CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Learner’s Book
answers
Unit 1 Getting started
1
a
144
b
9
c
125
d
4
2
a
512
b
128
3
a
157
b
153
4
a
4 and 3000 and 225
b
5
10
8
b
The square root of any integer between
144 and 169 is a possible answer.
a
14
b
6
10 a
i
9
All of them.
6
Exercise 1.1
1
2
3
4
5
7
1
ii
1
iii
2
b
( 5 + 1) × ( 5 − 1) = 4, and so on
c
( N + 1) × ( N − 1) = N − 1
d
Learner’s own answer.
a
integer 3
b
irrational
c
irrational
d
integer 7
e
irrational
a
1, 7 , −38 and − 2.25 are rational.
Reflection:
a
i
b
200 is the only irrational number.
b
a
integer
b
surd
c
surd
No. It might be a repeating pattern or it
might not.
d
integer
e
integer
f
surd
a
irrational because 2 is irrational
b
rational because it is equal to 4 = 2
c
irrational because 3 4 is irrational
d
rational because it is equal to 3 8 = 2
a
5
12
Learner’s own answer. For example:
2 and 2 − 2
i
4
ii
6
iii
10
iv
6
b
They are all positive integers.
c
Learner’s own answer.
d
Learner’s own answer.
a
7² = 49 and 8² = 64
b
4³ = 64 and 5³ = 125
11 a
b
aLearner’s own answer. For example:
2 and − 2 .
b
6
aThe square root of any integer between
16 and 25 is a possible answer.
3
No. It is not a repeating pattern.
Learner’s own answer.
true ii
true iii
false
Exercise 1.2
1
2
3
a
3 × 105
b
3.2 × 105
c
3.28 × 105
d
3.2871 × 105
a
6.3 × 107
b
4.88 × 108
c
3.04 × 106
d
5.2 × 1011
a
5400
b
1 410 000
c
23 370 000 000
d
87 250 000
4
Mercury 5.79 × 107 km; Mars 2.279 × 108;
Uranus 2.87 × 109
5
a
Russia
c
The largest country is approximately 9
times larger than the smallest country.
a
7 × 10−6
b
8.12 × 10−4
c
6.691 × 10−5
d
2.05 × 10−7
6
b
Indonesia
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
7
a
0.0015
b
0.000 012 34
c
0.000 000 079
d
0.000 900 3
8
a
30
b
9.11 × 10−25 kg
9
a
z
b y
10 a
a
36
8
a
1 b
81
9
a
i
2
b
i
x=5
10 a
65 is not between 1 and 10.
1
36
c
1
d
1
216
1
c
225
1
ii 4
4
1
d
1
400
ii
1
iii 9
9
x = 10
i
3
5
ii
39
iii
310
iv
36
i
3
ii
3−1
iv
3−2
v
3−3
b
6.5 × 105
c
4.83 × 107
11 a
1.5 × 10−2
b
2.73 × 10−3
c
5 × 10−8
12 a
6.1 × 106
b
6.17 × 105
11 a
56 b 52
c
5−2
c
1.75 × 105
12 a
6−1
b
73
13 a
7.6 × 10−6
c
11−10
d
4−4
b
8.02 × 10−5
13 a
x=4
b
x=6
c
1.6 × 10
c
x = −2
d
x=5
i
22
ii
43
iii
51 or 5
iv
23
14 a
b
−7
i
7 × 106
ii
3.4 × 107
iii
4.1 × 10−4
iv
1.37 × 10−3
b
To multiply a number in standard form by
10, you add 1 to the index.
c
To multiply a number in standard form
by 1000, you add 3 to the index. To divide
a number in standard form by 1000, you
subtract 3 from the index.
Reflection: You can compare them easily. You
can write the number without using a lot of zeros.
You can enter them in a calculator.
Exercise 1.3
1
a
d
1
4
1
216
b
1
8
e
1
f
10 000
c
1
81
1
32
2
3 , 2 and 4 are equal, 5 , 6
3
a
2−1
b
2−2
c
d
2
e
2
0
f 2
a
102
b
103
c
100
d
10−1
e
10−3
f
10−6
a
64−1
b
c
4−3
d
a
3 or 9 or 81
b
The three ways in part a.
4
5
6
2
b
7
−3
−4
−2
−6
−4
−2
−1
−1
Learner’s own answers.
d
Learner’s own answers.
14 a
b
Learner’s own answers.
c
Learner’s own answers.
32
d 5−6
15 a
6−3
b
9−1
c
15−4
d
10−5
16 a
25
b
87
c
5−6
d
122
17 a
26
b
2−6
c
36
d
3−6
e
93
f
9−3
Check your progress
1
0
26
c
iii
2
−3
a
rational
b
irrational
c
rational
d
irrational
e
rational
a
rational because it is equal to 25 = 5
b
irrational because it is 3 + 7 and 7 is
a surd
3
n=3
8−2
4
a
2−6
5
C, D, A, B
6
a
b
8.6 × 1010
1
49
b
1
81
6.45 × 10−6
c
1
128
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
7
a
53
b
c
8
a
65
b
12−5
c
4−6
d
152
50
5−2
4
b
x
+7
3
5
12
b
5
= 53
59
x2 + 2x
b
12y2 − 21yw
a
4(x + 3)
b
2x(2x + 7)
5
a
17
or 1 5
12
12
b
6
or 11
5
5
6
a
F = 25
b
a=
c
a=6
2
a
32 × 34 = 36
c
(7 ) = 710
3
a
4
6
F
m
a
b
x − 2y = 3 − 2 × 5
= 3 − 10
= −7
Learner’s own answer.
a
x = 1 and y = 14, x = 2 and y = 11, x = 3 and
y=6
b
Learner’s own answer. For example:
x = −4 and y = −1, x = −5 and y = −10,
x = −6 and y = −21
c
Learner’s own answer. For example:
x = −1 and y = 14, x = −2 and y = 11,
x = −3 and y = 6 or x = 4 and y = −1,
x = 5 and y = −10, x = 6 and y = −21
a
4( m + 2 p ) = 4( 2 + 2 × −4 )
= 4( 2 − 8)
= 4 × −6
= −24
b
p3 − 3mp = ( −4 )3 − 3 × 2 × −4
= −64 + 24
= −40
2 5
Exercise 2.1
1
Learner’s own answers.
For example: Part a is incorrect as −32
should be written as (−3)2, which is 9 and
not −9; part b is incorrect as (−2)3 is −8
and not 8.
Unit 2 Getting started
1
a
5
c
x3 + xy = 33 + 3 × 5
= 27 + 15
= ( −2 )5 − 64
= −32 − 64
= −96
= 42
c
y2 −
10 x
y
= (5 ) −
2
10 × 3
a
21
b
36
c
16
30
d
64
e
68
f
−18
5
g
14
h
−25
i
−7
= 25 − 6
j
82
= 25 −
7
5
= 19
2
3
3
5
3
3
 p
 −4 
 m  + ( p ) =   + ( −4 )
2
a
9
b
4
c
9
d
8
e
8
f
30
g
5
h
47
i
−30
j
−4
a
Learner’s own answers. For example:
i
a = 3, b = 10, c = 12, d = 2
ii
a = −3, b = −10, c = −12, d = −2
iii
a = 3, b = 4, c = −36, d = 3
b
Learner’s own answers.
c
Learner’s own answers.
Activity 2.1
Learner’s own answer.
8
9
Learner’s own counter-examples.
a
For example: When x = 2,
3x2 = 3 × 22 = 3 × 4 = 12, and
(3x)2 = (3 × 2)2 = 62 = 36, and 12 ≠ 36
b
For example: When y = 2, (−y)4 = (−2)4 = 16
and −y4 = −24 = −16, and 16 ≠ −16
c
For example: When x = 3 and y = 4,
2(x + y) = 2(3 + 4) = 2 × 7 = 14 and
2x + y = 2 × 3 + 4 = 10, and 14 ≠ 10
a
26
b
49
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
2
10 5a 2 − 9( b − a ) + 5 + 7ab = 5 × ( −2 )2 − 9( −1 − −2 ) +
c
b
2
( −1)
Width of rectangle = 2x = 2 × 3 = 6
+ 7 × −2 × −1
5
= 5 × 4 − 9 ×1+
2
−1
Perimeter = 2 × length + 2 × width =
2 × 8 + 2 × 6 = 28
+ 14
Area = length × width = 8 × 6 = 48
d
= 20 − 9 − 2 + 14
= 23
−5a
b
− 6a3 − ( ab )4 +
9
2
b −a
3
=
−5 × −2
−1
=
10
−1
5
9
2
( −1) − ( −2 )
− 6 × −8 − ( 2 )4 +
= −10 + 48 − 16 +
e
Learner’s own answer.
a
i
P = 2x + 10
ii
A = 3x + 6
iii
When x = 4, P = 18 and A = 18
i
P = 2y − 4
ii
A = 4y − 24
iii
When y = 10, P = 16 and A = 16
i
P = 4n + 8
ii
A = n2 + 4n
iii
When n = 6, P = 32 and A = 60
i
P = 2p2 + 8p
ii
A = 4p3
iii
When p = 2, P = 24 and A = 32
3
9
1+ 8
b
9
9
= 22 + 1
c
= 23
Reflection: Learner’s own answers.
