (supplementary angles)
Angle PNM + x = 180° (supplementary
angles)
x = 180° − 78°
= 112°
Angle AEC + 90° + 90° + 110° = 360°
(angles sum in a quadrilateral = 360°)
Angle AEC = 360° − 290°
= 70°
Angle ADE = 70° (base angles of an
isosceles triangle are equal)
x = 180° − 70° = 110° (angles on a
straight line add to 180°)
Angle PRQ = angle RPQ (base angles of
an isosceles triangle are equal)
2 × angle RPQ + 35° = 180° (angles in a
triangle add up to 180°)
4
a + 110° = 180°
a = 70°
a + b = 180°
b = 180° − 70°
= 110°
c + 100° + 100° + 60° = 360°
c = 100°
5
6
360°
Exterior angle = ____ = 60°
6
Interior angle = 180° − 60° = 120°
Sum of interior angles = 120° × 6 = 720°
M
x
x
N
P
Q
Let angle MQN = x
Triangle MPQ is isosceles ⇒ angle PMQ = x
(base angles are equal)
Angle MPQ = 180° − 2x (angles in a triangle
add up to 180°)
Therefore, angle MPN = 180° − (180° − 2x) = 2x
(angles on a straight line add up to 180°)
Angle PMN = 2x (base angles of an isosceles
triangle are equal)
So, angle NMQ = 2x + x = 3x = 3 × angle MQN
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7
a
360° (true for all polygons)
b
360°
____
= 24°
c
8
a
15
180° − 24° = 156°
9
Angle PSN = x (opposite angles in a
parallelogram are equal)
y is vertically opposite angle PSN so y = x
Check by measurement.
10 a
b
c
d
Check by measurement.
They all intersect at the same point.
Check by drawing.
The point of intersection is always the
centre of the circle passing through all
three points.
11 If n = number of sides, then
the total interior angle is 180°(n − 2) = 5400
n − 2 = 30
n = 32
So, the number of sides is 32
33
b
Angles in a triangle add to 180°
x x
Angle BAC = 180° − __ + __
(2 2)
= 180° − x
Angles on a straight line add up to 180°
so, angle CAD + (180° − x) = 180°
Therefore, angle CAD = y = x
b
12 a
360°
Exterior angle ____ = 72°
5
x = interior angle = 180° − 72° = 108°
Triangle ABC is isosceles
⇒ x + 2y = 180°
108° + 2y = 180°
2y = 72°
y = 36°
360°
13 3x = ____ = 36°
10
x = 12°
5
Therefore, exterior angle of B = __ × 12° = 20°
3
360°
____
= 18 sides
20°
14 a
b
c
d
Angle UVP is alternate with angle VPQ = a
so, angle UVP = a
Similarly, WVQ = c
Angles on a straight line add up to 180°
so, angle UVP + angle PVQ + angle WVQ
= 180°
⇒ a + b + c = 180°
angle RQV = 180° − c
a + b + c = 180°
so, 180° − c = a + b
Therefore, exterior angle RQV = sum of
two opposite interior angles
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Chapter 4
Getting started
1
Answers could include: observations, experiments and measurements, research using secondary
sources, questionnaires and sampling.
2
Some possible answers are:
Pictogram
Pie chart
Line graph
a
Number of medals that
each country won
b
Use the key and number
Read frequency
of circles to find number of from vertical axis
medals
and country name
from horizontal
axis
Use size of sector
to compare
countries
Read frequency
from vertical axis
and use slope of
line to see if the
number increases
or decreases for
different times (given
on horizontal axis)
c
Mostly for effect/
decorative
To compare data
sets
To compare data
sets
To show trends or
patterns
Used for discrete data in
categories
Used for discrete
data in categories
Used for discrete
data in categories
Data that changes
over time; best for
continuous data
(although this data is
discrete)
Use of different symbols
or different sized symbols
can make data difficult
to compare/interpret
(size and surface area of
symbols can be used to
mislead, especially with
circular symbols
Scale and/or axis
manipulation can
be misleading,
for example,
not starting at
0, reducing or
increasing intervals
% might not sum
to 100, circles
with different sizes
can give different
impressions, total
numbers may not
be given
Vertical scale can be
manipulated, may
not start at 0 and
can jump in different
sized steps, labelling
might not be clear
d
34
Bar chart
Total medal count
for some countries
Proportion of total Number of medals
medals awarded
each country won
to each continent over a period of time
Using 3D bars
can also make the
categories look
different
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Exercise 4.1
1
a, b Students’ answers will vary, below are possible answers.
Categorical data
Numerical data
Hair colour
Number of people in household
Eye colour
Hours spent doing homework
Favourite subject
Hours spent watching TV
Mode of transport to school
2
3
Brand of toothpaste used
Shoe size
Make of cell phone
Test scores
a
b
c
d
e
f
g
h
i
j
k
l
Continuous
Discrete
Continuous
Continuous
Discrete
Continuous
Continuous
Discrete
Continuous
Discrete
Discrete
Discrete
e
a
i
ii
iii
iv
i
ii
iii
iv
Experiment
Primary
Numerical
Discrete
Survey
Primary
Categorical
Discrete
h
i
ii
iii
iv
i
ii
iii
iv
Use existing data
Secondary
Numerical
Continuous
Survey
Primary
Categorical
Discrete
j
b
c
d
35
Number of books
read in a month
f
g
i
i
ii
iii
iv
i
ii
iii
iv
i
ii
iii
iv
i
ii
iii
iv
i
ii
iii
iv
Use existing data
Secondary
Numerical
Discrete
Experiment
Primary
Numerical
Discrete
Survey
Primary
Numerical
Continuous
Use existing data
Secondary
Categorical
Discrete
Use existing data
Secondary
Numerical
Discrete
i
ii
iii
iv
Survey
Primary
Numerical
Discrete
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Exercise 4.2
1
Score
Tally
Total
1
|||| |||
8
2
||||||||||
3
|||| ||
7
4
|||| |||
8
5
|||| |||
8
6
|||| ||
7
12
50
2
Students’ own answers.
3
a
b
c
d
7
2 and 12
It is impossible to score 1 with two dice.
There are three ways of getting each of
these scores.
Exercise 4.3
1
a
b
c
d
e
2
a
b
c
d
3
36
Number of coins
0
1
2
3
4
5
6
7
8
Frequency
6
2
6
4
4
2
4
1
1
8
2
None or two coins
30: add column and total the frequencies.
Amount ($)
0−9.99
10−19.99
20−29.99
30−39.99
40−49.99
50−59.99
Frequency
7
9
5
2
1
1
16
1
$10 – $19.99
Call length
Frequency
0−59 s
0
1 min−1 min 59 s
4
2 min−2 min 59 s
3
3 min−3 min 59 s
6
4 min−4 min 59 s
4
5 min−5 min 59 s
3
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Exercise 4.4
Exercise 4.5
1
1
a
b
c
d
2
Student’s own answers.
3
a
4 5899
5 33455566689
6 00378
Key
Key 4|5 represents 45 kg
2
a
Branch A
5
11
42
13
990
14
2
52
15
9
9864
16
059
9952
17
7
988600
18
056778888
980
19
0011368
100
20
000145
Key
Branch A 5 | 11 represents 115 pairs
b
37
b
c
Branch B: 205 pairs
Branch B, as the data are clustered round
the bottom of the diagram where the
higher values are located.
a
b
c
d
e
26
12 cm
57 cm
6
i
More data clustered round top of
diagram; possibly need to add 0 as a
stem.
ii Data clustered round bottom of the
diagram, possibly need to add more
stems (i.e. higher than 5).
a
b
c
d
Does not
check on
phone
Checks on
computer
5
3
Does not check
on computer
3
1
Most people surveyed check their email
on their phone (8 out of 12), of these five
of them check both. Only three people
don’t check email on their phones and
only one person does not check email on a
phone or a computer.
Exercise 4.6
1
a
Branch B 14 | 2 represents 142 pairs
4
Checks
on
phone
Branch B
12
3
15
33
Mostly right-handed
90
b
c
d
e
i
3695 miles
ii 8252 miles
iii 4586 miles
Istanbul to Montreal
21 128 miles
4 hours
Blanks match a city to itself so there is no
flight distance.
Exercise 4.7
1
a
250 000
b
500 000
c
125 000
d
375 000
7
101 beats per minute
142 beats per minute
Exercise raised the heart rate of everyone
in the group. Data moved down the stems
after exercise, indicating higher values for
all people.
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2
Answers may vary depending on the scale
students choose. For example:
France
Spain
USA
China
Italy
represents 25 000 000 arrivals
3
a
b
c
d
Reel deal
Fish tales
Golden rod – 210 fish;
Shark bait – 420 fish;
Fish tales – 140 fish;
Reel deal – 490 fish;
Bite-me – 175 fish
1435 fish
Exercise 4.8
1
a
Favourite take-away food
90
80
No. of people
70
60
50
40
30
20
10
0
Burgers
b
Noodles
Fried chicken Hot chips
Other
African countries with the highest HIV/AIDS
infection rates (2015)
38
25
20
15
10
Mozambique
Equatorial
Guinea
Malawi
Zambia
Namibia
South Africa
Zimbabwe
Lesotho
0
Botswana
5
Eswatini
% of adults (15–49)
infected
30
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2
a
Temperature (ºC)
Frequency
b
32−34
35−37
38−40
41−43
4
5
6
5
Average summer temperature in 20 Middle East cities
Temperature (°C)
41–43
38–40
35–37
32–34
0
1
2
3
4
5
6
7
Frequency
3
a
Local and international visitors on a
Caribbean island
45 000
40 000
No. of visitors
35 000
30 000
25 000
Regional visitors
20 000
International visitors
15 000
10 000
5000
0
b
Jan
Feb
Mar Apr
Month
May Jun
Local and international visitors on a
Caribbean island
55 000
50 000
45 000
No. of visitors
40 000
35 000
30 000
Regional visitors
25 000
International visitors
20 000
15 000
10 000
5000
0
39
Jan
Feb
Mar Apr
Month
May Jun
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Exercise 4.9
1
match they are even happier because their
team won; after that, the level of happiness
drops slowly, but remains high.
Students’ use of online support
3
Never used
Used in the past
Uses at present
2
Students draw their own graph to show their
happiness levels over a school day. If they
seem willing, they can share these with their
groups, but do not force them to do this.
Exercise 4.10
Home language of people passing
through an international airport
English
1
a
i
ii
b
It heats about 14 degrees in 30 seconds, so
assuming a 1 degree heating in 2 seconds,
it will reach 100 degrees in approximately
188 seconds.
2
a
b
2 °C
Between 07:00 and 09:00
3
a
62.5 bpm
b
At 6.30 a.m., as her pulse rate started
steadily increasing after a short rest.
c
150 bpm at 6.50 a.m.
d
It dropped fairly steadily, returning to the
starting rate after about 10 minutes.
Spanish
Chinese
Italian
French
German
Japanese
3
Land used on a farm to grow vegetables
Squashes
Pumpkins
Cabbages
Sweet potatoes
4
a
b
c
d
1
__
4
≈11%
0.25
i
225
ii 100
iii 200
iv 150
Practice questions
1
40
a
b
Graphs can tell a story
1 Students’ own discussions. Should indicate
that change in the graph is in response to
events and that lines sloping up indicate noise
levels increasing, while lines sloping down
indicate noise levels dropping.
2
approximately 33 °C
approximately 65 °C
Students’ own discussions. For example, the
spectators get happier as the game starts, then
their team scores a goal and the happiness
level goes up quickly and stays high, then the
other team scores a goal and the level drops
quickly, staying low. The other team scores
a second goal and the happiness level drops
really low and stays low until their team scores
a goal, then it increases to about the same
level as at the start of the match. It stays at
that level until their team scores a third goal
and then it goes up quickly. At the end of the
c
Primary data – it is data collected she
collected herself by counting.
Discrete data – the data can only take
certain values.
No. of broken
biscuits
Tally
Frequency
0
|||| |||| ||
12
1
|||| ||||
10
2
|||||||||
11
3
|||| |
6
4
|
1
40
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d
d
Number of broken biscuits
12
Frequency
10
8
5
6
4
e
4
28 (to nearest whole number)
f
83 (to the nearest whole number)
a
Pictogram
b
Each stick person represents 1 billion
people.
c
1 billion = 500 million
__
2
0
2
a
b
c
d
0
1
2
3
No. broken biscuits
Heathrow
15 397
Gatwick
Heathrow
London City
Luton
Stansted
4
24 000
40 000
6 000
11 000
15 000
6
1
__
d
2
200 years
e
2012
f
1 of a stick person.
9 full stick people and __
5
a
Football Tennis Bowling Total
Packthorpe
14
5
16
35
Gatwick
Rainbridge
21
13
11
45
Heathrow
Total
35
18
27
80
London city
b
Luton
45
Stansted
Key:
35
3
a
b
4980
District C – it has the highest percentage
of laptops.
c
Percentage (%)
41
Tennis
Bowling
% of people in four districts who own a laptop
and a mobile phone
100
90
80
70
60
50
40
30
20
10
0
Packthorpe
Rainbridge
Practice questions worked
solutions
1
A
B
District
C
Own a laptop
Own a mobile phone
4
Football
Frequency
= 10 000 flights
a
Sport played by students.
b
Five
c
Baseball
D
a
b
The data is collected directly through an
experiment, so it is PRIMARY data.
The data can only be numbers of biscuits
⇒ it can only take whole number values
⇒ it is discrete data.
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
c
No. of broken
biscuits
Tally
Frequency
0
|||| |||| ||
12
1
|||| ||||
10
2
|||||||| |
11
3
|||| |
6
4
|
1
4
5
d
Frequency
12
a
Discrete
b
5
c
Baseball
d
1 × 200 = 50
__
e
4
28
f
83
a
b
c
d
e
Pictogram
1 full symbol represents 1 billion people
1 billion
1930 − 1650 = 280 years
In 2012
1 of another
9 whole symbols and __
5
f
6
a
Tennis
Bowling
Total
2
Football
0
3
4
1
2
Number of broken biscuits
Packthorpe
14
5
16
35
d
Rainbridge
21
13
11
45
Airport
Total
35
18
27
80
a
b
c
Heathrow
15 397
24 000, 40 000, 6000, 11 000, 15 000
Gatwick
b
40
Heathrow
Football
London City
Tennis
Bowling
Luton
Stansted
Packthorpe Rainbridge
Means 4000
3
a
b
Past paper questions
Means 1000
0.83 × 6000 = 4980 people
District C because a large proportion of
people own laptops.
2
c
100
3
%
= Mobile phone
= Laptop
A
42
1
60 = 2 × 2 × 3 × 5
90 = 2 × 3 × 3 × 5
So 3 × 5 = 15
40 × 3 40 × 3
≃ _______2 = ______ = 30
4
20 − 4
a
Pentagon
b
360°
exterior angle = ____ = 20°
18
∴ interior angle = 180° − 20° = 160°
B C D
District
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4
a
1 = 2 −5
___
b
i
ii
5
7
25
318 − t = 36
18 − t = 6
t = 12
8 × 6 × w10 × w5 = 48w15
a = 590 (vertically opposite)
b = 370 (corresponding)
c = 1800 − 590 − 370 = 840 (angle sum in a
triangle)
6
180° − 38°
Angle ACB = __________ = 71°
2
Angle ACD = 180° − 71° = 109°
8
4p7q−1
9
a
b
c
Netball
d
e
Football
f
Hockey
10 a
b
Tennis
Key:
43
represents 4 people
i
ii
i
ii
iii
iv
17 × 4 = 68
61, 67, 71, 73
67
64
65
72
1
__
7
3.722
8=2×2×2
14 = 2 × 7
LCM = 2 × 2 × 2 × 7 = 56
i
12 °C
ii 17 °C
4
_ 3
_3
( √ 81 ) y 16 × 4 = 3 3 y 12 = 27y 12
2 3 = (2 2) p
⇒ 2 3 = 2 2p
2p = 3
3
p = __
2
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Chapter 5
Getting started
1
2
Exercise 5.1
A=F
B=Q
C=H
D=I
E=U
G=O
J=V
K=L
M=P
N=Y
R=T
W=X
S has no matching value.
1
b
c
d
Students’ own drawings. For example:
e
a
2
+50%
c
Start
+40%
+50%
a
b
−50%
b
a
+90%
d
e
f
g
44
15 ___
20
5 ___
10 ___
__
=
=
=
9 18 27 36
3 ___
9 ___
6 ___
__
=
= 12
=
7 14 21 28
8
6 __
12 = __
___
= 2 = ___
18 9 3 12
18 __
3
2 = __
___
= 1 = __
36 2 4 6
55 ____
165 ____
220
110 ___
____
=
=
=
128 64 192 256
1
__
3
1
__
3
3
__
4
3
__
5
1
__
5
2
__
3
3
___
10
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Fraction diagrams
1
a
Jay
b
Mia
1
4
1
4
1
4
1
4
Sam
1
2
1
4
1
8
1
8
1
4
Exercise 5.2
It is possible. For example:
1
3
4
1
4
c
2
a
10
___
b
3
___
c
2
__
d
1
__
a
1
__
24
1 × 2 × 1 = 1, area of entire rectangle is
__
2
1 of the
4 × 6 = 24, so shaded triangle is ___
24
rectangle)
2
1
4
1
2
5
20
1
8
1
16
13
16
45
1
12
1
6
1
6
2
12
1
3
3
9
1 2
=
4 8
1
5
=
2
10
9
4
c
d
92
22
___
= 2 ___
e
35
32
f
319
7
____
= 39 __
g
Students’ own work. There are many possible
solutions and using equivalent fractions will
produce more options. For example:
14
5
8
___
21
2
b
1 (area of the shaded triangle is
___
27
h
35
8
180
8
161
1 or ____
80 __
2
2
Exercise 5.3
1
a
2
__
b
5
__
c
1
__
d
13
4
___
= 1 __
e
11
___
f
1
___
g
7
__
h
1
2 ___
16
3
7
4
9
9
30
24
8
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
a
b
c
d
e
f
g
h
i
j
k
l
1
3 __
3
5
___
6
11
1
18 __
4
3
__
3
4
5
− __
6
11
12 ___
16
13
___
6
16
29
___
2
60
25
___
1
42
1
__
2
5
9 ___
12
7
3 ___
60
9
b
77
___
10
60
1
1
___
2
4
__
3
60
4
___
= 8 __
40
5
4
7
5
5
24
6
1
__
7
3
__
8
3
9 __
5
7
8
8
Exercise 5.6
2
Terminating: __
5
1
Non-terminating: __
3
5
__
3
___
16
8
3
__
7
4
__
9
1
90 people
2
4
___
3
21
98
4
3
__
7
5
1
__
3
Terminating: the denominators are all
products of powers of 2 and 5 only.
6
3
3 cups and 3 __ cups of water
4
4
They all have the same numbers in the same
order, but starting in different places.
Exercise 5.7
Exercise 5.4
1
2
3
4
5
6
7
8
46
7
___
Exercise 5.5
Fraction patterns
1 0.4, 0.333333…, 0.625, 0.428…, 0.444…,
0.1875
2
a
3
__
7
14
___
15
4
___
63
2
___
11
147
2
____
= 29 __
5
5
48
___
85
189
67
____
= 1 ____
122
122
13
___
14
1
4
a
7
___
b
3
__
c
1
__
d
9
___
e
3
___
f
1
___
g
43
___
h
33
___
i
47
___
10
4
5
25
20
40
20
25
40
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
j
k
l
2
271
____
250
1
____
400
1
______
50 000
a
b
c
d
e
60%
28%
85%
30%
4%
f
2%
416.67% = 416 __
3
Thinking about percentages
1 a Method D.
b There are different reasons why the others
won’t work, mostly because they all use
the incorrect method. Individual reasons
might include: A – no % included in the
calculation, B – multiplying by the answer
cannot give the answer, C − 0.013 is not
13% and E – 13 is given as fraction of 500
rather than 100.
2
a
b
c
d
3
Most will show 0.25 because 25% is
equivalent to 0.25.
12% × 650 or 650 × 12% (the other
methods involve converting the percentage
and don’t use the percentage function, for
example, 0.12 × 650, 12 ÷ 100 × 650,
650 ÷ 100 × 12).
You are dealing with percentages here, so
you include the sign.
Students will compare calculators and
probably discover that they don’t all work
in exactly the same way.
Students’ explanation of different methods for
calculating 15% of 500.
2
a
b
c
3
16 397 batteries (16396.8)
4
77.8 (3 s.f.)%
5
79.2 (3 s.f.)%
6
25%
7
0.025%
8
177.33%
Exercise 5.9
1
4%
2
21%
3
7%
4
19%
5
25%
6
44%
Exercise 5.10
1
a
b
c
d
e
44
46
50
42
41.6
2
a
b
c
d
e
79.5
97.52
60.208
112.36
53.265
3
a
b
c
d
e
111.6
105.4
86.8
119.04
115.32
4
a
b
c
d
e
3.62
23.3852
36.0914
0
36.019
Exercise 5.8
1
47
a
b
c
d
e
f
g
h
15
12
135
360
75
45
0.078 m
0.275 L
40%
25%
27.0% (3 s.f.)
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
33 h
6
$13.44
7
26 199
8
126 990
9
10 h 34 min
10 174.90%
11 No. 100% − 12% − 12% = 74.44% which is a
22.56% decrease
12 14.9
13 Students’ reasoning may vary.
a No, on day 8 the points will have doubled
but on day 7 they will have halved.
b Increasing by 40% and then 50% as the
increase is cumulative, so you are actually
increasing 140% by 50%.
c It is cheaper to take the 50% discount.
For example, if 400 MB of data cost $100,
then you would pay $50 for 400 MB, if
you take the extra 50% free, then you get
600 MB for $100 and this means you pay
$50 for only 300 MB.
e
f
g
h
i
j
$12.95
$37.54
$24.39
$105.90
$0.81
$0.66
6
a
b
40 students
33 students
7
$20
8
80 kg
9
210 litres (3 s.f.)
Exercise 5.12
1
a
b
c
d
e
f
g
h
i
j
k
l
3.8 × 102
4.2 × 106
4.56 × 1010
6.54 × 1013
2 × 101
1 × 101
1.03 × 101
5 × 100
4 × 10−3
5 × 10−5
3.2 × 10−5
5.64 × 10−8
2
a
b
c
d
e
f
g
h
i
2 400 000
310 000 000
10 500 000
9 900
71
0.000 36
0.000 000 016
0.000 000 203
0.0088
3
a
b
0.000 025 kg
2.5 × 10−5 kg
Exercise 5.11
1
175
2
362.857
3
1960
4
5
48
Sale price
($)
a
b
c
d
% reduction
Original
price ($)
52.00
10
57.78
185.00
10
205.56
4700.00
5
4 947.37
2.90
5
3.05
24.50
12
27.84
10.00
8
10.87
12.50
7
13.44
9.75
15
11.47
199.50
20
249.38
4
4.0208 × 1013 km
99.00
25
132.00
5
8.4 × 10−2 mm
$20.49
$163.93
$11.89
$19.66
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Investigation: Standard form on a calculator
1 Learners’ own answers.
2
a
b
i
1.09 × 105
ii 2.876 × 10−6
iii 4.012 × 109
iv 1.89 × 107
v 3.123 × 1013
vi 2.876 × 10−4
vii 9.02 × 1015
viii 8.076 × 10−12
ix 8.124 × 10−11
x 5.0234 × 1019
8.076 × 10−12
8.124 × 10−11
2.876 × 10−6
2.876 × 10−4
1.09 × 105
1.89 × 107
4.012 × 109
3.123 × 1013
9.02 × 1015
5.0234 × 1019
3
a
b
c
d
e
f
g
h
i
4
Volume of space is (3.27 × 10−7) × (2 × 10−7) ×
(1.16 × 10−4) = 7.5864 × 10−17.
7.927 × 10−17 . 7.5864 × 10−17, so it will not
fit.
5
a
b
2
49
Display will vary according to the calculator
used.
a 4.2 × 1012
b 0.000 018
c 2 700 000
d 0.0134
e 0.000 000 001
f 42 300 000
g 0.000 310 2
h 3 098 000 000
i
2.076 × 10−23
a
b
c
d
e
f
g
h
1.3607 × 1018
1.0274 × 10−15
1.0458 × 100
1.6184 × 1011
5.2132 × 1019
3.0224 × 10−16
2.3141 × 1012
1.5606 × 1017
1.07 × 109
1 × 1012
Exercise 5.14
1
a
b
c
d
e
f
g
h
i
j
k
l
8 × 1030
4.2 × 1012
2.25 × 1026
1.32 × 109
1.4 × 1032
3 × 101
2 × 101
3 × 103
3 × 1042
1.2 × 103
5 × 102
1.764 × 1015
2
a
b
c
d
e
3.4 × 104
3.7 × 106
5.627 × 105
7.057 × 109
5.7999973 × 109
3
a
b
c
d
e
f
g
h
8 × 10−10
6.4 × 10−12
3.15 × 10−9
3.3 × 10−2
2 × 1033
7 × 10−37
5 × 1012
1.65 × 101
Exercise 5.13
1
2.596 × 106
7.569 × 10−5
4.444 × 10−3
1.024 × 10−7
3.465 × 10−4
2.343 × 107
5.692 × 103
3.476 × 10−3
1.040 × 10−3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
a
b
c
d
2.731 × 10−2
2.88 × 10−1
7.01056 × 103
1.207 × 10−5
5
8.64 × 104 seconds
6
a
b
c
3 × 109 metres
6 × 109 metres
3.06 × 1010 metres
Practice questions worked
solutions
1
2
a
b
c
a
19 + 24 + 31 + 10 + 11 = 95
100% − 95% = 5% gained a grade U
b
6
24 = ___
____
c
3
29 975
4
7.5%
5
The 20% increase is not an increase on the
original salary, it is an increase on the larger
salary at the end of year 1. Timur’s salary
will be multiplied by 1.1 × 1.2 = 1.32. Timur’s
salary is increased by 32% overall.
6
n = 1.88 × 107
7
a
b
c
8
3 299 000
9
ab
____
× 10m + n + 2
10 a
b
c
27 500 × 1.09 = 29 975 cases
4
172 − 160
__________
× 100% = 7.5%
5
The second increase is 20% on an already
increased salary.
1.1 × 1.2 = 1.32 ⇒ 32% increase
6
3 × 10 8 × 2 × 10 7
n = ________________
3 × 10 8 + 2 × 10 7
6 × 10 15
= _______________________
300 000 000 + 20 000 000
6 × 10 15
= ___________
320 000 000
6 × 10 15
= _________8
3.2 × 10
160
= 1.875 × 107
= 1.88 × 10 7 (to 3 s.f.)
7
$4400 (2 s.f.)
$5000 (2 s.f.)
x n
V 1 − ____
(
100 )
11 On 1 January 2038 the painting is due to be
worth $2520. So the value of the painting will
first be worth more than $2500 during 2037.
100 25
19% of 93 800 = 0.19 × 93 800
= 17 822
3
9.46 × 1012
0.423
1.88 × 105
100
5 2 __
= __(__
+1
6 8 8)
5 31
= __ × __
8
26
5
= ___
16
2
16
5%
6
___
25
17 822
8)
5 __
1 + __
1
__
Practice questions
5
___
6(4
1
a
3.0 × 105 × 60 × 60 × 24 × 365
= 9.46 × 1012 km
b
4.0 × 10
___________
= 4.23 light years
c
3.0 × 105 × 0.625 = 1.88 × 105
miles per second
13
9.46 × 10 12
8
3.352 × 10
___________
= 3 299 000
9
xy = ab × 10m + n
6
1.016
ab
= ____ × 10 m + n + 2
100
3875
10 a _____ = $4403
0.88
b Divide by 0.88 again = $5004
100 − x
V ( ________)
100
n
c
50
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
11 $1800 × 1.02 = $1836
× 1.02 = $1873
× 1.02 = $1910
× 1.02 = $1948
× 1.02 = $1987
× 1.02 = $2027
× 1.02 = $2068
× 1.02 = $2109
× 1.02 = $2151
× 1.02 = $2194
× 1.02 = $2238
× 1.02 = $2283
× 1.02 = $2328
× 1.02 = $2375
× 1.02 = $2423
× 1.02 = $2471
× 1.02 = $2520
Therefore, the amount in 2038 is $2520.
51
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 6
h
5
p = − __
2
20
p = ___
13
x = −1
a
b
c
d
x = 11.5
x = 10.5
x = 16.7
x=3
e
f
1
x = − __
7
x = 10
a
x=3
Exercise 6.1
b
1
c
d
e
f
1
x = __
2
5
x = __
2
x = 0, so there is no solution
x=1
x = −13
a
x=1
b
1
x = __
3
c
3
x = − __
4
1
x = __
3
__
x=1
5
1
x = − __
6
Getting started
f
3
7
7
4
18
4
8
a+d
a
b
c
d
e
f
g
h
2
a
b
c
3
4
52
4
15
5
3a + c
a+b+c
a+b+d
x=7
x = −5
x=9
62
x = − ___
7
x=5
n = 11
q = 1.75
t = 0.5
e
f
x=2
x = −10
y = −3
11
x = ___
15
p=1
x = 60
a
b
c
d
e
x=2
p=3
t=1
m=5
n = 10
f
13
x = − ___
6
a
b
c
d
x=2
x=2
x = 12
x=1
e
15
x = ___
4
d
11
11
a+c
2a + c + d
g
7
b
4a + 4b + c
4b + a
a – 2d
−a + 4b
− c − 2d
−2a − c
− 2d
6
7
d
e
f
Exercise 6.2
1
a
b
c
d
e
f
g
h
i
j
k
l
3(x + 2)
3(5y − 4)
8(1 − 2z)
5(7 + 5t)
2(x − 2)
3x + 7
2(9k − 32)
11(3p + 2)
2(x + 2y)
3( p − 5q)
13(r − 2s)
2( p + 2q + 3r)
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
3
a
b
c
d
e
f
g
h
7(3u − 7v + 5w)
3x( y + 1)
3x(x + 1)
3p(5q + 7)
3m(3m − 11)
10m2(9m − 8)
12x3(3 + 2x2)
4pq(8p − q)
a
b
c
2m2n2(7 + 2mn)
abc(17 + 30b)
m2n2(49m + 6n)
d
1 (a + 3b)
__
f
g
h
i
j
2
1
__ x(6x 3 + 7)
8
8(x − 4)
(x + 1)2 (1 − 4x)
2x3(3 + x + 2x2)
7xy(x2 − 2xy + 3y)
( y + 3)(x + 2)
a
b
c
d
e
f
4(x + 3y)
15(3x + y)
a2b(3a − 4b)
3(17z3 + 7x2)
4x3y4(3 − 5x2y)
Fully factorised
a
b
c
d
e
f
g
h
i
3(3x + 4y)
3(3x − 4y)
10a(x + y)
(2x3 − 3)(5x2 − 2)
(x + y)(x − y − 2)
(x − 2m + 1)(x + 2m − 1)
(4x + 5y)(2a + 3b)
(abx + bcy)(1 + c)
(ax − by)(1 + k)
a
b
4(6x + 35)
240(2x + 5)
e
4
5
6
They are all divisible by 3.
3
a
b
4
n + n + 1 + n + 2 + n + 3 + n + 4 = 5n + 10
= 5(n + 2). Divisible by 5.
Seven numbers 7n + 21 = 7(n + 3).
Divisible by 7.
5
Sum of four numbers is divisible by 2, not 4.
Sum of six numbers is divisible by 3 not 6.
6
When the number of integers is p, where p is
prime, the answer is always divisible by p.
This doesn’t work for non-primes.
1
a
b
d
a=c−b
r=p−q
g
h = __
f
d−c
_____
b= a
e
a = bc
f
t+m
n = _____
a
a
m = an − t
b
t
a = ______
n−m
c
tz
x = __
y
d
x = bc + a
e
d
y = c − __
x
f
b=a−c
a
r = q(p − t)
b
x−a
b = _____
c
c
2
3
c
d
3(60 − x) − 4y
e
8
a
b
f
9003 × 10m
This is not in standard form because 9003
is greater than 10.
In standard form: 9.003 × 10m + 3.
Sums of consecutive numbers
1 Students’ choice of numbers.
n + 1 and n + 2.
n + n + 1 + n + 2 = 3n + 3 = 3(n + 1).
This is divisible by 3. This confirms the
answer to question 2.
Exercise 6.3
7
c
53
2
4
t
m = n − __
a
bc
a = ___
d
a = x − bc
xy
z = ___
t
a
b = c2
b
c2
b = ___
a
c
c 2
b = (__
a)
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
d
e
b = c2 − c
b = x − c2
f
x 2
y = (__
c)
a
P−l
w = __
2
w = 35.5 cm
b
6
7
a
(v − u)
a = ______
t
b
10.72 m/s
T2
l = g ____
( 4p 2 )
c
d
1
11 b = _______
(c − a) 2
12 2(2p2 + a)( pq3 + 2)
5
13 __
2
Practice questions worked
solutions
1
Practice questions
1
9
2
a
b
c
qx
y = ___
p
4
x2y(x2y2 + 7 − 3xy2)
5
5
− __
2
6
2n + 1 + 2n + 3 + 2n + 5 = 69
6n + 9 = 69
n = 10
Smallest number is 21.
7
8
a
b
a
b
c
9
a
b
c
d
e
f
10 a
b
54
2
a
b
14 − n
3n − (14 − n) = 4n − 14 = 22
n=9
c
Increase in height = 1300 − 500
= 800 m
800
____
= 4 °C
200
So, the new temperature = 23 °C − 4° °C
= 19 °C
increase in height
________________
=5
200
Therefore, increase in height = 5 × 200
= 1000 m
700 − 12
13 = ________
q
688
13 = ____
q
9T
F = ___ + 32
5
17.6 °F
−40 degrees
13q = 688
688
q = ____ = 52.9
13
3x + 42 cm
x + 3 cm
x + 7 cm
8x + 30 = 2(4x + 15) cm
8x + 30 = 3x + 42
x = 2.4 cm
3
Perimeter of each shape is 49.2 cm.
Area is 80.28 cm2.
5
12x + 13
0.5
3p − 5
T = ______
2
3p
−5
12 = ______
2
3p − 5 = 24
3p − 5 = 29
29
p = ___
3
Temperature will be 19 °C.
You will need to climb to 1500 m.
52.9
3
19 cm
4.27 cm
4
p
x __
__
y =q
py = qx
qx
y = ___
p
x4y3 + 7x2y − 3x3y3
= x2y(x2y3 + 7 − 3xy2)
0.8x + 3 = 2(0.6x + 2)
0.8x + 3 = 1.2x + 4
0.4x = −1
1 = −2.5
x = − ___
0.4
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
6
2n + 1 + 2n + 3 + 2n + 5 = 69
6n + 9 = 609
6n = 60
n = 10
The smallest number is 2n + 1 = 21
7
a
b
8
a
b
c
9
a
b
c
d
e
55
14 − n
3n + (14 − n)(−1)
= 3n − 14 + n
= 4n − 14
4n − 14 = 22
4n = 36
n=9
5
T = __ (F − 32)
9
9T
___
= F − 32
F
9T
Therefore, F = ___ + 32
5
9( )
__
F = − 8 + 32
5
72 160
= − ___ + ____
5
5
88
___
=
5
= 17.6 °C
5
x = __(x − 32)
9
9x = 5(x − 32)
9x = 5x − 160
x = − 40
So, −40 °F = −40 °C
x + 14 + x + 14 + x + 14
= 3x + 42
EF = 2x + 3 − x
=x+3
FG = 12 + 2x − (x + 5)
= 12 + 2x − x − 5
=x+7
x + 12 + 2x + 2x + 3 + x + 5 + x + 3 + x + 7
= 8x + 30
3x + 42 = 8x + 30
5x = 12
12 = 2.4
x = ___
5
f
Perimeter of ABC = 3x + 42
= 7.2 + 42
= 49.2 cm
Perimeter of DEFGHI = 8x + 30
= 19.2 + 30
= 49.2 cm
Area DEFGHI = (2x + 3)(12 + 2x)
− (x + 3)(x + 7)
= 7.8 × 16.8 − 5.4 × 9.4
= 80.28 cm2
10 a
3 + 2(x + 1) + 2(3x + 2) + 4(x + 1)
= 3 + 2x + 2 + 6x + 4 + 4x + 4
= 12x + 13
b
4(x + 1) = 3 + 2(x + 1)
4x + 4 = 3 + 2x + 2
2x = 1
1
x = __
2
3
7
c 3x + 2 = __ + 2 = __
2
2
3
4(x + 1) = 4 × __ = 6
2
7
Perimeter = 2(__ + 6)
2
= 7 + 12
= 19 cm
7
d Area of rectangle = __ × 6 = 21 cm2
2
Area of square × 1.15 = 21
21
Therefore, area of square = _____
1.15
_____
21 = 4.27 cm
Side of square = _____
1.15
1_ = c
11 a + __
√b
1_ = c − a
__
√b
√
_
√b
1
__
= _____
1
c−a
1
b = (_____
c − a)
2
12 4p3q4 + 8p2q + 2apq3 + 4a
= 2(2p3q4 + 4p2q + apq3 + 2a)
13
42x − 3 = 25 + 2(x − 3)
= 25 + 2x − 6
2
2x
−
3
(2 )
= 22x − 1
2(2x − 3) = 2x − 1
4x − 6 = 2x − 1
2x = 5
5
x = __
2
(22)2x − 3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 7
Getting started
a
b
c
d
e
f
1
___
9
14.14 cm × 14.14 cm
10 a
24
1
__
4
1
___
10
3
__
8
1
__
2
1
__
2
b
c
Exercise 7.1
1
a
b
12.5 cm
9 cm
2
a
b
c
25 cm
35 m
23 km
a
b
c
d
16 m2
42 m2
8 cm2
54 cm2
a
b
i
5850 cm2
ii 0.585 m2
360 cm
a
b
c
d
e
f
g
50 m2
52.29 m2
33.1 cm2 (3 s.f.)
37.8 cm2
36 cm2
145.16 cm2
55.7 cm2 (3 s.f.)
a
b
c
d
e
h = 6 cm
b = 17 cm
a = 2.86 cm (3 s.f.)
b = 5 cm
h = 10.2 cm (3 s.f.)
3
4
5
6
56
7
183 tiles
8
a
74.8 cm2
b
133
____
xy
0.3458 m2
1
__
2
Consider this as a red/black triangle on
top of a yellow/white triangle. The base
of each triangle is the same but the height
of the white/yellow triangle is twice the
height of the red/black triangle, so its area
will be twice as big. Therefore the area of
the red/black triangle is half of the area
of the yellow/white triangle which means
that the area of the yellow and white
region of the flag is the same as the area
of the red and black region.
11 Students’ answers will vary; the following are
just examples.
a
1 cm
4 cm2
4 cm
2 cm
6 cm2
3 cm
b
1 cm
6 cm2
6 cm
2 cm
6 cm2
3 cm
c
5 cm
3 cm
18 cm2
6 cm
5 cm
4 cm
24 cm2
6 cm
162
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d
6.6 cm
24 cm2
4 cm
6 cm
5 cm
6
Exercise 7.3
24 cm2
1
4 cm
6 cm
12 Area = 440 square units and
perimeter = 102 units
Increasing areas
1 4 cm2
2
Area of 10 cm square is
This is
25 times larger than the 2 cm square.
When we multiply the side length by k the area
is multiplied by k2.
5
14 cm2
6
The same result applies: multiplying the side
length by k means the area is multiplied by k2.
57
14p
C = ____ + 14 cm
2
12p
C = ____ + 12 cm
2
18.4p
C = ______ + 18.4 cm
2
49p
A = ____ cm 2
2
36p
A = ____ cm 2
2
84.64p
A = _______ cm 2
2
d
9p
A = ___ cm 2
2
3
a
b
12 cm
A = 144 − 36p cm2
4
A = 32p mm2
Exercise 7.4
1
a
b
c
d
e
f
g
h
A = 12.6 cm2
A = 25.1 cm2
A = 1.34 cm2
A = 116 cm2
A = 186 m2
A = 0.185 cm2
A = 36.3 cm2
A = 98.1 m2
P = 16.2 cm
P = 22.3 cm
P = 7.24 cm
P = 44.2 cm
P = 55.0 m
P = 1.88 cm
P = 24.6 cm
P = 43.4 m
2
a
b
c
d
A = 198 m2
A = 70.4 cm2
A = 94.7 cm2
A = 14.5 m2
l = 22.0 m
l = 17.2 cm
l = 29.6 cm
l = 9.69 m
3
a
b
A = 16.4 m2
A = 243 cm2
P = 6.54 m
P = 62.5 cm
4
1856 m
5
70.0 m
6
14 cm
7
9 cm
Exercise 7.2
2
d
C = 10p cm
C = 14p cm
A = 0.9025p cm2
100 cm2.
Answers are correct to 3 s.f.
a
b
c
d
e
f
A = 20.25p cm2
A = 1369p cm2
A = 3600p mm2
a
b
c
Area of 6 cm square is 36 cm2. This is nine
times larger than the 2 cm square.
4
1
C = 9p cm
C = 74p cm
C = 120p mm
f
14 72 cm
3
a
b
c
e
13 32 cm2
2
2 × 12 cm pizza ≈ 226.2 cm2 and 24 cm
pizza ≈ 452.4 cm2, so two small pizzas is not
the same amount of pizza as one large pizza.
A = 50.3 m2
A = 7.55 mm2
A = 0.503 m2
A = 0.785 cm2
A = 1.57 km2
A = 1.27 m2
C = 25.1 m
C = 9.74 mm
C = 2.51 m
C = 3.14 cm
C = 4.44 km
C = 4 m (exact)
Answers correct 3 s.f.
a A = 250 cm2
b A = 13.7 cm2
c A = 68.3 m2
d A = 55.4 cm2
e A = 154 m2
f A = 149 cm2
3
23 bags
4
White = 0.1 m2
5
0.03 m2
Red = 1.0 m2
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
8
25
A = 40 − ___p cm2
8
A = 90 − 4p cm2
a
b
4
a
b
768 cm3
816 cm2
5
3.39 m3 (3 s.f.)
6
d
25
5
A = 60 − ___p m2
P = 34 + __p m
3
6
2
A = 70.56 − 17.64p cm P = 16.8 + 8.4p cm
a
b
e
A = 324 − 81p m2
7
241 cm3 (3 s.f.)
a
b
c
d
e
P = 144 cm
P = 7.07 cm
P = 15.6 cm
P = 26.6 cm
P = 61.1 cm
8
a
b
c
9
Volume = 264 cm3
Surface area = 306 cm2
c
9
5
P = 21 + __p cm
2
P = 30 + 2p cm
P = 18p m
A = 1400 cm2
A = 3.63 cm2
A = 17.0 cm2
A = 32.6 cm2
A = 181 cm2
200 p cm3
542 cm2 (3 s.f.)
320 boxes
8.5 m2
48 m3
10 6.77 cm
Exercise 7.7.
Exercise 7.5
1
1
b
c
a
b
a
2
3
a
b
c
Trapezium-based prism
O and S
PQ = RQ = UV = VW
a is correct
4
1
4
2
5
8
a
1600p cm2 (3 s.f.)
b
32 000
______
p cm3 (3 s.f.)
3
2
5300 cm3 (3 s.f.)
3
549 000 000 000 km3 (3 s.f.)
4
2 600 000 m3 (3 s.f.)
5
a
b
6
1110 cm3 (3 s.f.)
7
a
b
8
2.29 cm (3 s.f.)
9
R = 3√ 2
__
340p m2 (3 s.f.)
725p m3 (3 s.f.)
754 cm3 (3 s.f.)
415 cm2 (3 s.f.)
__
r
10 a
3
5 mm
7
10.4 cm
6
5
a
b
344 cm2
0.000 24 m3
35 cm
Exercise 7.6
Volume = 66 cm3
2
a
b
3
58
b
1
i
ii
i
ii
720 cm3
548 cm2
13.8 mm3 (3 s.f.)
40.3 mm2 (3 s.f.)
Volume of metal in the tube
10.4
= (p × (_____) × 35)
2
Surface area = 144 cm2
2
10.4 − 1
− (p × (________) × 35) cm 3
2
2
c
544 cm3 (3 s.f.)
432 000 cm3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
d
Total surface area of tube = 2 × area of ring + area of outer tube + area of inner tube
(Note ‘ring’ is the 5 mm thick end of the cylinder.)
10.4
10.4 − 1
2 × [p × (_____) − p (________) ] + (p × 10.4 × 35) + [p × (10.4 − 1) × 35] cm 2
2
2
2
2
11 260 cm3
12 a
b
i
ii
i
ii
205 cm2
197.47 cm2
225 cm3
254.47 cm2
5
40
8 × p × 1.2 2 × ____
360
= 4.02 m2
6
a
Practice questions
b
1
33 900 mm (3 s.f.)
2
P = 32.3 cm (3 s.f.) Area = 47.7 cm2
3
2.31 m3 (3 s.f.)
4
550
____
p cm
5
4.02 m2
6
a
b
c
62.8 cm
125.7 cm2
37.5%
7
a
b
168 cm3
0.000 168 m3
8
3
4
_
1 × 8 × √ 48
Area of triangle = __
2
_
Therefore, total area = 40p + 2 √ 48
_
= 40p + 16 √ 3 cm2
59
c
11 = 1.375
Scale factor of lengths = ___
8
Therefore, increase in perimeter = 37.5%.
a
1×3×4×7
Volume = 6 × 5 × 7 − __
2
= 210 − 42
= 168 cm3
b
1 m3 = 100 × 100 × 100 cm3
= 1 000 000 cm3
Volume = 336 cm3
Surface area = 360 cm2
1
600 × p × 18 = 33 929 mm
= 33.9 m
2
1 (p + 4)
Perimeter = 9 + 6 + 9 + 1 + 1 + __
2
= 32.3 cm
1 p × 42
Area = 9 × 6 − __
2
= 28.9 cm2
4
82 – 42 = 48
8
Practice questions worked
solutions
3
60°
1 of
The triangle removes ____ = __
360°
6
each circle
5
Therefore, perimeter = 3 × __ × p × 8
6
= 20p cm
5
Area = 3 × __ × p × 4 2 + area of triangle
6
= 40p + area of triangle
7
168
Volume in m3 = _________ = 0.000 168 m3
1 000 000
1.4
Volume = p × (____) × 1.5 = 2.31 m3
2
1 × p × 5 2 × 12 + __
1 × __
4 × p × 53
__
3
2 3
250
= 100p + ____ p
3
550p
= _____ cm3
3
2
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
8
Each sloping face is a trapezium.
6
5
12
Therefore, surface area
12 + 6
=12 2 + 6 2 + 4 × (______) × 5
2
= 360 cm2
1 × 12 2 × 8 − __
1 × 62 × 4
Volume = __
3
3
= 336 cm2
60
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 8
Getting started
6
1, 2 Students’ individual work.
It is sensible to choose ‘higher’ for any number
where more than half the numbers are higher
(so any decimal between 0 and 0.5). It is
sensible to choose ‘lower’ for any number
where more than half the numbers are lower
(so any decimal between 0.5 and 1).
Talking about probability
1 Probability only tells us the expected, or ‘on
average’ proportion of ‘true’ answers. It does
not tell us exact proportions.
2
The three possible outcomes are not
necessarily equally likely.
3
Statistical experiments always give a range of
outcomes. Different outcomes are possible, so
they also have a non-zero probability.
4
Any random experiment like this can produce
unexpected outcomes, because each of the
combinations of red and blue socks is possible.
We should only give probabilities for outcomes
of experiments that have not already been
completed. It makes more sense to talk about
the probability that it rains tomorrow.
7
a
b
c
d
8
9 blue balls
9
a
b
c
d
2
a
1
b
a
1
___
b
c
d
3
4
61
10
3
___
20
131
____
260
141
____
260
2
b
a
80
b
7
___
c
5
b
235
____
= 0.783
300
233
a
20
30
13
1
__
4
1
__
2
3
___
13
a
7
___
50
0.14
1
___
Exercise 8.2
Exercise 8.1
1
77
76
___
77
4
__
9
5
__
9
0
__
9
1
First throw
Second
throw
5
b
1
___
i
1
__
ii
1
__
iii
3
__
iv
1
__
H
T
H
HH
TH
T
HT
TT
2
4
4
4
a
First dice
Second dice
3
a
×
1
2
3
4
5
6
1
1
2
3
4
5
6
2
2
4
6
8
10
12
3
3
6
9
12
15
18
4
4
8
12
16
20
24
5
5
10
15
20
25
30
6
6
12
18
24
30
36
5 750
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
b
1
___
i
ii
iii
iv
v
vi
a
b
Spinner
iii
iv
v
Second throw
ii
62
18
5
6
7
8
9
2
3
4
5
6
7
8
9
10
3
4
5
6
7
8
9
10 11
4
5
6
7
8
9
10 11 12
+
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
3
4
6
8
4
5
6
7
8
9
10
5
5
5
5
6
8
5
6
7
8
9
10 11
7
7
7
7
7
8
6
7
8
9
10 11 12
9
9
9
9
9
9
b
Set B, but set A is not far away from being
sensible
Exercise 8.3
1
a
b
c
d
2
4
1
4
6
2
6
2
6
3
6
10
2
2
10
2
5
2
12
4
6
2
12
3
12
15
1
3
5
3
15
3
24
4
6
2
12
3
24
vi
4
3
2
9
3
9
4
8
___
2
8
4
v
1
7
24
18
8
6
15
2
__
7
5
12
iv
6
4
10
17
___
5
3
6
iii 1
4
8
4
18
2
__
3
3
6
20 10
14 ___
7
___
=
20 10
9
___
20
9
___
20
8 __
___
=2
20 5
5
___
2
4
First throw
i
1
2
a
b
+
1
6 ___
3
___
=
i
ii
4
a Set A
Set B
Tetrahedral
dice
3
5
36
0
___
36
2
__
9
7
__
9
1
__
6
2
__
9
1
___
36
1
__
4
1
__
6
5
__
6
2
a
b
Red, red; red, blue; blue, red; blue, blue
i
58.3%
ii 41.7%
iii 24.3%
iv 51.4%
v 48.6%
vi 34.0%
vii 66.0%
3
a
1
____
b
c
d
e
f
169
1
_____
2704
1
___
52
3
__
8
1
__
2
1
__
2
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
a
b
c
d
e
0.24
0.24
0.36
0.76
0.52
Practice questions
a
2
__
b
3
__
c
0
2 a
1
__
b
0
c
5
__
d
1
__
1
3
a
5
5
3
6
3
Face
Probability
2
c
13
___
63
2
1
__
3
6
___
18
3
5
___
18
5
___
18
4
1
__
6
3
___
18
18
a
Josh
Soumik
4
b
1
2
__
9
4
___
18
b
6
___
c
18
___
d
25
___
+
$5
$1
$1
50c
20c
20c
20c
$5
$10
$6
$6
$5.50
$5.20
$5.20
$5.20
$5
$10
$6
$6
$5.50
$5.20
$5.20
$5.20
$5
$10
$6
$6
$5.50
$5.20
$5.20
$5.20
$2
$7
$3
$3
$2.50
$2.20
$2.20
$2.20
50c
$5.50
$1.50
$1.50
$1
70c
70c
70c
50c
$5.50
$1.50
$1.50
$1
70c
70c
70c
50c
$5.50
$1.50
$1.50
$1
70c
70c
70c
49
49
49
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
1
__
b
1
__
c
11
___
8
9
96
a
Square spinner
Pentagon
spinner
6
a
b
c
1
2
3
4
1
1, 1
2, 1
3, 1
4, 1
2
1, 2
2, 2
3, 2
4, 2
3
1, 3
2, 3
3, 3
4, 3
4
1, 4
2, 4
3, 4
4, 4
5
1, 5
2, 5
3, 5
4, 5
i
4 = __
1
___
ii
7
___
iii
1
2 = ___
___
iv
9
___
i
7
P(A or B) = ___
20
ii
20
5
20
20
10
20
7 ___
1 + ___
= 11
but P(A) + P(B) = __
5 20 20
Events A and B are not mutually
exclusive.
Practice questions worked
solutions
1
c
2
4 = __
P(banana) = ___
10 5
6 __
3
2 + 4 = ___
_____
=
10
10 5
0
a
2 = __
1
__
b
0
__
=0
c
5
__
d
2 = __
1
__
a
5 ___
3
4 , ___
4 , ___
___
,
a
b
2
3
64
6
3
6
6
6
3
b
18 16 18 18
Face 2
c
4 + 6 + 3 ___
13
5
13
_________
=
or 1 − ___ = ___
18
18
18
18
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
Josh: $1, 50c, $5, 20c, 20c, 20c
Soumik: $5, $5, $5, $2, 50c, 50c, 50c
a
Soumik
Josh
b
c
d
50c
20c
20c
20c
$6
$5.50
$5.20
$5.20
$5.20
$10
$6
$5.50
$5.20
$5.20
$5.20
$5
$10
$6
$5.50
$5.20
$5.20
$5.20
$2
$7
$3
$2.50
$2.20
$2.20
$2.20
50c
$5.50
$1.50
$1
70c
70c
70c
50c
$5.50
$1.50
$1
70c
70c
70c
50c
$5.50
$1.50
$1
70c
70c
70c
c
65
$10
$5
1
2
3
4
5
6
x
x
x
x
1
__
1
__
6
3
1 + __
1=1
4x + __
6 3
6 1 __
1
4x = __ − __
− 2 = __
6 6 6 2
1
x = __
8
1
P(rolling 3) = __
8
1 × __
1 = __
1
P(6 and 6) = __
3 3 9
P(2, 6) + P(6, 2) + P(3, 4) + P(4, 3)
1 × __
1 + __
1 × __
1 + __
1 × __
1 + __
1 × __
1
= __
8 3 3 8 8 8 8 8
1 + ___
1 + ___
1
1 + ___
= ___
24 24 64 64
1 + ___
1 = ___
11
= ___
12 32 96
a
Square
Pentagon
6
$5
$1
42 14
15
5
___ = ___
42 14
22 = ___
11
___
42 21
Probability
b
$5
3 ___
___
= 1
5
a
+
1
2
3
4
1
1, 1
2, 1
3, 1
4, 1
2
1, 2
2, 2
3, 2
4, 2
3
1, 3
2, 3
3, 3
4, 3
4
1, 4
2, 4
3, 4
4, 4
5
1, 5
2, 5
3, 5
4, 5
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
b
i
ii
iii
iv
c
i
ii
1
4 = __
P(total of the scores is 5) = ___
20 5
P(the scores have a difference of 1)
7
= ___
20
P(total of scores is 5 and scores have
a difference of 1)
= P(2, 3) + P(3, 2)
2 = ___
1
= ___
20 10
P(total of scores is 5 or scores have a
difference of 1 or both)
P(1, 4) + P(4, 1) + P(2, 3) + P(3, 2)
+ P(4, 3) + (3, 4) + P(4, 5) + P(5, 4)
+ P(1, 2) + P(2, 1)
10 1
= ___ = __
20 2
1
P(A or B) = __
2
4
P(A) = ___
20
7
P(B) = ___
20
11 ≠ __
1
P(A) + P(B) = ___
20 2
A and B overlap so P(A) + P(B)
counts some possibilities twice.
Past paper questions
1
9
a
6.05 × 10−2
b
4.0261 × 1011
1.6
10 a
b
c
d
11 a
b
c
d
12 a
b
c
201
5.6% = 0.056
0.065 = 0.065
5
___
= 0.056179
89
5 13
So 5.6%, ___, ____, 0.065
89 201
1 − 0.15 = 0.85
3
351
1 p × 4.5 2 × 10.4 = ____
__
p
13 a
5
a
b
1 − 0.38 = 0.62
0
6
a
b
6.4 × 105
6 × 10−4
7
a
isosceles
4.4 cm
1 × 5 × 4.4 = 11 cm2
__
2
1 × 5 × 4.4 × 6 = 66 cm3
__
2
6a + 4b
4 × 9 + (3 × −2) = 36 − 6 = 30
i
x = 80
ii 3x = 21
x=7
iii 10x + 5 = 27
x = 22
x = 2.2
3r = p + 5
p+5
r = _____
3
2
1.2
3 × 1.2 + p × (____) = 4.73 m2
2
4.73097 … × 0.2 = 0.946 m3
0.946 × 1000 = 946 litres
increase = 60.805 … litres
= 0.0608 … m3
0.0608
_______
= 0.0128 m = 1.29 cm
4.73
4pR 2
+ pR 2
p × 2.4 × 6.3 + p × 2.4 2 = _____
2
522p
3pR 2 = _____
25
522
R 2 = ______
25 × 3
√
5
3
7a(3a + 4b)
b
Colour
Blue
Red
Yellow
Green
Probability
0.15
0.2
0.22
0.43
b
66
1.2
____
× 100 = 7.5%
13
____
= 0.06467...
2
4
8
____
522
R = ____ = 2.64 cm
75
2
7.6
1 × p × 7.6 2 × 16 − __
1 × p × ____
__
( 4 ) ×4
3
3
= 953 cm3
200 × 0.2 = 40 times
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 9
Getting started
Exercise 9.1
1
1
a
b
c
d
2
a
b
c
3
a
b
c
d
Students’ own description of a number
sequence.
The term in position n, in other words, the
term in any position.
Next term is 11 because you subtract 4
each time.
Yes, once you know the rule for a
sequence, you can use it to work out
the value of the term in any position.
In this sequence: Tn = 27 − 4(n − 1),
so T20 = 27 − 4(20 − 1) = −49.
Even
Prime
12
6
14
3
18
4
10
2
19
c
d
e
11
16
9
15
17
1
+2
3
8
3
0.5
+2
13
+5
9
8
×3
2
+2
+5
243
5
+2
+5
×3
17
28
233
81
3.5
+2
+5
×3
15
13
18
27
f
13
–2
g
2
5
–3
...
+2
33
+5
+2
38
...
+5
+5
729 2187 6561 ...
×3
6.5
19
×3
8
×3
9.5
×3
11
...
6
–3
11
–3
9
–2
4.8
–1
–3
7
–2
3.6
–7
–4
–3
2.4
–3
3
5
–2
–2
1.2
–10 –13 ...
1
–2
0
–3
–3
–1
...
–2
–2
–1.2 –2.4
–1.2 –1.2 –1.2 –1.2 –1.2 –1.2 –1.2 –1.2
h
2.3
1.1
–0.1 –1.3 –2.5 –3.7 –4.9
–1.2 –1.2
...
–1.2 –1.2 –1.2 –1.2 –1.2
a
81, −243, 729; rule = multiply previous
term by −3.
b
Fr, Sa, Su; rule = days of the week.
c
u, b, j; rule = skip 1 extra letter of the
alphabet each time.
a
1 or any combination of an integer
x = __
2
and a half.
x = 5 or any number with 5 in the units
place.
x = any number ,0.
b
13
+2
11
9
+1.5 +1.5 +1.5 +1.5 +1.5 +1.5 +1.5 +1.5
7
8
7
×3
3
5
5
+5
2
4
67
b
Rational number can be written as
a
fractions in the form __ or the decimal
b
equivalent. Decimals must terminate or
be recurring. Irrational numbers cannot
be written as fractions and they result in
non-terminating, non-recurring decimals.
p is irrational because it cannot be
expressed as a fraction.
2 , so it is rational.
0.6666… = __
3
5
7
2
1
a
c
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 9.2
1
2
a
b
c
23, 27, 31
49, 64, 81
−17, −31, −47
a
b
1, −2, −5, …; −56
1, 0, −1, …; −18
c
1 , 2, 4.5, …; 200
__
d
2
0, 6, 24, …; 7980
e
3
3
1
__
, 1, __ …; __
f
3
a
b
c
d
e
f
g
h
4
a
b
c
d
7
2
4
2, 16, 54, …; 16 000
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
33
2n + 3
73
5n − 2
14 348 907
3n
21.5
1.5n − 1
−34
−3n + 11
−15
−2n + 15
−10.8
−1.2n + 7.2
450
2n2
4(2n − 1)
3996
30
Rule is 8n − 4, so 8n − 4 = 154 should give
integer value of n if 154 is a term:
8n − 4 = 158
8n = 158
n = 19.75
OR
19th term = 148 and 20th term
= 156, therefore 154 is not a term.
68
5
3, 6
6, 9
6
a
b
c
d
T6 = 48
n2 + 2n
T20 = 440
7
x = −2
8
n=5
9
a
2, 8, 18, 32, 50
b
i
2n2 + 1
ii
4n2
iii
6n2 + 1
c
n
1
2
3
Sequence
5
6
11
2
8
18
3
−2
−7
2n2
Sequence −
2n2
n
4
5
Sequence
20
33
2n2
32
50
−12
−17
Sequence − 2n2
d
e
−5n + 8
2n2 − 5n + 8
10 a
b
2n2 + n + 2
n2 + n + 11
11 a
It has a constant 2nd difference of 4, so
sequence is quadratic.
68
Tn = 2n2 − 4n − 2
4798
b
c
d
12 a
b
c
d
T5 = 30, T6 = 42
Tn = n2 + n
240
n = 10
13 a
b
c
2n − 1
263
3n − 1
1st difference: 5, 7, 9, 11
2nd difference: 2, 2, 2; so, sequence is
quadratic.
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Fibonacci patterns
1 a The number of clockwise and the number
of anticlockwise spirals will often
be consecutive terms of a Fibonacci
sequence.
b Students will need to physically count
and keep track of the sections they have
counted to find the pattern.
2
There are many examples including seeds on a
sunflower, sections on the skin of a pineapple,
the arrangement of leaves on the stems of
plants. An online search will give a selection
of answers.
3
a
b
An investigation will show that many
artists, including Salvador Dali and
Leonardo da Vinci, often produced work
using this ratio. It can also been seen in the
relationships between height and width
in buildings, for example in the Acropolis
and the Great Pyramid at Giza. The
golden ratio is also used to space out facial
features. Advertising logos and visually
appealing layouts often reflect the golden
ratio.
In simple terms, the golden ratio can
be worked out using the dimensions
of a rectangle where the ratio a : a + b
is equivalent to b : a. There are many
examples of diagrams that show this.
Exercise 9.3
a
b
c
d
69
Pattern
number n
1
2
3
4
5
6
n
300
Number of
matches m
4
7
10
13
16
19
m = 3n + 1
901
Pattern
number p
1
2
3
4
5
6
p
300
Number of
circles c
1
3
5
7
9
11
c = 2p − 1
599
Pattern
number p
1
2
3
4
5
6
p
300
Number of
triangles t
5
8
11
14
17
20
t = 3p + 2
902
Pattern
number p
1
2
3
4
5
6
p
300
Number of
squares s
5
10
15
20
25
30
s = 5p
1500
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 9.4
1
e
c
1 , 9 __
1 ,14 __
1 , … 124 __
1
4 __
2 2 2
2
−1, −3, −5, … −49
f
f
g
h
1 , 2, 2 __
1 , … 13 __
1
1 __
2
2
2
1, 7, 17, … 1 249
1, 4, 9, … 625
6, 7, 9, … 16 777 221
2
30 is T6 and 110 is T11.
3
T9
4
a
b
153
n=6
5
a
The subscript n + 1 means the term after
un, so this rule means that to find the term
in a sequence, you have to add 2, to the
current term (un). So, if the term is 7, then
un + 1 is 7 + 2 = 9
b
−8, −6, −4, −2, 0
Exercise 9.5
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
Rational
Rational
Rational
Rational
Irrational
Irrational
Rational
Rational
Rational
Rational
Rational
Rational
Rational
Irrational
Irrational
Irrational
a
6
__
b
c
70
427
______
5, 9, 13, … 101
−2, 1, 4, … 70
e
2
8
__
a
b
d
1
d
1
19
___
8
427
_____
1000
3
9
10 003
311
____
99
Possible answers include:
a 2
_
b √5
c
d
1
2
4
The set of rational numbers and the set of
irrational numbers are both infinite sets. But
the set of rational numbers is ‘countable’
whereas the set of irrational numbers is
‘uncountable’. This might suggest that there are
more irrational numbers than rational numbers.
The term ‘countable’ does not mean finite.
In this context we mean that, if you tried to
pair up every rational number with exactly
one irrational number, you would have a lot of
irrational numbers left over that you couldn’t
pair up but no rational numbers would be
upaired.
5
Students’ own answers. Example: An
‘imaginary number’ is a quantity of the
form ix, where x is a real number___
and i is__the
positive square root of −1, e.g. √ − 3 = i√ 3 .
Exercise 9.6
1
a
Let x = 0.6̇
Then 10x = 6.6̇
Subtracting:
10x = 6.6̇
−x = 0.6̇
9x = 6
6
So x = __
9
2
Simplify x = __
3
b
Let x = 0.1̇7̇
Then 100x = 17.1̇7̇
Subtracting:
100x = 17.7̇
−x = 0.17̇
So 99x = 17
17
x = ___
99
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
a
5
__
9
1
__
b
9
8
__
c
9
8
___
d
33
61
e ___
99
32
___
f
99
206
g ____
333
233
____
h
999
208
____
i
999
1
___
j
45
17
k ___
90
31
____
l
990
27
m ___
11
1034
_____
n
333
248
____
o
99
9990
_____
p
= 10
999
5994
q _____ = 6
999
8
__
r
9
999
____
=1
s
999
900
t ____ = 100
9
Recurring decimals
1 a i
0.1
ii 0.01
iii 0.001
iv 0.000 000 001
b As the number subtracted tends to 1, the
answer tends to 0. Yes it will reach 0.
c
d
e
f
71
2
g
As the fractions represent infinite 9 there
is no 1 at the end of the infinite 0 and so
0.999… = 1
a
b
4.41 . 4.1 but 4.1 , 4.5
Another 9 could be added to the end of
4.49999.
Yes. X = 4.49̇
10x = 44.9̇
9x = 40.5
c
40.5 9
x = _____ = __ = 4.5
2
9
No.
Exercise 9.7
1
2
3
2 √7
b
√7
c
10 √ 3
d
3 √ 11
e
2 √6
f
5 √ 10
g
15 √ 2
h
24 √ 2
i
−15 √ 7
j
−12 √ 6
k
14 √ 2
l
−10 √ 15
a
√ 54
b
√ 40
c
√ 98
d
− √ 24
e
√ 108
f
− √ 72
a
Student discussion; could include
__
changing them all to the form √ n .
_
_
_
A 3 √5 , 3 √3 , 2 √5
_
_
_
B 4 √ 6 , 3 √ 6 , √ 24
_
_
_
C 2 √ 15 , 3 √ 6 , 4 √ 3
Student discussion and ideas. Comparing
like surds first can mean you don’t need to
convert all of them.
b
2 , __
2
__
3 9
0.8̇
8
__
9
5
4 , __
__
, 0.9̇, 1
9 9
4
_
a
a
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
Possible errors are:
1: Added coefficients correctly but also
added roots.
2: Subtracted and added unlike surds.
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
b
5
b
7
8
9
_
Exercise 9.8
1
Any numerical examples to show the
expression are unequal. For example:
a
6
_
11 √ 3 + 3 √ 2
_
_
_
4 √5 + 3 √2 − 2 √3
1:
2:
_
_
_
_
+ √ 9 = 2 + 3 = 5 and √ 4 + 9 = √ 13
= 3.605...
_
_
_
_
√ 9 − √ 4 = 3 − 2 = 1 and √ 9 − 4 = √ 5
= 2.236...
√4
_
_
√ 33
c
24 √ 6
d
9 √ 30
e
− 4 √ 30
f
54
g
6 √6
h
6 √2
i
4 √ 10
a
√2
_
_
_
_
_
b
2 √ 11
c
−2 √2
d
6 √2
e
8 √5 − 4 √2
b
√6
f
8 √3 + 8 √2
c
− √ 10
a
2 √5 + √5 = 3 √5
b
2 √3 + 3 √3 = 5 √3
d
√2
1_ = ___
___
2
√2
c
4 √3 − 6 √3 = −2 √3
e
√ 15
d
6 √ 2 + 6 √ 2 = 12 √ 2
f
1
__
e
5 √3 − 8 √3 = −3 √3
f
9 √ 3 + 4 √ 3 = 13 √ 3
a
b
c
d
e
f
4+
+
_
_
_
_
_
+
3 √5
−
3 √5
_
_
− 2 √7
2
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
1 + 5 √3
_
3 √3 − 1
1
_
√3 + 5
8
_
_
3 √7 − 3 √2
_
2 √ 10
+
_
3
2 √5
__ 2
___
So, x = √ 28
__
__
= √4 × √7
__
= 2 √7
Checking for errors
72
b
a
3 √7
10 x 2 = (2 √ 35 ) − (4 √ 7 )
= 4(35) − 16(7)
= 140 − 112
= 28
b
√ 35
5 √3
___ 2
a
_
a
2D and 4C are incorrect.
2A and 4D are correct.
Possible mistakes are:
In 2D the student has added the numbers
instead of multiplying.
In 4C the student
__ has only multiplied the
numerator by √ 2 .
4
_
_
_
_
_
_
_
g
3 _
6 √3
h
−2 √3
i
4 √3
j
√5
___
_
_
_
k
3
_
3 √ 10
l
9
__
a
2 √5 + 8
b
3 √2 + 3
c
−2 √3 − 8
d
− 4 √ 5 − 24
e
6 √ 11 − 4
f
4 − 8 √3
g
2 √ 2 + √ 10
h
2 √3 + 3 √6
i
2 √5 − 8
j
4 √ 5 + 8 √ 15
k
2 √ 6 − 4 √ 21
l
5 √ 10 − 20
a
3 √2
b
8 √3
c
6 √2
2
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
d
1
e
5 √3
____
f
3
5
a
b
c
d
e
f
g
P = 6 √ 2 + 6 cm
c
6_ ___
6
1 × ___
__
× _ = 9 cm2
2 √2 √2
5
a
b
c
7p mm2
_
_
L = 50 + 2 √ 7 mm and W = 30 + 2 √ 7 mm
_
1528 + 160 √ 7
6
a
Length = 297 √ 2
_
2
_
5 √3
____
3
_
2 √ 11
_____
−
_11
√
15
____
5 _
2 − √2
______
2_
√5
3
− ____
10 _
4 + √6
______
2 _
√2
2
−
______
10
______________
_
= √ 297 2 × 3
_
= 297 √ 3
Alternatively,
_______________
_
2
d = √ (297 √ 2 ) + 297 2
_
= √ 264 627
_
____________________
i
12 − 3 √ 3
________
= √ 9 × 9 × 9 × 11 × 11 × 3
_
_
= (3 × 9 × 11) √ 3
13
√
2
3
+1
_______
11
_
= 297 √ 3
_
Exercise 9.10
_
k
8 + 3 √7
l
4 √6 − 2 √2
2 √2 − 4 √6
__________
or __________
_
_
6
_
− 11
_
_
_
1
11
3
c
d
Exercise 9.9
e
f
g
21 − √ 2
_______
439
_
1
20 − 4 √ 7 cm
2
a
b
4 cm
4 cm, 3 cm and 2.65 cm (students to
explain that they have rounded the
decimal in 2.65)
3
a
b
4 √ 59
944 cm2
4
a
h2 =
_
6_
6_
___
+ ___
(√2 )
(√2 )
2
36 36
= ___ + ___
2
2
72
= ___
2
= 36
_
So, h = √ 36 = 6 cm
73
a
b
10 √ 7 + 2 √ 12 − 2 √ 21 + 14
_______________________
_
7
_______________
_ 2
= √ 297 2 × 2 + 297 2
2 − √5
_
_
d = √ (297 √ 2 ) + 297 2
h
j
_
b
2
2
{Monday, Tuesday, Wednesday, Thursday,
Friday, Saturday, Sunday}
{Jan, Feb, Mar, Apr, May, Jun, Jul, Aug,
Sep, Oct, Nov, Dec}
{1, 2, 3, 4, 6, 9, 12, 18, 36}
{red, orange, yellow, green, blue,
indigo, violet}
{7, 14, 21, 28, 35, 42, 49}
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
{TOY, OYT, YTO, YOT, OTY, TYO}
Various answers are possible. Examples
include:
a hamster, mouse
b peas, beans
c Dublin, Amsterdam
d Rhine, Yangtze
e redwood, palm
f soccer, rugby
g Italy, Spain
h Aconcagua, Kilimanjaro
i
Bach, Puccini
j
lily, orchid
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
4
k
l
m
n
o
p
12, 15
flatback, Olive Ridley
Uranus, Neptune
surprised, mad
African, American
pentagon, quadrilateral
a
b
c
d
e
a
b
c
d
e
square numbers
continents of the world
even numbers less than 10
multiples of 2
factors of 12
false
true
true
false
true
Exercise 9.12
1
b
c
2
a
b
2
a
b
c
i
ii
i
ii
3
i
C ∩ D = {a, g, u, w, z}
ii C ∪ D = {a, b, g, h, u, w, x, y, z}
Yes, u is an element of C and D.
No, g is an element of both sets and will
be an element of the union of the sets.
a
b
Equilateral triangles have two sides equal.
All equilateral triangles are isosceles,
so F is entirely contained within G.
The intersection is simply F.
4
a
i
T ∪ W = {1, 2, 3, 6, 7, 9, 10}
ii T ∩ W = {1, 3}
Yes; 5 is not listed in T.
5
a
b
c
d
e
f
{cat, bird, turtle, aardvark}
{rabbit, emu, turtle, mouse, aardvark}
{rabbit, cat, bird, emu, turtle, mouse,
aardvark}
{ } or ∅
{rabbit, emu, mouse }
{rabbit, cat, bird, emu, turtle, mouse,
aardvark}
A = {6, 12, 18, 24} and
B = {4, 8, 12, 16, 20, 24}
A ∩ B = {12, 24}
A ∪ B = {4, 6, 8, 12, 16, 18, 20, 24}
i
P = {a, b, c, d, e, f }
ii Q = {e, f, g, h}
P ∩ Q = {e, f }
i
(P ∪ Q)9 = {i, j}
ii P ∩ Q9 = {a, b, c, d}
a
A
e
b
21
22 23
25
a
b
c
b
c
f
a
d
A
20
24
26 27
29
4
B
g
h
A ∩ B = {6, 8, 10}
A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10}
3
8
3
b
a
b
c
Exercise 9.11
1
a
28
B
30
32
36
35
31
33
34
x=6
n(V ) = 16
n(S )9 = 16
5
A
6
a
B
A
B
C
b
A
B
C
74
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
c
b
A
{equilateral triangle, square, regular
pentagon, regular hexagon}
{18, 21, 24, 27, 30}
B
c
C
d
A
4
{x : x is a multiple of 3 and 5}
5
a
B
C
e
i
{5}
ii {1, 2, 3, 4, 5}
iii {1, 2, 3, 4, 5}
iv {6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
v {1, 2, 3, 4, 5}
%
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17}
A
b
c
B
6
a
A = {x, y : y = 2x + 4} is the set of
ordered pairs on a straight line. The set is
infinite, so you cannot list all the points
on the line.
B = {x : x3 is negative} is the set of
negative cubes; any negative number
cubed will result in a negative cubed
number, so the set is infinite.
C
f
b
A
B
C
g
A
Practice questions
B
1
b
c
d
C
7
9
8
6
a
2
Pattern number (n)
1
2
Number of dots (d )
5
8
11 14
4
5
6
3
4
d = 3n + 2
182
29
a
Exercise 9.13
1
2
a
b
c
d
e
{x : x is a square number less than 101}
{x : x is a day of the week}
{x : x is an integer, x , 0}
{x : 2 , x , 10}
{x : x is a month of the year, x has 30
days}
a
b
d
e
{x : x is an integer, 1 , x , 9}
{x : x is a letter of the alphabet, x is a
vowel}
{x : x is a letter of the alphabet, x is a
letter in the name Nicholas}
{x : x is an even number, 1 , x , 21}
{x : x is a factor of 36}
a
{41, 42, 43, 44, 45, 46, 47, 48, 49}
c
3
75
b
c
d
e
3
Dots
1
2
Lines
4
7
3
10 13 16 19
298
3n + 1
28
_
_
_
_
5 √2 − 2 √8 = 5 √2 − 2 √4 × 2
_
_
_
= 5 √2 − 2 √4 × √2
_
_
= 5 √2 − 2 × 2 × √2
_
_
= 5 √2 − 4 √2
_
= √2
4
45
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
_
5
17 √ 3
_____
b
6
A
B
_
6
5 √ 6 cm
7
a
b
a = 29, b = 12
p = 29, q = 180
_
8
9
40 + 4 √ 7
________
a
b
c
d
e
f
g
9
1
3
{a, b, c, e, f, g}
{e, g}
{a, b, c, d, e, g}
{a, b, c, d, f }
{a, b, c, d, e, g}
17 a
b
c
1, 1, 2, 3, 5, 8, 13, 21
Fibonacci numbers
u13
18 a
b
c
1, 3, 6, 10, 15
Triangular numbers
n = 20
1 (n − 1)n
un−1 = __
2
d
10 4
5029
11 _____
1665
Practice questions worked
solutions
12 x = 0.999 999
10x = 9.999 99
9x = 9
x=1
1
a
8, 11, 14
b
8
3
1
13 __
3
c
14
Bus
Car
d
4
53
21
2
15 a
b
c
16 a
76
A = {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6}
2
(A ∩ B)9 = {integers NOT including 3 or 6}
A
B
3
d = 3n + 5
d(60) = 3 × 60 + 5
= 185
89 = 3n + 5
3n = 84
n = 28
O
b
c
82
21
29
14
a
29
a
b
c
11
O
O
13, 16, 19
If d = number of dots and L = number of
lines.