Exercise 2.2
1
2
3
4
n+5
b
5n − 5
c
n
+5
5
d
5(n + 5)
e
n −5
5
f
5−n
a
7x
b
20 − x
c
2x + 9
d
x
−4
6
e
x2
f
100
x
g
5(x − 7)
h
x3
k
(3x)2 + 7 or 9x2 + 7
l
(2x)3 − 100 or 8x3 − 100
a
i
2x + 2y
ii
xy
b
i
6x + 2y
ii
3xy
c
i
6x + 4y
ii
6xy
d
i
4x
ii
x2
e
i
8x
ii
4x2
f
i
2x2 + 4x
ii
2x3
j
a
i2 red + 2 yellow = 4 green;
both = 8x + 4
ii3 red + 3 yellow = 6 green;
both = 12x + 6
iii4 red + 4 yellow = 8 green;
both = 16x + 8
b
n red + n yellow = 2n green (or similar
explanation given in words)
c
i6 red + 2 yellow = 12 blue;
both = 12x + 12
x
aPerimeter = 2(x + 5) + 2(2x) =
2x + 10 + 4x = 6x + 10
Learner’s own answer.
6
x
3
i
b
4
d
a
Perimeter = 6x + 10 = 6 × 3 + 10 = 28
Area = 2x2 + 10x = 2 × 32 + 10 × 3 =
18 + 30 = 48
− 6( −2 )3 −
( −2 × −1)4 +
Length of rectangle = x + 5 = 3 + 5 = 8
ii9 red + 3 yellow = 18 blue;
both = 18x + 18
iii12 red + 4 yellow = 24 blue;
both = 24x + 24
7
d
3n red + n yellow = 6n blue (or similar
explanation given in words)
e
Learner’s own answer.
a(3w)2 = 36, 2v(3v − 2w) = 30, 5w(w + v) = 50
b
116
c
(3w)2 + 2v(3v – 2w) + 5w(w + v) =
9w2 + 6v2 − 4vw + 5w2 + 5vw =
14w2 + vw + 6v2
d
116
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
8
9
a3a2 − 7b = 61, 8b – 3a = 31, a2 + 6b = 37,
4(a + 3b) = 4
2
a
m14
b
n12
c
p7
d
q5
e
r3
f
t5
g
x21
h
y10
i
z12
j
5t7
k
5g2
l
−h9
a
Sofia is correct. x2 ÷ x2 = x2−2 = x0 = 1
b
133
c
3a − 7b + 8b − 3a + a + 6b + 4(a + 3b) =
4a2 + 7b − 3a + 4a + 12b = 4a2 + a + 19b
d
133
e
11
b
Learner’s own answer.
f
Not valid because although the perimeter
is positive, three of the side lengths are
negative, which is not possible.
c
x2 ÷ x2 = 1
d
All the answers are 1. Learner’s own
explanations. For example:
2
2
3
When simplified, all the expressions have
an index of 0, and anything to the power
of 0 = 1.
a2(3x2 + 4) + 2(5 − x2) or
3x2 + 4 + 3x2 + 4 + 5 − x2 + 5 − x2
b
2(3x2 + 4) + 2(5 − x2) =
6x2 + 8 + 10 − 2x2 = 4x2 + 18 = 2(2x2 + 9)
or 3x2 + 4 + 3x2 + 4 + 5 − x2 + 5 − x2 =
4x2 + 18 = 2(2x2 + 9)
c
or Any expression divided by itself, always
gives an answer of 1.
4
Arun is correct. Learner’s own
explanation.
For example: The variable x only appears
in the expression for the perimeter when
it is squared. When you square 2 and −2
you get the same answer.
5
a
6x5
b
12y9
c
30z7
d
4m7
e
4n13
f
8p3
a
Learner’s own answer.
b
Learner’s own answer.
c
Learner’s own answer.
Sasha’s method would be easiest to use to
simplify these expressions:
or: 2(2(−2)2 + 9) = 2(2 × 4 + 9) =
2(8 + 9) = 34
2
3
11 a
b
8y
Side length = 49 = 7 cm,
Perimeter = 4 × 7 = 28 cm
6
Perimeter = 4 × x or 4 x
Volume = x
3
7
Side length = 3 y
1
a
x ×x = x
5
=x
c
8
4+5
b
y ×y = y
2
d
u8 ÷ u 6 = u8− 6
(g ) = g
3 2
5
a
3q4
b
3r4
c
3t6
d
2u5
e
2v4
f
5w
a
D 1 x3
2+ 4
3× 2
f
5 12
5m3 + 3m3 = 8m3
h
8n2 − n2 = 7n2
A 2 y6
d
B 31
5
3
or (3x2)3 = 3x2 × 3x2 × 3x2 =
3 × 3 × 3 × x2 × x2 × x2 = 27 × x6 = 27x6
or (3x2)3 means everything inside the
bracket must be cubed. That means the 3
must be cubed as well as the x2.
5 ×12
= h60
b
(3x2)3 = 33 × (x2)3 = 27 × x6 = 27x6
w 5 ÷ w = w 5 −1
(h ) = h
2
5
C k
3
aArun is correct. Learner’s own
explanation. For example:
= w4
= g6
g
6 z9
z5
= .
4
6
36 z
= y6
9
= u2
e
4
2
6z9 ÷ 36z4 = 6
c
Exercise 2.3
4
7
12y7 ÷ 8y6 = 212 y6 = 3 y and
10 aSide length = 25 = 5 cm,
Perimeter = 4 × 5 = 20 cm
c
2
3
6x
and 2(2(2)2 + 9) = 2(2 × 4 + 9) =
2(8 + 9) = 34
b
5
4x5 ÷ 6x3 = 3 4 x3 = 2x ,
b
i
16x10
iii
16z28
ii
125y12
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Activity 2.3
4
a
y2 + 6y + 8
b
z2 + 14z + 48
a
Learner’s own spider diagram.
c
m2 + m − 12
d
a2 − 7a − 18
b
There are many possible expressions.
For example:
e
p2 − 11p + 30
f
n2 − 30n + 200
5
3x × 12x
2
10
4x8 × 9x4
36x14 ÷ x2
aThe plus at the end would change
to a minus and the 9 changes to a 1.
x2 + 1x − 20
b
The plus at the end would change to
a minus and the 9 changes to a −1.
x2 − 1x − 20
c
The plus in the middle would change to a
minus. x2 − 9x + 20
d
i
(x + A)(x + B) = x2 + Cx + D
ii
(x + A)(x − B) = x2 + Cx − D
iii
(x − A)(x + B) = x2 − Cx − D
iv
(x − A)(x − B) = x2 − Cx + D
72x20 ÷ 2x8
(6x6)2
36(x3)4
9
c
Learner’s own answers.
a
q−3 = 3
1
q
b
r−2 = 12
c
t
= 15
t
d
v
−5
r
−1
=1
v
10 a
A and iii, B and iv, C and i, D and vii,
E and vi, F and v.
b
6
7
a
C w2 + 12w + 27
b
A x2 + 2x − 35
c
B y2 − 2y − 48
d
A z2 − 9z + 20
a
(x + 2)2 = (x + 2)(x + 2)
Learner’s own answer. Any expression
= x2 + 2x + 2x + 4
1
that simplifies to give 7 .
6y
For example:
= x2 + 4x + 4
b
5 y2
30 y9
(x − 3)2 = (x − 3)(x − 3)
= x2 − 3x − 3x + 9
Reflection: Learner’s own answers.
Exercise 2.4
1
a
(x + 4)(x + 1)
3
6
8
a
= x + 1x + 4x + 4
2
i
y2 + 10y + 25
ii
z2 + 2z + 1
iii
m2 + 16m + 64
iv
a2 − 4a + 4
= x + 5x + 4
b
= x2 + 6x − 3x − 18
v
p2 − 8p + 16
= x2 + 3x − 18
vi
n2 − 18n + 81
c
= x2 − 8x + 2x − 16
(x − 3)(x + 6)
(x + 2)(x − 8)
2
9
b
(x + a)2 = x2 + 2ax + a2
a
(x + 3)(x − 3) = x2 + 3x − 3x − 9 = x2 − 9
b
i
x2 − 4
= x 2 − 6x − 16
d
= x2 − x − 4x + 4
ii
x2 − 25
= x2 − 5x + 4
iii
x2 – 49
(x − 4)(x − 1)
2
= x2 − 6x + 9
a
x2 + 10x + 21
b
x2 + 11x + 10
c
x2 + 2x − 15
d
x2 + 4x − 32
e
x2 − 9x + 14
f
x2 − 14x + 24
a
Learner’s own answers and explanations.
b
Learner’s own answers and explanations.
c
Learner’s own answer.
c
There is no term in x, and the number
term is a square number.
d
x2 − 100
e
x2 − a2
Activity 2.4
a
① 33 × 29 = 957, ② 28 × 34 = 952,
③ 957 − 952 = 5
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
① 16 × 12 = 192, ② 11 × 17 = 187,
③ 192 − 187 = 5
c
The answer is always 5.
d
e
n
n+1
n+5
n+6
5
a
1 2 3 2 5
+ = + =
2 6 6 6 6
b
1+ 2 3
1
= =1
2
2
2
c
5
1
≠1
6
2
d
① (n + 5)(n + 1) = n2 + 6n + 5,
② n(n + 6) = n2 + 6n,
She cannot cancel the 3 with the 6,
because the expression is 3x + y, all
divided by 6, not just 3x divided by 6.
x y 3x y 3x + y
+ =
+ =
2 6 6 6
6
③ n + 6n + 5 − (n + 6n) =
n2 + 6n + 5 − n2 − 6n = 5
e
Learner’s own answer.
The answer is always 5.
f
i
2
2
Learner’s own answer.
iiincorrect. Learners should show that
the correct answer is
Exercise 2.5
1
2
3
a
2x
5
b
4x
7
c
8
x
d
x
e
2x
5
f
4
x
a
2 y 3y 4 y 3y 7 y
+ = + =
5 10 10 10 10
b
2
1
10
1
9
−
=
−
=
5 y 25 y 25 y 25 y 25 y
c
3y
4
d
3y
8
e
11
9y
f
3y
14
a a 5a 2 a
+ = +
2 5 10 10
5a + 2 a
=
10
7a
=
10
b
5
2
25
14
+ =
+
7c 5c 35c 35c
25 + 14
=
35c
39
=
35c
d
7e 2e 21e 16e
− =
−
8
3
24
24
21e − 16e
=
24
5e
=
24
f
a
A, D, F
b
c
G; the answer is
a
c
e
4
7
correct
x
3
iii
correct
ivincorrect. Learners should show that
6
a
b
the correct answer is
9x − 8
20
i
a+b
5
ii
5a + 9b
12
iii
2a + 9
15
iv
ab + 12
4b
v
3ab + 40
10b
vi
8ab + 27
18b
Learner’s own checks.
Activity 2.5
Learner’s own answers.
7
b b 3b 4b
+ = +
4 3 12 12
3b + 4b
=
12
7b
=
12
5d 3d 25d 18d
− =
−
6
5
30
30
25d − 18d
=
30
7d
=
30
a
6 × 3 + 2 18 + 2 20
=
= = 10
2
2
2
b
3 × 3 + 1 = 9 + 1 = 10
c
10 = 10
d
Learner’s own explanation. For example:
He factorises the bracket to give
2 × bracket, which is then divided by 2.
The × 2 and ÷ 2 cancel each other out,
leaving just the bracket.
e
When x = 3, 6 × 3 + 1 = 18 + 1 = 19, 19 ≠ 10,
so the answer is wrong.
Learner’s own explanation. For example:
The expression shows that 6x + 2 must all
be divided by 2.
9
3
18
15
− =
−
20 f 20 f
10 f 4 f
18 − 15
=
20 f
3
=
20 f
B, C, E
4x − y
10
Arun has only divided the 2 in the
numerator by 2, and not the 6x by 2 as well.
8
f
Learner’s own answer.
a
2x + 1
b
x+2
c
2x − 3
d
2x − 5
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
9
4
6 x − 4 20 x + 25 2(3x − 2 ) 5( 4 x + 5 )
+
=
+
=
2
5
2
5
3x − 2 + 4x + 5 = 7x + 3
10 a
2(x + 3) = 2 × x + 2 × 3 = 2x + 6
5
3
T = 3x − 4
c
T = 53
d
T +4
x=
3
e
x = 22
a
v = 87
b
v = 125
c
i
2(x + 3) or 2x + 6
c
u = 27
d
u = 46
ii
2(x + 2) or 2x + 4
e
t = 10
f
a=2
iii
4(x − 3) or 4x – 12
a
20%
b
60%
iv
3(1 − 3x) or 3 − 9x
c
125%
a
65 kg
b
49.1 kg (1 d.p.)
c
95.9 kg (1 d.p.)
d
57.3 kg (1 d.p.)
a
i
B x=
y−z
2
2( y + 3h )
5
a
S = 60M
6
7
c
M=
a
i
b
m = , m = 12
c
F
a = , a = −1.75
m
8
b
S = 900
S
60
d
M = 22.5
ii
C x=
F = 60
ii
F = −78
iii
A x = 7k(y − 6)
iv
C x = 3ny + m
v
A x=
F
a
a
Number
of faces
Number Number
of
of
vertices edges
Cube
6
8
12
Cuboid
6
8
12
Triangular
prism
5
6
9
Triangularbased
pyramid
4
4
6
Square-based
pyramid
5
5
8
b
E = F + V − 2, or any equivalent version
c
V=E−F+2
i
ii
V=6
V=7
d
c i is a pentagonal-based pyramid and
c ii is a hexagonal-based pyramid
e
F = E − V + 2, F = 0, it is not possible
to have a shape with five edges and
seven vertices.
f
w−y
7
b
Learner’s own answer.
a
t=
m−9
7
b
t = 5(k + m)
c
t = pv − h
d
t=
10 a
A = a2 + bc
9
3D Shape
9q + w
5
b
A = 49.5
c
A = a2 + bc, A − bc = a2, a = A − bc
d
a=8
11 a
78.5 cm
c
6.25 cm
12 a
l = 3V
A
π
b
r=
b
2 cm
13 Sasha is correct as 30 °C = 86 °F and
86 °F > 82 °F (or 82 °F = 27.8 °C and
27.8 °C < 30 °C).
14 aShe is not underweight as her BMI is
20.05, which is greater than 18.5.
b
3.7 kg
Check your progress
1
Learner’s own answer.
2
8
b
Learner’s own choice and explanation.
Exercise 2.6
2
Ben’s age is x + 2, Alice’s age is x − 6
b
Reflection: Learner’s own answers.
1
a
a
39
c
12
b
161
perimeter = 16x + 8,
area = 5x(3x + 4) = 15x2 + 20x
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
3
4
5
6
a
x5
b
q6
c
h10
c
320 ÷ 101 = 320 ÷ 10 = 32
d
15m9
e
2u2
f
3p2
d
320 ÷ 100 = 320 ÷ 1 = 320
a
x2 + 7x + 10
b
x2 + x − 12
a
2.7
b
0.45
c
x − 3x − 54
d
x – 14x + 40
c
0.36
d
0.017
e
2
x − 64
f
2
x − 12x + 36
e
0.08
f
0.0248
a
2x
3
b
2y
15
g
9
h
0.0025
12 x − y
20
a
Learner’s own answer.
c
d
3x − 5
b
i
6.8 ÷ 10−3 = 6800
ii
0.07 ÷ 10−4 = 700
2
a
x = 31
b
c
y = ± x − 5z , y = ±6
6
2
z=
7
x−y
, z=6
5
2
Unit 3 Getting started
1
a
8
b
32.5
c
6
d
0.85
e
90
f
625
g
700
h
32
8
2
B
3
a
15.4
b
640
4
a
$345
b
$240
5
63.6 cm (3 s.f.)