L = 3d + 1
d
e
O
L(99) = 3 × 99 + 1
= 298
3n + 1
85 = 3n + 1
3n = 84
n = 28
There are 85 lines in the pattern with
28 dots.
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
4
_
_
_
_
_
_ _
5 √ 2 − 2 √ 8 = 5 √ 2_ − 2 √ 4_ √ 2
= 5_√ 2 − 4 √ 2
= √2
_ _
_
3 √ 3 × 5 √ 45 __________
3√ 3 × 5 √ 9 √ 5
__________
_
_ _
=
√3 √5
√ 15
=3×5×5×3
= 225
_
5
_
_
_
_
√3
√3
7 √3
7_ __
__
+ __
+ ____
=
_ 2
2
2
√3
(√ 3 )
√3
7 √3
= __ + ____
6
2
3
_
_
3 √ 3 + 14 √ 3
= __________
6
_
√
17 3
= _____
6
10
1 × 3 √_
_
Area = __
2 × ___
2
√3
_
10 √ 3
1 × 3 √_
2 × _____
= __
_ 2
2
(√ 3 )
_ _
= 5 √2 √3
_
= 5 √6
7
a
_ 2
_
(3 + 2 √ 5 ) = 9 + 12 √ 5 + 4 × 5
_
= 29 + 12 √ 5
Therefore, a = 29 and b = 12
b
_ 2
_
(3 + 2 √ 5 ) = 29 + 12 √ 5
_ _
12 √ 5 √ 45
_
= 29 + ________
√ 45
_
12 √ 225
_
= 29 + _______
√ 45
12 ×
15
_
= 29 + _______
√ 45
180
_
= 29 + ____
√ 45
Therefore, p = 29 and q = 180.
_
8
_
3(4 − √ 7 )
7(4 + √ 7 )
7 _ ______________
3 _ ______
______
______________
+
=
_
_ +
_
_
√
√
4 + 7 4 − 7 (4 + √ 7 )(4 − √ 7 ) (4 + √ 7 )(4 − √ 7 )
_
_
12 − 3 √ 7 + 28 + 7 √ 7
= __________________
_ 2
4 2 − (√ 7 )
_
40 + 4 √ 7
= ________
16 − 7
_
40 + 4 √ 7
= ________
9
77
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
9
10
a
b
c
d
e
f
g
M ∩ N = {f } ⇒ n(M ∩ N ) = 1
3
6
2
6
5
6
C
R
15 − x
x
14
53
7−x
30 174 5029
x = ______ = _____
9990
1665
12 x = 0.9̇
10 = 9.9̇
9x = 9
x=1
x
y
x + y + 53 = 78 (78 travelled by car or bus or
both)
53 + x + y + z = 107 (There are 107 in total)
so, z = 107 − 53 − 25 (Combining
both pieces of information)
= 29
y + z = 50 (50 did not travel by bus)
y = 21
x = 78 − 53 − 21
=4
a 53 + x + z = 53 + 4 + 29
= 86
b y = 21
c z = 29
5
3.02̇04̇ = x
10x = 30.2̇04̇
10 000x = 30 204.2̇04̇
9990x = 30 174
C
Z
15 − x + x + 7 − x + 5 = 24
27 − x = 24
x=3
n(R ∩ C′) = 7 − x = 7 − 3 = 4
11
B
15 a
b
c
16
A = {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6}
A ∩ B = {3, 6}
n(A ∩ B) = 2
The elements not in both A and B = the
integers that are not positive multiples of
3 between −4 and 7.
A
B
A
B
13 0.3̇003̇ + 0.0̇330̇
Let x = 0.3̇003̇
10 000x = 3003.3̇003̇
3003
9999x = 3003 ⇒ x = _____
9999
Similarly, 0.0̇330̇ = y
10 000y = 330.0̇330̇
330
Therefore, 9999y = 330 ⇒ y = _____
9999
3003
330
So, 0.3̇003̇ + 0.0̇330̇ = _____ + _____
9999 9999
3333 1
= _____ = __
9999 3
78
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
17 un+2 = un+1 + un
a u1 = u2 = 1
1, 1, 2, 3, 5, 8, 13, 21
b Th e Fibonacci sequence.
c
34
55
89
144
233
9th
10th
11th
12th
13th
It is in the 13th position.
1 n(n + 1)
18 un = __
2
79
a
b
1, 3, 6, 10, 15
Triangular numbers
c
1 n(n + 1)
210 = __
2
n(n + 1) = 420
2
n + n − 420 = 0
(n + 21)(n − 20) = 0
⇒ n = 20 or n = −21
But n . 0 so n = 20
d
1 (n − 1)(n − 1 + 1)
un−1 = __
2
1 n(n − 1)
= __
2
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 10
Getting started
1
b
y
6
x
−3
−2
−1
0
1
2
3
y
−1
0
1
2
3
4
5
5
5
3
4
2
3
1
−2 −1 O
2
3
a
b
y
6
4
1
2
2
3x
1
y-value is 4 more than the x-value.
y=x+4
−4 −3 −2 −1
−1
x
0
1
a
3
4
−2
Gradient = 3
y-intercept = −2
Equation of the line is y = 3x − 2
c
x
−3
−2
−1
0
1
2
3
y
−7
−5
−3
−1
1
3
5
Exercise 10.1
1
2
y
x
−3
−2
−1
0
1
2
3
6
y
−7
−4
−1
2
5
8
11
4
12
2
y
x
−4 −3
10
−2 −1
8
0
−2
1
3
2
4
−4
6
−6
4
−8
2
x
−3
−2
−1
0
−2
1
2
3
d
x
−3
−2 −1
0
1
2
3
y −19 −14 −9
−4
1
6
11
−4
15
y
−6
10
−8
5
x
−4
−3
−2
−1
0
1
2
3
4
−5
−10
−15
−20
80
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
e
x
−3
−2
−1
0
1
2
3
y
7
5
3
1
−1
−3
−5
h
x −3
−2
−1
0
1
2
3
y −8.5 −5.5 −2.5 0.5 3.5 6.5 9.5
y
y
10
8
6
5
4
2
x
x
−4
−3
−2
−1
0
−2
1
2
3
–4
4
−3
−2
−1
0
1
2
3
4
0
1
2
3
1
1.5
2
2.5
−5
−4
−6
−10
−8
f
x
−3
−2
−1
0
1
2
3
y
1
0
−1
−2
−3
−4
−5
i
x −3
−2 −1
y −0.5
0
0.5
y
2
y
3
1
x
−4
−3
−2
−1
0
−1
1
2
3
2
4
−2
1
−3
x
−4
−4 −3 −2 −1
0
1
2
3
4
−5
−1
−6
g
x
−3
−2
−1
0
1
2
3
y
9
8
7
6
5
4
3
j
x −3
−2
−1
0
1
2
3
y −12 −8
−4
0
4
8
12
y
10
15
9
10
8
5
y
x
7
−4
6
−3
−2
−1
0
−5
5
−10
4
−15
1
2
3
4
3
x
−4
81
−3
−2
−1
0
1
2
3
4
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
k
x
−3
−2
−1
0
1
2
3
y
−3
−3
−3
−3
−3
−3
−3
n
x
−3
−2
−1
0
1
2
3
y
−5
−4
−3
−2
−1
0
1
y
2
y
1
1
x
x
−3
−2
−1
0
1
2
3
−4
4
−3
−2
−1
−1
0
−1
1
2
3
4
−2
−2
−3
−3
−4
−5
l
x
−3
−2
−1
0
1
2
3
y
2
1
0
−1
−2
−3
−4
3
−6
o
y
x
−3
−2
−1
0
1
2
3
y
−3
−2
−1
0
1
2
3
2
4
1
3
x
−4
−3
−2
−1
0
1
2
y
2
3
−1
1
x
−2
−4
−3
−3
−2
−1
−4
m
0
−1
1
2
3
4
−2
x
−3
−2
−1
0
1
2
3
y
7
6
5
4
3
2
1
y
−3
−4
p
8
7
x
−3
−2
−1
0
1
2
3
y
3
2
1
0
−1
−2
−3
6
4
5
y
3
4
2
3
1
x
2
−4
1
x
−4
−3
−2
−1
0
1
2
3
4
−3
−2
−1
0
−1
1
2
3
4
−2
−3
−4
82
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
2x
−
y=
h
1
y=
2x
+2
2
g
y
2x
+1
y=
2x
3
y=
2
−3
−2
0
−1
1
x
3
2
Exercise 10.2
1
−1
−2
−3
a
b
x
−3
0
3
y=x+2
−1
2
5
x
−3
0
3
5
2
−1
y = −x + 2
c
a
b
c
d
e
f
x = −4
x=2
x=7
y=7
y=3
y = −6
2
The lines are parallel.
3
y = x + 2 is parallel to y = x − 2
y = −x + 2 is parallel to y = −x − 2
Same coefficients of x but different
constant values.
x
−3
0
3
y=x−2
−5
−2
1
x
−3
0
3
1
−2
−5
(h) x = – 7–
2
(i)
y (g) x = 12–
6
5
4
3
2
1
−6 −5 −4 −3 −2 −1−10
−2
−3
−4
−5
−6
(b) x = 3
(f) y = 4
(a) y = 3
x
1 2 3 4 5 6 (c) y = −1
(e) y = −3
(j) (d) x = −1
d
y = −x − 2
y
y=x+2
10
9
y = –x − 2
y=x−2
8
7
6
5
4
3
2
1
x
−10 −8 −6 −4 −2 0 1 2 3 4 5 6 7 8 9 10
−2
−3
−4
−5
−6
−7
−8
−9
−10
y = –x + 2
4
a
b
c
d
e
f
83
y = x + 2 cuts the x-axis at x = −2
y = −x + 2 cuts the x-axis at x = 2
y = x − 2 cuts the x-axis at x = 2
y = −x − 2 cuts the x-axis at x = −2
y = x + 2 and y = x − 2
−x + 2 and −x − 2
y = x + 2 and y = −x + 2
y = x − 2 and y = −x − 2
None of the graphs
Exercise 10.3
1
a
b
c
d
e
f
g
h
i
2
3
3
2
1
−3
−5
−1
1
__
3
2
__
3
1
− __
4
a
b
c
d
e
3
1
2
−3
−3
17
f ___
4
450 m
The gradient is equal to the coefficient of x in the
equation of the line.
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 10.4
1
a
4
d
y = −x + 3
3
3
2
2
1
−3 −2 −1 0
−1
2
3
−3
−3
−4
−4
Gradient = −1
−5
Gradient = 4
y-intercept = −5
y
y = 2x + 3
5
4
3
3
2
2
1
2
3
4
5
−5 −4 −3 −2 −1 0
−1
y=
y
1
x+2
3
x
1
2
3
4
5
−2
–3
–4
−3
–5
−4
1
3
y-intercept = 2
y-intercept = 3
−5
Gradient =
5
y
f
4
9
3
x
1
2
3
4
5
6
5
4
−2
3
−3
2
−4
1
−5
1
x
4
7
1
−5 −4 −3 −2 −1 0
−1
y=6−
y
8
2
84
5
1
x
–2
y-intercept = −2
4
−5
4
–5 –4 –3 –2 –1 0
–1
Gradient = −3
3
e
1
y = –3x −2
2
y-intercept = 3
5
Gradient = 2
1
−2
−2
b
x
−5 −4 −3 −2 −1 0
−1
x
1
y
4
y y = 4x − 5
1
c
5
−4 0
−1
x
4
8 12 16 20 24 28 32 36 40 42
−2
1
Gradient =
4
y-intercept = 6
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
a
y = –x + 4
5
e
y
8
7
6
5
4
3
2
1
4
3
2
1
x
−3 −2 −1 0
−1
1
2
3
4
5
−3
x-intercept = −1
y-intercept = 4
b
x + 2y = 4
3
x
−8−7 −6−5−4 −3−2−1 0
−2
y
x-intercept = 4
y
x= –+2
4
y
1 2 3 4 5 6 7 8
−2
−3
−4
−5
−6
−7
−8
y-intercept = −8
2
f
1
−3 −2 −1 0
−1
1
−2
x-intercept = −
y-intercept = 2
c
2
3
4
1
2
6
4
2
x
1
2
3
3
y-intercept = 6
x = 4y − 2
2
1
−2
–
x-intercept = 1
4
–
y-intercept = 1
2
y = mx + c
x
1
2
3
Gradient
1
__
2
y-intercept
−2
1
b
c
y = 2x + 4
2
4
d
y = 2x − 5
2
−5
e
y = 2x + 5
2
1
− __
3
5
g
1x + 2
y = − __
3
y = 3x − 2
h
f
−2
2
3
−2
y = −4x + 2
−4
2
i
y = 2x + 4
2
4
j
y = 6x − 12
6
1
__
8
−12
−12
6
k
l
85
x
1x − 2
y = __
2
y = −2x + 1
a
y
−3 −2 −1 0
−1
2x − 3y = − 9
y-intercept = 3
4
x-intercept = −2
d
y
−8−7 −6−5−4 −3−2−1 0 1 2 3 4 5 6 7 8
−2
−3
−4
−5
−6
−7
−8
x-intercept = −4.5
y
y x+ 2 =3
0
8
7
6
5
4
3
2
1
x
1x − 3
y = __
8
y = −12x + 6
−3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
5
a
b
c
d
e
f
y = 2x + 3
y = −3x − 2
y = 3x − 1
y = 0.75x − 0.75
y = −2
y=4
a
3
y = − __ x − 0.5
2
3
y = − __ x + 2
4
1x−3
y = __
2
b
c
6
a
y = −4x − 1
b
1x+1
y = __
3
y = −3x + 2
y = 5x + 2
y = 3x + 1
y = −x + 2
y = 2x − 3
c
d
e
f
g
h
i
7
a
b
y = 4x − 5
y = −3x + 17
c
9
6
y = __ x − __
5
5
71
17
y = ___ x − ___
4
4
d
8
c
d
e
f
c
y = −3x − 5
y = 2x + 13
x
y = __ − 3
2
y = −x − 4
x = −8
y=6
2 , e.g. y = __
2x−5
Any line with gradient __
3
3
Any line with same y-intercept,
e.g. y = 2x + 3
y=3
Investigation
No. Imagine a straight road with a straight
footbridge crossing above it. Both the road and
the bridge follow a straight line path, but they do
not cross because there is vertical distance between
them.
Exercise 10.5
1
y = −5x + 8
2
a
Gradient AB = −2;
1 ; −2 × __
1 = −1,
Gradient PQ = __
2
2
so AB is perpendicular to PQ.
b
1 ; __
1 × −2 = −1,
Gradient MN = __
2 2
so MN is perpendicular to AB.
3
4
5
6
1x + 5
y = − __
3
1 or x + 2y − 1 = 0
1 x + __
a y = − __
2
2
b y=x−3
1:
Gradient A = −2, gradient B = __
2
1 = −1, so A is perpendicular to B.
− 2 × __
2
10
Gradient AB = ___; gradient AC = −1, so AB
9
is not perpendicular to AC and figure cannot
be a rectangle.
7
a
b
c
8
2x − y + 6 = 0
y = 5x − 18
y= x− 4
y=1
a, c
10 a
b
c
d
86
b
Any line with the same gradient, e.g.
a
b
9
2x−1
y = __
3
__
y= 1x−2
4
11 a
y = 2x − 2
y = 2x
y = 2x − 4
1
y = 2x + __
2
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 10.6
1
a
12
d
y
5
y
y = −5x + 10
y = 4x + 2
4
3
11
2
10
1
9
x
8
−2 −1 0
−1
7
1
−2
6
−3
x-intercept = –0.5
y-intercept = 2
5
4
3
e
2
3
1
−1 0
−1
1
2
3
4
1
5
x
−3 −2 −1 0
−1
−2
x-intercept = 2
y-intercept = 10
4
f
2
3
x
−1 0
−1
1
2
3
4
1
x
−1 0
−1
1
2
3
−2
x-intercept = 2
y-intercept = 2
y y = −3x + 6
g
y y = 2x − 3
6
2
5
1
x
4
−2
3
2
−1 0
−1
1
2
3
−2
1
x
1
2
−2
x-intercept = 2
y-intercept = 6
87
y y = −x + 2
5
−2
x-intercept = 3
y-intercept = −1
−1 0
−1
3
2
1
7
2
−3
1
x-intercept = − −
3
y-intercept = 1
3
c
1
−2
y=x
– −1
3
y
y y = 3x + 1
2
x
b
2
3
−3
−4
−5
−6
x-intercept = 1.5
y-intercept = –3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
h
2
l
2x − 1
y y=−
3
x
−2 −1 0
−1
1
2
4
3
3
2
−2
1
−3
x-intercept = 1.5
y-intercept = −1
1
−2 −1 0
−1
x
0
−1
–−2
y =x
4
y
1
2
3
4
5
6
7
8
x
9 10
−2
−3
x-intercept = 8
j
–+1
y = 2x
5
2
y-intercept = −2
3
y
1
x
−4 −3 −2 −1 0
−1
1
2
1
a
b
c
d
e
f
g
h
c=2
c = −4
c = −9
c = −8
c=4
c=3
c = −2
c=2
3
Exercise 10.7
−2
x-intercept = −2.5
y-intercept = 1
k
8
7
6
5
4
3
2
1
−8 −7 −6 −5 −4 −3 −2 −1 0
x-intercept = −8
1
2
x-intercept = 0.5
y-intercept = 6
2
88
y y = −12x + 6
5
1
i
6
1
a
b
c
d
e
f
g
h
i
2
AB = 5.39 midpoint = (3, 4.5)
CD = 4.47 midpoint = (−4, 6)
EF = 8.60 midpoint = (−2.5, 2.5)
GH = 7.07 midpoint = (3.5, 0.5)
IJ = 5.10 midpoint = (2.5, −3.5)
KL = 12.6 midpoint = (1, −3)
MN = 5.39 midpoint = (−3.5, −2)
OP = 7.81 midpoint = (−4.5, −4)
3
5.83
4
B
x
y= –+2
4
y
x
1 2 3 4 5 6
−2
−3
−4
−5
−6
−7
−8
y-intercept = 2
Length = 8.49 midpoint = (6, 9)
Length = 4.47 midpoint = (3, 8)
Length = 5.66 midpoint = (6, 5)
Length = 3.16 midpoint = (4.5, 9.5)
Length = 5 midpoint = (2.5, 5)
Length = 1.41 midpoint = (11.5, 3.5)
Length = 5 midpoint = (1, 3.5)
Length = 6.08 midpoint = (4.5, 2)
Length = 11.05 midpoint = (−2.5, 1.5)
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
B
6
AB = 6.40
AC = 4.24
BC = 7.28
7
a=7
8
E = (−6, −2)
Exercise 10.8
1
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
2
a
b
c
d
e
f
12 − 7x + x2
3 + 7x − 6x2
6m2 − 17m + 7
−8x2 + 2x + 3
8a2 − 2b2
−8m2 − 2mn + 3n2
g
i
j
k
l
3
1
x 2 + __ x + __
4
8
2
1
2x 2 − __ x − __
3
6
2
−36b − 26b + 42
6x2 + 9x − 15
6x3 + 9x2 + 2x + 3
15x4 − 18x2 + 3
a
b
2x2 + 9x + 9
3y2 + 10y + 7
h
3
89
x2 + 4x + 3
x2 + 10x + 24
x2 + 19x + 90
x2 + 15x + 36
x2 + 2x + 1
x2 + 9x + 20
x2− 3x − 28
x2 + 5x − 24
x2 − 1
x2 − x − 72
x2 − 13x + 42
x2 − 9x − 52
y2 − 11y − 42
z2 − 64
t2 + 13t − 68
h2 − 6h + 9
1g−2
g 2 + 3 __
2
9
d 2 − ___
16
c
d
e
f
g
h
i
7z2 + 15z + 2
4t2 + 17t − 15
2w2 − 23w + 56
16g2 − 1
72x2 + 23x − 4
360c2 − 134c + 12
−2m2 + 10m − 12
4
a
b
c
−2x4 + 6x2y − 4y2
−4x4 + 2xy2 − 4x3 y + 2y3
6x3 + 9x2y − 2xy − 3y2
5
a
b
c
d
e
f
g
h
i
15x3 + 21x2 − 24x − 12
x3 − 5x2 − 25x + 125
12x3 + x2 − 9x + 2
4x3 + 32x2 + 80x + 64
12x3 − 32x2 + 25x − 6
18x3 − 33x2 + 20x − 4
x3 + 6x2 + 12x + 8
8x3 − 24x2 + 24x − 8
x4y4 − x4
j
x
x
1 − ___
___
+ ___
6
a
b
c
2
81
4
18
16
1 (x − 2) 2 cm 3
V = (2x + __
2)
2x3 − 7.5x2 + 6x + 2
0.196 cm3
Exercise 10.9
1
a
b
c
d
e
f
g
h
i
x2 − 2xy + y2
a2 + 2ab + b2
4x2 + 12xy + 9y2
9x2 −12xy + 4y2
x2 + 4xy + 4y2
y2 − 8x2 y + 16x4
x4 − 2x2y2 + y4
4 + 4y3 + y6
4x2 + 16xy2 + 16y4
j
1
1 − ____
1 + _____
____
k
3xy y
9x
____
− ____ + ___
4x 2
4xy
16y 2
2
2
4
4
4
b
l
a 2 + ab + ___
4
2
2
m a b + 2abc4 + c8
n 9x4y2 − 6x2y + 1
16
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
o
2
3
16xy
4x
____
+ _____ + 16 y 2
2
p
3
9
x2 − 6x + 9
a
b
c
4x − 12
2x2 + 2x − 19
2y2 + 8x2
d
8x
x
___
+ ___ − 2
2
e
3
2
2
6x + 13.8x + 3.6
f
g
h
i
j
−16x2 + 8xy + 2x − 2y2
−x2 + 3x − 22
4x2 − 12xy − 19y2
−2x3 − x2 − 17x
4x2 − 13x − 1
a
b
c
d
e
f
−49
9
66
36
0
321
Discussion
The two numbers still multiply to give the constant
term, but no longer add to give the coefficient of x
because they will be multiplied by the coefficients
of x in the brackets.
For example, (2x + 1)(3x + 2) = 6x2 + 4x + 3x + 2
= 6x2 + 7x + 2. The 1 and 2 multiply to give the
constant term of 2, but they do not add to the give
the coefficient of x (which is 7). This is because the
1 and 2 are multiplied by 2x and 3x respectively.
3
2
90
a
b
c
d
e
f
g
h
i
j
k
l
(x + 12)(x + 2)
(x + 2)(x + 1)
(x + 4)(x + 3)
(x + 7)(x + 5)
(x + 9)(x + 3)
(x + 6)(x + 1)
(x + 6)(x + 5)
(x + 8)(x + 2)
(x + 10)(x + 1)
(x + 7)(x + 1)
(x + 20)(x + 4)
(x + 7)(x + 6)
a
b
(x − 6)(x − 2)
(x − 4)(x − 5)
(x − 4)(x − 3)
(x − 4)(x − 2)
(x − 8)(x − 4)
(x − 7)(x − 7)
(x − 10)(x + 2)
(x − 9)(x + 2)
(x − 8)(x + 4)
(x + 3)(x − 2)
(x + 11)(x − 3)
(x + 12)(x − 2)
a
b
c
d
e
f
g
h
i
j
k
l
( y + 17)( y − 10)
( p − 6)( p + 14)
(x − 12)(x − 12)
(t + 18)(t − 2)
(v + 15)(v + 5)
3(x + 4)(x + 3)
5(x + 1)(x − 2)
3(x − 5)(x + 2)
3(x − 1)2
2(x − 9)(x + 2)
−2(x + 3)(x + 4)
(x − 10)(x + 10)
Exercise 10.11
1
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
(x + 6)(x − 6)
( p + 9)( p − 9)
(w + 4)(w − 4)
(q + 3)(q − 3)
(k + 20)(k − 20)
(t + 11)(t − 11)
(x + y)(x − y)
(9h + 4g)(9h − 4g)
4(2p + 3q)(2p − 3q)
(12s + c)(12s − c)
(8h + 7g)(8h − 7g)
3(3x + 4y)(3x − 4y)
2(10q + 7p)(10q − 7p)
5(2d + 5e)(2d − 5e)
(x2 + y2)(x2 − y2)
x( y − x)( y + x)
2
a
b
c
d
(x + 3)2
(x + 2)2
(x − 7)2
(x − 9)2
Exercise 10.10
1
c
d
e
f
g
h
i
j
k
l
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
e
f
g
h
i
(6 + x)2
(7 + x)2
(2 − x)2
(5 − x)2
(2x + 5)2
1
x 2 − ___
x2
3
a
b
4x2 − 2x + 25
2x2 + 8
4
a
7 + 4 √3
_
_
3
71
b
4 + 2 √3
4
6
c
2 + √3
______
Exercise 10.12
1
2
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
x = 0 or x = 9
x = 0 or x = −7
x = 0 or x = 21
x = 4 or x = 5
x = −7 or x = −1
x = −3 or x = 2
x = −2 or x = −1
x = −10 or x = −1
x = 3 or x = 4
x = 6 or x = 2
x = 10 or x = −10
t = −18 or t = 2
y = −17 or y = 10
p = −14 or p = 6
w = 12
a
b
c
d
e
f
g
h
i
x = −5 or x = 2
x = −2 or x = 1
x = 1 or x = −10
x = 4 or x = −4
x = −9 or x = 4
x = −4 or x = 4
x = 3 or x = 1
x = 12 or x = 2
x = 2 or x = 1
Practice questions
1
91
2
a
b
c
d
15x2 + 2x − 8
x2 + 20x + 36
4x2 − 9
12y 4 − 5y2 − 3
5
a
b
c
d
e
_
2
y = 1.5x + 3
y=3
y = −4x − 4
x
1
y = − ___ − 3 __
2
10
x=3
g
2x
y = ___ − 3
3
y = −x
a
y = 4x + 4
b
y = −3x + 13
c
y = 0.5x + 0.9
7
a
b
c
(m − n)(m + n)
10 000 − 9
(100 − 3)(100 + 3) = 97 × 103
8
a
b
(x − 17)(x + 2)
(4x − 7y)(4x + 7y)
9
x = 2 or x = −14
f
6
10 27x3 + 54x2 + 36x + 8
11 a
b
i
ii
iii
6x(2x − 1)
( y − 6)( y − 7)
(d + 14)(d − 14)
i
1
x = 0 or x = __
2
y = 6 or y = 7
d = 14 or d = −14
ii
iii
12 20 cm or 380 cm
13 2 or 4
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
_ 2
14 x = 7 or x = −2
15 a
b
c
d
e
(x − 21)(x − 29)
x = 21 or x = 29
AD = 100 − 2x m
Area = 100x − 2x2
100x − 2x2 = 1218
x = 21 or x = 29
If x = 21, width = 21 m and length = 58 m
If x = 29, width = 29 m and length = 42 m
b=6
b
(7, 4)
c
y = −2x + 8
5
_
16 − 4 × 3
4 + 2 √3
= _______
4
_
2 + √3
= ______
5
a
b
c
units2
d
1
f
2
3
(x + 2)(x + 18)
= x2 + 2x + 18x + 36
= x2 + 20x + 36
c
(2x + 3)(2x − 3)
= 4x2 + 6x − 6x − 9
= 4x2 − 9
g
6
a
b
(2x − 5)2 = (2x − 5)(2x − 5)
= 4x2 − 10x − 10x + 25
= 4x2 − 20x + 25
7
(x + 2)2 + (x − 2)2
= x2 + 4x + 4 + x2 − 4x + 4
= 2x2 + 8
_
_
_
_
(2 + √ 3 )(2 + √ 3 ) = 4 + 2 √ 3 + 2 √ 3 + 3
_
_
_
= 7 + 4 √3
_
_
(1 + √ 3 )(1 + √ 3 ) = 1 + √ 3 + √ 3 + 3
_
= 4 + 2 √3
3
y = __ x + 3
2
y=3
y = −4x − 4
7
1 x − __
y = − ___
2
10
x=3
2x − 3
y = __
3
y = −x
a
y = 4x + c
At x = 3, y = 16,
16 = 12 + c
c=4
So, y = 4x + 4
b
y = −3x + c
At x = 7, y = −8,
−8 = −21 + c
c = 13
So, y = −3x + 13
c
y = 0.5x + c
At x = 3, y = 2.4,
2.4 = 1.5 + c
c = 0.9
So, y = 0.5x + 0.9
(4y2 − 3)(3y2 + 1)
= 12y4 − 9y2 + 4y2 − 3
= 12y4 − 5y2 − 3
1
1
1
__
__
___
2
(x − x )(x + x ) = x − 1 + 1 − 2
x
1
= x 2 − ___
x2
b
4
(3x − 2)(5x + 4)
= 15x2 − 10x + 12x − 8
= 15x2 + 2x − 8
b
a
_
_
e
a
_
28 − 14 √ 3 + 16 √ 3 − 8 × 3
= _______________________
Practice questions worked
solutions
d
92
_
(7 + 4 √ 3 )(4 − 2 √ 3 )
= ________________
_
_
(4 + 2 √ 3 )(4 − 2 √ 3 )
2
16 a
d
c
_
(2 + √ 3 ) _______
7 + 4 √3
________
_
_ 2 =
(1 + √ 3 ) 4 + 2 √ 3
AB2 = (9 − 2)2 + (6 − 10)2
= 49 + 16
= 65
AC 2 = (−6 − 2)2 + (−4 − 10)2
= 64 + 196
= 260
BC 2 = (−6 − 9)2 + (−4 − 6)2
= 225 + 100
= 325
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AB2 + AC 2 = 65 + 260
= 325
= BC 2
Therefore the lengths satisfy Pythagoras’
theorem
⇒ the triangle is right angled.
8
9
a
m2 − n2 = (m + n)(m − n)
b
9991 = 10 000 − 9
= 1002 − 32
c
9991 =
−
= (100 + 3)(100 − 3)
= 103 × 97
a
b
(x − 17)(x + 2)
(4x)2 − (7y)2 = (4x + 7y)(4x − 7y)
1002
32
10 x2 + 12x = 28
x2 + 12x − 28 = 0
(x + 14)(x − 2) = 0
x = −14 or x = 2
11 (3x + 2)3 = (3x + 2)(3x + 2)(3x + 2)
= (9x2 + 6x + 6x + 4)(3x + 2)
= (9x2 + 12x + 4)(3x + 2)
= 27x3 + 54x2 + 36x + 8
12 a
b
i
ii
iii
12x2 − 6x = 6x(2x − 1)
y2 − 13y + 42 = ( y − 7)( y − 6)
d 2 − 196 = d 2 − 142
= (d + 14)(d − 14)
Using previous answers:
i
6x(2x − 1) = 0
x = 0 or 2x − 1 = 0
1
x = 0 or x = __
2
ii ( y − 7)( y − 6) = 0
y = 6 or y = 7
iii
(d + 14)(d − 14) = 0
d = −14 or d = 14
13 Square ⇒ all sides are equal in length
So, 2x2 + 3x − 9 = x2 − 3x + 7
x2 + 6x − 16 = 0
(x + 8)(x − 2) = 0
x = −8 or x = 2
If x = −8, x2 − 3x + 7 = 64 +24 + 7 = 95 ⇒
square perimeter = 4 × 95
= 380 cm
93
If x = 2, x2 − 3x + 7 = 4 − 6 + 7 = 5 ⇒ square
perimeter = 4 × 5
= 20 cm
14 AB2 = (a − 3)2 + (4 − 3)2
= a2 − 6a + 9 + 1
= a2 − 6a + 10
_
_ 2
But AB = √ 2 ⇒ a 2 − 6a + 10 = (√ 2 )
=2
Therefore, a2 − 6a + 8 = 0
(a − 4)(a − 2) = 0
a = 4 or a = 2
15 5 x(x + 3) = (5 2) x 2 − x − 7
Therefore, x(x + 3) = 2(x 2 − x − 7)
x 2 + 3x = 2x 2 − 2x − 14
So, x 2 − 5x − 14 = 0
(x − 7)(x + 2) = 0
x = 7 or x = − 2
16 a
x 2 − 50x + 609 = (x − 29)(x − 21)
b
2x 2 − 100x + 1218 = 0
x 2 − 50x + 609 = 0
(x − 29)(x − 21) = 0
x = 29 or x = 21
c
2AD = 100 − 2x
so AD = 50 − x
d
Area = x(50 − x)
e
x(50 − x) = 609
50x − 50x2 = 609
x2 − 50x − 609 = 0
Therefore, x = 29 or x = 21
So, the rectangle is 21 × (50 − 21) = 21 × 29
(or 29 × (50 − 21) = 29 × 21)
17 a
b
b − 2 __
______
=1
11 − 3 2
2b − 4 = 8
2b = 12
b=6
(
2 )
6+2
3 + 11 _____
______
,
, i.e. (7, 4)
2
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c
Gradient of perpendicular line
1 = −2
= − ____
1
__
(2)
y = −2x + c passes through (7, 4)
so, 4 = − 14 + c ⇒ c = 18
Therefore, y = −2x + 18
y
B(11, 6)
A(3, 2)
C
–1
D
9
x
Line through A and B has equation
1x+c
y = __
2
and passes through (3, z),
3
1
so 2 = __ + c ⇒ c = __
2
2
1 x + __
1
Therefore, y = __
2
2
This passes through x-axis. When y = 0,
1 x + __
1 = 0 ⇒ x = −1
__
2
2
Line through A perpendicular to AB
passes through the x-axis when
−2x + 18 = 0
Therefore, x = 9.
1 AD × AC
Area = __
2
__________________
1 √________________
(9 − 3) 2 + (2 − 0) 2 √ (3 − − 1) 2 + (2 − 0) 2
= __
2
_ _
1
= __ √ 40 √ 20
2
_ _ _ _
_
1
= __ √ 8 √ 5 √ 4 √ 5 = 10 √ 2
2
94
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Chapter 11
Getting started
1
Student activity
2
a
b
c
d
e
f
4
3
Student investigation into Pythagorean triples.
4
a
b
c
a −1
______
2
a
a +1
______
2
i
3
2
4
2
5
ii
5
12
13
iii
7
24
25
iv
9
40
41
All the sets of numbers are Pythagorean
triples. They all satisfy the relationship
a2 + b2 = c2.
Other odd values of a also generate
Pythagorean triples.
When a = 1 the other values are 0 and 1.
These values satisfy the relationship
a2 + b2 = c2, but they are not a
Pythagorean triple because 0 is not a
positive integer.
k = 10.4 cm
h = 8.06 cm
d = 6.08 m
f = 13 m
a
b
c
d
e
Right-angled
Not right-angled
Not right-angled
Right-angled
Right-angled
Exercise 11.2
1
53.2 inches
2
3.03 m
3
277 m
4
3.6 m
5
0.841 m
6
a
b
c
d
7
P = 42.4 cm
8
6.02 cm
9
Height = 13.9 cm and area = 111 cm2
5.39
3.16
9.90
10.30
10 23.4 m
Exercise 11.1
11 Area is 45.0 cm2, so there is enough paint.
1
2
3
95
11
2.5
2.38
7
26
27.78
e
f
g
h
a
b
c
d
e
x = 10 cm
y = 13.4 cm
h = 2.59 cm
p = 1.62 cm
t = 7.21 m
12 4.24 cm
a
b
c
d
e
x = 7.42 m
y = 3.63 cm
t = 8.66 cm
p = 12 m
a = 6 cm
4
a
b
c
d
x = 2.80 cm
y = 4.47 cm
h = 4.28 cm
p = 8.54 km
Gradients and triangles
1–3
y
6
5
3
2
1
−2 −1 O
−1
1
2
3x
−2
−3
−4
−5
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6
Gradient of AB = __ = 3
2
3
Gradient of BC = __ = 3
1
4
Exercise 11.4
1
The vertical length is always 3 × the horizontal
length. The gradient of the line is 3. They are
the same.
5
It doesn’t matter where you draw the triangles,
the gradient of the line is the same all the way
along.
6
The length of the hypotenuse divided by the
shortest side is always equal to 3.16 (to 2 d.p.).
7
The ratios of corresponding sides are the same
for all the triangles and the internal ratios of
sides are the same for all the triangles.
b
Exercise 11.3
1
2
96
a
6
__
= 1.2
5
2
The ratio of corresponding sides are not
the same so the shapes are not similar.
All sides of shape 1 have length x and all
sides of shape 2 have length y so the ratio
of corresponding sides will be equal and
the shapes are similar.
4=2
__
5
__
= 1.25
c
4 = 1.3̇
__
4
3
Ratios not equal, so not similar.
d
80
___
= 1.3̇
60
___
= 1.3̇
45
60
Ratios of corresponding sides equal,
therefore they are similar.
e
9
__
= 1.5
8
6
Ratios of corresponding sides equal,
therefore they are similar.
They are not similar because not all
corresponding angles are equal.
12 = 1.5
___
a
b
c
d
e
f
g
h
i
j
Similar; all angles equal
Similar; sides in proportion
Not similar; angles not equal
Not similar; sides not in proportion
Similar; angles equal
Similar; sides in proportion
Not similar; sides not in proportion
Similar; sides in proportion
Similar; angles equal
Similar; all angles equal
2
a
b
c
d
e
f
x = 12
y=5
p = 12
a = 12
b = 5.25
c = 5.14
a
b
c
d
e
f
g
h
3
x = 10
4
a
3
AC = 8.75 cm
4
Angle BAC = Angle EDC (alternate angles)
Angle ABC = Angle DEC (alternate angles)
Angle ACB = Angle DCE (vertically opposite
angles)
All three angles are equal so the triangles are
similar.
CE = 4.51 cm
5
BC = 2.97 m
6
Lighthouse = 192 m
7
r=8
8
x = 60
f
b
c
d
x=9
y = 14
p = 3.30
y = 7.46
x = 50, y = 16
x = 22.4, y = 16.8
x = 7.5, y = 12.5
x = 178
All the angles on any square are 90°, so all
corresponding angles are equal.
All squares have four equal sides, so the
ratio of corresponding sides will always be
equal.
The ratio of corresponding sides of
different rectangles will not always be
equal.
Circles may be different sizes, but they
are all identical in shape, so are therefore
similar to each other.
All regular shapes are similar. For
example, all regular pentagons are similar
to each other. Irregular shapes do not
behave in the same way.
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Exercise 11.5
Exercise 11.7
1
1
a
b
c
d
421.88 cm2
78.1 m2
1562.5 m2
375 cm2
a
b
c
d
x = 24 cm
x = 30 m
x = 2.5 cm
x = 15 cm
3
a
b
c
Area will be 4 times larger.