2
Exercise 3.1
Learner’s own answer.
d
Learner’s own answer. For example: An
alternative method is to realise that ÷ by
10−x and × by 10x are the same. So, in this
case 2.6 ÷ 10−2 = 2.6 × 102
e
Learner’s own answer.
a
3.2 ÷ 103 = 3.2 ÷ 1000 = 0.0032
b
3.2 ÷ 102 = 3.2 ÷ 100 = 0.032
c
3.2 ÷ 101 = 3.2 ÷ 10 = 0.32
d
3.2 ÷ 100 = 3.2 ÷ 1 = 3.2
e
3.2 ÷ 10−1 = 3.2 × 10 = 32
f
3.2 ÷ 10−2 = 3.2 × 100 = 320
g
3.2 ÷ 10−3 = 3.2 × 1000 = 3200
h
3.2 ÷ 10−4 = 3.2 × 10 000 = 32 000
a
Yes. Learner’s own explanation.
b
i
greater
iii
smaller
1
a, D and ii; b, A and v; c, E and iv; d, C and i;
e, B and iii
2
a
3.2 × 103 = 3.2 × 1000 = 3200
b
3.2 × 102 = 3.2 × 100 = 320
c
3.2 × 101 = 3.2 × 10 = 32
d
3.2 × 100 = 3.2 × 1 = 3.2
e
3.2 × 10−1 = 3.2 ÷ 10 = 0.32
f
3.2 × 10−2 = 3.2 ÷ 100 = 0.032
11 Do not tell anyone the secret!
g
3.2 × 10−3 = 3.2 ÷ 1000 = 0.0032
h
3.2 × 10−4 = 3.2 ÷ 10 000 = 0.000 32
12 a
a
Yes. Learner’s own explanation.
b
i
smaller
iii
greater
3
4
5
9
c
ii
9
the same
ii
the same
10 a
2.5
b
47 600
c
70
d
8.5
i
400
ii
40
iii
4
iv
0.4
v
0.04
vi
0.004
b
Smaller
c
Smaller
d
i
0.12
ii
1.2
a
1300
b
7800
c
240
d
85 500
e
65
f
8000
iii
12
iv
120
g
17
h
0.8
i
0.085
v
1200
vi
12 000
j
0.45
k
0.032
l
1.25
a
b
e
Larger
320 ÷ 103 = 320 ÷ 1000 = 0.32
f
Larger
320 ÷ 102 = 320 ÷ 100 = 3.2
g
Learner’s own answer.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
13 a
c
0.8 × 101
8 ÷ 100
0.08 ÷ 10–2
=8
80 × 10–1
0.008 × 103
800 ÷ 102
32 ÷ 102
0.32 × 100
8
9
b
3.2 ÷ 101
320 ÷ 103
= 0.32
1.6
b
−5.6
c
−5.4
d
6
e
0.3
f
−0.66
g
3.6
h
−0.44
iv
15
v
12
vi
10
d
i
Smaller
e
Learner’s own answer.
a
False
b
True
c
False
d
True
ii
c
300
d
40
11 a
A and iv, B and v, C and vi, D and vii,
E and iii, F and i
b
Learner’s own answer. Any question that
gives an answer of 0.024. For example:
0.03 × 400 × 0.002
c
Learner’s own answer.
15.96 ÷ 0.57 = 28, 159.6 ÷ 0.57 = 280,
15.96 ÷ 28 = 0.57, 15.96 ÷ 280 = 0.057
13 a
b
8 × 0.2 = 1.6
0.08 × 0.2 = 0.016
b
0.4 × 0.007
4 × 7 = 28
4 × 0.007 = 0.028
0.4 × 0.007 = 0.0028
123 × 57 = 7011
i
701.1 ii
701.1 iii
70.11
iv
7.011
v
7.011 vi
0.070 11
14 a
Learner’s own answer.
b
Learner’s own answer.
c
iEstimate: 4 × 30 = 120
Accurate: 119.625
3
C, D, I, K (0.015); A, F, H, J (0.15);
B, G, L (1.5); E (15)
4
a
20
b
−50
c
−30
d
600
iiEstimate: 10 ÷ 0.2 = 50
Accurate: 62
e
40
f
−400
iiiEstimate:
g
200
h
−300
Accurate: 19 200
a
0.81 × 100 = 81 = 9
0.09 × 100
9
b
6.4 × 1000 = 6400 = 1600
0.004 × 1000
4
6
a
D
7
a
i
0.8
ii
iv
5.6
v
i
Larger
b
b
15 a
c
B
d
D
2.4
iii
4
7.2
vi
8.8
C
ii
Larger
He has made a mistake. The denominator is
0.12, not 1.2; he wrote the answer with only
one decimal place. Answer = 50.
120
8 × 2 = 16
5
10
20
For example: 28 × 0.057 = 1.596,
2.8 × 0.57 = 1.596, 28 × 5.7 = 159.6,
2.8 × 5.7 = 15.96
a
a0.08 × 0.2
iii
b
Exercise 3.2
2
30
12 Learner’s own answers and discussions.
Reflection: Learner’s own answers.
1
ii
200
Activity 3.1
Learner’s own answers.
60
10 a
3.2 × 10–1
32 × 10–2
i
Smaller
60 × 4
= 24 000
0.01
0.2 ÷ 0.4 = 0.5 m
b
0.45 m
c
Learner’s own answer.
Exercise 3.3
1
a
200 × 1.1 = $220
220 × 1.15 = $253
b
200 × 0.9 = $180
180 × 0.85 = $153
c
200 × 1.2 = $240
240 × 0.95 = $228
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
2
aLearner’s choice of who they think is
correct, with reason.
b
The coin is now worth less than $800.
Learner’s explanation. For example: The
10% decrease will be $80, but the 10%
increase will be less than $80 as it is 10%
of a smaller amount than $800.
d
Learner’s own answer.
a
i
b
=
c
i
Five years. 10 000 × 0.94 = 6561,
10 000 × 0.95 = 5904.9
e
10 000 × 0.9n
Learner’s own answers.
Exercise 3.4
1
57.6
ii
57.6
=
ii
=
a
b
c
i
25, 26, 27, 28, 29, 30, 31, 32, 33, 34
ii
25
iii
34
i
85, 86, 87, 88, 89, 90, 91, 92, 93, 94
ii
85
iii
94
i265, 266, 267, 268, 269, 270, 271, 272,
273, 274
ii
265
iii
274
4
a–e Learner’s own answers.
5
a
i
195
b
i
630 ii 108.864
6
a
1.1235
7
a
i
72 ii
b
i
285
8
a
0.7216 b
$4618.24
b
11.5
9
a
A and iii, B and iv, C and i, E and ii,
F and v
c
12.4
a
i54.5, 54.6, 54.7, 54.8, 54.9, 55.0, 55.1,
55.2, 55.3, 55.4
b
ii
b
ii
64.4
d
$67.41
52.8
48.412
D and 0.81
2
3
10 aZara is correct. 1.04 × 1.04
is the same as (1.04)2, so
5000 × 1.04 × 1.04 = 5000 × (1.04)2
b
5000 × (1.04)3
c
5000 × (1.04)4
d
8. The power on the 1.04 is the number of
years.
e
i
5000 × (1.04)12
ii
5000 × (1.04)20
iii
5000 × (1.04)n
f
11 a
b
11
d
Activity 3.3
$800 − $80 = $720, $720 + $72 = $792.
3
The population after 10 years.
Sofia is correct.
Learner’s explanation. For example: 10%
of $800 is $80, so the value goes up to $880.
10% of $880 is $88, so the value goes down
to $792. The 10% decrease is greater than
the 10% increase. It is not the same value.
c
c
15 years
i
10 000 × 0.9
ii
10 000 × 0.92
iii
10 000 × 0.93
The population after 5 years.
i845, 846, 847, 848, 849, 850, 851, 852,
853, 854
ii
845
iii
854
a11.5, 11.6, 11.7, 11.8, 11.9, 12.0, 12.1,
12.2, 12.3, 12.4
b
ii
54.5
iii
55.4
42 × 1.3 = 54.6 = $55
4
a–c Learner’s own answers.
5
a–c Learner’s own answers and discussions.
6
a
3.5 ⩽ x < 4.5
b
11.5 ⩽ x < 12.5
c
355.5 ⩽ x < 356.5
d
669.5 ⩽ x < 670.5
a
15 ⩽ x < 25
b
335 ⩽ x < 345
c
4745 ⩽ x < 4755
d
6295 ⩽ x < 6305
a
250 ⩽ x < 350
b
1850 ⩽ x < 1950
c
4650 ⩽ x < 4750
d
7950 ⩽ x < 8050
7
8
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
9
Exercise 4.1
Learner’s own answers and discussions.
a
i
b
The lower and upper bounds of a
rounded number will always be +/− half
of the degree of accuracy.
10 a
b
11 a
b
i
ii
0.5
iii
5
1555 cm
ii
1565 cm
ii
172.5 cm
50
1
a
171.5 cm
c
171.5 cm ⩽ x < 172.5 cm
b
12
c
0.046
d
59
e
0.0728
f
g
37
a
d
= 16
2
= 24
15
y=
5
x = −3
5
c
y=4
d
y=8
h
18
e
a = −6
f
a = −1
−1.6
b
3.6
g
x=2
h
z=4
c
−0.0028
d
600
e
300
f
9
g
7.5
h
0.11
a
i
20 000 × 1.08
ii
20 000 × (1.08)2
iii
20 000 × (1.08)3
b
The value of the painting after 5 years.
c
The value of the painting after 20 years.
d
6 years. 20 000 × (1.08)5 = 29 386.561 54,
20 000 × (1.08)6 = 31 737.486 46
e
20 000 × (1.08)n
a
i
b
7150 m2 ⩽ x < 7250 m2
ii
7150 m2
b
x=9
c
y = 25
d
y = 25
2
a
5
3
a
2x > 10
b
4x < 36
c
y + 5 ⩾ 13
d
y − 5 ⩽ −11
b
c
7
3
2
3
Learner’s own answers.
Learner’s own answers and explanations. For
example:
a
Substitute x = 26 back into the original
equation and check that left hand
side = right hand side.
b
When he expanded the bracket on the lefthand side he didn’t multiply the 8 by 2.
When he brought +8 to the right-hand
side he forgot to make it −8.
c
2 x + 16 = 18 − 3 x
5 x + 16 = 18
5x = 2
x=
x=5
5
When he brought the −3x to the left-hand
side he forgot to make it +3x.
7250 m2
a
3
a, b x = 15
c
4
=
9
y =1
b
3
−10
y=
x = −11
4
2
48
−6
6 y + 3 y = 22 − 7
9 y = 15
2 y = 16 × 3
2 y = 48
Unit 4 Getting started
12
x=
a
$265.20
1
8
= 11 + 5
y=
74 500
3
5
3
3
Check your progress
2
2y
2y
a
15 − 10 x = 9
−10 x = 9 − 15
−10 x = −6
x = −2
12 A, i and e; B, i and f; C, ii and b; D, iii and a;
E, ii and c; F, iii and d
1
b
−16
x=
1555 cm ⩽ x < 1565 cm
i
8 x = −30 + 14
8 x = −16
2
5
= 0.4
Check: When x = 0.4,
2(0.4 + 8) = 2 × 8.4 = 16.8 and
3(6 − 0.4) = 3 × 5.6 = 16.8
5, 6, 7
d
5
Learner’s own answer.
a, b x = 13
c
Learner’s own answers.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
6
a
42
=7
c
42 = 7c
42
=c
7
i, ii and iii Learner’s answers and discussions.
12 = 15d
12
c=6
15
a
10x − 8 = 5x + 12, x = 4
b
12(x − 5) = 4(x + 1), x = 8
4
c
5x − 4 = 2x + 20, x = 8
5
d
5=
=d
12
d=
c
Activity 4.1
12
= 15
d
b
15
=
21
=7
e+2
e
21 = 7(e + 2)
21
7
= e+2
3= e+2
3− 2 = e
e =1
3
a
8
a, b, c and e Learner’s own answers and
explanations.
9
a = 27 b
c
b=7
1
4
d
i
x =14 ii
a
i
A + 10
b
A + 10 = 2(A − 6)
c
A = 22
c = 3 d
x =6
3
5
iii
54 =
b
x=9
c
54 °, 54 °, 72 °
ii
iiThe two shorter sides of a rectangle
have side lengths of 6(3a − 4) and
3(4a − 3). Work out the value of a.
1
5
A−6
iiiThere are x sweets in bag A. There are
five fewer sweets in bag B than bag A.
The sweets in bag B are shared between
180 people. Each person gets 15 sweets.
How many sweets are in bag A?
10 a2(x + 3) + 7x − 5 + 5(7 − x) = 48 OR
4x + 36 = 48
b
x=3
c
12 cm, 16 cm, 20 cm
11 a
9a = 4a + 20
b
i
x=6
ii
a = 2.5
iii
x = 17
b
a=4
Exercise 4.2
c
Triangle sides 12 cm, rectangle sides 7 cm
and 11 cm
1
12 a
b
B and D
1
; B x = 15; C x = 8640;
15
1
D x = 15; E x =
15
13 a
c
d
1 Work out x. 5 x − 3 = 2 x + 15
5 x − 2 x = 15 + 3
3 x = 18
A x=
x=
85
=5
y
b
18
3
=6
2 Work out y. y = 5x − 3
= 5×6−3
= 30 − 3
= 27
3 Check values are correct. y = 2x + 15
There are 15 sectors in the pie chart.
152
=8
y+2
85
85
= 5 → y = = 17 and
y
5
152
152
= 8→
= y + 2 → 19 = y + 2 → y = 17
y+2
8
= 2 × 6 + 15
= 12 + 15
= 27
4 Write the answers: x = 6 and y = 27
Learner’s own answer.
2
13
Learner’s own problem. For example:
iA quadrilateral has sides of length
x cm, 2(x + 1) cm, 3(x + 2) cm, and
4(x + 3) cm. The perimeter is 80 cm.
Work out the value of x.
d = 11
x=−
270
x−4
14 a
15 a
7
75
, x=8
x+7
126
9 = , x=7
2x
x = 5, y = 9
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
3
x = 4, y = 13
4
x = 7, y = −5
5
a
y = 3x + 1
y=x+9
b
6
7
8
9
11 a
1 Add the two
equations.
x
0
3
y
1
10 19
x
0
3
6
x − y =   4
3x + 0y = 54
y
9
12 15
3x = 54, x =
6
2x +   y = 50
+
y
20
y = 3x + 1
18
16
y=x+9
14
12
10
8
6
4
2
0
x
0 1 2 3 4 5 6
c
(4, 13)
d
The coordinates give the solution of the
equations; x = 4 and y = 13
e
Learner’s own answer. For example: The
solution of simultaneous equations is the
point of intersection of the straight-line
graphs.
a
i
x = 2, y = 6
ii
x = 2, y = 6
b
x = 2, y = 6
c
Learner’s own answers and explanations.
a
i
x = 2, y = 7
ii
x = 6, y = 2
Substitute x = 18
into first equation
2 × 18 + y = 50
y = 50 − 36
= 14
54
3
= 18
3
Check in second
equation
18 − 14 = 4
4
x = 18 and y = 14
2
Substitute y = 9
into first equation
b
1 Subtract the two
equations.
x + 4y = 41
−
x + 4 × 9 = 41
x + 2y = 23
0x + 2y = 18
2y = 18, y =
x = 41 − 36
=5
18
2
=9
3
Check in second
equation
5 + 2 × 9 = 23
4
x = 5 and y = 9
2
Substitute y = 4
into first equation
c
1 Subtract the two
equations.
3x + 2y = 38
3x + 2 × 4 = 38
− 3x −   y = 26
3x = 38 − 8
0x + 3y = 12
3y = 12, y =
3x = 30, x =
12
3
=4
3
30
3
= 10
Check in second
equation
3 × 10 − 4 = 26
4
b
Learner’s own answers.
a
i
x = 9, y = 4
ii
x = 10, y = 8
b
12 a
i
x = 2, y = 3
ii
x = 4, y = 8
x = 10 and y = 4
Learner’s own answer.
iYou can add or subtract. If you add,
you eliminate the ys, if you subtract
you eliminate the xs.
a
x = 5, y = 2 b
x = 16, y = 3
ii
Subtract to eliminate the xs.
c
x = 7, y = 4
d
x = 3, y = 6
iii
Add to eliminate the ys.
iv
Subtract to eliminate the ys.