Area will be 9 times larger.
Area will be smaller by a factor of 4.
4
8:3
2
c
i
SM
ii PQ
iii BC
i
MSR
ii EFG
iii OPQ
ABCDEFG is congruent to SMNOPQR
2
a
b
c
d
A, C
D, F
B, G
E, H, L
3
a
b
c
d
DEF similar GHI
ABCD similar EFGH
MNOP congruent STQR
ABCDEFGH congruent PIJKLMNO and
both similar to WXQRSTUV
ABC similar MON
b
Exercise 11.6
1
k2; k3
2
a
b
c
e
4
216 cm2
4
172 cm2
5
a
b
c
16 mm
157.9 cm2
83.2 cm3
6
a
b
c
d
20.83̇ cm 3
21.3̇ mm 3
0.75 m3
56.64 m3
7
a
b
c
d
525 cm2
6860 cm3
36 cm
14.15 cm
9
97
4
16 : 1
64 : 1
3
8
a
Height
13 cm
11 cm
9 cm
Surface
area
x cm2
121x cm 2
_____
81x
____
cm 2
Volume
y cm3
1331y
______
cm 3
729y
_____
cm 3
x = 3.72
a
169
2197
169
2197
b
c
5
Since triangle FAB and FED are congruent:
Angle FAB = angle FED and that makes
triangle CAE a right angled isosceles triangle.
It follows that AC − BC = EC − DC,
so BC = CD.
BF = DF (corr sides of congruent triangles)
Therefore BFCD is a kite (two pairs of
adjacent equal sides).
6
The two turns are the same so all the
corresponding angles in the triangles must be
the same. Since they both walk 1000 metres
before turning then all three corresponding
side lengths must be the same and so the
triangles are congruent.
Practice questions
1
215 m further
2
4.21 m
3
a
b
35 cm
37 cm
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4
a
b
5
3.7 km
6
4.5 cm
7
a
b
b
a2 + b2 = c2
(7x) 2 + (24x) 2 = 150 2
49x 2 + 576x 2 = 22 500
625x 2 = 22 500
x 2 = 36
336 cm
c
⇒b+1=c
p 2 − 12
b = _______
2
Let angle CAB = x
Then angle ABC = 90° − x (angles in
triangle sum to 180)
Angle CAB = angle DCB = angle DAC = x
Angle BCA = angle BDC = angle CDA =
90°
Angle ABC = angle CBD = angle ACD =
90° − x
All corresponding angles are equal, so
triangles ABC, CBD and ACD are similar.
p 2 + 12
c = _______
2
The difference between b and c is:
c−b
p2 + 12 p2 − 12
= _______ − _______
2
2
p2 1
p2 1
− (__ − __
= __ + __
2 2
2 2)
By comparing ratios of corresponding
sides in triangles ABC and CBD:
=1
BC
AB = ___
___
CB
BD
c __
a
__
=
a e
ce = a 2
By comparing ratios of corresponding
sides in triangles ABC and ACD:
AC
AB = ___
___
AC AD
c __
b
__
=
b d
cd = b 2
c
a 2 + b 2 = ce + cd
= c(d + e)
= c2
8
a
b
16.8 cm
103 cm2
9
a
a2 + b2 = c
___________
_
1
HC = √ 521 2 − 320 2 = √ 169 041
_
Difference = 350 + √ 169 041 − 521
= 240.1 m
2
√ 4. 5 2 − 1. 6 2 = 4.21 m
3
a
b
___________
a
_
√ 21 2 + 28 2 = 35 cm = AC
AD2 + AC 2 = 122 + 352
_
AD = √ 1369 = 37 cm
(7x)2 + (24x)2 = 1502
49x2 + 576x2 = 22 500
625x2 = 22 500
22 500
x = _______ = 36
625
2
98
Practice questions worked
solutions
4
u2 − v2
u2 + v2
+ (_______) = (_______)
2
2
u2 + 2 u2 v2 + v2
u 4 − 2 u 2 v 2 + v 4 _______________
_______________
2
2
=
⇒ u v +
4
4
⇒ 4 u2 v2 + u4 − 2 u2 v2 + v4 = u4 + 2 u2 v2 + v4
⇒ u4 + 2 u2 v2 + v4 = u4 + 2 u2 v2 + v4
(uv) 2
17, 144, 145
(If a = 17 then uv = 1 × 17 because 17 is
a prime number and its only factors are 1
and itself. If you substitute the values of 1
and 17 into the formulae for b and c then
you get b = 144 and c = 145.)
Let a = prime ( p), then a = p × 1
(so u = p, v = 1)
If b and c differ by 1 then:
2
b
_
x = √ 36 = 6
So, perimeter = 7 × 6 + 24 × 6 + 150
= 42 + 144 + 150
= 336 cm
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5
b
S
17 2 − 1 2
b = ________ = 144
2
2
17 + 1 2
c = ________ = 145
2
1452 = 21 025 = 1442 + 172
so (17, 144, 145) is a Pythagorean triple.
5.6
J
4.2
___________
√ 5.6 2 − 4.2 2
6
c
= 3.70 km
3 ___
__
= r
8
12
3 × 12 9
So r = ______ = __ cm
2
8
7
a
b
All three have the same angles.
a __c
__
2
e = a ⇒ a = ce
b __c
__
= ⇒ b2 = cd
a b
c
a 2 + b2 = ce + cd
= c(e + d )
= c2
a
3
uv = 17 so u = 17 and v = 1
a = 17
If a is prime, then a = uv and u and v are 1
and a or a and 1
To make b positive, u . 1, so u = a
and v = 1.
a2 + 1
a2 − 1
b = ______ and c = ______
2
2
2
2
a +1 a −1
c − b = ______ − ______
2
2
2
2
a +1−a +1
= ______________
2
2
__
=
2
=1
____
8
b
343
× 12 = 16.8 cm
√____
125
____ 2
343
√____
( 125 )
3
× A = 201.6
A = 103 cm2
9
a
u2 − v2
a 2 + b 2 = (uv) 2 + (_______)
2
2
4 (uv) 2 + (u 2 − v 2) 2
=_________________
4
(u 2) 2 + 2 (uv) 2 + (v 2) 2
=___________________
4
u2 + v2
= (_______) = c 2
2
2
So, c 2 = a 2 + b 2
99
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Chapter 12
Getting started
4
For example, 1, 2, 3, 4, 15
1
He has used the ‘middle’ value of the ordered
list. This is called the median.
5
2
Most of the numbers are fairly close to the
value given by Rohan, so it is reasonable.
Mode = none; mean = 96.4; median = 103
He will choose the median because it’s the
highest.
6
4451.6 cm
3
The actual values for each of the 7 weeks are
lost. Instead, they are represented by a single
value, which cannot give the full picture.
7
2.38 kg
8
91.26̇ °C
Jess has not arranged the times in order but
has picked the middle value in the list. She has
considered the position only and not how the
value compares with the other values in the list.
9
For example, 3, 4, 4, 6, 8
4
5
Write the list in order and then take the
middle value:
1
2.5
3
5.5
7
7.5
9.5
The middle value is 5.5 hours.
10 For example, 2, 3, 4, 7, 9
mX + nY
11 ________
m+n
Units for averages
1
Mean = 3.22 metres
Range = 2.8 metres
6
It would not change, because the middle
number has not changed.
2
320 cm
160 cm
7
You can add the numbers up and divide by 7,
for example. This is called the mean. This can
be useful if you want the average to change
when any of the values change.
3
Mean = 322 cm
Range = 280 cm
They have been multiplied by 100.
The mean and range are both multiplied by
the same number as the original data.
4
Changing the units changes the numerical
value of both the mean and range. This means
that you can make any mean or range bigger
or smaller than any other just by changing
the units of either value. If you fix the units
for both sets of data, then you are comparing
‘like-for-like’.
5
If you add m to each value, you also add m to
the mean BUT the range remains unchanged.
Because the values are all m units larger, then
the overall mean will be the same m units
larger. But the range measures the difference
between the largest and smallest values, both
of which have increased by the same value and
so are the same distance apart.
Exercise 12.1
1
a
b
c
d
i
ii
iii
i
ii
iii
i
ii
iii
i
ii
iii
Mode = 12
Median = 9
Mean = 8
Mode = 8
Median = 6
Mean = 5.7
Mode = 2.1 and 8.2
Median = 4.15
Mean = 4.79
Mode = 12
Median = 9
Mean = 11.7
2
Mean increased from 8 to 11.7 because of the
extreme value of 43 in(d). No change to mode
or median.
3
a
b
100
i
ii
i
ii
280 cm
440 cm
410 cm
54
48.5
84.25
98.875
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Exercise 12.2
1
a
b
c
2
a
b
c
d
3
Ru
i
ii
Oli
i
ii
Ru
Oli
b
Mean = 0.152
Range = 0.089
6
a
Mean = 0.139
Range = 0.059
b
c
d
e
Backlights. Footlights has the best mean
but the range is large, whereas Backlights
and Brightlights have the same range but
Backlights has a higher mean.
2
a
b
c
d
7
a
Mean = 4.5
Median = 4
Mode = 4 and 5
Range = 8
a
Frequency
Total
$6.50
180
$1170
$8
215
$1720
$10
124
$1240
b
c
3
4
5
101
a
b
c
d
Mode = no letters
Median = 1 letter
Mean = 0.85 letters
Range = 5
a
b
c
Mode = 1 child
Median = 2 children
Mean = 2.12 children
a
i
ii
iii
Mode = 8 marks
Median = 6.5 marks
Mean = 6.03̇ marks
6
5
0045789
6
0112336689
7
04
12
The data has many modes.
74 − 46 = 28
61 kg
Stem Leaf
12
156688899
13
01233468
14
002236
15
0
29
132.5
Exercise 12.4
1
a
b
141.7 cm
140 , h < 145
2
a
b
c
5.28 min
5 min 17 s
2,t<4
3
57.36 °C
4
a
$4130
$7.96
4
Key
12 | 1 represents 121 components
per hour
Price
b
Stem Leaf
Key
4 | 6 represents 46 kilograms
Archimedes median = 13
Bernoulli median = 15
Archimedes range = 16
Bernoulli range = 17
Bernoulli
Archimedes
Exercise 12.3
1
If she wants to suggest that the class is
doing better than it really is, she would
use the mode and say something like:
most students got 8 of 10.
b
Hawks: 76.7 kg
Eagles: 78.4 kg
75 < M , 85 for both
5
39.2 cm
6
42.23 years
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Exercise 12.5
1
a
b
c
d
e
2
a
b
c
Median = 6, Q1 = 4, Q3 = 9, IQR = 5
Median = 17, Q1 = 12, Q3 = 21, IQR = 9
Median = 14, Q1 = 5, Q3 = 18, IQR = 13
Median = 3.4, Q1 = 2.45, Q3 = 4.95,
IQR = 2.5
Median = 15.65, Q1 = 13.9, Q3 =
18.42, IQR = 4.53
40.25 − 2.35 = 37.9 hectares
Q1 = 3.55, Q2 = 7.2 and Q3 = 23.83
In this case, the high range shows that the
data is very spread out but this is skewed
by one high value, so the IQR is more
representative of the spread of sizes.
3
Median = 6, Q1 = 4, Q3 = 8, IQR = 4
4
a
Summer: median = 18.5, Q1 = 15.5,
Q3 = 23.5
Winter: median = 11.5, Q1 = 9.25,
Q3 = 12.75
5
b
Summer: IQR = 8
c
The lower IQR in winter shows that
car numbers are more consistent. In
poor weather people either use their
own transport or take transport more
consistently.
a
b
c
6
a
b
c
d
102
massive difference in consumption given
that the difference between the two IQRs
is only 0.375.
7
b
Test 1: range = 55; IQR = 82 − 39.5
= 42.5; median = 60.5
Test 2: range = 69; IQR = 81.5 − 45
= 36.5; median = 61.5
Test 3: range = 44; IQR = 71.5 − 45 = 26.5;
median = 62
Interpretations will vary, but generally the
students performed worst on Test 3.
Practice questions
1
a
b
c
2
5
3
a
b
Winter: IQR = 3.5
294 g
15.2 g
5
38
Sunshine
8 3
8
7 6 4 2
9 5
Julia: median = 23, Q1 = 13, Q3 = 24
Aneesh: median = 18, Q1 = 14, Q3 = 20
Julia: IQR = 11 Aneesh: IQR = 6
The IQR for the Algebraist is more
consistent than that for the Statistician
and is therefore more likely to have a
particular audience while the variation is
greater for the Statistician and therefore
could appeal to a varying audience.
i
6.5
ii 5.9
i
10.85
ii 14.05
i
3.275
ii 3.65
At first glance it seems like country
driving gets much better fuel consumption
as it appears that the data is distributed
more towards the higher end of the stems.
However, the smaller interval and the
decimal nature of the data mean that
when you look at IQR, there is not such a
a
c
d
e
f
g
h
i
j
4
5
2
3
1
4
Shade
4
1
1
0
3
1
0
1
2
3
4
5
2
0
0
4
5
3
2
5
9
5
6
2 2 4 9 9
8 9
9
Key
Sunshine 9|4 represents 4.9g
Shade
3|4 represents 3.4g
3.1 g
1.9 g
5.1 − 0.4 = 4.7 g
4.9 − 0.2 = 4.7 g
Those collected in sunshine were, on
average, larger, but both were equally
spread out.
4.3 − 1.8 = 2.5 g
3.4 − 1.2 = 2.2 g
There are no extreme values, which means
that the range gives a sensible measure of
spread as well as the interquartile range.
a
1 and 21
b
135 + 5n
________
c
d
8
2
a
b
c
24
25 , T < 30
30
42 + n
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Practice questions worked
solutions
1
a
14.7 × 20 = 294 g
b
Total = 294 + 30 × 15.6
= 762
c
2
762
Mean = ____ = 15.24 g
50
762 + 184
__________
= 17.2
50 + n
762 + 184
so 50 + n = __________
17.2
n = 55 − 50 = 5 sweets
a
b
19 + 19 = 38 insects
Sunshine
8 3
8
7 6 4 2
9 5
c
d
e
f
g
h
i
j
103
a
2
3
1
4
5
Shade
4
1
1
0
3
1
0
1
2
3
4
5
2
0
0
4
5
3
2
5
9
5
6
2 2 4 9 9
8 9
9
Key
Sunshine 9|4 represents 4.9g
Shade
3|4 represents 3.4g
d
135 + 5n = 147 + 3.5n
1.5n = 12
n=8
Size 2
a
b
16 + 5 + 2 + 1 = 24
25 , T < 30
c
16 × 27.5 + 5 × 32.5 + 2 × 37.5 + 1 × 42.5
____________________________________
= 30 °C
19 + 1
Need the ______ = 5th and
4
19 + 1
______
3×
= 15th
4
43 − 18 = 25 g
34 − 12 = 22 g
No real outliers and data is fairly
symmetrical.
Both IQRs are similar and both ranges are
similar.
24
Past paper questions
1
a
b
2
x2 − 7x − 5x + 35 = x2 − 12x + 35
x
___
= sin 35°
12
x = 12 sin 35°
= 6.88 cm
19 + 1
______
= 10th is 31 g
2
19 g
51 − 4 = 47 g
49 − 2 = 47 g
On average, those collected in the
sunshine are heavier.
Total frequency = 42 + n
2 + n + 5 + 2 + 1 , 32
i.e. n , 22
and 11 + 2 + n + 5 + 2 + 1 . 21
n.0
So 1 < n < 21
b
2 × 21 + 3 × 11 + 4 × 2 + 5n + 6 × 5 + 7 × 2 + 8 × 1
____________________________________________
42 + n
135 + 5n
________
=
42 + n
135 + 5n
c ________ = 3.5
42 + n
Total tremors = 18 × 4.5 = 81
Total in first 10 years = 10 × 4.1 = 41
81 − 41
In the last 8 years, mean = _______
8
= 5 per year
3
4
3
4
5
2, 8, 14
6k − 4 = 422
6k = 426
k = 71
a
(−3, −1)
b
3
__
c
3
y = __ x − 1 or y = 1.5x − 1
2
a
All three angles are the same.
b
AC ___
6
____
=
2
27
18
1 = 9 cm
AC = 27 × __
3
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6
a
b
17
3n + 2
7
a
(−2, 4)
4
1
i
− __ = − __
2
8
1x + 3
ii y = − __
2
i and ii
b
c
y
L
6
5
A
4
3
2
1
B
C
–6
–5
–4
–3
–2
–1 0
–1
1
2
3
4
5
6
7
8
x
–2
iii
8
a
9.25 units
i
20
18
16
Frequency
14
12
10
8
6
4
2
0
Infant
Child
Adult
Senior
Type of ticket
ii
iii
iv
b
104
17 − 12 = 5
adult
3
12 = ___
___
= 0.3
40 10
i
ii
104 − 18 = 86
18 27 31 45
median is 45
iii
____ = 51
60
72
104
∑x
7
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
9
10
a
b
14 a
4, 10, 18, 28
−7n + 32
%
.
A
x = 0.47
.
10x = 4.7
.
100x = 47.7
90x = 43
43
x = ___
90
11 a
b
48
47
43
_
_
50
_
√ 4 2 + 8 2 = √ 16 + 64 = √ 80 = 4 √ 5 = 8.94
7a(3a + 4b)
5(4x2 − 9y2) = 5(2x − 3y)(2x + 3y)
13 a
20°
6.75
BE = _____
___
⇒ BE = 4.5 cm
5.2
7.8
780
volume multiplier = ____
32
c
B
42
44
12 a
b
b
45
41
46
−5 + 3
5 + 9 ______
_____
( 2 , 2 ) = (7, −1)
_
49
√
b
i
ii
iii
C
{41, 43, 47}
{44, 46, 49, 50}
∅
____
780
height multiplier = ____
32
3
√
____
3
780
height = 2 × ____ = 5.80 cm
32
105
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 13
Getting started
1
2
× 1 000 000 = 106
× 1000 = 103
Kilograms
3
× 1000 = 103
Grams
÷ 1000
3
÷ 1000
Kilo = thousands
Milli = thousandths
Centi = hundredths
Students’ conversion diagrams (examples
could include: tonnes–kilograms−grams, or
amps–milliamps)
Exercise 13.1
1
106
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
4000 g
5000 m
3.5 cm
8.1 cm
7300 mg
5.760 t
210 cm
2000 kg
1.40 m
2.024 kg
0.121 g
23 000 mm
35 mm
8036 m
9.077 g
5
a
b
c
d
e
f
14 230 mm, 0.014 23 km
19 060 mg, 0.000 019 06 t
2750 ml, 275 cl
4 000 000 mm2, 0.0004 ha
1300 mm2, 0.000 000 13 ha
10 000 mm3, 0.000 01 m3
6
a
b
c
27 m3
27 000 000 cm3
2.7 × 1010 mm3
7
a
b
c
1.09 × 1012 km3
1.09 × 1021 m3
1.09 × 1030 mm3
8
a
b
c
1.13 × 102 cm3
1.13 × 105 cm3
1.13 × 10−13 km3
9
a
b
6
20 g
10 a
b
c
No
No
Yes
11 a
Computers use a binary number system
and ‘mega’ stands for 220. This is equal to
1 048 576.
Examples include:
Pico – a millionth of a millionth (÷1012)
Nano – a thousandth of a millionth (÷109)
Deca – 10 times
Hecto – 100 times
Giga – a thousand million times (×109)
Milligrams
÷ 1 000 000
2
4
2 m , 324 cm
3.22 m , 3 __
9
1 litre < 0.65 litres , 780 ml
125 ml , __
2
60
b
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Babylonian mathematics
1 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
2
3
Exercise 13.3
Answers will depend on students’ research.
Possible reasons are that 60 is a convenient
number because it has a lot of factors. It is
possible to count to 12 using the bones in
the fingers of one hand. So a base 60 system
might have come about by multiplying the
12 on one hand by the 5 fingers on the other.
Another possibility is that the number 60
comes from the length of a growing season
for certain crops. So, a year would be
approximately 6 × 60 days.
History appears to suggest that there
are 360 degrees in a circle because that is
approximately the number of days in a full
year. In reality, 360 is a very convenient
number because it has so many whole number
factors. This means a circle can be divided into
smaller parts without having to use fractions
in any way.
1
a
b
c
20 02
45 min
23 min
2
a
1 h 7 min
b
Aville
11 10
Beeston
11 45
Crossway
11 59
Darby
12 17
3
c
14 25
a
b
c
d
00 17
12 h 40 min
5 h 46 min
i
01 29 or 13 34
ii unlikely to be 01 29 because it is in the
middle of the night – in the dark.
i
1– 6 February (Wed–Mon)
ii 1– 4 February (Wed–Sat)
e
Exercise 13.2
1
a
b
i
22 30 to 23 30
ii 09 15 to 10 45
iii 19 45 to 21 10
09 30
Exercise 13.4
2
3 h 39 min
3
9 min 47 s
4
Monday 10 February 02 30
5
a
b
c
d
107
Day
Mon
Tues
Wed
Total
time
worked
7h
55 min
7h
55 min
7h
25 min
Day
Thurs
Fri
Total
time
worked
7h
53 min
8h
24 min
39 h 32 min
$223.36
i
yes
ii He entered 5 [DMS] instead of
17 [DMS] and then subtracted 12°459
and got a negative time for the
afternoon.
1
a
b
c
d
e
f
11.5 < 12 , 12.5
7.5 < 8 , 8.5
99.5 < 100 , 100.5
8.5 < 9 , 9.5
71.5 < 72 , 72.5
126.5 < 127 , 127.5
2
a
b
c
d
e
f
2.65 < 2.7 , 2.75
34.35 < 34.4 , 34.45
4.95 < 5.0 , 5.05
1.05 < 1.1 , 1.15
−2.35 < −2.3 , −2.25
−7.25 < −7.2 , −7.15
3
a
b
c
d
e
f
g
h
i
j
131.5 < 132 , 132.5
250 < 300 , 350
402.5 < 405 , 407.5
14.5 million < 15 million , 15.5 million
32.25 < 32.3 , 32.35
26.65 < 26.7 , 26.75
0.45 < 0.5 , 0.55
12.335 < 12.34 , 12.345
131.5 < 132 , 132.5
0.1335 < 0.134 , 0.1345
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
250 kg < 300 kg , 350 kg
5
a
b
6
4.45 m < L , 4.55 m
a
b
c
d
e
f
g
h
i
j
30.8 < a2 , 31.9
13 900 < b3 , 14 100
5.43 < cd 3 , 5.97
609 < (a2 + b2) , 615
c
0.248 , ___2 , 0.252
b
ab
2.66 , ___ , 2.82
cd
c __
b
− 43.5 , __
a − d , − 46.5
a c
2.66 , __ ÷ __ , 2.82
(d b)
__
a
48.9 , dc + __ , 50.7
(
b)
__
a
47.9 , dc − __ , 49.7
(
b)
13 Upper bound is 0.910 m/s
Lower bound is 0.769 m/s
Exercise 13.6
2
a
b
c
d
e
4 lb
4 kg
36 kg
132 lbs
i
Correct
ii 18 lb = 8 kg
iii 60 lb = 27 kg
iv Correct
3.605 cm < length , 3.615 cm;
2.565 cm < width , 2.575 cm
9.246825 cm2 < area , 9.308 625 cm2
9.25 cm2 < area , 9.31 cm2
3
a
b
c
d
$16
$64
£36
£24
511 105 787 km2 < surface area
511 266 084 km2
1.086 525 72 × 1012 km3 < volume of
Earth , 1.087 036 906 × 1012 km3
4
a
b
c
165 min
4.8 kg
(40m) + 30 = 25
⇒ m = −0.125 kg
You cannot have a negative mass of meat.
As the graph assumes it will always take
at least 30 minutes to cook any piece of
meat, you cannot use this graph for meat
with a very small mass that will take less
than 30 minutes to cook.
√
√
a
b
Max 5.32 and min 4.86
Only 1 can be used. The value of a is
5 to 1 s.f. If we find the maximum and
minimum values to 2 s.f. we get 5.3
and 4.9. This doesn’t tell us any more
than the answer is 5 to 1 s.f.
140 °F
60 °F
−16 °C
38 °C
3
a
i
ii
a
b
c
d
37 kg < mass left , 39 kg
4
Max = 232.875; min = 128.625
1
2
b
c
b
99.5 m < 100 m , 100.5 m
15.25 seconds < 15.3 seconds
, 15.35 seconds
Exercise 13.5
1
12 a
5
The smallest number of cupfuls is 426.4, and
the largest is 433.6.
6
Maximum gradient = 0.0739
Minimum gradient = 0.06
7
a
b
8
63.4° < x° , 63.6°
9
45
45.2 % < (___ × 100) , 46.7%
98
8.1 cm2 < area of 4 , 8.5 cm2
5.76 cm < hypotenuse , 5.90 cm
10 332 kg < mean mass , 335 kg (3 s.f.)
11 117.36 < number of 5s , 117.84
108
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
2
a
35
a
b
c
d
Monday: 8 hours 25 minutes
Tuesday: 8 hours 43 minutes
Wednesday: 8 hours 42 minutes
Thursday: 8 hours 38 minutes
Friday: 10 hours 17 minutes
44 hours 45 mins
$400.51
$352.45
3
a
b
c
d
33.5, 34.5
550, 650
12.685, 12.695
665, 675
y
30
Feet
25
20
15
10
5
0
x
1
2
3
4
5
6
7
8
9
10
Metres
3600 ft (answer may vary +/− 100 foot)
4
12.25 kg, 12.75 kg
c
1050 m (answer may vary slightly if
answer to (b) varies from that shown)
5
a
a
50
Sploges in hundreds
6
b
b
i
7540
ii 49 692
iii 9238.50
iv 25 426.50
1232.61
a
3.78 × 1011
b
c
378 500 < area of Japan , 377 500
335 people per square kilometre
7
a
b
c
Approximately 18 litres
Approximately 6 gallons
L = 4.5G. The formula is not exact
because the values read from the graph
are approximate.
8
a
b
c
−1.55
1.53
0.62
9
a
4.116 × 103 cm3 < volume of cube
, 4.038 × 103 cm3
4.116 × 106 mm3 < volume of cube
, 4.038 × 106 mm3
y
40
30
6
20
10
0
x
10
15
20
5
Squidges in hundreds
25
b
625 squidges (answer may vary)
c
224 000 ploggs (answer may vary:
220 000 − 228 000)
Exercise 13.7
1
$18.50
2
$4163.00
3
£7960
4
$384.52
5
$2589.20
6
$113.77
b
Practice questions
1
109
a
b
104 km/h
69 mph
10 a
b
c
11.94 cm2
7.09 cm
0.89
11 a
b
759 cm3
4.47 cm < a , 4.95 cm
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Practice questions worked
solutions
1
a
Conversion graph, km/h to mph
80
mph
60
40
20
0
20
b
40
60
km/h
80
100
120
100
120
Conversion graph, km/h to mph
80
mph
60
40
20
0
2
a
40
60
km/h
c
d
8.25
8.43
8.42
8.38
10.17
42 + 2 + 45 minutes = 44 hours and
45 minutes
44.75 × $8.95 = $400.51
0.88 × $400.51 = $352.45
a
b
c
d
upper bound
34.5
650
12.695
675
b
3
110
20
80
lower bound
33.5
550
12.685
665
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
Greatest = 12.75 kg
Least = 12.25 kg
5
a
i
ii
iii
iv
6
1 US $ = 75.40 Indian Rupees
100 US $ = 100 × 75.40 Indian Rupees
= 7540 Indian Rupees
€1 = 82.82 Indian Rupees
€600 = 600 × 82.82 Indian Rupees
= 49 692 Indian Rupees
1 Dhs = 20.53 Indian Rupees
450 Dhs = 450 × 20.53 Indian Rupees
= 9238.5 Indian Rupees
1 SR = 20.10 Indian Rupees
1265 SR = 1265 × 20.10 Indian Rupees
= 25 426.50 Indian Rupees
b
1 Australian dollar = 56.79 Indian Rupees
1 Australian dollar
1 Indian Rupee = ______
56.79
14 000
14 000 Indian Rupees = ______ Australian dollars
56.79
= 246.52 Australian dollars
a
378 000 × 1000 × 1000 = 3.78 × 1011 m2
b
Upper bound = 378 500
Lower bound = 377 500
c
126 500
126 500 000
___________
= 335
9
a
b
377 500
7
8
a
b
c
a
2.55 + 4.35 − 8.45 = −1.55
b
2.55 2
_____
= 1.53
4.25
2.55
__________
= 0.59
8.55 − 4.25
c
111
18.5 litres
5.95 gallons
L = kG
18.5 = k × 4
18.5
k = ____ = 4.625
4
so, L is approximately equal to 4.625G
The value of k was calculated using
estimates from the graph.
10 a
Upper bound = 14.55 × 13.25 × 21.25
= 4116 cm3
Lower bound = 14.45 × 13.15 × 21.25
= 4038 cm3
Upper bound = 145.5 × 132.5 × 213.5
= 4 116 013 mm3
Lower bound = 144.5 × 131.5 × 212.5
= 4 037 872 mm3
1 × 5.25 × 4.55 = 11.94 cm2
__
2_____________
b
√ 5.35 2 + 4. 65 2 = 7.09 cm
c
4.65
____
= 0.89
11 a
b
5.25
7.52 × 13.5 = 759.375 cm3
√
________
volume
A = ________
h
√
√
______
325
Upper bound = _____ = 4.95 cm
13.25
______
275
Lower bound = _____ = 4.47 cm
13.75
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 14
Getting started
10
2x
=
8
y
y
+
1, 2, 4
c
10
6
5
4
4
3
2
−
5x
0
1
(3, 4)
1
4
−
y=
2
−1
1
−1 0
−1
−2
1
2
3
4
5
a
b
c
d
x = −2, y = −2
x = 3, y = 3
x = 3, y = −2
x = −1, y = 6
2
−5
e
−6
f
1 , y = −2
x = __
7
4 , y = __
4
x = __
3
3
9
1
x = ___ , y = − ___
11
11
5
3
x = __ , y = − __
4
4
7
x = __ , y = 1
4
25
22
___
x = , y = ___
17
17
−3
−4
3
3 square units
12 square units
3 < A , 12
a
b
Exercise 14.1
c
1
d
a
10
4
+y
a
=1
0
6
y
2x
8
b
(3, 4)
4
x+
2
2y
=
0
2
4
6
Solution is x = 3 and y = 4
b
8
11
10
1
2x
+y
=4
x
2
−
=
6x + 4y = −2
−1
1
1
2
3
4x
−2
−3
x
−1
3x + 2y = 7
2
−5 −4 −3 −2 −1 0
−1
(1, 2)
1
−2
x
y
5
3
y
5
When is there a solution?
x
12
4
3
4
The scale can sometimes make it difficult
to read off certain values, such as
fractions, accurately.
The equations must be solved
algebraically.
y
4
3
Solution is x = 3 and y = 4
x
−2
3
5
6
2
0
1
2
3
−4
Solution is x = 1 and y = 2
112
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
3
4
m x = 5, y = −7
c
d
solutions. If they ARE the same, then __ and __
Exercise 14.2
1
a
b
c
x = 2, y = 5
x = 3, y = −2
x = −10, y = 6
d
10
4 , y = ___
x = __
−
3
3
x = −2, y = 4
e
f
g
h
2
3
113
n
b
o
4
11 , y = 17
x = − ___
3
1
1
__
x = , y = __
2
2
10
19
___
x = , y = ___
17
17
a
b
c
d
e
f
g
h
i
j
k
l
x = 4, y = 4
x = 2, y = 6
x = 1, y = 2
x = 5, y = −1
x = 3, y = 4
x = 1, y = 3
x = 6, y = 3
x = 5, y = 4
x = 4, y = 3
x = 4, y = 6
x = 6, y = 6
x = 4, y = 2
a
b
c
d
x = 2, y = 4
x = 4, y = 3
x = −5, y = −10
x = 5, y = 5
e
f
7
9
x = __ , y = __
4
4
x = 5, y = 3
g
6
9
x = __ , y = ___
5
10
x = −1, y = −4
j
c
d
__
If __
a is not the same as b there will be
4.
must both be equal to __
3
l
i
The coefficients of x and y in the first equation
have been multiplied by the same number to
get the second equation, BUT this is not true
for the right-hand sides.
a
k
7
6
x = __ , y = − ___
3
13
118
5
x = − ____ , y = − ___
55
11
35
29
x = ___ , y = ___
4
12
x = 1, y = −4
h
The lines are parallel and never meet, so
there will be no solutions that work in both
equations at the same time.
a
b
c
d
e
f
g
h
i
j
k
3
7
x = − __ , y = __
3
2
3
29
x = __ , y = ___
5
5
x = 3, y = 4
x = 2, y = 4
x = −3, y = 5
x = 6, y = 3
x = 3, y = 5
x = 3, y = −4
x = 5, y = 3
x = 2, y = 4
x = 2, y = 3
x = −2, y = 1
x = −3, y = −2
1, y = 2
x = __
2
1, y = 3
m x = − __
2
n x = −3, y = 4
o x = 5, y = 8
l
b
c
d
e
f
504
112 , y = ____
x = ____
−
25
25
x = 3, y = −2
x = −8, y = −2
x = 6, y = −18
x = −0.739, y = −8.217
x = 5.928, y = −15.985 (3 d.p.)
6
a
b
c
d
90 and 30
−14.5 and −19.5
31.5 and 20.5
14 and 20
7
Pen drive $10 and hard drive $25
8
48 blocks (36 of 450 seats and 12 of 400 seats)
9
Students will create their own problem for
each other.
5
a
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 14.3
1
a
x
−5
−4
−3
−2
−1
0
1
2
3
4
5
−5
−4
−3
−2
−1
0
1
2
3
4
5
−3
−2
−1
0
1
2
3
4
5
6
7
−9
−8
−7
−6
−5
−4
−3
−2
−1
0
1
−8
−7
−6
−5
−4
−3
−2
−1
0
1
−6
−5
−4
−3
−2
−1
0
1
1.0
1.1
1.2
1.3
3.4
3.5
3.6
3.7
–3.4
–3.3
–3.2
–3.1
2.8
2.9
3.0
−4.6
−4.5
−4.4
−3.2
−3.1
−3.0
b
x
c
p
d
y
e
q
f
x
g
x
h
x
i
2
a
b
c
d
e
f
g
h
i
k
{4, 5, …, 31, 32}
{8, 9, …, 18, 19}
{18, 19, …, 26, 27}
{−3, −2, −1}
{−3, −2, − 1, 0}
{3, 4, …, 10, 11}
{−6, −5, −4}
{4, 5, 6}
{3, 4}
2
Exercise 14.4
1
a
b
x,2
x.3
c
14
y < ___
15
y . −2
c>2
x < −1
x,6
p.3
d
e
f
g
h
114
3
i
j
k
x . −15
g>4
w,8
l
7
k , ___
10
a
b
c
d
y . 30
y < 30
z . 62
k . 33
e
10
r , ___
3
a
q , 12
b
11
g < ___
2
3
h > − __
2
h , 19
c
d
e
f
44
y > − ___
3
n , 48
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
b
c
13
v < − ___
6
31
___
e.
28
1
t . 9 __
4
109
t . ____
4
763
____
t.
4
11 , x , −2
− ___
2
−3 , x < 9
6 , x < 13
d
16
___
,x<7
g
h
i
j
k
4
a
e
5
3
4
3
2
1
−3 −2 −1 0
−1
4
a
4
5
x
−5
b
y
5
4
3
2
1
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
x
−2
−3
1
2
3
4
3x − 2y > 6
−4
x
−5
−2
c
−3
6
5
y
4
3
y
x<5
2
1
3
2
1
115
3
−4
y
−3 −2 −1 0
−1
−3
2
−3
1
−2
1
−2
3
2y − 3x > 6
2
−3 −2 −1 0
−1
y . 3 − 3x
−5 −4 −3 −2 −1 0
−1
Exercise 14.5
x + 2y , 4
y
5
1
78 + 28b < 630
19 boxes (cannot load a fraction of a box)
4
x
2
p > 1 700 000
2
4
3
p
4
3
4
3
16
4
− ___ , x < − ___
5
17
1
2
−3
1 700 000
6
x−y>0
1
−2
a
b
y
1
2
3
4
x
–6 –5 –4 –3 –2 –1 0
–1
1
2
3
4 5
6
x
–2
–3
–4
–5
–6
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
g
y
5
5
y.3
4
4
3
3
2
2
1
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
1
x
−10 −8 −6 −4 −2 0
−1
−2
e
−3
−2
−4
−3
−5
−4
4
6
8
x
10
−5
7
5
a
b
c
Above
Below
Above and below
6
a
b
y < 4x + 5
x+y,3
c
1x+1
y > __
3
3
y < − __ x
2
5
4
3
2
1
−10 −8 −6 −4 −2 0
−1
2
4
6
8
10
x
d
−2
x + 3y < 10
−3
Shading the wanted region
Shading the wanted region for a single inequality
works well, but when there is more than one
inequality then it is more difficult because it
is hard to see the region that satisfies all the
inequalities since shaded regions overlap. Only the
region that is shaded for all the inequalities will
count as the wanted region.