10 Sofia is correct, x = −3 and y = 6. Zara got the
signs round the wrong way.
14
2
b
Learner’s own answer.
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
c
Learner’s own answer. For example:
Subtract to eliminate one of the letters
when the coefficients of that letter are the
same number and both positive or both
negative. Add to eliminate one of the
letters when the coefficients of that letter
are the same number and one positive and
one negative.
d
i
x = 9, y = 6
ii
x = −3, y = 2
iii
x = 8, y = 3
iv
x = 9, y = 5
5
b
x = 5, y = −2
c
x = 2, y = 4
d
x = 7, y = 1
14 a
x = 2, y = 2
7
a
2
3
4
–5
–4
–3
–2
–1
0
–4
–3
–2
–1
0
5
x<3
aHe has multiplied out the bracket
incorrectly.
b
i
x⩽2
b
x > −2
c
x ⩾ 10
d
x < −20
e
a
−2 ⩽ x < 2
f
−10 < x ⩽ 15
1
2
3
4
–5
–4
–3
–2
–1
–2
–1
0
1
True
ii
iii
0
–4 –3 –2 –1
0
x = −10
3(−10 + 2) ⩽ 2 × −10 − 5
False
–5
0
5
6
4
For x ⩽ −11 the substitutions give values
that are true and when x > −11 it gives a
false value.
7
8
9
8
a
4 ( 2 y + 3) − 5 y < 18 − y
8 y + 12 − 5 y < 18 − y
8 y − 5 y + y < 18 − 12
f
1
3(−11 + 2) ⩽ 2 × −11 − 5
−24 ⩽ −25
e
3
x = −11
True
d
2
3(−12 + 2) ⩽ 2 × −12 − 5
−27 ⩽ −27
0
–20 –15 –10
x = −12
−30 ⩽ −29
a
c
15
1
x ⩽ −11
b
4
0
3x − 2x ⩽ −5 − 6
Exercise 4.3
3
–1
3x + 6 ⩽ 2x − 5
Reflection: Learner’s own answers.
1
4
3(x + 2) ⩽ 2x − 5
3 × 2 + 2 = 6 + 2 = 8 and
4 × 2 + 2 × 2 = 8 + 4 = 12
0
3
b, c Learner’s own answers.
x = 9, y = 4
–1
2
d
6
13 a
2
1
c
All answers should be x = 6, y = 18
1
0
b
Activity 4.2
b
a
2
3
b
4
5
4y < 6
y < 1.5
6
a
7
c
−2, −1, 0 or 1
a
x>2
b
x⩽4
15 < 17
c
x < −3
d
x ⩾ −3
True
−4
b
i
y=1
4(2 × 1 + 3) − 5 × 1 < 18 − 1
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
ii
c
y = 1.5
When x = 5, 3 × 5 − 7 < 4 × 5 − 118 < 9
True
4(2 × 1.5 + 3) − 5 × 1.5 < 18 − 1.5
When x = 4, 3 × 4 − 7 < 4 × 4 − 115 < 5
False
16.5 < 16.5
False
iii
4(2 × 2 + 3) − 5 × 2 < 18 − 2
9
2<x⩽5
14 a
y = 2
18 < 16
0
False
b
a
a < 3.5
b
b ⩾ 11
c
c⩽6
d
d > −27
c
5n + 5 ⩽ 30
c
5, 12 and 13
b
2
d
b
Learner’s own answers.
c
Learner’s own answer. For example:
3
4
5
6
5
10
15
20
25
30
5
6
8
9
10 11
5
6
7
3<n<9
n⩽5
11 aLearner’s own answer. For example: To
make the x positive, Sergey adds x to both
sides and subtracts six from both sides.
He then rewrites the final inequality with
the x on the left and so he has to change
the < to >. To make the x positive, Natalia
divides both sides by −1, but this has the
effect of changing the < to >.
2
5 ⩽ y ⩽ 20
0
Learner’s checks for each solution.
10 a
1
3
4
7
−3 < m < 6
–4 –3 –2 –1 0
1
2
3
4
8
Check your progress
1
a
x = −4
b
a = −2.5
c
x = 2.4
d
y=9
e
m = 16
f
n = 10
Learner’s own checks for each solution.
2
x = 5, y = 19
2(x − 8) ⩾ 4x − 26
3
x = 19, y = 7
2x − 16 ⩾ 4x − 26
4
a
a<2
b
b⩾5
c
c > −1
d
d ⩾ −5
2x − 4x ⩾ −26 + 16
Learner’s own checks for each solution.
− 2x ⩾ −10
10 ⩾ 2x
5
−1 < x ⩽ 2
a
5⩾x
x ⩽ 5
12 a
–2
x > −4 or −4 < x
0
1
2
3
4
–2
–1
0
1
−4 < n < 1
b
x ⩾ 5 or 5 ⩽ x
c
x > 6 or 6 < x
d
x ⩽ −13 or −13 ⩾ x
e
x < 4 or 4 > x
Unit 5 Getting started
f
x ⩾ −2 or −2 ⩽ x
1
140 °
13 a
3x − 7 < 4x − 11
2
62 °
3
a
a and d OR b and e OR c and f
b
c and d
c
a and c OR d and f
–5
16
b
–1
b
For example:
3 x − 7 < 4 x − 11
−7 + 11 < 4 x − 3 x
4< x
x>4
–4
–3
2
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
4
5
The angle next to a = c (alternate angles);
the third angle at the same point is b
(corresponding angles); the 3 angles on a line
have a sum of 180 °.
a
Learner’s own diagram.
b
Each angle should be 37.5 °.
c
Learner’s own check.
8
9
a
Six triangles; 6 × 180 ° = 1080 °
b
Eight triangles; 8 × 180 ° = 1440 °
a
Exercise 5.1
1
60 °, 25 °, 95 °
2
a
x = 36, y = 50
c
A + B + C + D = 116 ° + 72 °
+ 122 ° + 50 ° = 360 °
b
122 °
Polygon
Number
of sides
Sum of
interior angles
triangle
3
180 °
quadrilateral
4
360 °
pentagon
5
540 °
hexagon
6
720 °
octagon
8
1080 °
decagon
10
1440 °
b
The sum of the angles = (n − 2) × 180 °
c
7 × 180 ° = 1260 °; correct because there are
seven triangles.
3
a = 40 °, b = 30 °, c = 70 °, d = 120 °
4
75
5
a
Trapezium. One pair of parallel sides.
10 a
b
A = 60 °, B = 120 °, C = 135 °, D = 45 °
11 144 °
6
C = 40 °, B = D = 100 °, A = 120 °
7
a
54 ° (angle of isosceles triangle AOB)
b
36 ° (angle BOC is 108 ° and triangle OBC
is isosceles)
c
90 ° = 54 ° + 36 °
8
9
b
c
45 ° + 51 ° = 96 °
A + B + C + D = 96 ° + 65 ° + 127 ° + 72 ° = 360 °
Exercise 5.2
17
There is no other way. Either the two
squares are adjacent or they have one
triangle between them on one side and
two triangles between them on the
other side. This way will look different
if it is reflected, but it is still the same
arrangement.
13 aLearner’s own diagram of a regular
arrangement of triangles.
1
110 °
2
40 °
3
136 °
4
a
103 °
b
128 °
5
a
88 °
b
128 °
6
a, bLearner’s own diagram of a hexagon split
into four triangles.
7
135 °
The second way could be drawn in a
reflected form.
Reflection: Learner’s own answer
10 a
b
12 a, b There are two ways:
x = 65 ° (angles on a straight line);
y = 45 ° = 115 ° (corresponding angles) − 70 °
(alternate angles)
105 °
100 °
c
4 × 180 ° = 720 °
d
120 °
a
109 °
b
100
b
Learner’s own diagram of a regular
arrangement of hexagons.
c
Because 108 ° is not a factor of 360 °.
d
Learner’s tessellations based on the two
drawings in Question 12.
e
Learner’s own diagram: two octagons
(135 ° angle) and one square (90 °) at every
point.
f
Learner’s own answer.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Reflection: In this case, subtract the 360 ° at the
centre. 5 × 180 − 360 = 540 gives the same answer.
Exercise 5.3
1
a–c Learner’s own diagram and explanation.
The explanation is the same as for a pentagon.
Walking round the hexagon you turn through
each angle in turn and the total is 360 °.
2
a = 99 °; b = 112 °; c = 125 °
3
a
Yes, vertically opposite angles.
b
Yes. They are not all on the same side, but
the vertically opposite angles will be the
same as you walk round the quadrilateral.
4
a
120 °
5
a
360 °
6
a
b
c
90 °
b
360 ÷ 8 = 45 °
Sides
Exterior
angle
Equilateral triangle
3
120 °
Square
4
90 °
Regular pentagon
5
72 °
Regular hexagon
6
60 °
Regular octagon
8
45 °
Regular decagon
10
36 °
b
The exterior angle = 360 ÷ n degrees
c
i
7
a
9
8
a
i
150 °
ii
160 °
iii 170 °
b
i
12
ii
18
iii
9
30 °
ii
18 °
b
140 °
The answers to all the questions in this exercise are
diagrams. Each question asks the learner to check
their accuracy either by measuring themselves or
by asking a partner to measure.
Question 12 asks learners to think about
whether there are different ways to complete the
construction. They should be able to decide which
method is easier or more likely to give an accurate
drawing.
1
a
10 cm
b
13 cm
c
17 cm
2
a
4.3 cm
b
12.1 cm
c
14.2 cm
3
a
12 cm
b
4.8 m
c
75 mm
4
a
6.6 cm
b
5.0 cm
c
13.5 m
5
a
2
b
3
c
4 =2
d
Learner’s own diagram. A continuation of
the spiral pattern.
e
The 3rd hypotenuse is 2, the 8th hypotenuse
is 3 and the 15th hypotenuse is 4.
a
392 + 70 2 = 80 cm to the nearest cm.
b
1052 + 582 = 120 cm to the nearest cm.
6
10 a
8
b
11 a
360 − 2 × 135 = 90
12
c
20
3.50 2 − 0.912 = 3.38 m to the nearest cm.
7
8
36
15 sides
b
d
a
Learner’s drawing.
b
5.12 + 6.82 = 8.52, so it is a right-angled
triangle.
c
5.12 + 6.82 = 72.25 = 8.52. The triangle
satisfies Pythagoras’ theorem, and so is
right-angled.
24
Learner’s own diagram.
12 (360 − 60) ÷ 2 = 150 ° is the interior angle. The
exterior angle is 180 − 150 = 30 °. The number
of sides is 360 ÷ 30 = 12.
13 Interior angle 168 ° means exterior angle 12 °
and 360 ÷ 12 = 30 so it has 30 sides. Interior
angle 170 ° means exterior angle 10 ° and
360 ÷ 10 = 36 so it has 36 sides. But interior
angle 169 ° means exterior angle 11 ° and 11 is
not a factor of 360 so that is not possible.
18
Exercise 5.4
Exercise 5.5
72 °
Regular polygon
Reflection: Yes they do. Check with some values
for n. It is easier to see if you write (n − 2) × 180 ÷ n
as (180n − 360) ÷ n
9
Either 152 + 20 2 = 25 cm or
20 2 − 152 = 13.2 cm to 1 d.p.
10 a
b
90 + 40 = 130 m
130 − (90 2 + 40 2 ) = 31.5 m to 1 d.p.
11 aSquare perimeter = 4 × 25 = 100 mm,
rectangle perimeter = (2 × 20) + (2 × 30) =
40 + 60 = 100 mm
b
Diagonal of square = 35.4 mm; diagonal
of rectangle = 36.1 mm
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
c
Learner’s diagram and value.
3
d
The values so far support Sofia’s
conjecture and any further values should
too. The square has the minimum diagonal
for a given perimeter. All the examples
here are for a perimeter of 100 mm, but it
is true for any given perimeter.
Exercise 6.1
12 There are two possible answers. Either the two
shorter sides are 1 and 4 OR the hypotenuse is
9 and one of the other sides is 8.
These are suggested answers but there are many
other possibilities. It is not possible to give a
complete list of answers.
1
13 a7.52 + 5.52 = 86.5 and so length of
diagonal = 86.5 .
b
x2 + 5.52 = x2 + 30.25 and so length of
diagonal = x 2 + 30.25 .
c
d =
x 2 + y2
i
7 +7 =
ii
98 =
14 a
b
2
2
x2 + x2 =
A number is assigned to each person. 50
numbers between 1 and 632 are generated.
Any number that is a repeat is ignored.
Learner’s own answers.
a
For example: Can boys estimate more
accurately than girls? Can learners
estimate acute angles more accurately
than obtuse angles? Can learners
accurately estimate how long one
minute is?
b
For example: Girls can estimate the length
of a short line more accurately than boys.
Older learners can estimate an obtuse
angle more accurately than younger
learners. Learners tend to underestimate
one minute of time.
c
Learner’s own answers. This will depend
on the predictions. For example: Methods
could take names from a hat or use
random numbers. The method could
take learners from different groups in the
school.
98
49 × 2 =
72 × 2 = 7 2
2x 2 = x 2
Check your progress
1
a = 65. The reason could use corresponding
angles and the exterior angle of a triangle.
2
116 ° (x = 106)
3
10 sides
d
Learner’s own answer and explanation.
4
a
Learner’s own diagram.
e
Learner’s own answer.
b
Each side should be 8.5 cm.
f
Learner’s own generalisation, depending
on their data.
5
35 m or 35.3 m or 35.36 m are possible answers.
6
x = 10 and y = 24
2
Learner’s own answers.
a
For example: Are lessons too long? Are
there too many lessons in a day? Should
school start earlier in the day?
b
For example: Learners want longer
lessons. Learners want fewer lessons in a
day. Learners would prefer to start school
one hour later.
c
Learner’s own answers. This will depend
on the predictions. For example: The
method could take learners from different
groups in the school.
Unit 6 Getting started
In many questions these are suggested answers and
there are many other possibilities. It is not possible
to give a complete list of answers.
1
2
19
Learner’s own answers.
a
For example: length or width.
b
For example: number of doors or
passenger seats.
c
For example: colour or manufacturer.
d
Learner’s own answer and explanation.
Learner’s own answer. For example: Using
random numbers of position on the register.
It could include a specific number from each
year group.
e
Learner’s own answer.
f
Learner’s own generalisation, depending
on their data.
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
3
7
Learner’s own answers.
a, b For example: Questions and predictions
could be about lengths of words, lengths of
sentences, lengths of articles or vocabulary
used.
a
iiWhat do you think is the cause of
global warming?
b
c, d, e Learner’s own answers. This will
depend on the predictions.
f
c
Reflection: Learner’s own suggestions about
making predictions and choosing a sample to
test them.
aTo encourage people to buy Supremo
Shampoo.
b
3
4
5
6
For example: Sample choice, asking a
question suggesting a particular answer,
people giving an answer they think the
questioner wants.
aFor example: It is cheap. It is quick. It
gives a large sample.
b
For example: Many people do not use
social media. Many people will not
reply. People who reply might only do so
because they have a strong opinion.
a
8
c
Learner’s own explanation. For example:
The vertical axis starts at 30 and not
at zero.
d
Learner’s own diagram. The vertical axis
should start at 0, and they should use a
uniform scale.
a
30%
b
The people who reply might all have a
similar opinion and not be representative.
b
26
aThe questioner is suggesting the answer
they want, i.e. ‘yes’.
b
20
d
2
For example: Do not let the person know
which drink is the new recipe. Ask ‘Which
drink do you prefer?’. Arrange for half
the people to have the original drink first
and for half of the people to have the
original drink second.
iPeople will not want to admit they are
overweight.
iiThe question is too personal. A better
question would be, for example, ‘Do
you weigh less than …’ and give a
particular value.
Exercise 6.2
17 girls and 13 boys
iPeople are likely to say ‘yes’.
iiWhat is a fair price for entry to this
exhibition?
Learner’s own generalisation, depending
on their data.
1
iIf you ask people to agree with you,
they might do so just to avoid conflict.
iPeople might not know what ‘enough
exercise’ is. They might say they do
enough exercise when they do not.
iiHow many times a week do you
take exercise, such as walking for 30
minutes, cycling or going to a gym?
8
People are more likely to reply if they have a
complaint.