−4
−5
5
y
4
3
2
1
−10 −8 −6 −4 −2 0
−1
−3 , x , 5
2
4
6
8
x
10
Exercise 14.6
1
y
7
−2
6
−3
5
−4
4
−5
3
x=4
x+
2y
=
2
1
−5 −4 −3 −2 −1 0
−1
116
2
y
6
f
y
0<x<2
d
y
=
1
6
x
2
3
4
5
x
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
6
d
e
f
5
g
8
y
7
l
5
21
__
___
(x + 2 ) − 4
72
57
__
x
+
− ___
(
2 ) 4
2
3
21
__
___
(x − 2 ) − 4
81
72
__
___
(x + 2 ) − 4
2
13
165
___
x
−
− ____
(
2)
4
(x − 10)2 + 300
2
a
b
c
d
e
f
x = 0.74 or −6.74
x = −0.54 or −7.46
x = 3.41 or 0.59
x = 1.14 or −6.14
x = 2 or 1
x = 11.92 or 0.08
3
a
b
c
d
e
f
g
h
i
x = 3.70 or −2.70
x = 1.37 or −4.37
x = 0.16 or −6.16
x = 2 or 4
x = 1.89 or 0.11
x = 5.37 or −0.37
x = 1.30 or −2.30
x = 3 or −1
x = 1.62 or −0.62
4
Students’ own choice of question.
x
+
4
y
h
=
5
3
y=2
2
i
1
y=0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
x
j
k
3
8
y
7
x=4
6
5
x
+
4
y
=
y=3
5
3
R
2
1
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
x
4
y < −x + 4, y . 2x + 1, x < 2
5
(3, 0), (2, 0), (2, 1), (1, 1), (1, 2), (1, 0), (0, 3),
(0, 2), (−1, 4)
6
5
y
y=4 4
3
4
−5 −4 −3 −2 −1 0
−1
1
−2
−3
−4
−5
(0, 4) (1, 4) (2, 4) (1, 3)
Exercise 14.7
1
117
a
b
c
(x + 3)2 + 5
(x + 4)2 − 15
(x + 6)2 − 16
2
Exercise 14.8
y=
1
=
+2
3x +
2
y
x
(x + 3)2 − 4
(x − 2)2 + 8
(x − 1)2 + 16
2
3
4
5
x
1
a
b
c
3(x + 1)2 + 11
2(x + 2)2 − 7
6(x + 1)2 + 14
e
3
1
2 (x + __) + __
2
2
4(x − 1)2 + 7
f
2(x − 1)2 + 15
g
5(x + 1)2 − 4
d
h
i
2
7
73
3(x + __) − ___
12
6
2
3
33
2(x − __) − ___
4
8
2
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
a
b
c
d
e
f
7
209
5(x + ___) − ____
20
10
2
13
161
2(x − ___) − ____
4
8
2
10
1100
3(x − ___) + _____
3
3
108
120
135
144
150
165.6
a
b
c
d
2340
360
156
24
j
k
l
2
3
2
3
a
b
c
d
e
f
x = 3.73 or 0.268
x = 3.30 or −0.303
x = 0.896 or −1.40
x = −0.851 or 2.35
x = −1.37 or 0.366
x = 0.681 or −0.881
4
a
b
c
d
e
f
x = 4.79 or 0.209
x = 0.631 or 0.227
x = 0.879 or −0.379
x = 1.35 or −2.95
x = −2.84 or −9.16
x = 6.85 or 0.146
5
a
b
Exercise 14.9
1
2
118
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
x = −3 or −4
x = −6 or −2
x = −7 or −4
x = −5 or 1
x = −8 or 2
x = 8 or −20
x = 4 or 2
x = 7 or −4
x = 8 or −3
x = 8 or 4
x = 11 or −9
x = 12 or −3
x = 6 or 4
x = 5 or 7
x = −3 or 12
a
b
c
d
e
f
g
h
i
j
k
l
x = 0.162 or −6.16
x = −1.38 or −3.62
x = −2.38 or −4.62
x = −0.586 or −3.41
x = 3.30 or −0.303
x = 3.41 or 0.586
x = 7.16 or 0.838
x = 2.73 or −0.732
x = 6.61 or −0.606
x = 8.24 or −0.243
x = 8.14 or −0.860
x = −0.678 or −10.3
c
_
_
5 − √ 21
5 + √ 21
x = _______ or x = _______
2
2
_
_
√
−3 + 65
−3 − √ 65
________
________
x=
or x =
4
4
_
_
√
√
5
+
5
−
37
37
x = _______ or x = _______
6
6
_
_
_
_
d
x = − 2 + √ 3 or x = −2 − √ 3
e
x = − 4 + √ 7 or x = −4 − √ 7
f
5 − √ 15
5 + √ 15
x = _______ or x = _______
2
2
_
_
6
x = 1.61 cm (−5.61 is not a solution because
length cannot be negative)
7
a
b
4.53 metres
248 months
Exercise 14.10
1
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
(3x + 2)(x + 4)
(2x + 3)(x − 1)
(3x + 2)(2x − 1)
(3x + 8)(x + 2)
(2x − 5)(x + 2)
(4x − 1)(4x + 9)
(3x + 1)(x + 5)
(4x − 1)(2x + 1)
(2x + 3)(x − 2)
(2x + 3)(x + 3)
(3x + 8)(x − 2)
(5x − 3)(2x + 1)
(5x + 1)(x + 1)
(2x − 1)(x − 9)
(6x − 5)(2x + 3)
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 14.11
h
1
As Exercise 14.10
i
2
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
j
(3x − 7)(2x + 3)
−(2x + 3)(x + 5)
(2x + 3y)(2x + 3y)
(3x + y)(2x − 7y)
x(x − 9)(x − 4)
2(3x − 4y)(x − 5y)
(3x + 2)(2x + 1)
(3x − 4)(x − 3)
3(x − 5)(x − 8)
p(x + 3)(x + 4)
x(5x − 6)(x − 2)
3x(4 − x)2
(x − 1)(x − 2)
4(x − 2)(x − 1)
(2x)(6x + 13)
3
a
c
d
2
y
__
4
5
10
e
t
__
b
f
g
h
i
j
2
a
b
c
d
e
f
g
119
x
__
6
u
__
3
t
___
10
y
__
2
3z
___
4
4t
__
3
xy
___
3
x
___
4y
1
__
2
y
__
2
5x
3b
2x
___
3y
c
b
__
d
ac
___
e
abc
____
f
9b
___
a
4
2
j
a
18
_____
h
i
4
2
4c
(abc)2
3y
___
4x
4x 2z
_____
3y
9
g
Exercise 14.12
1
b
3b
2
____
3de
1
____
4 b2
a
___
5b
ab
i
17z 3
x 3z
____
2y
3v
_______
7u 2w 4
x+3
_____
x+4
x
_____
x+4
y3
_____
y+1
x−6
_____
x−4
x+5
_____
x−3
8
j
3x + 2
______
b
c
d
e
f
g
h
3x − 2
x+3
k _____
x+8
2x − 3
______
l
x+1
7x
−1
m ______
x−4
5y − 4
n ______
y−7
3x
−7
o ______
5x − 4
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
a
3x − 4
______
b
7x + 1
x2 + y2
c
1
__
d
e
x2 + 1
1
f
Exercise 14.14
1
x
√
_
x3 + y3
Exercise 14.13
1
a
___
e
4
3
y2
____
14
3z 2
____
14
t__2
3
1
f
1
__
b
c
d
b
8t
___
c
12u
____
d
z
___
e
5(x + y)
_______
f
3x
___
g
h
i
j
i
j
1
____
d
k
2d
___
e
l
r
____
f
h
a
b
c
2
a
b
c
2y 2
7c
2pq
3z 2 t 2
______
x3
2xt
___
3
3
____
4xy
3
a
b
c
d
64t y
_______
d
e
3
_______________
e
f
1
________
g
(√ z 2 + t 2 )
____________
4 4
27
4(x + y) 5(x − y)
f
4(a − b)
_ 3
h
120
3y
___
6
3f
___
2e
gh 2
____
32
2
g
2
x2
a
432(x 2 + y 2)
z−t
_____
z−y
4
a
b
4
15
35
14
12
2
11y
____
8
a
___
40
a
__
2
7x + 18y
________
63
19(x + 1)
__________
2
56
29pqr
______
136
93p
____
70
71x
____
84
62x 2
_____
63
33 − 5x
_______
18
x+3
_____
a
23
____
12a
19x
____
6y
3a + 2
______
a2
17
___
6x
7
___
5e
2x + 5
_____________
(x + 1)(x + 4)
5x − 7
_____________
(x − 1)(x − 2)
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
c
7x + 39
_____________
f
(x + 2)(x + 7)
5
___
2x
7
____
6xy
2 + x2
______
x
g
x + 2x + 5
__________
h
(x − 1)(27y − 14)
_________________
i
2y − x
_______
j
+ 4xy −
−
____________________
d
e
k
l
c
50x − x2
= −(x2 − 50x)
= −{(x − 25)2 − 625}
= 625 − (x − 25)2
x = 25
625 m2
7 (x + 4) 2 − 33
___
x = −4 ± √ 33
3
2x 2y
4x 2y
yz 2
z3
12xyz 2
4
4, 5, 6, 7, 8, 9
c
d
12 a
b
5
6
7
8
9
x
0
3
9
y
9
6
0
x
0
1
5
y
13
11
3
10 11 12 13
x
10
2
16
16
x = 40 or x = 60
40 cm
(3x − 1)(x − 2)
(x − y)(x + y − 2z)
x−7
15 _____
x+3
16 a
b
17 i
x+y=9
2
5(x − 3) 2 > 0
5(x − 3) 2 + 6 > 6 . 0
d
4
(100 − x)
x
___
+ _________ = 325
14 5(x − 3) 2 + 6
8
6
100 − x
2
(100 − x) 2
x 2 ___
x2
100 − x
__
and (_______) = _________
=
(4)
4
16
16
5+x
13 _____________
(x + 2)(x − 2)
c
2x + y = 13
12
9
b
3
y
14
x = 1.41 or x = −6.41
11 a
18x − 4 , 88
18x , 92
x , 5.11
432 cm2
3
8
6y 3
10 ____
x2
1
_____
x+3
2
_____
x+2
a
c
50 − x
x(50 − x) = 50x − x2
d
63y 2
2
b
a
b
2
x = 2, y = 3
a
6
2(x + 1)
1
4
14
2
Practice questions
b
5
ii
1−p
p − p2
3
1 or p = __
p = __
4
4
15
___
16
___
√ 37
5
2
0
d
121
0
2
4
6
8
10 x
x = 4, y = 5
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
18 a
y
6
5x
+
5 6y
4
4
=
y
30
=
x
1– x
y= 2
3
+1
b
2
b
0
−1
1
2
3
4
5
6 x
5
2
Greatest value for x + 2y = 8 ___
11
(occurs at intersection of x = y and
5x + 6y = 30)
Practice questions worked
solutions
1
6x − 5y = −3
6
①
5x + 4y = 22
②
4 × ① → 24x − 20y = −12
5 × ② →25x + 20y = 110
a
b
0
3
9
y
9
6
0
x
0
1
5
y
13
11
3
graph showing x + y = 9 and 2x + y = 13
for 0 < x < 10, 0 < y < 15
d x = 4, y = 5
1 n(n + 1) = 105
__
2
n(n + 1) = 210
n2 + n − 210 = 0
(n + 15)(n − 14) = 0
⇒ n = 14 or −15
So, n = 14 because n . 0.
a
y
③
④
x
2x + 2y = 100
2y = 100 − 2x
y = 50 − x
Width = 50 − x
③ + ④ → 49x = 98
98
x = ___ = 2
49
① → 6(2) − 5y = −3
12 + 3 = 5y
5y = 15
y=3
Check in ②: 5(2) + 4(3) = 10 + 12 = 22 
So, x = 2 and y = 3.
2
x
c
1
−1
a
b
Area = xy = x(50 − x) = 50x − x2
c
50x − x2 = −(x2 − 50x)
= −{(x − 25)2 − 625}
= 625 − (x − 25)2
Maximum when x = 25 m
Area = 25 × 25 = 625 m2
d
2(3x + 1 + 6x − 3)
2(9x − 2)
9x − 2
9x
, 88
, 88
, 44
, 46
46
x , ___
9
46
1
1
___
__
__
= 5 and x , 5
9
9
9
Largest possible x = 5 if x is an integer.
Largest area = (3(5) + 1)(6(5) − 3)
= 16 × 27
= 432 cm2
7
x2 + 8x − 17
= (x + 4)2 − 16 − 17
= (x + 4)2 − 33
x2 + 8x − 17 = 0
(x + 4)2 − 33 = 0
(x + 4)2 = 33 __
x + 4 = ± √ 33 _
x = −4 ± √ 33
8
x2 + 5x − 9 = 0
ax2 + bx + c = 0
a = 1, b = 5, c = −9
_
3
2a
0
4
12
_
_
5 √ 61
So, x = − __ ± ____
2
122
_______________
− b ± √ b 2 − 4ac − 5 ± √ 25 2 − 4 × 1 × − 9
x = ______________ = ____________________
2
− 5 ± √ 61
= _________
2
2
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9
1 (3x − 2) . 2
__
4
3x − 2 . 8
3x . 10
10
x . ___
3
3x − 12 , 17
3x , 29
29
x , ___
3
10
29
___
So,
, x , ___
3
3
2
x
4x ÷ 3____
10 ___
3y
9y 4
1 9 y 43
4 x1
_____
×
= _____
13 y1
1 3 x 22
4 y3
= ____
x2
11 a
b
c
d
12 a
b
100 − x
100 − x
x
Squares have side __ and _______
4
4
2
2
x
100 − x
so areas are __ and (_______)
(4)
4
(100 − x) 2
x 2 __________
___
+
= 325
16
16
x2 + x2 − 200x + 10 000 = 5200
2x2 − 200x + 4800 = 0
x2 − 100x + 2400 = 0
(x − 60)(x − 40) = 0
x = 60 or x = 40
Smaller square: x = 40 cm
= total wire used
= perimeter
3x2 − 7x + 2
= (3x − 1)(x − 2)
x2 − y2 = (x − y)(x + y)
2yz − 2xz = 2z( y − x)
So, x2 − y2 − 2xz + 2yz
= (x − y)(x + y) + 2z(y − x)
= (x − y)(x + y) − 2z(x − y)
= (x − y)(x + y − 2z)
3
1
13 _____________ + _____
(x − 2)(x + 2)
x−2
3
x+2
= _____________ + _____________
(x − 2)(x + 2)
(x − 2)(x + 2)
x+5
= _____________
(x − 2)(x + 2)
123
14 5x2 − 30x + 51
= 5(x2 − 6x) + 51
= 5{(x − 3)2 − 9} + 51
= 5(x − 3)2 − 45 + 51
= 5(x − 3)2 + 6
So, 5(x − 3)2 > 0 for all x
Therefore, 5(x − 3)2 + 6 > 6 . 0 so it is always
positive.
(x − 4)(x − 3)
(x − 7)(x + 7)
15 _____________ × _____________
(x − 3)(x + 3)
(x + 7)(x − 4)
(x + 7)
= _______
(x + 3)
16 a
P(tails) = P(not heads)
= 1 − P(heads)
=1−p
b p(1 − p)
3
c
p(1 − p) = ___
16
3
p − p2 = ___
16
16p − 16p2 = 3
16p2 − 16p + 3 = 0
(4p − 1)(4p − 3) = 0
3
1 or p = __
p = __
4
4
3
d p = __
4
P(at least one head) = 1 − P(no heads)
= 1 − P(tails) × P(tails)
1 × __
1
= 1 − __
4 4
1
= 1 − ___
16
15
= ___
16
17 x2 − 5x − 3 = 0
__________________
− (− 5) ± √ (− 5) 2 − 4 × 1 × (− 3)
__________________________
x=
2
_
√
5
±
25
+
12
= ____________
2
_
5 ± √ 37
= _______
2
_
_
5 + √ 37
5 − √ 37
_______
_______
So, a =
and b =
2
2
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_
i
_
5 √37
5 √37
a − b = __ + ____ − ( __ − ____ )
2
2
2
2
_
= √37
ii
_
_
5 √37
5 √37
a + b = __ + ____ + __ − ____
2
2
=5
2
18 a
2
y=x
y
y=
6
1
x+1
2
R
O
6
x
5x+6y=30
Region is triangle R
b
124
6
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 15
Getting started
1
2
3
a
b
3
Student C is correct.
Student A’s solution suggests that 1 cm
1 cm in the real world.
on the map is ______
50 000
This would mean the real world is smaller
than the map. Instead, student A needed
to use the fact that 1 cm on the map is
50 000 cm in the real world. Student A has
also converted cm to km incorrectly.
Student B found the distance 1 950 000 mm
correctly, but has converted to km
incorrectly. There are 1000 m in each km
and 1000 mm in each m. This means there
are 1 000 000 mm in 1 km, so the answer is
1.95 km.
It would also have 3000 seats, they would just
be smaller! It’s important to understand that
a scale model has exactly the same features in
the same numbers. They are just a different
size.
14.4
0.32 × _____ = 57.6 cm
0.08
Exercise 15.1
1
6.8 m × 5.2 m
2
a
b
3 cm
2.4 cm
3
a
b
5.6 cm
15°
1
a
b
c
270°
135°
045°
2
a
b
082°
315°
3
a
b
c
d
e
110°
050°
230°
025°
280°
4
a
b
c
108°
288°
147 km
5
a
b
9.6 km
090°
6
a
b
121° to 123°
471.7 m
Exercise 15.4
1
a
C
D
100 m
90 m
A
2
125
20 m
34.8 m
Exercise 15.3
Exercise 15.2
1
a
b
80°
75°
120 m
b
c
BCD = 92°; ADC = 113°
80 m
a
b
20°
3.4 m
B
2
Hypotenuse Opposite A Adjacent A
a
c
a
b
b
y
z
x
c
p
q
r
d
l
n
m
e
c
d
e
f
e
f
g
a
b
c
opp(30°) = 5.7 cm
opp(40°) = x cm
adj(50°) = x cm
opp(65°) = q m or adj(25°)
opp(25°) = p m or adj(65°)
hypotenuse = r m
Lengths and angles in triangles
1–3 Students’ own drawings and measurements.
4–5 The answers are all the same (0.577).
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
6
7
8
For angles of 30° and 60°, the value
opp(30°)
of _________ is 0.577 for all triangles.
adj(30°)
For right-angled triangles with angles A and B,
opp(A)
the value of _______ will be the same for any
adj(A)
angle A whatever the side lengths of the
triangle.
If you divide a different pair of sides the value
will be different, but for a given angle and a
given pair of sides the value will be the same
whatever the side lengths of the triangle.
Exercise 15.5
1
2
a
b
c
d
e
f
g
h
0.700
1.04
0.325
1
0.279
0.323
0.00873
0
a
1
tan A = __
2
3
tan A = __
2
__
tan A = 1
4
tan B = 4
b
c
d
e
3
4
126
e
f
g
h
i
40.6 cm
115 m
2.61 m
95.8 km
39.8 m
5
a
b
1.0724
32.2 m
6
32.3 m
7
a
b
8
0.45 m
9
36.18 m
1.73
2
Exercise 15.6
3
tan x = __
2
n
tan z = __
m
m
tan y = __
n
f
g
tan C = a
tan D = p2
a
b
c
d
e
f
5.20 cm
4.62 m
35.7 m
3.54 km
18 cm
10.3 cm
a
b
c
d
20.8 cm
16.1 cm
9.17 cm
7.85 cm
1
a
b
c
d
40.4°
60.0°
74.3°
84.3°
2
a
b
c
d
22°
38°
38°
70°
3
a
b
c
a = 35.0°
b = 77.5°
c = 38.7°
d = 51.3°
e = 18.4°
f = 30°
d
e
4
71.8° (1 d.p.)
5
21.2° (1 d.p.)
6
a
b
7
AB = 6.32 (3 s.f.)
ACB = 64.6° (1 d.p.)
13.3 (3 s.f.)
26.7 (3 s.f.)
Exercise 15.7
1
a
i
b
i
c
i
4
__
5
7
___
25
12
___
13
ii
ii
ii
3
__
5
24
___
25
5
___
13
iii
iii
iii
4
__
3
7
___
24
12
___
5
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
3
e
i
f
i
g
i
a
b
c
d
e
f
g
h
0.0872
0.9962
0.5000
0.8660
0.8660
0.5000
0.9962
0.0872
a
g
cos 42° = __
ii
ii
ii
h
a
b
c
d
e
f
g
h
i
j
k
l
0.845 m
4.5 m
10.6 km
4.54 cm
10.6 cm
9.57 cm
14.1 cm
106 cm
4.98 cm
42.9 m
2.75 m
137 m
a
b
c
d
81.9°
57.1°
22.0°
30°
d
e
f
g
127
29
8
___
17
4
__
5
13
___
85
e
c
sin 60° = __
a
RQ
cos 25° = ___
RP
y
sin u = __r
q
cos 48 = __r
e
sin 30° __
f
HI
cos 35° = ___
JI
x
__
cos u = r
c
5
ii
i
b
4
20
___
d
21
___
29
15
___
17
3
__
5
84
___
85
iii
iii
iii
iii
20
___
21
8
___
15
4
__
3
13
___
84
6
a
b
c
d
e
f
7
1.93 m (2 d.p.)
8
a
b
10.1 km (3 s.f.)
14.9 km (3 s.f.)
9
a
b
c
d
2.11 km
5.87 km
054°
7.98 km
25.9°
44.9°
69.5°
79.6°
26.9°
11.5°
10 a
b
473 m
1608 m
11 a
b
14.1 m (3 s.f.)
5.13 m (3 s.f.)
12 552 m (3 s.f.)
13 a
b
c
d
x = 14.82 cm
y = 10.09 cm
z = 44.99 m
a = 29.52 cm
b = 52.80 cm
14 a
i
ii
i
ii
i
ii
i
ii
b
c
d
0.577
0.577
1.11
1.11
−1.73
−1.73
0.249
0.249
sin x
∴ tan x = _____
cos x
15 a
b
c
i
1
ii 1
1
sin2 x + cos2 x = 1
16 Students’ posters
Exercise 15.8
1
a
b
c
2 cm
9 cm
8 cm
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
a
b
c
16 cm
_
5 √3
5_ ____
___
cm
=
3
√3
_
20 √ 3
20 _____
___
_
√3
=
3
103.68 cm2
4
24 √ 3 m
cm
3
The two angles add up to 360 degrees. Cos x and
cos (360 − x) are the same each time.
Tan 30 and tan 210 = 0.577
Tan 60 and tan 240 = 1.73
Tan 15 and tan 195 = 0.268
Tan 100 and tan 280 = −5.67
The two angles differ by 180 degrees. Tan x and
tan (x + 180) are the same each time.
_
Exercise 15.9
Exercise 15.10
1
a
b
1
2
AB = 13.856 cm (3 d.p.)
3
a
b
c
4
ABC = ACB = 38.9° and BAC = 102.1°
5
a
b
c
020°
281.9 m
98 668 m2
a
b
c
d
e
f
g
h
i
j
−cos 60°
sin 145°
−cos 44°
sin 10°
−cos 92°
cos 40°
sin 59°
sin 81°
cos 135°
cos 30°
6
a
b
3.5 m (1 d.p.)
DE = 6.1 m (1 d.p.)
2
7
QT = 16 cm
8
a
b
c
d
e
a
b
c
d
e
f
g
h
i
30, 150
90
45, 315
78.7, 258.7
150, 210
191.5, 348,5
109.5, 250.5
60, 240
104, 284
9
77.255 cm2
3
a
b
c
d
e
f
g
h
i
45
120
55
45
270
120
270
90
696, 384
4
30, 150, 210, 330
ABC = 16.2°
BC = 17.9 m
ABC = 59.0°
AB = 1.749 (3 d.p.)
Capacity = 4.05 m3
AOE = 72°
AOM = 36°
OM = 1.376 cm (3 d.p.)
0.688 cm2
6.882 cm2 (3 d.p.)
10 6.882a2 cm2
na 2
11 __________
360°
tan(_____)
2n
Investigation
1
Sin 30 and sin 150 = __
2
Sin 10 and sin 170 = 0.174
Sin 60 and sin 120 = 0.866
Sin 5 and sin 175 = 0.0872
The two angles add up to 180 degrees. Sin x and
sin (180 − x) are the same each time.
Cos 30 and cos 330 = 0.866
1
Cos 60 and cos 300 = __
2
Cos 50 and cos 310 = 0.643
Cos 15 and cos 345 = 0.0644
128
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5
6
a
b
c
d
e
f
30°, 150°
120°, 240°
18.4°, 198.4°
No solutions
60°, 300°
30°, 150°, 330°
2
3
a
b
c
d
6.96
8.58
25.3
38.8°
a
b
c
d
e
f
10.6 cm
5.73 cm
4.42 cm
5.32 cm
6.46 cm
155 mm
a
b
c
d
e
f
54.7°
66.8° or 113.2°
69.8° or 110.2°
25.3° or 154.7°
52.7° or 127.3°
50.5°
4
C = 63°
AC = 15.9 cm
CB = 21.3 cm
5
F = 25°
DE = 9.80
EF = 14.9 cm
6
R = 32.2°
P = 27.8°
QR = 7.0 cm
7
a
8
129
1
AC = 8.62 cm
2
DE = 22.3 cm
3
P = 53.8°
4
a
b
c
TU = 18.7 m
U = 32.1°
T = 52.9°
5
a
b
c
X = 60°
Y = 32.2°
Z = 87.8°
6
a
b
Return = 14.4 km
296°
7
51.6 m on a bearing of 274°
41.4, 60, 300, 318.6
Exercise 15.11
1
Exercise 15.12
Exercise 15.13
b
c
Y is opposite a side shorter than X, so
Y , X and therefore ,40°.
Y = 30.9° and Z = 109.1°
XY = 22.1 cm
a
b
c
ACB = 51°
ABC = 52°
AC = 32.25 mm
1
a
b
c
d
e
f
2
108 cm2
3
0.69 m2
4
42.1 cm2
5
a
b
c
30.6 cm2
325.9 cm2
1.74 m2
6
a
b
174 cm2
8.7 cm and 21.5 cm
7
a
b
Q = 22.6°
P = 53.1°
10.0 cm2
15.0 cm2
52.0 cm2
17.2 cm2
22.7 cm2
24.2 cm2
Exercise 15.14
1
a
b
c
AC = 25 cm
EC = 13.0 cm
27.5°
2
a
b
c
EG = √ 50 m
_
AG = √ 75 m
AGE = 35.3°
_
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
a
b
c
d
e
AC B = 53.1°
BC = 5 m
CD = 4.2 m
BM = 4.5 m
BCD = 65°
4
a
b
c
14.9 cm
15.2 cm
u = 11.4°
a
_
x2 + y2
5
b
c
d
√
13 a
b
c
d
e
f
Practice questions worked
solutions
1
90°
_______
√x 2 + z 2
___________
x2 + y2 + z2
√
1
2.9 cm
2
AC = 9.8 m, BC = 6.9 m
3
DAB = 47.9°
4
9.9 m
5
a
b
X = 10.1 m (to 3 s.f.)
y = 20.6°
a
b
i
ii
i
ii
iii
a
b
5.16 m
3.11 m2
8
a
b
7 cm
51.1°
9
a
b
c
(90°, 1)
−1
2 solutions
7
Window measures 250 cm by 350 cm
350
250
Drawing measures ____ cm by ____ cm
150
150
√
= 2.87 cm
2
12 cos 35° = 9.83 m
12 sin 35° = 6.88 cm
3
25
72
QX = 60 tan 4° = 50.3 m
78.3 m
250.3 m
257.4 m
077°
b
i
ii
iii
i
ii
90 – 25
= 65
72
72
tan x = ___ ⇒ x = tan −1(___) = 47.9°
65
65
4
5
AB = 107.3 km
PAB = 66.6
143.4°
5h
12 km/h
_
√3
1_ or ___
12 ____
12
4 √3
12 tan 35° + 1.5 = 9.90 m
12 = tan 50°
a ___
x
12
x = _________
= 10.07 m
tan 50°
b
Horizontal distance RS = 72 − 30 − 10.07
= 31.93 m
12
tan y = _____
31.93
12
⇒ y = tan −1( _____
= 20.60°
31.93 )
6
a
i
ii
b
60 × tan 40° = 50.3 m
60
_________
= 78.3 m
cos 40°
i
250.3 m
ii
√ 60 2 + 250. 3 2 = 257.4 m
iii
130
72
x
10 30, 150
11 a
________________
2
2
250
350
Diagonal length = (____) + (____)
150
150
Practice questions
6
30°, 330°
88.4°, 268.4°
153.4°, 333.4°
180°
30°, 210°, 150°, 330°
30°, 41.8°, 138.2°, 150°
____________
250.3
tan −1(_____) = 77° to nearest degree
60
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
7
a
b
8
AB2 = 32 + 82 − 2 × 3 × 8 × cos 15°
AB = 5.16 m
1 × 3 × 8 × sin 15° = 3.11 m2
__
2
a
8
d
20
5
2
1_
= ____
4 √3
_
= √ 32
b
_ 2
h 2 = 9 2 − (√ 32 )
= 81 − 32
= 49
_
h = √ 49 = 7 cm
b
7
tan −1(__) = 37.9°
9
a
b
c
(90°, 1)
−1
There are two, at x = 210° and 330°.
1
10 cos 300° = __
2
⇒ x = 30° or 180° − 30°
= 30° or 150°
2
2 tan x = 72
tan x = 36
x = 88.4°
2(tan x + 2) = 3
3
tan x + 2 = __
2
1
tan x = − __
2
x = 153.4°
d
e
cos x = −1
x = 180°
1_
tan x = ± ___
√3
x = −30° + 180° or 30°
= 30° or 150°
f
(3 sin x − 2)(2 sin x − 1) = 0
h
d = 32
√3
cos x = ___ ⇒ x = 30°
c
x
i
ii
60
___
= 12 km/h
√2
1 × ___
2
1 × __
__
__
(
2)
2
2
__________
_____
_
12
= _4
√3 × 1
√3
13 a
_
√42 + 42
9
11 a
100
____
= 5 hours
_
8
9
i
_
5
4
d2 =
b
2 or sin x = __
1
sin x = __
3
2
x = 41.8°. 180° − 41.8° or x = 30°,
180° − 30°
so, x = 41.8°, 138.2°, 30°, 150°
_____________________________
AB = √ 100 2 + 60 2 − 2 × 100 × 60 × cos 80°
= 107.3 km
ii
sin 80°
sin PAB _________
________
=
100
107.3
100 sin 80°
PAB = sin −1(____________)
107.3
= 66.6°
iii
131
360° − 150° − 66.6° = 143.4°
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 16
2
3
It is likely to show positive correlation,
because temperatures are lower at higher
altitudes, so there would be greater snowfall
near the top of the mountain. There are lots
of other factors, such as the direction of the
prevailing wind, the gradient of each slope
and the overall climate of the region. The
correlation may be very weak; the points could
be scattered much more.
Students will come up with their own ideas,
but possible examples include:
The heights and masses of 100 people. This is
likely to be a positive correlation.
Temperature and sales of rain jackets. This
is likely to be a negative correlation, because
higher temperatures usually go with drier
weather.
Taking all of the schools in Japan, you could
plot the number of classrooms against the
number of miles from Mount Fuji. It is very
unlikely that these two values would show any
correlation.
a
Positive; weak
b
Zero correlation
c
Negative; weak
d
Negative; strong
Relationship between width
and length of leaves
200
180
Length (cm)
160
140
120
100
80
60
40
20
0
10 20 30 40 50 60 70 80
Width (cm)
b
d
3
Strong positive correlation.
40 cm
a
Relationships between mass of a dog
and duration of morning walk
40
35
Just because there is a correlation between two
variables, it does not mean that one of them
‘causes’ the other to change. In this situation
it is likely that higher temperatures will mean
that ice creams sell better, but also that more
people are on the beaches, leading to shark
attacks.
Exercise 16.1
1
a, c
Mass of dog (kg)
1
2
30
25
20
15
10
5
0
b
4
20
40
60
80
Duration of walk (min)
Zero correlation
a, c
300
Relationship between number of
assistants and queuing time
250
Waiting time (seconds)
Getting started
200
150
100
50
0
132
10
20
30
No. of sales assistants
40
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Strong negative correlation.
d
Value is outside the range of the collected
data and waiting time will be negative
time!
a
b
c
A = 122, B = 92, C = 56, D = 28, E = 200
Strong negative correlation.
Scatter diagram showing the relationship
between time watching TV and maths score
100
Maths score (%)
5
b
d
e
80
60
40
20
0
0
50
100
150
200
250
Time spent watching TV (min)
105 mins
The correlation is strong and Aneesh’s
score is within the range of the collected
data. This means the estimate is likely to
be reliable. It can never be exact, but it is
expected to be close to the actual value.
Practice questions
1
a, c
Relationship between price (£) and area
12 000
10 000
Price (£)
8000
6000
4000
2000
0
0
b
d
e
133
0.5
1.0
1.5
2.0
Area (m2)
2.5
3.0
3.5
4.0
Painting E because other paintings of a
similar size are much cheaper.
$6400
Value is outside the range of the collected
data.
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2
a, c
Comparison of 1st and 2nd year maintenance
Repairs in second year (y) (minutes)
5000
4000
3000
2000
1000
(f )
0
b
d
e
f
20
40
100
80
60
Maintenance hours (x)
120
140
Strong negative correlation.
1600 minutes
Repair time is a negative number – value
is outside the range of the collected data.
Approximately 130 hours – this is an
extrapolated value so might not be
accurate.
Practice questions worked
solutions
1
a
y
12000
10000
8000
Price ($)
6000
4000
2000
0
b
c
d
e
134
0
0.5
1
1.5
2 2.25 2.5
Area (m2)
3
3.5
x
E (circled above)
See graph.
1.5 × 1.5 = 2.25 m2 is approximately equal
to $6100
2.1 × 2.1 . 4, which is outside the range
of the data, where the pattern may change.
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
a
y
6000
5000
4000
Repairs
(minutes) 3000
2000
1000
0
0
20
40
60
80
100
120
x
Maintenance (hours)
b
c
d
e
f
Strong, negative, linear correlation
See graph
approximately 1750
You would need to extend the line, but it
would then predict a negative repair time.
approximately 12.2 hours
Past paper questions
1
5.665 < l , 5.675
2
10x + 5y = 15
x − 5y = 40
11x = 55
x=5
10 + y = 3
y = −7
x = 5, y = −7
135
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3
8
a, c, d
Area in cm = 32 × 24 0002
= 1.8432 × 1010 cm2
1 km2 = (1000 × 100) × (1000 × 100) cm2 = 1 ×
1010 cm2
Area = 1.8432 km2
y
60
55
50
45
40
Written test
area factor = 24 0002
9
35
x−5
3x
− 1 − 2x − 4
______________
= ______________
(x + 2)(3x − 1)
(x + 2)(3x − 1)
30
25
20
15
10
5
0
0
5
10
15
20
25
30
35
40
45
50 x
Speaking test
4
b
e
strong positive linear correlation
38
a
c
correct position of town B 9 cm from A at
angle of 140° from A
i
triangle constructed using points A
and B from part a and C 7 cm from A
and 5 cm from B, bearing of C from
A less than 140°
ii between 38° and 42°
2c − 3d
12x = 30
5
x = __
2
27 − 9x
d
× 38.64
77.28
2A = 2_________
_____
= ______ = 8.4
e
i
b
5
a
b
a+b
ii
5.5 + 3.7
9.2
3x = 5y
2y = x + 4
3x = 5y
3x = 6y − 12
6y − 12 = 5y
y = 12, x = 20
6
Pupper = 2(12.5 + 4.5) = 2 × 17 = 34
7
(x + 2)2 − 4 − 9
a = 2 and b = −13
136
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10 a
y
0
b
11 a
b
c
12 a
b
c
137
90°
180°
270°
1
tan x = __
3
x = 11.3°, 11.3° + 180°
= 11.3° or 191.3°
AC 2 = 6.42 + 10.62 − 2 × 6.4 × 10.6
× cos 102°
AC = 13.5 cm
10.6
BX = ______
______
⇒ BX = 8.6 cm
sin 44 ∘ sin 58 ∘
1 × (8.68 + 6.4) × 10.6 sin 78° = 78.2 cm2
__
2
_
1
13 × 24 + 20 × 24 + √ 231 × 24 + 2 × __
2
_
× √ 231 × 13 = 1354 cm2
150
1 × √_
__
f = cos −1(____
=
28.1°
231 × 13
2
170 )
× 24 = 2371 cm3
24.6°
360°
13 a
b
c
d
e
14 a
b
c
75.6
i
2a − 3b
3
ii __
4
i
x = −5
ii 20 − 12x = 23
12x = −3
1
x = − __
4
32x6 = 9x6
6x2 + 3xy − 10xy − 5y2 = 6x2 − 7xy − 5y2
130 + 80 + 170 + 50 = 530 m
100 2 + 150 2 − 120 2
cos u = _________________ ⇒ u = 52.9°
2 × 100 × 150
150
i
f = cos −1(____
= 28.1°
170 )
ii
d
x
360 − 28.1 = 332°
14 982 m2 = 1.498 hectares
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 17
Getting started
Exercise 17.2
1
1
2
3
Student answers will depend on how much
they already know. They can use the key
words in this chapter, the glossary at the end
of the book, a dictionary or online sources
(such as Investopedia) to find the meanings of
words they don’t know.
a
b
40 + 5% = 42 or 1.05 × 40 = 42
40 − 10% = 36 or 0.9 × 40 = 36
a
100I
P = ____
RT
100I
R = ____
PT
100I
T = ____
PR
b
c
a Net
income ($)
net
b % _____
( gross )
B Willis
317.00
47
M Freeman
158.89
35
J Malkovich
557.20
43
H Mirren
383.13
42
M Parker
363.64
43
Employee
2
a
b
c
Mean weekly earnings: $836.63
Median weekly earnings: $853.30
Range of earnings: $832.50
3
a
Difference between gross and net income:
M Badru: 3954.52
B Singh: 724.79
Percentage of gross income that each
takes home as net pay:
M Badru: 69.3%
B Singh: 57%
Exercise 17.1
1
$49.50
2
$428.75
3
a
b
c
d
e
b
$13.50
$6.45
9.35
$12.15
$13.68
Exercise 17.3
1
Taxable
income
Annual tax
Monthly
tax
4
$2085.75
a
$98 000.00
$21 149.25
$1762.44
5
$474.30
b
$120 000.00
$27 309.25
$2275.77
6
$8250
c
$129 000.00
$29 829.25
$2485.77
7
Annie
$319.20
d
$135 000.00
$31 509.25
$2625.77
Bonnie
$315.00
e
$178 000.00
$43 856.75
$3654.73
Connie
$300.30
Donny
$403.20
Elizabeth
$248.85
2
a
b
c
138
i
ii
iii
Yes
No – he pays $6181.25
$6181.25 = $4681.25
+ (40 000 − 34 000) × 0.25
$67 616.75
i
She owes additional tax.
ii $238.25
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Types of tax
Investigation
Answers will vary from country to country, but the
basic definitions of terms are:
a Value added tax – tax added at each stage of
a product’s lifespan as value is added, in other
words a consumption tax levied on a product
repeatedly at every point of sale.
b General sales tax – tax levied on goods when
they are sold at retail outlets, calculated as a
percentage of the value of the goods.
c Customs and excise duties – customs duties
are also called import taxes, they are levied
on good imported to a country to try and
protect local industry that makes the same
items; excise duties are indirect taxes on the
sale of locally made goods or services, such as
alcohol, tobacco and energy.
d Capital gains tax – a tax on profit from the
sale of assets.
e Estate duties – a tax on wealth or assets that
are inherited.
f Property taxes – rates and other taxes levied
on real estate.
g Air passenger tax – a charge levied on
passengers (often older than 16) who fly out
of different airports, varies from place.
h Corporate tax – direct tax on the income or
capital of some businesses.
h
i
$1370.00
$190 500.00
3
4 years
4
7% p.a.
5
33 years 4 months
6
a
b
c
$32
$96
i
$40.80
ii $136.80
7
a
b
c
$11 700
£3700
15.4% (1 d.p.)
Exercise 17.5
1
a
b
c
2
$2850
3
a
b
$141.83
$2072
4
a
b
£301
33.5% (1 d.p.)
5
a
b
$3657.80
13.09% (2 d.p.)
$100
$60
$460
Exercise 17.4
Exercise 17.6
1
1
a
b
2
$88 814.66
3
$380 059.62 (2 d.p.)
4
a
b
2
139
a
b
c
d
e
f
g
h
i
$15.00
$12.19
$62.50
$312.00
$144.38
$108.00
$190.04
$72.00
$21 375.00
a
b
c
d
e
f
g
545.00
715.00
$1120.00
$1416.00
$1071.88
$1305.00
$803.85
$10 035.20
$9920.00
$4998.09
$5077.92
Personal finances
Students’ own discussion.
Exercise 17.7
1
a
b
c
8.207 billion
8.642 billion
10.629 billion
2
a
b
1882
1721
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
a
Time (days)
0
1
2
3
4
5
6
7
8
Total number of
microbes (millions)
1
2
4
8
16
32
64
128
256
b
y
Number of microbes (millions)
60
50
40
30
20
10
0
c
4
1
2 2.5 3 3.5 4
5
Time (days)
d
i
Approximately 6 million
ii Approximately 12 million
Just over 4 days
a
b
6.5 minutes
12 grams
5
$27 085.85
6
a
b
c
d
6
7
d
$2903.70
8
a
b
1
9
15 hours
a
b
c
2
Is it worth it?
a Around 20%
b Students’ discussion, could include model and
make of car, colour, mileage done, whether it
is maintained and serviced or not, whether or
not it is in accident.
c $12 600
a
b
c
d
3
140
Students’ own answers. Good advice is to
buy a low mileage one year old car. This has
already depreciated by about 20%, so you get
good value for your money.
d
7 137 564
10 years
x
Exercise 17.8
$10 120
$8565.57
$5645.41
$11 000(0.92)n
7
8
i
ii
i
ii
i
ii
i
ii
$5.00
25.00%
$50.00
10.00%
$0.30
20.00%
$0.05
16.67%
i
ii
i
ii
i
ii
i
ii
$100.00
25.00%
$0.10
13.33%
$0.25
5.00%
$0.65
10.00%
Percentage profit = 66.67%
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 17.9
1
a
b
c
d
$156.00
$400.00
$399.15
$500
4
a
b
5
$33.60
6
$647.51
7
a
b
c
$18.20
71.2%
$30 000.00
$2 977.53
$2 307.59
2
$840
3
$3225
4
$360
8
28.07%
5
$220.80
9
11%
6
$433.55 for 10 and $43.36 each
7
28%
10 a
b
$24 300
25.9%
8
a
b
11 a
b
619 173.64
13 years
12 a
Decreases by 16% every 2 minutes, so
exponential decay.
$41.32
37%
Exercise 17.10
Original
price ($)
%
discount
89.99
5
4.50
85.49
125.99
10
12.60
113.39
599.00
12
71.88
527.12
22.50
7.5
1.69
20.81
65.80
2.5
1.65
64.16
10 000.00
23
Original
price ($)
Sale
price ($)
89.99
79.99
11
125.99
120.00
5
599.00
450.00
25
22.50
18.50
18
65.80
58.99
10
10000.00 9500.00
5
2
Practice questions
1
a
b
$451.95
2.75 hours
2
a
b
$12
$14.40
3
8.5%
141
b
Savings
Sale
($)
price ($)
y
100
80
Temperature (°c)
1
2300.00 7700.00
60
40
20
%
discount
0
c
0
2
4 6 8 10 12 14 x
Time (minutes)
7.25 minutes
Practice questions worked
solutions
1
2
a
36 × 10.48 + 1.5 × 10.48 × 4.75 = $451.95
b
420.75 − 36 × 10.48
3
__________________
≈ 2 __ hours
a
$15 × 0.8 = $12
b
15 × 0.8 × 1.2 = $14.4
1.5 × 10.48
4
3
102
____
= 1.085 ⇒ 8.5% increase
4
a
35 × 1.25 − 25.55 = $18.20
b
18.20
_____
= 0.712… so 71% profit
94
25.55
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
(160 × 0.07 × 3) = $33.60
6
500 × 1.093 = $647.51
7
a
c
35 730.48
_________
= $30 000
1.06 3
35 730.48
_________
= $2977.54
12
35 730.48
_________
× 0.775 = $2307.59
12
8
14 875 − 10 700
______________
× 100 = 28.1%
9
2200 − 1950
___________
× 100 = 11.4%
11 a
b
142
y
100
80
2200
b
96
80.6 − 67.7
__________
× 100 = 16.00%
80.6
Same decay (16%) each time.
b
14 875
10 a
96 − 80.6
_________
× 100 = 16.04%
Temperature (°c)
b
12 a
18 000 × 1.35 = 24 300
60
40
20
14 300 − 18 000
______________
× 100 = 25.9%
24 300
0
100 000 × 1.210 = $619 173.64
100 000 × 1.213 = $1 069 932 after 13 years
c
0
2
4 6 8 10 12 14 x
Time (minutes)
7.8, 7.9 minutes
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 18
Getting started
(b)
12
1
(a)
10
a
b
c
d
e
f
2
a
b
c
d
AB: Penguin moving downwards through
air towards surface of the water; BC:
Enters water and continues downwards;
CD: At C, penguin turns and starts to
swim back upwards towards the surface.
6m
1 second
3.5 metres below the surface
5.5 seconds
Students’ answers will vary. However, it is
likely to curve up above the x-axis again
as the bird surfaces and then curve back
down as it dives under the surface again.
8
(c)
(a) y = x2 + 1
(b) y = x2 + 3
(c) y = x2 – 2
(d) y = –x2 + 1
(e) y = 3 – x2
6
4
2
−3
−2
0
−1
1
2
3
−2
−4
Set A: all symmetrical about y, all pass
through origin (0, 0) and all ∪-shaped.
Set B: all symmetrical about y, all pass
through origin (0, 0) and all ∩-shaped.
Set C: all symmetrical about y and all
∪-shaped, all have a y-intercept.
Set A and B: main difference is the width
of the graph, Set C, main difference is the
y-intercept and that affects whether or not
graph has x-intercepts.
If the coefficient of x is . 0 (positive)
the graph is ∪-shaped, if the coefficient
is , 0 (negative) the graph is ∩ shaped.
The value of the coefficient also affects
the width of the graph. As the coefficient
increases, the graph becomes narrower.
i
a is the y-intercept.
ii For a . 0, the graph is shifted
vertically upwards by a units and the
turning point is above the x-axis.
If a , 0, the graph is moved a units
down and the turning point is below
the x-axis.
2
(e)
−6
(d)
−8
f
When the value of the constant term
changes the graph moves up or down the
y-axis.
a
b
c
d
e
C
B
A
D
E
Exercise 18.1
1
143
x
−3
−2
−1
0
1
2
3
a
y = x2 + 1
10
5
2
1
2
5
10
b
y=
x2
+3
12
7
4
3
4
7
12
c
y=
x2
−2
7
2
−1
−2
−1
2
7
d
y = −x2 + 1
−8
−3
0
1
0
−3
−8
e
y = 3 − x2
−6
−1
2
3
2
−1
−6
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 18.2
1
12
x
x2
y=
− 2x + 2
−1
0
1
2
3
5
2
1
2
5
y
10
y = x2 − 5x − 4
8
y
y = x3 –2x + 2
6
6
4
5
2
4
−4
0
−2
3
−4
1
−6
0
4
6
8
−3
−2
−1
0
−3
−4
0
1
2
−3
0
5
1
2
x
−2
2
−1
2
1
2
−8
x
3
−10
2
x
x2
−5x
−4
y=
x
x2
−5x
−4
x2
− 5x − 4
−2
−1
4
1
0
10
5
0
−4
−4
−4
−4
10
2
−4
−8 −10
3
9
0
1
−12
2
1
4
−5 −10
4
5
6
16
25
36
−4
3
x
y=
x2
+ 2x − 3
x
y = x2 + 2x − 3
y
5
y = x2 + 2x − 3
−15 −20 −25 −30
4
−4
−4
−4
−4
3
y = x2 − 5x − 4 −10
−8
−4
2
2
1
−4
−3
−2
−1 0
−1
x
3
4
−2
−3
−4
−5
144
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4
x
y=
−x2
− 4x
0
1
0
−5
y
2
3
4
−12 −21 −32
Exercise 18.3
1
a
y
y = –x2 – 4x
0 0
1
2
3
4
(0, 0)
x
–5
x
(1, −3)
–10
y = −3x2
–15
–20
–25
b
–30
y
y = 3x2
–35
(1, 3)
5
x
−6
−5
−4
−3
y = −x2 − 6x − 5
−5
0
3
4
x
−2
−1
0
y = −x2 − 6x − 5
3
0
y = x2 − 6x − 5
c
y
1
y = x2
2
−5
5
x
(0, 0)
y
(1, 0.5)
4
x
(0, 0)
3
d
2
y
1
−8
−6
−4
−2
0
y = x2 − 4
x
(−2, 0)
(2, 0)
−1
x
−2
−3
(0, −4)
−4
−5
6
a
b
c
d
e
6m
2 seconds
3 seconds
4.5 m
The water surface is at h = 0.
e
y
(0, 9)
(−3, 0)
(3, 0)
x
y = −x2 + 9
145
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f
b
y
y
y = 3x2 − 12
(−2, 0)
2.5
x = −1
2.0
(2, 0)
x
1.5
1.0
0.5
(0, −12)
g
y
y = −2x2 +
( 1 , 0)
2
1
2
x
−1.0
1
(0, )
2
1
(− , 0)
2
0.5 1.0
−1.5 −2.0 −1.5 −1.0 −0.5
−0.5
−1.5
x
−2.0
−2.5
c
2
a
y
x = −1
y
x = −3
4
6
3
4
2
2
1
x
−10 −8
−6
−4
−2
2
x
4
−5
−2
−4
−3
−2
−1
1
−1
−4
2
−2
−6
−8
d
y
4
−10
2
−12
−14
x
−6
−5
−4
−3
−2
−1
−2
1
−4
−6
−8
x = −3
146
−10
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e
h
y
y
30
x = −2
25
−5
−4
−3
x
−2
−1
−1
20
−2
15
−3
10
−4
5
−5
x
−4
−2
−5
2
4
6
8
10
−6
−10
3
a
b
c
d
y = −x2 − 4x + 5
y = 4 − x2
y = x2 − 3x − 4
y = x2 − 2x − 3
4
a
b
c
d
(20, 0)
0 < x < 20
−10 < h < 0
x=3
−15
f
y
5
x=2
4
3
2
h
0
1
−1
1
2
3
4
20
30
40
50
−2
x
−2
x
10
5
−4
−1
−6
−2
−8
−3
−10
−4
−12
g
y
2.0
e
f
x = 0.5
1.5
Exercise 18.4
1.0
1
0.5
x
−1.0 −0.5
−0.5
−1.0
147
Width = 40 m
Max height = 10 m
0.5
1.0
1.5
2.0
2.5
a
x
2
y = __
x
−6
1
− __
3
−4
x
1
2
2
y = __
x
2
1
−0.5
−3
2
− __
3
−2
−1
−1
−2
3
2
__
3
4
6
1
__
3
0.5
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2
y
1 23
1 13
1
2
3
1
3
x
–6 –5 –4 –3 –2 –11 0
–
1
2
3
4
5
6
3
–2
3
–1
–113
–123
–2
b
x
−5
−4
1
y = − __
x
0.2
0.25
−3
1
__
3
−2
−1
1
2
0.5
1
−1
−0.5
3
1
− __
3
4
5
−0.25 −0.2
y
1
y=− x
1.00
0.75
0.50
0.25
x
−5 −4 −3 −2 −1 0
0.25
1
2
3
4
5
0.50
0.75
−1.00
c
x
6
y = − __
x
−6
−4
−3
−2
−1
1
2
3
4
6
1
1.5
2
3
6
−6
−3
−2
−1.5
−1
6
y
6
y=− x
5
4
3
2
1
−6 −5 −4 −3 −2 −1 0
−1
x
1
2
3
4
5
6
−2
−3
−4
−5
−6
148
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d
x
4
y = __
x
−6
2
− __
3
−4
−3
1
−1 __
3
−1
4
−2
−1
1
2
−2
−4
4
2
3
1
1 __
3
4
6
2
__
3
1
c
y
y
12
10
3
8
4
y= x
2
6
4
1
−6
−4
2
0
−2
2
4
6
x
−2.0 −1.5 −1.0 −0.5
−2
−1
0.5 1.0 1.5
2
x
−4
−2
d
y
−3
6
5
−4
2
4
a
8
y
3
6
2
4
1
2
x
−3 −2
x
−8
−6
−4
2
−2
4
6
1
3
4
−2
−4
e
y
−6
8
−8
6
8
2
−1
8
−2
b
−1
4
y
2
6
x
4
−15
−10
−5
5
10
15
−2
2
x
−8
−6
−4
−2
−2
2
4
6
8
−4
−6
−8
149
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f
b
y
y
6
6
4
5
2
4
x
−15
−10 −5
−2
5
10
3
15
2
−4
1
−6
x
−4
−8
a
b
x
20
40
60
80
100 120
y
12
6
4
3
2.4
y
2
240
y= x
14
−2
12
1
2
3
4
c
Graph is still disjoint but both curves are
above the x-axis on opposite sides of the
y-axis.
d
Division by 0 is undefined.
e
y = 0 (the x-axis) and x = 0 (the y-axis)
f
x = 0 and y = 3
g
i
y
0.5
10
−6
8
−4
6
−2
−0.5
2
4
x
6
−1.0
4
−1.5
2
0
−1
−1
−10
3
−3
−2.0
0
20
40
60
80
100 120 140
x
−2.5
−3.0
c
4
a
240
y = ____
x
ii
y
9
x
−4
−3
−2
−1
1
− __
2
y
1
___
1
__
1
__
1
4
x
1
__
1
2
3
4
5
y
4
1
1
__
1
__
1
___
4
16
2
9
4
4
8
7
6
9
16
3
2
x
−4
150
−3
−2
−1
0
1
2
3
4
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
h
Suggested answer: The equation y = ax−1
can be written with positive indices
a
as y = __
x . This is the standard form of
a reciprocal graph and it will give a
hyperbola with two curves in opposite
quadrants. The equation y = ax−2 can be
a
written with positive indices as y = ___2 .
x
The range for this function is all positive
numbers, so the two curves of the graph
will be above the x-axis. If y = −ax−2,
a
the equation becomes y = − ___2 and the
x
range will be negative numbers, meaning
that the two curves will be below the
x-axis.
3
a
x = −1.3 and x = 2.3
9
y
8
7
6
5
4
3
y = x2 − x − 3
2
1
Exercise 18.5
1
2
x
−3
a
x = −1 and x = 2
b
x = −2.4 and x = 3.4
c
x = −2 and x = 3
a
−1
0
1
−2
−1
y = −x2 − x + 1
−5
−1
1
x
0
1
2
y = −x2 − x + 1
1
−1
−5
−4
b
x = −2.6 and x = −0.4
y = x2 + 3x + 1
−2
−1
5
y
4
3
2
y = −x 2 − x + 1
1
x
−3
4
−3
y
0
3
−2
−3
1
2
−1
x
b
−2
1
−1
−2
−3
2
3
4
x
−4
−3
−2
−1
0
−1
1
−2
−3
−4
−4
−5
c
151
x = −1.6 and x = 0.6
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
a
6
3
y
5
y = 4 − x2 + 2x
4
a
b
c
x = 9.1 and x = 0.9
x = −2 and x = 4
x = 3.8 and x = −1.8
4
2
7
x
−1 0
−1
1
3
2
6
5
4
4
3
2
−2
−3
1
−4
b
5
i
ii
a
−5 −4 −3 −2 −1 0
−1
−2
−3
−4
x = −1.2 and x = 3.2
x = 0 or x = 2
5
y
4
5
a
b
c
x = −1, y = 9 and x = 7, y = 17
x = −1.64, y = −0.27 and x = 2.14, y = 7.27
x = 2, y = 5
6
a
(1, 6) and (5, 14)
b
( 3
1
1
2
3
−2
−3
−4
−5
−6
i
ii
iii
x = −1.2 and x = 3.2
x = −1.8 and x = 3.8
x = −1 and x = 3
Exercise 18.6
1
a
b
c
d
2
Students’ own graphs.
a (0, 0) and (3, 9)
b (−1.4, −1.4) and (1.4, 1.4)
c (2, 0)
152
y = −4
There are no points of intersection.
2
−3 −2 −1 0
−1
x
1 2 3 4 5
−5
y = x2 − 2x − 4
3
b
y = x2 + 2x + 3
9
8
1
−2
y
11
10
3
x = 2 and x = −1
x = 2 and x = −2
x = −2 and x = 1
x = 1.2 and x = −0.4
4
x
5
_
c
_
√ 14 _________
8 + 5 √ 14
____
,
(
3
) and (
_
_
8 − 5 √ 14
− √ 14 _________
_____
,
3
3
)
_
)
2
_
_
1 − 3 √5
________
, 5 − 3 √5 )
(
2
_
1 + 3 √5
________
, 5 + 3 √5
and
Plotting simple cubic graphs
Students should find that increasing the value of a
makes the graph narrower and the value of d is the
y-intercept. There is only ever one x-intercept and
√
____
d
this is at the point − __
a.
3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 18.7
1
x
a
y = 2x3
b
y = −3x3
c
x3
d
y=
−3
−1
−54 −16
−2
0
2
24
3
0
−3
−29 −10
−3
−2
−1
6
25
−51 −13
1
3
5
19
57
−45 −16
−3
0
−1
0
9
9
43
81
−2
y=3+
−2
2x3
0
1
e
y=
f
y = 2x3 − 4x + 1
−41
−7
3
1
−1
g
y = −x3 + x2 − 9
27
3
−7
−9
−9
−44 −15
−2
1
0
h
y=
x3
x3
a
−
−
2x2
2x2
60
+1
2
3
16
54
−24 −81
−13 −27
1
10
c
y
40
40
30
y = 2x3
20
20
−3
−2
y
10
x
0
1
−1
−20
2
y = x3 − 2
3
−3
−2
−40
−1 0
−10
1
x
3
2
−20
−60
−30
b
100
y
−40
80
d
60
60
y = −3x3
40
50
20
−3 −2
−1 0
−20
y
40
1
2
x
3
30
y = 3 + 2x3
20
−40
10
−60
−80
−100
−3
−2
−1 0
−10
1
2
3
x
−20
−30
−40
−50
−60
153
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e
h
y
10
−3
−2
y
−1 0
−10
1
2
y = x3 − 2x2 + 1
10
y = x3 − 2x2
3
x
−3
−1 0
−10
1
x
2
3
−1 −0.5
0
0.5
1
+ 8x −15 −5.6
0
2.6
3
1.5
2
2.5
3
3.5
1.9
0
−2
−20
−20
−30
−30
−40
−40
−50
−50
f
2
y
50
x
y=
40
30
−3
−2
a
y = 2x3 − 4x + 1
x3
−
6x2
−
6x2
x
20
y=
10
x
4
4.5
5
y = x3 − 6x2 + 8x
0
5.6
15
−1 0
−10
x
1
2
3
x3
b
20
−20
−30
10
−40
5
g
40
−2
y
1
2
3
4
x
6
5
−20
c
10
−1 0
−10
−1 0
−5
y = x3 − 6x2 + 8x
−15
20
1
2
3
x
3
i
x = 0, x = 2 and x = 4
ii
x = 0.7, 1, and x = 4.3
a
−20
x
−4
−3
−2
−30
x3
y = ___
10
−6.4
−2.7
−0.8 −0.1
−40
y = 6x − x2 −40
x
−27
−16
2
3
4
0.8
2.7
6.4
y = 6x − x2 8.1
9.1
8.1
x3
y = ___
10
154
y
−10
30
y = −x3 + x2 − 9
−2
−1.9 −3 −2.6
15
−50
−3
+ 8x
−1
−7
5
0
1
0
0.1
0
5
6
12.5 21.6
5.1
0
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
10
y
y = −x2 + 6x
8
6
4
2
−10 −8 −6 −4 −2 0
−2
2
4
6
8
x
10
−4
3
y= x
10
−6
−8
−10
4
b
x = 0 and x = 4.2
a
The square root of a negative number is
undefined, so x cannot be negative and
the domain is x > 0.
b
i
y
12
10
y=4 x
8
y=3 x
6
y=2 x
4
2
−1 0
−2
ii
iii
c
1
2
3
4
5
6
7
8
9x
As a increases, the curve moves
further from the x-axis.
Negative values of a mean that the
curve is reflected in the x-axis.
Students’ sketches and notes will vary, but
the graph will be a reciprocal graph in the
first quadrant. Larger values of a move
the curve further from the origin and
negative values of a produce a reflection
in the x-axis.
Exercise 18.8
1
a
b
c
d
x
−3
2
y = 3 + x2 − __
x
1
__
y = 3x − x
2
y = − x + x2 + __
x
3
y = −x − 2x + 1
−2
−1
−0.5
−0.2
12.7
8
6
7.3
13.0
N/A −7.0 −0.8
2
6.5
11.3
−8.7
−5.5
−2
0.5
4.4
N/A −4.4 −0.5
2
5.5
8.7
11.3
5
0
−3.3
−9.8
N/A
3.8
2
3.5
6.7
13
4
2.1
1.4
0.6 −0.1
−2
34
0
1
0.2
9.8
0.5
1
2
3
−11.5 −32
Note: The y-values are rounded to 1 decimal place.
155
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
a
2
y = 3 + x 2 − __
x
Exercise 18.9
1
25
a, b
30
20
25
15
10
5
−15 −10 −5 0
−5
−10
y = 3−x
10 15
5
2
y = 3 + x2 − x
20
10
5
1
y = 3x − __
x
−4
1
y = 3x –x
8
2
4
−1
−3 −2 −1 0
−2
1
2
3
x
0
−5
1
3
2
x
a
i
ii
2
0.8
b
10
−4
y
8
−6
−8
y =10
x
6
−10
2
y = − x + x 2 + __
x
4
2
y
10
x
8
–0.2
6
0
0.2
0.4
0.6
0.8
1
x = 0.67 (but allow 0.66 − 0.68)
4
2
2
y = −x + x2 + x
x
1 2 3
−3 −2 −1 0
−2
3
−4
−6
a
b
c
d
2
5.3 hours
64
20 hours
4
−8
−10
400
y = −x3 − 2x + 1
40
y
30
20
10 y = −x3 − 2x + 1
−3 −2 −1 0
−10
1
2
3
x
Temp (°C)
d
4
The two graphs are symmetrical about the
y-axis.
2
12
−2
c
6
c
−3
y
10
y = 3x
15
−15
b
y
300
200
−20
−30
100
−40
0
1
2
3
4
Time (min)
156
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5
a
iii
18 000
y
1.0
16 000
Population
14 000
0.5
12 000
10 000
−3
8000
6000
2000
0
1
2
3
Time (months)
4
5
3
−2.0
−2.5
b
c
3 months
64 000
a
Instructions will vary, but should include
determining whether the graph is
increasing or decreasing using the value of
a. If a is positive, the graph is decreasing;
if a is negative, the graph is increasing.
Use a + q to determine the y-intercept.
Work out the asymptote by finding the
line y = q.
If a , 0, the graph is below the asymptote
and if a . 0, the graph is above the
asymptote.
i
y
−3.0
c
i
ii
Many possible answers. For example,
y = 4 × 2(x + 0) + (−6), with y-intercept
at −2.
Greatest possible intercept for these
values is 9. Many possible equations,
including y = y = 4 × 2(x + 0) + 5
Exercise 18.10
1
a
9
−2
y
y = x2
8
2
−3
2
−1.5
0
b
1
−1.0
4000
6
x
0
−1
−0.5
−2
7
x
−1 0
−2
1
2
6
3
5
4
−4
(ii)
−6
(i)
3
2
−8
1
−10
ii
−3 −2 −1
y
6
4
b
0
1
2
3
x
i
4
ii −1.75
(−1.5, 2.25)
2
−3
−2
−1
0
x
1
2
3
−2
−4
−6
157
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
a
300
Population
2
y
200
100
0
1930
b
3
1940
1950
1960
1970
Year
1980
1990
f
g
h
i
49x6
−16x3
84x11
−80x4
2
a
b
c
d
e
f
6
3
32
−8
−36
960
3
(3, 27)
The gradient at point (1950, 170) is −4.4
people per year.
Rate of change of population in the
village in 1950.
a
10
y
8
y = x3 + 1
6
4
2
−3
−2
−1
0
A
1
2
3
x
Exercise 18.12
1
a
b
c
d
4x3 + 5x4
9x2 − 20x3
42x5 + 18x
x2 − 28x6
−6
e
−8
f
32
30x 4 − ___ x 3
11
−14x + 18x5
−2
−4
b
g
3
h
i
Gradients of tangents
1 For the curve x2 the gradient of the tangent at
any point is twice the value of x at that point.
2
For the curve x3 the gradient of the tangent at
any point is three times the value of x2 at that
point.
Exercise 18.11
1
158
a
b
c
d
e
4x3
6x5
9x8
12x2
24x
x
2010
2000
j
16
36x 2 + ___ x 7
3
11
−120x − 80x9
8x − 36x2 + 20x3
32
6
3
− ___ x 3 + __ x 2 − __ x
7
2
11
93
52
12
2
a
b
c
3
(1, 5) and (−2, −4)
4
_
9
(0, 0) or (√ 3 , −__)
4
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 18.13
1
2
a
b
c
d
e
f
g
h
i
5
−4
0
7
−3
8x − 4
21x2 + 2
x2 + x
m
a
b
c
d
e
f
g
h
i
j
k
l
m
n
2x + 2
5x4 + 8x3
2x − 1
2x − 9
16x3 + 24x2
−10x + 20
4x + 5
6x − 7
24x + 23
12x − 13
42x − 44
2x + 6
8x + 4
18x − 12
o
3 2 __
6
__
x + x
p
q
r
67
4
(3 3)
9
1
y = ___ x − ___
20
16
2
( 19 , 0)
3
0.25
4
26
___
( 9 , 2)
34
___
1
a
b
c
d
e
(2, −3) min
(−3, −13) min
(4, 14) max
(2, −8) min
(1, −1) max
f
(− 2 , − 4 ) min
3
__
13
___
g
( 10 , 20 ) max
h
i
j
k
(−2, 15) max and (2, −17) min
(0, 3) min and (4, 35) max
(2, −4) min
(0, −25) min
l
( 4 , − 8 ) min
89
3 ___
___
3
__
9
__
m (3, 81) max and (4, 80) min
n (0, 0) min and (2, 4) max
2
m
y
(3, 81)
(4, 80)
x
0
( 3 , −3)
6
(1, 5) and (2, −4)
7
(2, 11) and (−2, −5)
8
a
1
__
b
a
b
n
y
a = 2, gradient at x = 4 is 92
dy
___
at x = −3 is 50
dx
Exercise 18.14
159
e
2 , __
1
__
5
1
y = 56x − 144
y = 18.25x − 19.25
Exercise 18.15
5
5
14 x 6 + x 5
___
3
10x − 20
2x
3
c
d
(2, 4)
0
3
x
y = 6x − 9
y = −4x − 4
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
a
b
dh
___
= 7 − 10t
3
dt
2.45 m
a
b
p = −10, q = 6.3, r = 9.2
x
0.6
1
1.5
2
2.5
4
a
54 thousand
y
5
a
Length = 2 − 2x and
width = 1 − 2x
V = length x width x
depth = x(2 − 2x)(1 − 2x)
The width is only 1 m and we are
subtracting two lots of x from this length.
So we can only subtract something less
than 0.5.
x = 0.211, V = 0.192
x
3
3.5
4
4.5
5
y
0.3
1.9
3.8
6.3
9.2
3
4
b
c
a
b
y
8
6
4
2
2
A: x = −2
B: y = −x
C: y = x2 − 2
D: y = 2x + 1
i
ii
−5.9 −3.7 −2.3 −1.1
10
Practice questions
1
−10
0
1
2
5
x
−2
(−2, 2)
(3, 7) and (−1, −1)
−4
c
(− 3 , 3 )
d
D
e
C
a
x
−2
y
7
x
0.5
1
1.5
2
y
3.25
4
5.25
7
1 __
1
__
−6
−8
b
y=x
2
−1.5
−1
5.25
9
−0.5
4
y
3.25
−10
0
3
4
c
d
x = 2.9
Gradient = 6
a
b
c
d
vi
ii
i
iv
8
7
y = x2 + 3
6
5
4
3
2
1
−3 −2 −1 0
c
d
160
1
2
3
x
No, x2 will never equal x2 + 3
i
x = +2.4 or −2.4
ii x = +1.7 or −1.7
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
a
i
ii
Practice questions worked
solutions
p = 160, q = 10, r = 2.5
160
M
1
140
a
120
100
b
80
c
60
40
20
0
b
t
0
1
2
x = 2.31 and y = 2.77
x = −0.65 and y = 1.78
7
a
b
c
y = 3(x +
x = −1
(−1, −4)
a
b
c
d
17
6x + 2
14
y = 14x −11
e
11
___
( 14 , 0)
9
4
5
iii Rate of change = 28.2
t=1
6
8
3
1)2
6
7
d
e
2
A: x = −2
B: y = −x
C: x2 − 2
D: y = 2x + 1
i
(−2, 2)
ii (−1, −1) and (3, 7)
2x + 1 = −x
3x = −1
1
x = __
3
1
1
__ __
(− 3 , 3 )
Graph D
C is symmetrical about the y-axis
a Missing values are 7, 3.25
b, d y = x2
−4
11
10
y=x
2
9
8
7
5
4
y = 2a − x
Area = x(2a − x)
A = x(2a − x) = 2ax − x 2
3
2
dA
___
= 2a − 2x
dx
2a − 2x = 0 ⇒ x = a
y = 2a − x = a
So all sides have length a and the
rectangle is a square.
161
y=6
6
−x − 21
y = _______
4
10 a
b
c
y = x2 + 3
12
1
−3
c
d
−2
−1
0
1
2
3x
Meet when x2 + 3 = x2
3 = 0 which never occurs
Solutions are:
i
2.4
ii 1.7
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
a
b
c
b
0.3 3
6
p = ____ − ___ = −9.982 = −10.00 to 1 d.p.
12
0.6
4.5 3
6
q = ____ − ___ = 6.3
4.5
12
53 6
r = ___ − __ = 9.2
12 5
5
a
ii
y
10
8
120
7
100
6
80
5
60
4
40
3
20
2
0
iii
1
2
3
4
5x
140
−2
120
−3
100
−4
80
−5
60
−6
40
−7
20
−8
0
b
−10
d
Point of intersection with x-axis is when
x = 2.9
Gradient is approximately equal to 6.3
a
b
c
d
(vi) Negative quadratic curve
(ii) Exponential growth curve
(i) Cubic graph with positive y-intercept
y-values are always negative
0
1
2
3
4
5
6
7x
0
1
2
3
4
5
6
7t
m
160
−1
−9
162
y
160
140
0
4
160
p = ____
= 160
20
160 ____
160
= 10
q = ____
=
16
24
160 ____
5
160 __
r = ____
=
=
6
2
64
2
9
1
c
i
Estimated gradient = −26
m = 160 − M
⇒ 160 − M = M
2M = 160
M = 80
160
⇒ 2t = 2
So, 80 = ____
2t
⇒t=1
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
6
y = 2x2 − 3x − 1 ①
3y − x = 6
②
Substituting ① in ②
3(2x2 − 3x − 1) − x = 6
6x2 − 9x − 3 − x = 6
6x2 − 10x − 9 = 0
_______________
10 ± √ 100 − 4 × 6 × − 9
x = _____________________
12
_
√
79
5
= __ ± ____
6
6
_
√
79
41
___
____
y=
±
18
18
7
8
a
b
y = 3x2 + 6x − 1
= 3(x2 + 2x) − 1
= 3[(x + 1)2 − 1] − 1
= 3(x + 1)2 − 4
x = −1
c
(−1, −4)
a
x=2⇒y=3×4+2×2+1
= 17
dy
___ = 6x + 2
dx
dy
___
at x = 2 = 6 × 2 + 2 = 14
dx
b
c
d
e
163
Gradient 14 at (2, 17)
y = mx + c
y = 14x + c
17 = 28 + c
c = −11
y = 0 ⇒ 14x = 11
11
x = ___
14
11, 0
so, (___
14 )
9
x = 5 ⇒ y = (5 − 3)2 = 4
Now y = (x − 3)2 = x2 − 6x + 9
dy
___ = 2x − 6
dx
dy
At x = 5, ___ = 10 − 6 = 4
dx
Gradient of tangent = 4
1
Gradient of perpendicular normal = − __
4
1=1
because 4 × − __
4
1x+c
y = − __
4
5
16 5 21
__
4 = − + c ⇒ c = ___ + __ = ___
4
4
4
4
1
21
So, y = − __ x + ___
4
4
10 a Perimeter = 2x + 2y = 4a
So, 2y = 4a − 2x
y = 2a − x
b
Area = xy = 2(2a − x) = A
A = 2ax − x2
c
dA
___
= 2a − 2x = 0 when A is maximum
dx
So, x = a when A is maximum
x = a and y = 2a − x
= 2a − a
=a
x = y = a and the rectangle is a square.
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 19
Getting started
1
It fits onto itself as it turns round a point, so it
has rotational symmetry.
3
a
b
c
d
2
2
Possible solution:
1
Possible solution:
Shape
Number of lines
of symmetry
Square
4
Rectangle
2
Equilateral triangle
3
Isosceles triangle
1
Scalene triangle
0
Kite
1
Parallelogram
0
Rhombus
2
Regular pentagon
5
Regular hexagon
6
Regular octagon
8
3
e
f
Answers will vary; check each other’s
answers.
Many options, but simplest solution is:
Exercise 19.1
1
164
a
b
c
d
e
f
g
h
None
CD, HG
CD, HG
AB
AB, EF
AB, CD
CD
AB, CD, GH
4
5
Students’ own answers.
Exercise 19.2
1
a
b
c
d
e
f
g
h
2
5
2
6
2
1
1
1
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2 b
c
d
Regular
polygon
Lines of
symmetry
Order of
rotational
symmetry
Triangle
3
3
Quadrilateral
4
4
Pentagon
5
5
Hexagon
6
6
Octagon
8
8
Decagon
10
10
Lines of symmetry = order of rotational
symmetry in regular polygons
Number of sides = lines of symmetry
= order of rotational symmetry in regular
polygons
3
Students’ own answers.
4
There may be some variation, depending on
the font chosen, but the most likely answers
are:
a ABCDEMUVWY
b HIOX
c HINOSX
5
2
1
Each has a rotational symmetry of 2
2
a
b
c
d
e
f
Infinite
1
2
8
Infinite
1
Exercise 19.5
1
a
b
c
2
Join OP and construct a line at right angles to
OP that will be the chord.
3
x = 43°
4
13.5 cm
5
AO = 9 cm
Area AOCB = 108 cm2
6
O is the centre of both concentric circles.
Construct OX perpendicular to AD.
∴ X is the mid-point of AD and BC
∴ BX = XC and AX = XD
AB = AX − BX = XD − XC = CD
7
a
b
c
8
10 √ 3 ≈ 17.3 cm
1
165
4
Infinite
Infinite
2 if base is a right-angled, isosceles triangle
2
2
Infinite
7
2
Exercise 19.4
Students’ own answers.
Exercise 19.3
a
b
c
d
e
f
g
h
i
AB = 5 cm
AB = 30 cm
AB = 2.4 m
17.3 cm
4.25 m
31.1 mm
__
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 19.6
1
a
b
c
d
x = 43°, y = 43°, z = 94°
x = 124°, y = 34°
x = 35°
x = 48°
2
a
b
x = 41.5
x = 38°
3
a
Tangents subtended from the same point
are equal in length.
i
CAB = 70°
ii DAC = 20°
iii ADC = 70°
b
Understanding the alternate segment theorem
1 Students should notice that the angle between
the chord and the angle in the alternate
segment is equal. They can compare their
diagrams to the one given to check they used
the correct angles.
2
b
c
d
Exercise 19.7
1
a
b
c
d
e
f
g
p = 50°, q = 65°, r = 65°
b = 80°
c = 30°, d = 55°, e = 45°, f = 45°
p = 85°, q = 105°
b = 60°
x = 94°, y = 62°, z = 24°
p = 85°, q = 65°
a
b
c
AOB = 2x
OAB = 90° − x
BAT = x
3
a
b
c
a = 70°
b = 125°
c = 60°, d = 60°, e = 80°, f = 40°
4
a
b
c
90° − x
180° − 2x
2x − 90°
5
a
Draw the chords AD and BC. ADX and
BCX are angles in the same segment, so
they are equal. Similarly, angle DAX is the
same as angle CBX. AXD and BXC are
vertically opposite angles, so they are the
same, too. This means that both triangles
contain the same three angle and so they
are similar.
AX . You can then
DX = ___
Using similarity, ___
CX BX
multiply through by CX and BX.
2
b
166
a
Any two angles drawn in the same
segment are equal, this means that you
can draw another angle in the alternate
segment, using the diameter as one of the
lines forming the angle and know that it is
still equal to y.
Triangle PAB is right angled as it is the
angle in a semicircle.
Angle APB must be 90 − y.
The angle between the diameter and a
tangent is 90, so 90 − y + x = 90, so x = y.
Exercise 19.8
1
a
b
c
d
e
f
2
Angle BTC = 180° − 30° − (180° − 60°) = 30°
because angles in a triangle add up to 180°.
So angle TDC = 30° by the alternate segment
theorem.
CTD = 180° − 60° − 30° = 90° (angle sum in a
triangle)
So CD is diameter because the angle in the
segment is 90°.
3
CTD = 90°
So TDC = 180° − 90° − x = 90° − x
So by the alternate segment theorem
CTB = 90° − x
But BCT = 180° − x
So y + 180° − x + 90° − x = 180°
So 2x − y = 90°
4
103°
120°
85°
80°
120°
90°
90°
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Practice questions
1
a and e
2
Order 3
3
a
7
x = 26
Angle at centre = 2 × angle at circumference
360 − 4x = 2(3x + 50)
360 − 4x = 6x + 100
10x = 260
x = 26
8
a
For example:
b
c
d
9
b
a
b
c
For example:
d
angle QSP = 60° (alternate segment
theorem)
angle SQP = 60° (angle sum of triangle)
angle PBQ = 60° (angle sum of triangle)
angle QRS = 140° (PQRS is cyclic
quadrilateral)
They are vertically opposite angles.
Angles in the same segment
From parts (a) and (b), angle BXA =
angle DXC and angle XAB = angle XDC.
This means that angle ABX = angle DCX.
So all three angles are the same and the
triangles are similar.
Triangles ABX and DCX are similar so
the ratio of sides AX and BX is equal to
the ratio of sides DX and CX.
DX
AX = ____
____
BX CX
so (AX )(CX ) = (BX )(DX )
10 a
i
ii
b
4
a = 90°, b = 53°, c = 90°, d = 53°
5
a
b
c
6
6
7
7
Angle BCD = 180° − 43° = 137°
Opposite angles of a cyclic
quadrilateral add up to 180°.
Practice questions worked
solutions
OPX = OQX = 90° (tangent perpendicular to
radius)
POQ = 150°
So, PXQ = 360° − 90° − 90° − 150° (angle sum
of quad)
Angle PXQ = 30°
1
a
b
c
d
e
2
167
i
ii
1 × angle = BOD =
Angle BAD = __
2
1 × 86° = 43°
__
2
Angle at centre = 2 × angle at
circumference.
Yes. The dotted lines are lines of
symmetry. There is rotational symmetry
of order 2 about the intersection of the
lines of symmetry.
Neither
Rotational symmetry only
Reflection symmetry only
Yes. Exactly the same as a. but with
rotational symmetry of order 4.
3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
a
7
Reflex angle AOC = 360° − 4x° because angles
at a point add up to 360°.
Angle at centre = 2 × angle at circumference
360° − 4x = 2(4x − 50°)
360° − 4x = 8x + 100°
12x = 260°
260°
x = _____ = 21.7°
12
8
a
b
60° by the alternate segment theorem.
180° − 40° − 60° = 80°
(Angles in a triangle add up to 180°.)
c
QPB = 60° (Base angles of an isosceles
triangle are equal.)
PBQ = 180° − 60° − 60° = 60° (Angles in
a triangle add up to 180°.)
180° − 40° = 140° (Opposite angles in a
cyclic quadrilateral add up to 180°.)
b
d
9
4
a
b
c
d
5
a
b
c
6
90°, because a diameter and tangent
always meet at a right angle.
90° − 37° = 53°
90°, because the angle in a semicircle is
always 90°.
180° − 37° − 90° = 53°
6
One through end faces.
One through centre of each rectangular
face.
6+1=7
7
a
b
c
Vertically opposite angles
Vertically opposite angles
ABX = DCX (Angles in the same
segment are equal.)
and
BAX = CDX
So triangles ABX and DCX have the same
angles.
d
BX because ratios of
AX = ____
____
10 a
DX CX
corresponding sides are equal.
So (AX )(CX ) = (BX )(DX )
i
ii
b
i
ii
1 × 86° = 43°
BAD = __
2
Angle at centre = 2 × angle at
circumference
BCD + 43° = 180°
BCD = 137°
Opposite angles in a cyclic
quadrilateral adds up to 180°.
XPQ = XQP = 90° because a radius meets a
tangent at 90°.
Total angles in OXPQ = 360°
So PXQ = 360° − 90° − 90° − 150°
= 30°
168
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Chapter 20
Getting started
b
c
d
e
f
g
2
fx means number of passengers (x)
multiplied by the frequency ( f ).
Find the product of x and f for each row.
32
Sum of fx
1.625
1
2
Answers will vary but should include that
the diagram shows a frequency diagram
(histogram) and a frequency polygon. The
histogram shows non-overlapping class
intervals on the horizontal axis and the
frequency on the vertical axis. The frequency
polygon is plotted at the class midpoints and
shows the shape of the distribution.
Frequency density
a
100 < n , 200
18
100
0.18
200 < n , 250
18
50
0.36
250 < n , 300
32
50
0.64
300 < n , 350
31
50
0.62
350 < n , 400
21
50
0.42
400 < n , 500
20
100
0.2
b
Number of sweets in jar
Freqency density
0.6
0.4
100
200
300
400
Number of sweets (n)
500
3
2
1
6
9
12
15 18 21
Mass, m (kg)
24
27
30
Mass of actors
a
b
c
a
b
c
d
0.2
0
169
4
5
0.8
4
3
a
No. of sweets Frequency Class Frequency
(n)
(f )
width
density
5
0
Exercise 20.1
1
Mass of children
6
Frequency density
1
2
e
18
16
14
12
10
8
6
4
2
0
60
65
Mass, m (kg)
70
Students’ ideas.
They both show the shape of the
distribution. In the histogram the larger
the area the greater the frequency for that
class interval, and in the stem-and-leaf
diagram the longer the leaves the greater
the frequency for that stem value.
When you have too many individual items
of data to list separately on a stem-andleaf diagram.
80
73
7
Body fat is too low for intense physical
activity.
No − the expectation is that soldiers are
physically active and therefore keep their
body fat at a satisfactory level.
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7
a
b
Age (a) in years
Frequency
0 , a < 15
12
4.0
15 , a < 25
66
3.5
25 , a < 35
90
35 , a < 45
90
45 , a < 70
50
Frequency density
6
b
156
a
No − frequency density and not frequency
given.
Yes − most of the bars are with the
boundaries of the speed limits.
b
c
0 < s , 50
240
50
4.8
50 < s , 65
320
15
21.3
65 < s , 80
500
15
33.3
80 < s , 95
780
15
52
95 < s , 110
960
15
64
110 < s , 125
819
15
54.6
125 < s , 180
638
55
11.6
8
170
a
2.5
2.0
1.5
1.0
0
140
c
d
9
Speed (km/h) Frequency Class Frequency
width
density
d
3.0
0.5
i
ii
Heights of students
150
160
170
180
190
Height (h cm)
200
210
150 –160
75.7
Answers will vary depending on the data that
students collect.
Check that students measure time in seconds
and collect the raw data by experiment before
they organise it into a frequency distribution.
A suitable scale might have a wider class at the
start (0–15 seconds) and end (.50 seconds)
with varied intervals between those values.
Histograms could be drawn on graph paper to
make it easier to work with frequency density.
240 below the minimum speed limit
15%
Height (h cm)
Frequency
140 < h , 150
15
150 < h , 160
35
160 < h , 165
20
165 < h , 170
18
170 < h , 180
22
180 < h , 190
12
190 < h , 210
12
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Exercise 20.2
a
b
c
Height in cm
5 < h , 15
15 < h , 20 20 < h , 25 25 < h , 40
Number of plants
3
7
10
5
Cumulative frequency
3
10
20
25
21−25 cm
Cumulative frequency
1
Exercise 20.3
30
1
a
b
c
d
e
30.0 cm
27.5 cm
33.5 cm
6 cm
29.5 cm
2
a
i
Paper 1: 48%
ii Paper 1: 28%
iii Paper 1: 52%
Paper 1: .66%
20
10
0
10
30
20
Height (cm)
40
Median = 21 cm
$36.25
p = 12, q = 24, r = 35
a
40
171
b
4
30
10
10
30
20
40
50
Amount spent ($)
60
b
c
d
10
20
30
40
Mass in kg
200
180
160
140
120
100
80
60
40
20
Speed
10 20 30 40 50 60 70 80 90100110120 130140
Speed (km/h)
Masses of children
50
45
40
35
30
25
20
15
10
5
i
45 kg
ii 330 girls
10%
a
20
0
b
c
a
Median amount spent $37
Cumulative frequency
3
3
Amount spent on books
0
d
b
Cumulative frequency
a
b
c
Cumulative frequency
2
Paper 2: 60%
Paper 2: 28%
Paper 2: 65%
Paper 2: .79%
50
Median = 102 km/h
Q1 = 92 km/h
Q3 = 116
IQR = 24 km/h
14.5%
60
19 kg
7
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Practice questions
26
24
22
20
18
16
14
12
10
8
6
4
2
0
2
a
Time taken by home owners to
complete a questionnaire
a
We say we are 29 right up until the day we
become 30.
b
y
80
Cumulative frequency
Frequency density
1
3
60
40
20
0
2
4
6
8
10
Time taken, t (min)
c
d
e
12
6
0
20 40 60 80 100 x
Age (year)
Approximately 43
Approximately 56 − 40 = 18
Approximately 12.5%
Practice questions worked
solutions
5
1
Time
Number
0<t,2
2
Frequency
3
density
2<t,3
18
2
3<t,4
25
4<t,6
12
6<t,9
5
9 < t , 15
2
4
0
b
c
172
0
5
10
15 20 25 30
Height (in cm)
35
Height (h cm)
Frequency
0 < h , 10
4
10 < h , 15
12
15 < h , 20
18
20 < h , 25
27
25 < h , 30
16
30 < h , 40
16
22.5 cm
40
Frequency density
1
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
Frequency
density
2
__ = 1
2
18
___
= 18
1
25
___
= 25
1
12 = 6
___
2
5
__
= 1.67
3
2 = 0.33
__
6
25
18
6
1.67
1
0
2
4
6
8
10
Time
0.33
12
14
16
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2
3
a
Additional bars have heights 0.4, 2.4, 3.6
b
5.4 × 5 = 22
3.2 × 5 = 16
1.6 × 10 = 16
c
4 × 5 + 12 × 12.5 + 18 × 17.5 + 22 × 22.5 + 16 × 27.5 + 16 × 35
_______________________________________________________
= 22.5
a
We know there are no further ages until past 30.
88
b
Age
Cumulative
frequency
0 < t , 30
2
0 < t , 40
20
0 < t , 50
47
0 < t , 60
65
0 < t , 70
77
0 < t , 90
80
90
Cumulative frequency
80
70
60
50
40
30
20
10
0
c
d
e
0
30 40 50 60 70 80 90
Age
approximately 52
approximately 25
approximately 28%
Past paper questions
1
a
b
2
0.892€ to the dollar
3
8300 × 1.0566 = $11 509.64
173
4
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4
a
i
ii
b
5
a
rhombus
x
−2
−1
0
1
2
3
4
y
−1
4
7
8
7
4
−1
b
y
9
8
7
6
5
2
4
3
2
1
−2
0
−1
1
2
3
4x
−1
−2
c
d
174
x=1
x = −1.8, 3.8
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6
7
a
b
c
d
e
f
2.35 + 4.45 = 6.80
10 − 3.4 − 2 × 0.85 = 4.90
i
34 × $8.25 = $280.50
ii $280.25 + 1.5 × $8.25 × 8 = $379.50
3.5 + 7 + 8 + 10.5 = 33 hours
$85.20
$13 891.50
a
x=2
b
i
ii
c
i
k
4 = __ ⇒ k = 8
2
8
8
y = __ = ____ = 0.032
x 250
x
−8
−4
−2
−1
y
−1
−2
−4
−8
ii
y
8
6
4
2
−8
−6
−4
−2
0
2
4
6
8x
−2
−4
−6
−8
8
d
y=x
a
180 − 2 × 62 = 180 − 124 = 56°
b
d
360
180 − ____ = 144°
10
x = 90° − 58° = 32°
y = 90° − x = 58°
28° CED is alternate with the given 28° angle
e
√ 212 + 282 = 35 cm
c
175
_______
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9
a
b
i
14a + 4a = 18a
ii 14a2
6, 9, 14
c
i
x
−6
−4
−3
−2
−1
1
2
3
4
6
y
−2
−3
−4
−6
−12
12
6
4
3
2
ii and iii
y
12
10
8
6
4
2
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6x
−2
−4
−6
−8
−10
−12
iv
x = 1.5
110°
10 x = _____ = 55°y = 24°
2
11 x2 − 7x + 5 = 0
x2 − 2x + 1 = 5x − 4
y = 5x − 4
12 a
i
Mass (k kg)
Cumulative frequency
176
k < 10
k < 25
k < 35
k < 40
k < 50
3
22
43
48
50
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ii
50
Cumulative frequency
40
30
20
10
0
0
10
20
30
40
50 k
Mass (kg)
iii
13 a
i
≈42
y
1
0
90°
180°
270°
360° x
−1
ii
b
c
rotational symmetry order 2 about (180, 0)
3
sin x = __ ⇒ x = 48.6°, 180 − 48.6° = 131.4°
4
i
(x + 5)2 − 25 + 14 = (x + 5)2 − 11
ii
y
14
O
x
(−5, −11)
177
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14 a
x
−3.5
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
y
−4.1
2
5.1
6
5.4
4
2.6
2
2.9
6
12.1
b
y
15
10
5
−3.5 −3 −2.5 −2 −1.5 −1 −0.5
0
0.5 1 1.5 x
−5
c
d
e
x = −3.3
x3 + 3x2 + 2 = −2x
x = 2.2
2,k,6
b
2.65
_____ × 100 − 100 = 6%
2.5
500 × 1.0157 = $554.92
c
1.10620 = 1.3736… so 37.4 % increase
15 a
_____
d
16 a
b
c
r=
2607
× 100 = 96%
√_____
6400
22
i
range = 27 − 20 = 7, mode = 21, mean = 22.7
ii
3
___
= 0.2143
14
nx − (n − 1)(x + 1) = nx − (nx + n − x − 1) = −n + x + 1
i
16.54
ii
25
Frequency density
20
15
10
5
0
0
5
10
15
20
25
30
35 t
Temperature (°C)
178
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17 a
(x − 5)(x + 5) _____
x+5
_____________
=
(x − 5)(x + 4)
b
(x + 5)(x − 1) + x(x + 8) _____________
2x + 12x − 5
_______________________
=
c
i
2
x(x + 1)
ii
iii
179
x+4
x(x + 1)
dy
___
= 6x 2 − 8x
dx
6(42) − 8(4) = 6 × 16 − 32 = 64
6x2 − 8x = 0
2x(3x − 4) = 0
4
x = 0 or x = __
3
98
y = 6 or y = ___
27
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Chapter 21
Getting started
1
a
b
c
3:2
4:9
3 : 10
2
a
36°, 72°, 108° and 144°
108
____
= 30%
360
Answers will vary but could include: cost
of flooring or building; water flow/usages;
salary or wages earned; cost of hiring;
speed of wind or athlete; cricket scoring;
heart rate for fitness; pressure exerted;
exchange rates.
Examples could include: litres/km;
pressure per square inch (psi); words per
minute; gigabytes per second and so on.
b
3
a
b
4
a
b
a = 1, b = 20, c = 12, d = 6 and e = 3.5
y = 4x
5
a
b
c
d
e
f
1 : 10
1 : 100
100 : 1
1 : 1 000
1 000 : 1
1 : 60
6
a
b
c
d
e
f
g
h
1:2
1:8
3:8
3 : 25
3 : 200
1 : 20
8:5
2 : 15
7
a
b
c
d
Length : width of a screen
19.5 : 9 = 39 : 18; 16 : 10 = 8 : 5; 21 : 9 = 7 : 3
24 cm
Students’ own investigation and
measurements.
Exercise 21.1
1
a
b
c
d
e
f
g
1:1
1:5
25 : 3
3 : 10
3 : 20
1:5
10 : 4 : 8 = 5 : 2 : 4
2
a
b
12 : 5
5 : 12
3
a
b
c
d
2:3
3:4
11 : 16
1:2
4
a
b
c
d
e
f
1 : 12
1:2
1:8
7:6
10 : 3
5 : 12
180
Exercise 21.2
1
2
a x=9
b
c
d
e
f
g
h
i
j
k
l
y = 24
y=2
x=6
x = 176
y = 65
x = 35
y = 180
y = 1 400
x = 105
x = 1.25
y=4
a
b
c
d
e
f
g
h
x = 15
x=8
y = 20
x = 2.4
x = 0.6
y = 3.25
x = 5.6
y = 7.2
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3
a
e
False: The order matters, so you cannot
just reverse the ratios.
True
20 4
3
False: ___ = __
and not __
4
15 3
False: 48 is 6 times 8, so the daughter’s
age must be 6.
True
a
b
c
d
1g
1.33 g
7:5
3:5
a
b
c
18 : 25 : 5
1.67 g
4.17 g
a
b
20 ml
2.5 ml
b
c
d
4
5
6
7
8
22.5 cm
9
b
40 : 160
1 200 : 300
15 : 35
12 : 48
150 : 450
22 : 16
220 : 80
230 : 460 : 1 610
0.3 l = 300 ml
3
Josh gets 27, Ahmed gets 18
4
Annie gets $50, Andrew gets $66.67 and
Amina gets $83.33
5
Students should draw a 16 cm line with 6 cm
marked and 10 cm marked.
7
181
4 p r 3 : 4p r 2
__
3
= 4p r 3 : 12p r 2
= 4p r 2 × r : 4p r 2 × 3
=r:3
1
2
6
pr 2 : 2pr
= pr × r : pr × 2
=r:2
Exercise 21.4
15 750 kg
a
b
c
d
e
f
g
h
1200 people
10 a
Exercise 21.3
1
37.5 cm
(i)
a
1 : 200
0.005 : 1
b
1 : 250
0.004 : 1
c
1 : 25 000
0.00 004 : 1
d
1 : 200 000
0.000 005 : 1
e
1 : 28.6
0.035 : 1
f
1 : 16 700 000
0.000 000 06 : 1
2
a
b
c
d
4m
6m
14 m
48 m
3
a
b
c
d
0.0012 m = 0.12 cm = 1.2 mm
0.0003 km = 300 mm
0.0024 km = 2400 mm
0.00151 km = 1510 mm
4
a
Rectangle with dimensions:
100 mm × 250 mm
Rectangle with dimensions:
80 mm × 200 mm
b
N (kg)
P (kg)
K (kg)
5
5.5 mm
a
0.25
0.375
0.375
6
12 : 1
b
1.25
1.875
1.875
7
0.9625 mm
c
5
7.5
7.5
8
d
6.25
9.375
9.375
a
b
c
1.8 m : 2.25 m : 1.35 m
(ii)
1740 km
1640 km
1520 km
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9
a
b
c
d
e
0.9 m
i
5.65 m
ii 4.05 m
iii 1.6 m
iv 1.98 m
i
11.34 m2
ii 9.3312 m2
iii 8.019 m2
1.44 m2
$59.39
conditions become terrible, such as in the
event of a natural disaster or war when
people will move away and population
density may be reduced.
3
Population density
Students may find information online, in social
studies textbooks or in an atlas.
1
a
b
c
2
a
b
182
Greenland is the least densely populated
area on Earth with 0.0259 people/km2,
the Sahel area in Africa has a population
density of 2 people/km2. The most densely
populated places on Earth are usually
small islands. Macau, for example, has
19 737 people/km2 and Singapore has
8000 people/km2.
Small areas with lots of people will
automatically have a high population
density. Generally, though, high
population densities are linked to ‘pull
factors’ that attract people to live in a
particular area.
Population density is a simple relative
measure that doesn’t tell you how spread
out people are in the area mentioned,
so social scientists consider distribution
patterns as well. For example, Nepal has a
population density of 203 people per km2,
but Nepal is in the Himalaya, so people
are not evenly spread throughout the
country. In fact, the capital Kathmandu
has a population density of 19 726 people
per km2.
Examples will vary. A healthy coral reef
would be densely populated with marine
animals, but if the coral bleached and
died or there was a tsunami, the animals
that lived on the reef would move away.
Not generally in the modern world as
people tend to be limited in their choices
of moving or not. But worldwide, there
is a trend towards urbanisation as people
move to cities because they think the
conditions there will be good for them.
Similarly, there are examples where
Answers will vary.
a Ecologists study how increasing or
decreasing population density impacts
biodiversity and use of resources.
b Epidemiologists may study how quickly
infectious diseases spread in areas of high
population density.
c Planners may use population density to
make decisions about energy use and
supply, fibre and data connections needed
and transport links.
Exercise 21.5
1
a
b
c
d
e
f
2.4 kg/$
0.12 l/km
$105/night
0.25 km/min
27 students/teacher
3 hours/hole dug
2
a
b
9600 t
48 000 t
3
a
b
120 l
840 l
4
7.4 minutes
5
12.75 km
6
a
b
805 km
76.67 km
7
a
b
3000 km
312.5 km
8
a
b
c
d
e
2 hours 40 minutes
2 hours 30 minutes
4 hours 26 minutes 40 seconds
1 hour 40 minutes
34 days 17 hours 20 minutes
9
110 km/h
10 18.7 km/h
11 a
b
37.578 km/h
40.236 s
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Exercise 21.6
1
a
b
c
d
700 m
7 min
09 07 and 09 21
Going to the library
2
a
b
c
45 min
17 54
17 15
3
a
Distance (km)
8
6
4
2
2
0
20 40 60 80 100 120 140 160 180 x
Time (minutes)
18 minutes
16 minutes
a
b
c
d
10 km/h
3 min
3.33 m/s
a
For the first 50 minutes the taxi travelled
a distance of 10 km at 12 km/h, then was
stationary for 50 minutes, then took 20
minutes to return to starting point at
30 km/h. The taxi was then stationary
for 40 minutes, then travelled 5 km in 40
minutes at a speed of 7.5 km/h and was
then stationary for 40 minutes.
130 minutes – the graph is horizontal.
25 km
i 12 km/h
ii 10 km/h
iii 6 km/h
iv 6.25 km/h
30
25
20
15
10
5
0
b
c
3
a 6 min
Distance–time graph
Distance (metres)
4
Neo’s journey
y
10
0
b
c
iv The object is moving very quickly in the
direction of y at a constant speed, then
stops and is stationary for a while, then
continues in the same direction at the
same speed as before, then stops and
is stationary again. Example: a train
travelling from Valladolid to Madrid,
stopping at Segovia on the way.
v The object travels slowly at first, then very
quickly, then slowly again in the direction
of y. Example: an Olympic runner doing
interval training.
vi The object is moving at a constant speed in
the opposite direction to y, then it suddenly
changes direction and travels at a slightly
faster speed in the direction of y.
b
c
d
5 10 15 20 25 30 35 40 45 50 55 60
Time (seconds)
15 m
5m
Exercise 21.7
1
183
a and b
Answers will vary, examples:
i The object is moving in the direction of y
at a constant speed. Example: a heliumfilled children’s balloon released in a large
hall (with no breeze).
ii The object is stationary. Example: a
parked car.
iii The object is moving in the direction of y
at a constant speed, then suddenly changes
direction, moving at a much faster speed.
Example: a squash ball travelling towards
the court wall, hitting it then bouncing
back.
4
a
b
c
d
1500 m
2 m/s
He was stationary.
0.5 m/s
5
a
Other questions are possible, these are just
examples:
What is the total time taken to attain a
height of 16 m?
When was the helicopter descending?
When was the helicopter ascending?
During what time period was the vertical
speed the greatest?
At what speed was the helicopter
travelling between 2 and 4 seconds?
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Exercise 21.8
1
a
d
i
0 km/h
ii 3 km/h
iii 8 km/h
iv 2 km/h
From 10 00 until 10 20 the speed remains
constant at 8 km/h.
From 10 20 to 10 30 the speed drops
uniformly to 2 km/h.
i
0.375 km
ii 0.833 km
iii 7.58 km
3.64 km/h
2
a
b
c
2 m/s2
35 m
3.5 m/s
3
a
b
c
1 m/s2
100 m
15 m/s
b
c
1
c, d, e, f, i
2
a
a = 6, b = 15
b
6
y = __ x
5
Exercise 21.10
$6.75
2
60 min
3
70 s
4
172.5 kg
5
10.5 km
6
a
b
d
Exercise 21.11
1
a
b
2
i
100
ii 25
iii 8
250 cm
Number of
people
120 150 200 300 400
Days the water
will last
3
a
b
4
722.86 km/h
5
3 h 36 min
40
32
24
16
12
8 days
2 days
320 g flour, 64 g sultanas, 80 g margarine,
99 ml milk, 32 g sugar, 16 g salt
4:1
1
a
b
c
2
p and q are not inversely proportional because p ×
q is not constant.
3
a
4.5
y = ____
x
b
62.5
y = _____
x
c
2
y = __
x
d
0.28
y = _____
x
e
4.8
y = ____
x
4
a
b
c
d
k = 5120
y = 10
y = 23.70
x = 5.98
x
7
250 g
8
a
b
550 km
27 litres
5
9
a
b
13 ft
i
4m
ii 6.5 m
6
184
i
30 ft
ii 6.59 m
6.49 m
Exercise 21.12
Exercise 21.9
1
c
1.5
15
8
0.1
0.25
0.5
1
y
25
4
x
25
100
y
10
5
3.70
26
0.0625
64
1
50
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7
8
9
a
b
c
2.5
1000
0.125
a
864
F = ____
b
p=4
c
F = 32
8
p3
18 km/h
f
17.5 km
a
20
d = ___ t 2
9
35.6 m
1.25 s
b
c
9
400
a
b
10 6.4
c
d
11 60
12 a
b
c
e
False
False
True
e
13 5 h
14 16 666. 7N (16.7 kN to 3 s.f.)
f
15 a 2 °C
b As temperature varies inversely it will
never reach −1 °C.
16 a
b
40
6
10 a
x = 0.06 m
b
x = 0.72 m
c
m = 9 kg
d
Px 2
E = ____
h2
m = 116.7h kg
17 25%
Practice questions
1
Raja receives $40
2
300 cm = 3 m
3
a
b
4
9 cups
5
a
b
e
1.6 kg raisins
1.2 kg dates
15 5
10 2
start and ___ = __
of the set at the end.
25 5
8% increase
The number of cards must be divisible by
both 4 + 6 + 5 = 15 and 6 + 10 + 9 = 25.
The LCM of 15 and 25 is 3 × 5 × 5 = 75.
6
a
b
1.31 × 10−6 metres.
1.31 micrometres.
7
a
b
c
90 km/h
1080 km/h2
15 km
d
1 min
2 __
2
185
11 a
6
2 of the set at the
Bahram. He has ___ = __
c
F, because it is a straight line through the
origin.
B, because the y-coordinate of every point
on the line is the same.
E.
A. y decreases as x increases and it is a
curve. The answer is not C because this
graph shows a straight line.
Also A. y decreases as x increases and it is
a curve. The answer is not C because this
graph shows a straight line.
C could be e.g. y = 4 − 3x. It will be an
equation in the form y = mx + c, where c
is positive and m is negative.
D could be y = x2 + 3 or y = xn + c where
c is positive and m . 1.
1.536
R = _____
d2
b
R = 0.0423
c
1.536
d = ______
R
226 mm
1.15 ohms
d
e
√
__________
Practice questions worked
solutions
1
7 + 5 = 12
$96
____ = $8
12
Raja receives 5 × 8 = $40
2
25 × 12 = 300 cm
= 3m
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3
4
4 + 5 + 3 = 12
4.8
___ = 0.4
12
a 4 × 0. 4 = 1.6 kg
b 3 × 0.4 = 1.2 kg
×3 3 : 4 ×3
9 : 12
b
c
9
a
b
c
9 cups wholemeal
5
a
b
6
Initially distribution in fractions:
6 ___
5
4 : __
2 : __
4 : ___
1
___
:
i.e. ___
15 15 15
15 5 3
Final distribution in fractions:
10 ___
9
6
6 ___
9
2 : ___
___
:
:
i.e. ___ : __
25 25 25
25 5 25
Bahram has the same fraction in both.
9
___
25
___
= 1.08, i.e. 8% increase
1
__
c
3
Must be divisible by 15 and 25.
LCM of 15 and 25 = 75 cards.
a
118
_______
= 0.00131 mm
9 × 10 4
= 0.000 001 31 m
= 1.31 × 10−6 m
7
d
1.31 × 10−6 × 10 6 = 1.31 micrometres
a
1.5 km/minute = 1.5 × 60
= 90 km/h
1.5
___ = 0.3 km/minute/minute
5
15 + 5
______
( 2 ) × 1.5 = 15 km
1 minutes
2 __
2
4 × 0.5 = 0.4 km/minute
__
5
7.5 + 2.5
________
( 2 ) × 0.5 + 15 = 2.5 km + 15 km
= 17.5 km
b
c
d
e
f
8
a
d ∝ t2
d = kt 2
20 = k × 9
20
k = ___
9
20
⇒ d = ___ t 2
9
d
e
f
20
20 × 16 320
d = ___ × 4 2 = _______ = ____ m
9
9
9
_______
9 × 3.5
20 2
___
t = 3.5 ⇒ t = _______ = 1.25 seconds
20
9
√
F Straight line through (0, 0).
B y-coordinate the same for every point.
E y ∝ x 2 ⇒ y = kx 2
x = 0 ⇒ y = 0, so E
k
1
__
y ∝ __
x or y = x is a curve.
As x increases, y decreases, so A.
k
y = ___3 Also A.
x
C y = ax + b, with a . 0 and b , 0.
D y = bx2 + c, with b . 0 and c . 0.
10 x = km
30 = k × 5 ⇒ k = 6
a x = 6m
b x = 12 × 6 = 72 cm
c 0.54 = 6m
0.54
⇒ m = ____ = 0.09 kg
6
d E = kx2
P
P = kh2 ⇒ k = ___
h2
P x2
So ⇒ E = ____
h2
P x2
e 49P = ____
⇒ x 2 = 49h 2
h2
so x = 7h metres
7h
6m = 7h so m = ___ kg
6
k
11 R = ___2
d
k
a 0.096 = ___
16
k = 16 × 0.096
= 1.536
1.536
so R = _____
d2
1.536
b R = _____ = 0.0427 ohms
36
______
1.536
1.536
c d 2 = _____ ⇒ d = _____
R
R
________
1.536
d d = ________
= 226 mm
3 × 10 −5
1.536
e R = _____
⇒ R 3 = 1.536
R2
√
√
3
_
R = √ 1.536 = 1.15 ohms
186
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Chapter 22
Getting started
1
a
b
c
d
2
3
1 of your
It means you should keep ___
10
speed + 2 vehicle lengths between you and
the vehicle in front.
11 truck lengths
10 km/hr
There is a partial variation; the faster you
go, the more distance you need between
you and the vehicle in front.
a
b
c
Cost of the USB port
1.8 is the cost of 1 metre of cable
12.25 m
a
h = 3a − 12
b
12 + h
a = ______
3
h = 10.5 m
3m
c
d
3
a
b
c
d
x = 13
x=9
x=2
x = 11
4
a
i
ii
i
ii
b
5
a
b
a
b
c
d
e
f
g
h
2
187
a
b
c
d
4x = 32
x=8
12x = 96
x=8
x + 12 = 55
x = 43
x + 13 = 25
x = 12
x − 6 = 14
x = 20
9 − x = −5
x = 14
2
Students’ own heart rates.
3
a
b
c
As R increases, C increases; as R decreases
C decreases.
5
a
b
c
d
7
x = 17.5
28
___
=4
y=3
y = 12
y = 46
y = 70
Answers will vary. But for a HR of 70,
C = 70 × 80 = 5600 ml; The norm for C
is between 4 and 7 litres, so check that
students get an answer in that range.
Students’ own answers
85.7 beats per minute
4
x
__
= 2.5
x
x=7
x
7x
x
t = ____ + ____ = ____
320 240 960
52.5 minutes
Heart rates
1 Each value compares two different quantities:
ml/min, ml/beat, bpm.
Exercise 22.1
1
(s + 2q) cm
(4p + 3r) cm
3rs cm2
(4ps + 8pq + 6rq) cm2
Inversely proportional; as one value
increases the other decreases.
Artery: 0.0016
Venae cavae: 0.000 316
Arteriole: 1
As the diameter decreases the resistance
increases. The power of four means that if
the diameter is halved the resistance will
becomes 16 times as great.
When the arteries become narrower the
resistance increases, which means that
blood flow is reduced and the risk of
heart-related health problems increases.
Exercise 22.2
1
Child = 15.5 years and parent = 46.5 years
2
Silvia has 70 marbles; Jess has 350 marbles.
3
Kofi has $51.25 and Soumik has $46.25
4
$250 and $500
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5
9 years
d
6
Width = 15 cm and length = 22 cm
7
48 km
8
Pam = 12 years and Amira = 24 years
9
6.30 p.m.
e
f
g
h
10 50 km
i
Exercise 22.3
j
1
−8 and −5 or 5 and 8
2
t = 2 seconds
3
12
4
4 and 7
5
6 cm
c
6
8 cm
d
7
a
b
12 sides
n not an integer when the equation is
solved
e
8
a
Width of smaller rectangle = (x − 1) cm
Since two rectangles are similar:
1 : x = (x − 1) : 1
x(x − 1) = 1
x2 − x − 1 = 0
x = 1.62 or x = −0.62
Negative solution can’t work as a length
must be positive
Perimeter = 5.24 cm
b
c
d
9
2
b
11 7 or −2
3
b
c
d
e
f
a
d
14 6 or −4
e
15 2.75 cm
Exercise 22.4
x = m − bp
x = pr − n
m
x = __
4
c+b
x = _____
a
d − 2b − c
x = __________
m
x = 3by
p
x = __
m
np
x = ___
m
mk
x = ___
2
20
___
x= p
m − 3y
x = _______
3
4t − c
______
x=
4
y + 15
______
x=
3
5
__
x=
2
m
x = ___ + y
4c
a
x = 2r − __
pr
E
m = ___
c2
100I
R = _____
PT
2k
___
m= 2
v
2A
___
−a
b=
h
3V
h = ___
A
3V
____
h= 2
pr
__
m
x = √ __
a
_
x = √m + y
_
x = √n − m
_
x = √ ay
__
ac
x = ___
b
√
_
j
x = √a + b 2
__
n
x = √ __
m
m2
x = ___
y
2
a
x = ___
5
x = y2 + z
k
x = ( y + z)2
f
16 7 and 8
c
4
c
13 1.96 seconds
a
b
a
b
12 3 cm by 8 cm
188
f
0.836 seconds
10 (−7, −6, −5), (4, 5, 6)
1
a
√
____
g
h
i
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2
c
x = _____
(a − b)
a−m 2
m x = ______
( b )
y2 + 1
n x = ______
3
y − a2
o x = ______
2
2 + by 2
a
p x = ________
4y 2
a 38 °C
b 100 °C
c 0 °C
l
5
6
a
b
c
2.11
6.18
0.40
a
b
c
d
e
f
b−x
a = _____
1−x
L
a = _________
B+1+C
5b
a = _____
b−1
x(y + 1)
a = _______
y−1
3
−y
a = _____
y−1
_____
2
a = ______
m−n
√
√
__
2
E
c = __
m
3
a = √c 2 − b2
4
a
6
b
c
4x + z
______
2(b − a)
y = ________
3
2y
a = _____
1−y
7
189
√
x = ___
k
a
V = 2 010 619 cm3
b
V
r = ___
ph
√
√
___
___
b
A = 1.13 m2
c
11 a
b
√
___
4A
d = ___
p
SA = pr2 + 2prh
SA 1 ( 2
Sealant = ___ = __
pr + 2prh)
6
6
5.608 litres
Shadow maths
1 It means that as the time of day changes, the
length of the shadow (L) changes.
H , where a is
The function for this is L = _____
tan a
the angle of the light (usually the Sun)
2
Students may remember how to do this from
primary science lessons, but they can find out
how to use a shadow stick (basically a vertical
ruler) to develop a ratio that they can apply to
other objects. They may also discover online
calculators that use coordinates, time and
shadow length to determine heights.
Exercise 22.6
i f(2) = ii f(−2) = iii f(0.5) = iv f(0) =
a
8
−4
3.5
2
b
8
−12
0.5
−2
c
3
−5
0
−1
d
11
11
3.5
3
e
0
8
−0.75
0
f
6
−10
2
a
b
c
d
−5
−1
5
−17
3
a
b
c
d
0
−4
5
−3.9375
___
___
yz2
2E
v = ___
m
A = 1.13 m2
3
√
b
1
2x
y = ___ + 2
3
y = 3x − c
2A
r = ___
u
E = 49
10 a
_
d
5
9
a
c
Exercise 22.5
1
8
−1.875
−2
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4
a
b
c
d
0
−9
−2
5
5
a
b
c
16
16
1
6
4
x = __
3
1
x = __
3
a x=6
7
8
b
9
a
b
2a
2a + 4
8a
8a
11 a
b
9
x=2
12 a
b
c
15
3
1
2
3
190
4
gh(4) = 5, hg(4) = __
5
a −56 + 16x2 − x4
b 56 − 16x2 + x4
c −56 + 16x2 − x4
d 56 − 16x2 + x4
7
a
e
−25
3
__
2
7
− ___
34
1
__
3
−15
a
(x 2 + 36) 2
b
√ x 8 + 36
c
0
d
√ 76
b
c
9
fg(x) = x + 3; gf(x) = x + 3
b
fg(x) =
− 15x + 1;
gf(x) = 10 x 2 − 15x + 5
c
fg(x) = 27 x 2 − 48x + 22;
gf(x) = 9 x 2 − 12x + 4
50 x 2
a
b
c
d
4x2 − 36
fg(x) = ________ ;
3
16x2
gf(x) = _____ − 9
9
−2x
−4
16
−2
a
b
c
9x + 4
18x2 + 1
3456
25
26
7
26
29
5
8
a
d
726
____
d
Exercise 22.7
1
e
a
b
c
d
x = −2 or 3
x = −6
10 a
b
c
d
150
4
6
1
Domain: x > − __
4
Range: y > 0
d
_
_
1 , which is undefined.
hgf(1) = __
0
Exercise 22.8
1
a
x
__
7
___
1
√___
7x
b
3
c
3
d
e
f
g
h
i
j
√
_
x
x−3
_____
4
2(x − 5)
2x − 2
x
__
+2
3
2x − 9
______
2
4x − 2
______
2+x
_
√x − 5
3
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k
2
3
x −8
______
2
3
x+1
_____
l
x−1
g −1 (x) = 3(x + 44)
x
a i
f −1(x) = __
5
ii ff−1(x) = x
iii f−1f(x) = x
b
i
ii
iii
c
i
ii
iii
d
e
f −1 (x) = √ x + 1
ff−1(x) = x
f−1f(x) = x
iii
g
4
5
6
f−1(x) = g(x)
f−1(x) = g(x)
f−1(x) ≠ g(x)
f−1(x) = g(x)
a
b
c
8
20
11
a
−10
b
5x + 2
______
d
191
3
a
b
c
d
c
$2
2
16 5c coins and 34 10c coins
3
a
12T
b
11 hours
___
c
11 − T
Time = ___
6
20
x = 1.54
d
_
i
ii
iii
ii
f
3
i
ii
iii
i
1
f−1(x) = x − 4
ff−1(x) = x
f−1f(x) = x
x+7
f −1 (x) = _____
2
ff−1(x) = x
f−1f(x) = x
f −1 (x) = √ x − 2
ff−1(x) = x
f−1f(x) = x
x2 + 1
f −1 (x ) = ______
2
ff−1(x) = x
f−1f(x) = x
9
f −1 (x) = __
x
ff−1(x) = x
f−1f(x) = x
i
ii
iii
Practice questions
_
11 − T
Distance = 48(___
)
6
11 − T = 64
12T + 48(___
)
6
2 = 40 minutes
T = __
3
Lana cycled 8 km
4
9 cm × 13 cm
5
16
6
a
b
30
30
___
− 2 = _____
x
x+4
30(x + 4) − 2x(x + 4) = 30x
30x + 120 − 2x 2 − 8x = 30x
2x 2 + 8x − 120 = 0
x 2 + 4x − 60 = 0
2x + 4 = 16
7
84 cm
8
1_ − 1
x = ___
√y
z 2 (y + 1) 2
x = _________
(y − 1) 2
9
10 a
b
c
d
False
True
True
False
11 a
b
c
d
14
x = 1.26 or −0.26
x = 1.76 or −0.76
x=1
e
2
i − 56 __
5
ii 3
4
iii − 7 __
5
6
f
g
h
4−x
_____
3
63x2 − 99x − 32
25
x = ___
9
4
18x − 36x3 − 39x2 + 57x + 40
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12 a
36T = 24
2 = 40 minutes
24 = __
T = ___
36 3
2 = 8 km
Distance to station = 12 × __
3
1−x
1 − _____
1
+x
ff(x) = ________
1
−x
1 + _____
1+x
1+x−1+x
___________
1+x
____________
=
1+x+1−x
___________
4
y
1+x
b
c
13 a
b
c
f −1 (x)
xy = 13 × 9
2(x + y) = 44 ⇒ x + y = 22
x = 13 and y = 9
Rectangle is 13 m by 9 m
1−x
= _____
1+x
1−x
fffff(x) = _____
1+x
7
3−x
_____
4
4
5
n2 + (n + 1)2 = 545
n2 + n2 +2n + 1 = 545
2n2 + 2n − 544 = 0
n2 + n − 272 = 0
(n + 17)(n − 16) = 0 ⇒ n = 16
6
30
30
_____
= ___ − 2
3
14 − __
4
Practice questions worked
solutions
1
2
3
6x + 3(x + 2) = 24
6x + 3x + 6 = 24
9x = 18
x=2
White paint costs $2 per litre.
Let x = number of 5c coins.
Let y = number of 10c coins.
x + y = 50
⇒ 5x + 5y = 250
5x + 10y = 420 ⇒ 5x + 10y = 420
②−①
5y = 170
y = 34
x = 16
a
b
c
d
192
Distance = speed × time
= 12T
5
11 hours
1 __ hours = ___
6
6
Remaining distance = 64 − 12T
11 − T
Time taken = ___
6
64
−
12T
48 = _________
11 − T
___
6
so 88 − 48T = 64 − 12T
x
117 m2
2x
= ___
2
=x
x+4 x
a ⇒30x = 30(x + 4) − 2x(x + 4)
⇒30x = 30x + 120 − 2x2 − 8x
2x2 + 8x − 120 = 0
x2 + 4x − 60 = 0
b
①
②
7
(x + 10)(x − 6) = 0 ⇒ x = 6 or x = −10
so, x = 6 because x . 0
Therefore, total = 6 + 6 + 4
= 16
1 (x + 1)(3x + 2)
(2x − 1)(x − 1) = __
2
__
2
2x − 3x + 1 = 1 (3x2 + 5x + 2)
2
4x2 − 6x + 1 = 3x2 + 5x + 2
x2 − 11x − 1 = 0
_______________
11 ± √ 121 − 4 × 1 × − 1
x = _____________________
_
11 ± √ 125
= _________
2
2
_
√
11
−
125
But _________ , 0 ⇒ x − 1 would be
2
negative
_
11 + √ 125
So, x = _________
2
Perimeter = x + 1 + 3x + 2
__________________
+ √ (x + 1) 2 + (3x + 2) 2
= 84.6 cm
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8
√
1_ = ______
1
_ _
x + ___
√x
√x √y
_
_
(√ x ) 2
d
g(x) = 2g(x) − 1
g(x) = 1
4 − 3x = 1
3x = 3
x=1
e
y = 4 − 3x
3x = 4 − y
4−y
x = _____
3
x↔y
4−x
So f−1(x) = _____
3
f
3(4 − 3x)2 − 3(4 − 3x) − 4 − 4{4 − 3(3x2 −
3x − 4)}
= 3(16 − 24x + 9x2) − 12 + 9x − 4 − 16 +
36x2 − 36x − 48
= 63x2 − 99x − 32
g
h
4 − 3(4 − 3x) = 17
4 − 12 + 9x = 17
9x = 25
25
x = ___
9
3(3x2 − 3x − 4)2 − 3(3x2 − 3x − 4) − 4 −
(3x2 − 3x − 4)2
= 2(3x2 − 3x − 4)2 − 9x2 + 9x + 12 − 4
= 2(9x4 + 9x2 + 16 − 18x3 − 24x2 + 24x) −
9x2 + 9x + 12 − 4
= 18x4 − 36x3 − 39x2 + 57x + 40
12 a
1−x
1 − _____
1
+ x 1 + x − 1 + x 2x
ff(x) = _________ = ____________ = ___ = x
2
1
−
x 1+x+1−x
1 + _____
1+x
1_
+ 1 = ___
√y
1_
x + 1 = ___
√y
1_ − 1
x = ___
√y
_
9
√x + z
_
y = ______
√x − z
_
_
(√ x − z)y = √ x + z
_
_
y √ x − yz = √ x + z
_
_
y √ x − √ x = z + yz
_
√ x (y − 1) = z(1 + y)
√
z(1 + y)
x = _______
y−1
_
z 2 (1 + y) 2
x = _________
(y − 1) 2
10 a
b
c
d
11 a
b
fg(x) = (5 − x) − 5
= −x ≠ x
−1
So, f (x) ≠ g(x)
g(5 − x) = 5 − (5 − x)
=5−5+x
=x
−1
g (x) → 5 − x
fg(x) = −x
gf(x) = 5 − (x − 5)
= 10 − x ≠ fg(x)
f(−2) = 3(−2)2 − 3(−2) − 4
= 12 + 6 − 4
= 14
3x2 − 3x − 4 = −3
3x2 − 3x − 3 = 0
_____________
c
3 ± √9 − 4 × 3 × −1
x = _________________
_ 6
√
3
±
21
= _______
6
x = −0.264 or 1.264
3x2 − 3x − 4 = 0
_____________
3 ± √9 − 4 × 3 × −4
x = _________________
6
_
√
3
±
57
= _______
6
x = − 0.758 or 1.758
193
b
1−x
f−1(x) = _____
1+x
c
1−x
fffff(x) = f(x) = _____ because ff(x) = x
1+x
13 f(x) = 3 − 4x
a f(−1) = 3 − 4(−1) = 7
b y = 3 − 4x
4x = 3 − y
3−y
x = _____
4
x↔y
3−y
So, f−1(x) = _____
4
c ff−1(4) = 4
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
14 ______ = − 2
2x − 1
5 = −4x + 2
4x = −3
3
x = − __
4
194
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Chapter 23
Getting started
1
a
b
c
2
A – flipped (reflected) across the y-axis
B – rotated 90° clockwise about the origin
C – moved right and down; D – enlarged
A – reflection
B – rotation
C – translation
D – enlargement
i
A, B and C
ii D
a
b
c
2
a
y
4
2
Direction of
movement
y=0x
0
x
y
If number is
positive
−2
right
up
−4
If number is
negative
left
down
2
b
(4)
4
6
y
x=2
4
2
D
C
C′
D′
B
B′
A′
2
c
A
Q
M
−2
0
2
4
x
6
−2
N
−4
P
c
Exercise 23.1
1
y
4
a
2
y=1
x
−2
0
2
4
6
−2
−4
b
3
a
y
C′
D′
B′
3
2
1
E′
−3 −2 −1 0
195
B
C
A
D
E
1
x
2
3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
b
c
4
c
B9 = (−1, 3)
A is invariant – A and A9 are the same
point.
2
a and b
y
−2
x=1
P
Q
R"
Q"
0
−2
−2
S′
R′
Q′
2
R
y=2
S"
2
4
P"
a
b
c
d
e
90° clockwise about (−6, 2.5)
180° about (3.5, 2)
90° clockwise about (4, 0)
180° about (0, 0)
90° clockwise about (−4, −1)
3
a
Centre of rotation A; angle of rotation
90° clockwise
Centre of rotation point on line AC; angle
of rotation 180°
Centre of rotation point on line AC; angle
of rotation 90° clockwise
6
y
2
D D′
(a)
2
F′
E E′
b
y=1
1
E"
−3
b
−1
−2
F"
0
(c)
−1
1
2
c
x
3
4
D"
F is at (−2, 3)
F9 is at (2, 3)
a
b
c
No
No
Yes
Exercise 23.3
1
Exercise 23.2
1
2
x
a and c
F
x
0
P′
4
S
5
y
a
Image
a
Object
y
b
4
B
C
B′
Object
A
C′
Image
2
0
–2
b
2
x
2
a
A → B(− 6) A → C(3)
0
6
b
A → B( 0) A → C(− 6)
−7
1
c
A → B(0) A → C( 6)
5
−3
y
4
2
C
3
A
B
B′
−2
0
−2
A′
2
4
6
c
x
A
C′
B
b
C
a
d
196
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
b
y
9
8
7
6
5
C 4
(−1, 4)
3
2
1
A"
C"
A (3, 5)
B (2, 1)
A′ B"
x
C′
−6 −5 −4−3 −2 −1 0 1 2 3 4 5 6
−1
−2
B′
−3
5
6
3
y
7
6
5
a and b
4
y
9
N′ (2, 8)
8
6
1
M
N
4
O′ (2, 5)
4
5
P
O
1
−4 −3 −2 −1 0
2
3
4
5
6
Exercise 23.4
1
a
b
c
d
2
Scale factor 2; centre of enlargement
= (8, 0)
Scale factor 2; centre of enlargement
= (3, −2)
Scale factor 2; centre of enlargement (−3, 4)
1 ; centre of enlargement (0, 0)
Scale factor __
2
a
y
O
197
A
2
3
4
5
x
6
7
8 9
y
15
14
13
12
11
10
9
8
7
6
5
4
3
2 D′
1
x
1
1
1 ; centre of enlargement (0, −1)
Scale factor __
2
Scale factor 1.5; centre of enlargement (4, 2)
6
2
B
P
0
5
3
C
3
2
7
P′ (−2, 5)
x
O
X 9 (7, −1)
Y 9 (6, 4)
Z 9 (3, −7)
M′ (−2, 8)
y
0
A
D
A′
B
C
B′
C′
x
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
7
a
b
c
d
9.6 cm wide
Length will be tripled.
No; the image will not be in proportion.
2.5 cm long and 1.5 cm wide
8
a
b
Scale factor is 0.75
1.78 times smaller
x
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
9
a
b
Answers will vary depending on how
students view the shapes. There are
translations – the large coloured stars
are moved to the right (or left) along
a line. There are enlargements of both
the star shapes and the frames. There
are reflections – for example, the large
coloured stars could be reflections of each
other. The smaller shapes containing the
blue, orange, green and turquoise stars
bottom centre of the design could also be
rotations around the centre of the space in
the middle.
Answers will vary, but could include
translating the larger shapes with the
coloured stars down to fit the spaces in
the bottom row, or reflecting that entire
row so it fits into the open space.
b
y
5
4
3
2
1
−5 −4 −3 −2 −1 0
−1
2
a
b
c
2
3
4
5
x
−2
−3
−4
−5
−6
−7
3
a–c
y
Exercise 23.5
1
1
5
y = −x
y=x−1
y=2−x
B′
4
3
2
a
y
1
5
−5 −4 −3 −2 −1 0
−1
4
3
−2
A′
−3
2
C
1
−5 −4 −3 −2 −1 0
−1
−2
−3
−4
−5
1
2
3
4
5
x
B
C′
4
5
D
E
P
1 2 3 4 5 6
x
1
2
3
x
A
−4
−5
4
y
3
2
1
0
−7−6 −5−4−3−2−1
−1
F
−2
−3
−4
−5
−6
−7
198
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
8
y
y
5
6
J
A
4
5
4
3
K
N
2
C
B
2
1
−5 −4 −3 −2 −1 0
–1
B′
1
2
3
4
5
x
L
M
−3
2 M′3
4
5
x
N′
J′
−3
−4
−4
−5
−5
6
y
9
5
y
13
12 A′ A"
11
10
9 A
8
7
6
C" 5
C′
4 B′ B B" C
(2, 5)
3
2
1
F′
4
3
2
1
−5 −4 −3 −2 −1 0
−1
G′
E′
1
2
−2
3
4
5
x
−9−8−7−6−5−4−3−2−10 1 2 3 4 5 6 7 8 9
x
10
D′
13
12
11
10
P′′ 98
7
6
5
M′′ 4
3
2
1
−3
−4
O′′
−5
N′′
−6
7
1
−6 −5 −4 −3 −2 −1 0 L′ 1
−1
K′
−2
C−2
′
A′
3
y
y
Mʹ
M
N
Nʹ
Pʹ
P
O Oʹ
x
−9−8−7−6−5−4−3−2−10 1 2 3 4 5 6 7 8 9
5
4
3
M
Exercise 23.6
N
1
2
P′
1
Q
−5 −4 −3 −2 −1 0
−1
O′
O
1Q′ 2
−2
−3
−4
N−5
′
199
M′
P
3
4
5
x
a
(6)
b
(2)
c
( 2)
d
( 2)
e
(− 4)
f
(4)
g
(4)
h
(− 2)
4
4
−4
−4
6
0
8
4
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
P
M
M
e
B
Q
A
U
K
V
T
W
Q
a
b
4
⟶
AB = (4)
0
⟶
ii BC = (1)
3
They are equal.
i
a
(2)
b
(− 1)
c
(− 1)
d
(− 3)
e
−4
( 3)
f
(2)
a
b
c
d
e
f
2
a
b
c
200
E
N
J
3
L
a
(8)
b
(21)
2
9
F
L
⟶
DC = (4)
0
⟶
AD = (1)
3
c
(10.5)
d
(
e
( 6)
f
(− 84)
g
( 6)
4.5
0.75
3)
1.5
− 36
4
1.5
⎜ ⎟
5
⎛ __
5⎞
−
3
35
___
⎝− 9 ⎠
h
6
0
Exercise 23.8
1
5
Exercise 23.7
1
T
P
C
3
f
R
D
⟶ ⟶
2GO = 1_2GC
⟶
⟶
3DG = 1CL
⟶
⟶
6BE = 2CL
d
S
(− 21)
⎜ ⎟
b
(− 5)
12
3
(− 7)
3
a
( 8)
b
(24)
c
(− 12)
d
(0)
e
(12)
f
(21)
−6
( 14)
( 7)
−3
⎜ ⎟
⎛ __
9⎞
−
4
21
___
⎝ 4⎠
4.5
(− 10.5)
⟶
⟶
DF = 2JK
⟶ →
JQ = 1_4JF
⟶ ⟶
HP = 1_2HF
(− 6)
2
9
⎛ __
3⎞
2
7
__
⎝− 2 ⎠
a
4
12
12
8
−4
2
0
16
g
( 9)
h
(− 7)
a
b
c
2q − 2p
2p + q
p−q
10
−2
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
5
6
a
x+y
b
3
__
(x + y)
c
3
1 x + __
− __
y
4
4
a
2a + 3b
3b
a + ___
2
b
b
a + __
2
b
c
d
4
Exercise 23.9
1
2
a
b
c
d
e
f
g
4.12
3.61
4.24
5
4.47
5
5.83
13 a
a
b
c
d
10.30
13.04
5
10
3
a
b
c
5
13
17
4
a
A(4, 2), B(−1, 3), C(6,−2)
⟶
⟶
⟶
AB = (− 5), CB = (− 7), AC = ( 2)
5
−4
1
10
8.60
b
b
5
a
b
6
100 km/h
7
6.71 km/h (3 s.f.)
8
a
b
c
9
i
b−a
ii
1 (b − a)
__
iii
1 b − __
2a
__
iv
1 b − __
2a
__
3
3
6
3
6
⟶ ⟶
1 b − __
2 a) and
MN = NC (they are both __
3
6
they share a common point N, so the
points M, N and C are collinear.
Perpendicular vectors
1 Students can draw any perpendicular vectors
to use.
a
c
2 If your vectors are ( ) and ( ) work
b
d
out ac and bd. You will notice that for all
perpendicular pairs ac = − bd. You can also
write this as ac + bd = 0. The quantity
ac + bd is known as the scalar product of the
two vectors.
Practice questions
1
a
5
y
−p + q
b
2 −p + q
__
(
)
C
4
3
B
A
2
1
−4 −3 −2 −10 1
−1
a
c
201
b−a
3
⟶ ⟶ ⟶
CD = CA + AD
⟶
So CD = − 2a + 3b − a = 3b − 3a = 3AB
So CD is parallel to AB, so the triangles
are similar.
3
2
1
___ + ___
3 3
1p
q + __
2
a
__
10 a
2
b
b −__
2
a−b
c _____
2
3a + 3b
_______
d
4
11 (4, 5)
⟶
12 a AB = b − a
⟶ 1
b AM = __
(b − a)
2
⟶ 1
c ON = __
a
3
d
−2
x
2
3
4
5
6
7
D
−3
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
2
A: reflection about y = 0 (x-axis)
6
B: translation (− 3)
2
C: enlargement scale factor 2, centre origin
D: rotation 90° anticlockwise about the origin
3
a
a
B'(4, 16)
y
16
14
i
(− 3)
12
ii
( 3)
10
1
−6
b
B(2, 8)
A'(4, 10)
8
m
C'(10, 8)
A(2, 11)
C(5, 7)
6
4
2
4
5
a
b
(−1, 2)
Scale factor −2
0
b
3
(− 2)
Rotation 180° about centre (6, 0)
c
i
a
10 y
9
8
7
6
5
4
3
2
1
b
c
d
7
0
4
6
8
12 x
10
(0, 6)
2
4
a
10
y
9
8
7
6
x
−10 −8 −6 −4 −2 0 1 2 3 4 5 6 7 8 9 10
−2
−3
−4
−5
−6
−7
ii
2
C(9,5)
5
4
3
B(3,2)
2
1
4:1
–2 –1
b
a
A(3,1)
0 1 2 3
–1 D(0,–1)
4
5
6
7
8
x
9 10
–2
202
b
a−b
c
|a| = 3.16
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
8
a
i
ii
Translation (7)
3
Enlargement scale factor 3,
centre origin
Practice questions worked
solutions
1
y
iii
b
9
Rotation 90°, centre (2, 1) and
translation (− 3)
1
iv Enlargement scale factor −2, centre
(0, 4)
Shapes B, D
a
7
6
5
4
b
c
d
0
x
D
A′(3, 4)
(4, −7)
l (b)(i)
A(5, 0) x
−7 −6 −5 −4 −3 −2 −1
1 2 3 4 5 6
−1
(a)(ii) B′(3, 1)
−2
C′(1, −2)
−3
−4
−5
−6
−7
10 a
A
B
y
B(1, 3)
3
2 (a)(i)
C(–1, 2)
1
b
C
x
y = __
2
b+c
ii
b+c+d
⟶
⟶
DE = 2b and EC = − (b + c)
So, following the path from B to A to D to
E to C,
c = −b + (b + c + d) + 2b − (b + c)
c=d+b
⟶
AD = b + c + d
= c + c = 2c
So AD is parallel to BC and the
quadrilateral is a trapezium.
2
A: Reflection in x-axis
3
a
B: Translation with column vector (− 3)
2
C: Enlargement, scale factor 2, centre (0, 0)
D: Rotation 90° anticlockwise about (0, 0)
i
m + n = ( 3 − 2 ) = ( 1)
−3
−4 + 1
ii
3n = (3 × − 2 ) = (− 6 )
3×1
3
b
3
4
4
a
b
Centre is (−1, 2).
Scale factor is −2.
5
a
(− 2)
b
Rotation, centre (6, 0), 180°
c
i
3
y
(2, 2)
(−4, 2)
(10, 2)
R
S 0
(−4, −2)
ii
203
3
6
10
(−2, 2)
22 = 4
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
6
a
ii
A1 (4, 16)
y
iii
A (2, 11)
B1
(4, 10)
iv
C1
(10, 8)
B (2, 8)
(0, 6)
C (5, 7)
9
Enlargement, scale factor 3,
centre (0, 0).
Rotation 90° anticlockwise
about (0, 0).
2, centre
Enlargement, scale factor −__
3
(0, 7).
b
B, D
a
i and ii.
y
4
l
x
b
c
d
7
2
Centre is (0, 6), where the ray lines meet.
2
22 = 4
a
−2
0
2
4
6
x
−2
y
C(9, 5)
−4
10 a
b
B(3, 2)
1x
Equation of l is y = __
2
b+c
b+c+d
E
b
A(3, 1)
a
0
b
c
8
x
C
a − b = (3) − (3) = ( 0 )
2
1
−1
|a| =
_
√32 + 11
a
c
d
_
= √ 10 = 3.16
B
D
y
b
12
11
10
9
E
8
7
6
5
4
3
2
D 1
–3 –2 –1 0
i
204
A
C
B
d
A
1 2 3 4 5 6 7 8 9 10 11
Translation (7)
3
x
⟶
⟶
DE = −d + b + c = 2AB = 2b
2b = −d + b + c
So, a + b = c
⟶
AD = b + c + d
= c + (b + d)
=c+c
⟶
= 2c = 2BC
So, BC is parallel to AD.
⇒ ABCD is a parallelogram.
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
b
c
d
e
Chapter 24
Getting started
1
2
Statement A: Not correct. You cannot assume
that both outcomes are equally likely. Think
about what factors affect the weather.
Statement B: Not correct. It is a
misconception that a probability gives the
proportion of outcomes that will actually
happen. You can experiment to check this
statement, but you can also think about a
simpler example, the probability of heads is
0.5, but this does not mean you will get 10
heads if you toss a coin 20 times.
Statement C: Not correct. The probability of
1 . It is no harder
each number on a dice is __
6
to roll a 6 than any other number. You may
never have rolled four 6s in a row personally,
but you cannot base probability on personal
experience.
Statement D: Not correct. Probability is not
based on patterns of recent events.
a
b
c
d
e
3
a
b
n(A): number of elements in set A
A ∪ B: union of set A and B
A ∩ B: intersection of set A and B
n(A): number of elements in region A and
any intersections
A ∪ B: combine elements of set A and
set B with none repeated
Exercise 24.1
a
1st
draw
r
205
a
2nd
draw
r
b
g
b
r
b
g
g
r
b
g
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
b
c
AA
AB
AC
AD
BA
BB
BC
BD
CA
CB
CC
CD
DA
DB
DC
DD
16
1
___
16
Exercise 24.2
1
1
2
% symbol for universal set (sample space)
Elements only in set A
Elements shared by set A and set B
Elements not in A or B but contained in
the universal set
Elements only in set B
A ∩ B: elements in the overlapping region
1
2
9 possible outcomes
3
5
4
1
2
1
2
H HH
1
2
T HT
1
2
H TH
1
2
T TT
H
4 outcomes
T
P(TT or HH) =
2
a
2
4
=
8
10
8
10
2
10
B
BB
8
8
×
10
10
R
BR
2
8
× 10
10
B
RB
2
8
×
10
10
R RR
2
2
×
10
10
B
2
10
8
10
R
2
10
b
1
2
i
1
P(RR) = ___
25
ii
8
P(RB) + P(BR) = ___
25
16
P(BB) = ___
25
iii
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
a
7
12
7
12
W
5
12
R
7
12
W
W
25
P(RR) = ____
144
49
P(WW ) = ____
144
i
ii
4
a
b
c
d
e
0.49
0.09
0.21
0.42
0.51
5
a
4
b
4
__
c
1
__
d
R
R
5
12
b
5
12
9
9
He is equally likely either to buy two
birds, or to buy one of each.
Frequency trees
1 a
83
Patients
Cold
Covid-19
Self-diagnosis
52
31
Cold
Covid-19
18
13
b
c
2
Actual
20
32
Cold
COVID-19
Cold
18
13
COVID-19
20
32
Self
206
Covid-19
Actual diagnosis
(after testing)
It clearly shows the outcomes.
A frequency tree shows the actual data
values and a probability tree shows the
probabilities.
a
b
Cold
Students’ own opinion with justification.
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
3
a
b
5
0.62
41.94%
a
29
Exercise 24.3
1
a
b
c
d
e
1
__
b
iii
iv
6
L
3
1
5
11
c
3
i
ii
B
b
4
__
5
1
__
4
11
___
20
a
b
c
d
12
3
21
12
e
7
___
f
7
65
93
____
130
27
____
130
37
____
130
12
12
___
19
a
P
T
1
12
3
7
9
11
8
12
5
G
a
C
5
8
b
8
14
3
__
P
207
N
11
4
b
37
32
___
a
H
4
27
37
2
2
__
3
1
__
6
1
__
3
1
2
a
V
P
i
5
___
ii
5
__
iii
1
__
7
8
8
b
c
7
14
d
3
12 = ___
___
a
b
x=9
i
102
ii 17
ii 23
c
17
____
28
7
4
80
20
130
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Exercise 24.4
1
a
1st card 2nd card
12
51
13
51
13
51
13
51
13
52
13
52
iii
iv
2
1
4
1
4
1
4
1
4
i
1
___
ii
1
___
iii
208
24
= 25
A
B
C
D
1
3
1
3
1
3
1
3
T
6
9
8
2
4
a
3
__
b
9
___
5
17
a
= 100
C
3
13 12 ___
=
P(♥♥) = ___ × ___
52 51 51
13 12 ___
3
P(♣♣) = ___ × ___
=
52 51 51
26 26 13
P(red, black) = ___ × ___ = ___
52 51 51
12 = ___
4
P(♥♥ given first card is ♥) = ___
51 17
a
b
4
C
13
51
13
51
13
51
12
51
ii
d
1
___
3
13
51
13
51
12
51
13
51
13
52
i
1
__
13
51
12
51
13
51
13
51
13
52
b
c
B
1
2
C
1
2
D
1
2
A
1
2
C
1
2
D
1
2
A
1
2
B
1
2
D
1
2
A
1
2
B
1
2
C
1
2
C
D
B
D
B
C
C
D
A
D
A
C
B
D
A
D
A
B
B
C
A
C
A
B
A
58
b
i
ii
5
a
b
c
d
22
20
0.58
11 or 0.275
___
40
60
____
= 0.247
243
37
___
= 0.617
60
48
____
= 0.350
137
71
____
= 0.444
160
6
2n − 1
4n
2n
4n + 1
2n + 1
4n + 1
B
B
G
B
G
2n
4n
G
24
24
0
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
P(2 counters same colour)
= P(BB or GG)
2n − 1 2n + 1 2n
2n
= ______ × ______ + ______ × ___
4n
4n + 1
4n + 1 4n
2
2
4 n − 2n 4 n + 2n
= _________ + _________
4n(n + 1) 4n(n + 1)
8 n2
= _________
4n(n + 1)
2n
= _____
n+1
7
a
S
K
A
O
K
A
O 12 outcomes
S
A
O
S
K
O
S
K
A
b
c
8
1
___
1
6
1
5
L
1
5
M
1
5
1
6
1
6
1
6
1
6
1
6
R
B
Sm
1
___
30
1
5
1
5
1
5
L
M
R
B
Sm
Sn
M
R
B
Sm
Sn
L
R
B
Sm
Sn
L
M
B
Sm
Sn
L
M
R
Sm
Sn
L
M
R
B
Locker 2
Locker 1
1
3
1
3
1
3
b
1
2
Sam
1
2
Kerry
Locker 3
1
Kerry
Kerry
1
Sam
Raju
1
Kerry
Kerry
1
Raju
Raju
1
Sam
Sam
1
Raju
Sam
1
2
Raju
Conditional – once the first name is
chosen it cannot be chosen again, so the
second choice depends on the first, and so
on.
1 way
6 ways
1
__
6
c
d
e
4
10 ___
15
11 a
Friday
0.21
Sn
209
a
12
1
___
12
a
b
9
0.79
b
i
ii
Saturday
0.83
Rain
0.17
No rain
0.3
Rain
0.7
No rain
Rain
No rain
0.1743
0.4113
1
12 P(rain both days) = ___
50
96
P(sun both days) = ____
125
53
P(1 day sun and 1 day rain) = ____
250
13 a
3
4
1
4
Good wind
5
8
Goes windsurfing
3
8
Does not go windsurfing
1
16
Not a good wind
b
15
___
c
33
___
d
31
____
15
16
Goes windsurfing
Does not go windsurfing
32
64
128
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Practice questions
1
a
6
Starter
Main
Dessert
R
F
V
R
F
V
b
2
3
a
b
c
a
b
4
16
10
P(maths) = 0.84
P(maths or physics) = 0.96
i
4
b
S
19
a
i
ii
b
84 __
____
=2
n
_____
n+5
n2 − n
____________
or equivalent
2
n + 9n + 20
7
n −n
____________
= ___
2
n2
+ 9n + 20
22
⇒ 22(n2 − n) = 7(n2 + 9n + 20)
210 5
148 ____
74
____
=
ii
210 105
1320 ___
_____
= 44
2730 91
a
3
a
7
1
___
a+b−4
7
9
L
I
L
I
L
I
L
I
C
Salad
6
L
I
L
I
L
I
L
I
C
Soup
G
H
⇒ 22n2 − 22n = 7n2 + 63n + 140
⇒ 22n2 − 22n − 7n2 − 63n − 140 = 0
⇒ 15n2 − 85n − 140 = 0
Music
⇒ 3n2 − 17n − 28 = 0
Maths
c
25
12
n=7
35
P(exactly one black and one white) = ___
66
78
25
b
c
5
P(music) = 0.264
P(music given maths) = 0.243
a
A
B
b
h
a
o
i
m
k
210
i
ii
iii
e
f
r
u
x
s
b
j
c
p
g
t l
n
C
q
d
v
w
y
{a}
{b, c, p}
{p}
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
Practice questions worked
solutions
1
a
1
2
1
4
Curry
1
4
Soup
1
2
Roti
Fish
1
4
Vegetarian
1
2
Salad
1
4
Roti
Lassi
Ice cream
Lassi
Ice cream
Lassi
Ice cream
Lassi
Ice cream
a
1 × __
1 × __
1 = ___
1
P(soup, curry, ice cream) = __
2 4 2 16
10
b
130 + 80 ____
210 ___
________
=
= 21
c
240 ___
____
= 24
a
i
b
3
Ice cream
1
2
Fish
1
4
Vegetarian
2
Lassi
Curry
1
4
Lassi
Ice cream
1
2
1
4
Ice cream
Ice cream
Lassi
1
4
1
2
Lassi
250
250
250
25
25
P(at least one green) = 1 − P(no green)
8
9
= 1 − ___ × ___
15 14
72
= 1 − ____
210
148
74
= ____ = ____
210 105
3 ___
6 ___
5
5
4 + ___
4 × ___
1 − P(same colour) = 1 − (___ × ___
+
×
15 14 15 14 15 14 )
20 + 12 + 30
= 1 − ____________
210
148
74
= ____ = ____
210 105
10
4 × ___
11 × ___
P(brown, not brown, not brown) × 3 = 3 × ___
15 14 13
44
= ___
91
ii
b
211
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
a
Ma
7
Mu
x
103 − x
37 − x
a
i
n
_____
ii
n(n − 1)
n
n−1
_____
× _____ = ____________
n + 5 n + 4 (n + 5)(n + 4)
n+5
n(n − 1)
7
____________
= ___
(n + 5)(n + 4) 22
b
22n(n − 1) = 7(n + 4)(n + 5)
25
b
c
5
22n 2 − 22n = 7n 2 + 63n + 140
15n 2 − 85n − 140 = 0
3n 2 − 17n − 28 = 0
25 + 103 − x + x + 37 − x = 140
154 − x = 140
x = 25
25 + 12 = 37
25
____
103
i
c
P
h
A
j
4 or 7 but n . 0
n = − __
3
so, n = 7
Therefore, P(exactly one of each)
= P(W, B) + P(B, W)
5
5
7
7
= ___ × ___ + ___ × ___
12 11 12 11
70
= ____
121
e
f
l
t
n
r
v
B
b
a
i
w
x
(3n + 4)(n − 7) = 0
S
Z
k
q
c
g
Past paper questions
d
m
1
a
b
stationary
The student is travelling fastest between
the time 1300 and 1320 because the graph
is steepest then.
2
a
i
C
u
y
ii
6
a {a}
b {b, c, h, p}
c {p}
%
E
G
H
c
7
10
2
b
6
3
12
9
4
4
9
M
5
8
3
1
7
a
S
d
a+d=b+c−2
a + b + c + d + 29 = 59 (Total number)
a + b + c + d = 30
We need b + c = a + d
= 32 − (b + 2)
b + c = 16
212
ii
b
3
2
1
2 = __
iii ___
12 6
No. 2 is prime and even.
Maxi travels 20 km before Pippa starts.
Now 110 km apart.
110
Both travel ____ = 55 km further.
2
20 + 55 = 75 km
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
4
a
i
enlargement, scale factor 4, centre of
enlargement (−4, −5)
ii rotation through 90° clockwise about
origin
b
i and ii
y
4
3
C
2
1
A'
−9 −8 −7 −6 −5 −4 −3 −2 −1 0
−1
1
−2
3
4
5x
B
−3
A
2
−4
−5
b
−6
−7
c
d
8
tan b = __ = 2 ⇒ b = tan −1 (2) = 63.4°
4
Yes, they have the same three angles.
90° − 63.4°
=26.6
D
B
63.4°
63.4°
5
a
b
c
d
213
E
i
1.125 × 152 = 171
3
ii 152 + 171 + __ × 152 = 380
8
152 : 171 : 57
8 rows
i
$6 is 4 parts so 1 part is $1.50.
7 × 1.5 = $10.50
ii 9 × 1.5 = $13.50
iii 120 × $13.50 + 136 × $10.50 + 30 × 6
= $3228
3228
iv _____ = 71.7%
4500
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
6
a
rotation 900 anticlockwise about (0, 0)
b
enlargement, scale factor 3, centre (5, −7)
c and d
y
9
8
7
B'
6
C
5
4
3
2
1
−9 −8 −7 −6 −5 −4 −3 −2 −1 0
−1
1
2
3
4
−2
−3
B
5
6
7
8
9x
A'
A
−4
−5
−6
−7
−8
−9
7
a
b
(− 6 )
14
( 21)
− 12
_
8
20 = k √ 25 = 5k ⇒ k = 4
_
y = 4 √ 36 = 24
9
a
i
ii
214
rotation 900 anticlockwise about
(0, −1)
1 centre
enlargement, scale factor __
3
(6, 6)
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
b
y
10
8
6
4
A'
2
−8
−6
−4
−2
0
C
2
4
6
8
10 x
−2
−4
−6
B
A
−8
−10
⟶ ⟶ ⟶
10 OS = OP + PS
⟶
4 (PQ)
= a + __
9
4 (−a + b)
= a + __
9
5
4b
= __ a + __
9
9
1
1 × __
1 = ___
1 × 2 = ___
11 2 × __
5 4 20
10
12 a
b
fg(x) = 4(5x − 4) + 3 = 20x − 13
p = −13
5x − 1
y = ______
3
3y = 5x − 1
3y + 1
x = ______
5
3x + 1
h −1(x) = ______
5
215
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CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
k
13 m = _______
( p − 1) 2
45
m = _______2
( p − 1)
45
m = ___
25
9
= __
5
14 a
k ∴ k = 45
5 = __
9
3m =(15)
21
i
_
b
15 a
b
c
d
e
16 a
b
_
ii √ 10 2 + 24 2 = √ 676 = 26
⟶ ⟶ ⟶
3
OE = OA + AE = p+ __ q
4
i
$7680
ii $34 240
$5306.04
26.7%
36 515
______
= $33 500
1.09
9
5
___ (x + 290) = ___
x
22
12
x = 15 660
5
∴ Arun paid ___ × 15 660 = $6525
12
9
9 m s−1 in 50 s ___ = 0.18 m/s2
50
1000
deceleration = 1944 × ___________
3600 × 3600
= 0.15 m/s2
9
time = _____ = 60 s
0.15
c
Speed (m/s)
y
9
0
d
216
0
50
100
150
Time (s)
200
250
300 x
1 × 50 × 9 + (130 × 9) + __
1 × 60 × 9 = 225 + 1395 + 270 = 1890 m
distance = __
2
2
time = 240 s
1890
average speed = _____ = 7.88 m/s
240
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
17 a
b
%
P
Q
M
C
%
34 − x
=6
……
34 − x
x
……
= 28
= 11
……
5
……
73 − x + 5 = 50
x = 28
c
i
ii
iii
8
11
29
(C ∩ S ) ∩ B9
19
v ___
30
3 ____
4 × ___
2
vi ___
= 12 = ___
19 18 342 57
8+7 1
15 1
vii p(A) = ___ = __
p(B) = _____ = __
Same probability means they are equally likely.
30 2
30
2
iv
18 a
i
ii
b
i
ii
iii
217
1.991 × 103
2(s − ut)
1 at 2 = s − ut ⇒ a = ________
__
2
t2
(2x + 3)(x − 1) − (x + 1)(x − 2) = 62
2x 2 + x − 3 − (x 2 − x − 2) = 62
x 2 + 2x − 1 = 62
x 2 + 2x − 63 = 0
(x + 9)(x − 7)
x = −9 or x = 7
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
19 a
b
_______________________________
BC = √ 80 2 + 115 2 − 2 × 80 × 115 × cos 72° = 118.1 m
sin ABC ______
sin 72
________
=
115
118.1
ABC = 67.8°
c
d
e
i
2550
ii 7.20
11.8 km/h
distance
________
= sin 72
80
distance = 80 sin 72 = 76.1
76.1 m
20 a
b
c
(32)2 + 1 = 82
x+2
_____
7
gg(x) = (x2 + 1)2 + 1 = x4 + 2x2 + 2
a = 1, b = 2, c = 2
d
3 7x−2 = 81
3 7x−2 = 3 4
7x − 2 = 4
7x = 8
8
x = __
7
21 a
i
ii
1.5 , h < 1.6
1.62 m
b
i
7
14 = ___
____
ii
c
i
120 60
7
6
7
21
3(___ × ____ × ____) = ______
60 119 118
20 060
Height (h metres)
Cumulative frequency
218
h < 1.4
h < 1.5
h < 1.6
h < 1.7
h < 1.8
h < 1.9
7
25
55
79
106
120
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023
CAMBRIDGE IGCSE™ MATHEMATICS: CORE & EXTENDED COURSEBOOK
ii
y
120
110
100
Cumulative frequency
90
80
70
60
50
40
30
20
10
0
1.3
1.4
1.5
1.6
1.7
1.8
1.9 h
Height (m)
d
219
i
ii
approximately equal to 1.63 m
1.59 m
Cambridge IGCSE™ Mathematics – Morrison, Hamshaw © Cambridge University Press & Assessment 2023