9
A good survey would choose men and women
of different ages in the correct proportions
questioned at different times of the day. These
are the numbers required:
Men
Women
Under 30
15
15
30 or more
45
45
Ask the first question about age. When the
required number has been reached, do not ask
any more people in that particular category.
10 aNo. Learner’s own explanation. For
example: The sample is too small to make
a valid conclusion.
b
Learner’s own explanation. For example:
The scale does not start at zero, which
makes the proportional differences
between men and women look greater
than they really are.
c
Learner’s own diagram. The vertical axis
should start at 0, and they should use a
uniform scale.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Check your progress
1
3
a
diameter = 16 cm
aWhich cake do you think tastes best?
Which cake looks most attractive?
Do you dislike any of the cakes?
b
2
2
A = πr2
= 3.142 × 82
= 3.142 × 64
People will prefer type A. Type A looks
most attractive. Most people dislike Type A.
= 201.09 cm 2 (2 d.p.)
Learner’s own answer. For example: Including
random numbers or using registers and a
particular number from each year.
b
diameter = 9 cm
aIt will be biased towards people travelling
to work.
b
= 3.142 × 4.52
= 3.142 × 20.25
Unit 7 Getting started
44 m2
diameter = 2.6 m
a
2
4.8 cm or 48 mm
3
a
4
Group 1: A, D, G, H; Group 2: B, F;
Group 3: C, E
= 3.142 × 1.32
= 3.142 × 1.69
5
a
320 000
b
560 000 000
= 5.31 m 2 (2 d.p.)
c
6.82
d
4.5
34 cm2
b
c
= 63.63 cm 2 (2 d.p.)
r = d ÷2
= 2.6 ÷ 2
= 1.3 m
1
37.70 cm
21.99 m
r = d ÷2
= 9÷2
= 4.5 cm
A = πr2
Choose people on trains on different days
and at different times of day.
b
r = d ÷2
= 16 ÷ 2
= 8 cm
A =πr 2
3
Exercise 7.1
1
a
153.938 cm2
b
i
153.86 cm2 ii 153.958 cm2
iii
154 cm2
i
0.05% ii
iii
0.04%
A = πr2
radius = 2 cm
= 3.14 × 22
= 3.14 × 4
c
= 12.6 cm (1 d.p.)
d
π = 3.142
e
Learner’s own answers and explanations.
For example: It is best to use the π button
for the most accurate answer, but if
you have to use an approximation, then
π = 3.142 is the best to choose as it gives
an approximate answer closest to the
accurate answer.
a
113 cm2
b
56.7 m2
c
415 cm2
d
18.1 m2
2
b
a
A = πr
radius = 9 cm
2
= 3.14 × 9
= 3.14 × 81
2
= 254.3 cm 2 (1 d.p.)
c
radius = 4.2 m
A = πr2
= 3.14 × 4.22
= 3.14 × 17.64
4
= 55.4 m 2 (1 d.p.)
5
0.01%
aLearner’s own answers and explanations.
For example: Ellie has made the mistake
of multiplying the radius by pi and then
squaring, rather than squaring the radius
and then multiplying by pi.
Hans has made the mistake of multiplying
the radius by 2, rather than squaring the
radius.
21
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
15 a
3.14 × 1.7 2 = 3.14 × 2.89
= 9.0746
b
Area = 9.07 m 2 (3 s.f.)
6
A = π 
2
a
i
A = 98.5 cm2
ii
C = 35.2 cm
i
A = 804.2 mm
ii
C = 100.5 mm
d
 2 
7
b
8
c
Learner’s own answers.
a
or A =
45π cm
iv
400π cm2
i
A = πr 2 =
1
× π × 144
1
1
2
2
× π × 122 =
2
1
1
2
2
× π × 24 + 24 = 12π + 24 m
Exercise 7.2
1
1
2
2
1
× π × 6.22 = 60.38 cm 2
2
1
× 3 × 36 = 54 cm 2 ;
2
1
× π × 14.852 = 346.40 m 2
b
1
1
2
1
2
d
r = 9.64 m;
2
2
Accurate: A =
1
× π × 9.642 = 145.97 m 2
2
i
A = 245.4 m2 ii P = 64.3 m
b
i
A = 831.0 mm2
ii
2
3
Learner’s own answers.
Area of semicircle = 10.618 cm , Area of
quarter-circle = 9.0792 cm2 and 10.618 > 9.0792.
2
Learner’s own answers and explanations.
b
Learner’s own answers and explanations.
c
i
ii 2.4 m
× 12 × 6 = 36
Area A = l × w = 5 × 12 = 60
1
1
2
2
× π × 62 = 56.55
Area rectangle = l × w = 4 × 1.5 = 6
i
3 cm
ii
68 cm2
b
i
7 cm, 8 cm
ii
98 cm2
c
i
7 cm
ii
138 cm2
a
i
7 × 4 + 0.5 × 7 × 5 = 45.5 cm2
ii
48.1 cm2
i
3 × 3 + 0.5 × 3 × 1.52 = 12.375 m2
ii
10 m2
i
0.5 × 4 × 10 + 0.5 × 3 × 52 = 57.5 cm2
ii
50.5 cm2
c
d
iii 9.0 mm
12 a, b A and v, B and i, C and vi, D and iii,
E and iv, F and ii
2
a
b
10 Marcus is correct.
11 a
1
Shaded area = 28.27 − 6 = 22.27 cm2
P = 118.3 mm
Activity 7.1
×b×h =
Area circle = πr2 = π × 32 = 28.27
× 3 × 100 = 150 m 2 ;
a
2
Total area = 60 + 56.55 = 116.55 cm2
× π × 7.352 = 84.86 cm 2
1
1
Area B = π r 2 =
× 3 × 49 = 73.5 cm 2 ;
1
Area A =
Total area = 36 + 36 = 72 cm2
c
2
Area A = l × w = 5 × 4 = 20
Area B = l × w = 12 × 3 = 36
r = 7.35 cm;
3.3 cm
a
Total area = 20 + 22 = 42 cm2
× 3 × 225 = 337.5 m 2 ;
2
2
1
Area B = l × w = 11 × 2 = 22
1
Estimate: A ≈ × 3 × 102 =
i0.5 × 3 × 302 + 0.5 × 3 × 152 = 1687.5 mm2.
The following could be accepted as
an alternative:
0.5 × 3 × 302 + 0.5 × 3 × 152 = 1687.5
ii
4
14 84 m2
5
22
iii
Reflection: Learner’s own answers.
Estimate: A ≈ × 3 × 152 =
13 16.44 m
144π mm2
ii P = π d + d =
Estimate: A ≈ × 3 × 62 =
Accurate: A =
9
ii
= 72π m 2
Estimate: A ≈ × 3 × 7 2 =
d
25π mm
2
Accurate: A =
c
i
πd 2
4
Accurate: A =
b
Learner’s own answers and explanations.
1539.4 mm2
a
Learner’s own answer.
b
Learner’s own answers and explanations.
c
Learner’s own discussions.
a
34 cm2 b 34.365 cm2 c
187.56 mm2
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
6
You can also say that there are one billion
nanometres in a metre or 1 nanometre is
one billionth of a metre.
Sofia is correct, the two shaded areas are
approximately the same size.
Area of 1st shape = 86.31 cm2, Area of 2nd
shape = 87.96 cm2
2
Activity 7.2
Learner’s own answers.
7
a
b
8
9
i
18(π − 2) cm2 ii
50(π − 2) cm2
iii
72(π − 2) cm2
iv
4.5(π − 2) cm2
Learner’s own answer. For example: The
answer is always a number times the
bracket π − 2. The number outside the
bracket is always half of the square of the
radius.
c
1
d
Learner’s own discussions.
2
aA kilolitre is a very large measure of
capacity. It is represented by the letters kL.
1 kilolitre = 1000 litres which is the same
as 1 kL = 1 × 103 L.
You can also say that there are one
thousand litres in a kilolitre or 1 litre is
one thousandth of a kilolitre.
b A gigametre is a very large measure of
length. It is represented by the letters Gm.
1 gigametre = 1 000 000 000 metres which is
the same as 1 Gm = 1 × 109 metres.
You can also say that there are one billion
metres in a gigametre or that 1 metre is
one billionth of a gigametre.
r 2 ( π − 2)
Learner’s own answers and explanations. For
example: The shaded areas are the same as
they are both ‘Area of square of side length
10 cm − Area of circle of radius 5 cm’. The
areas of both are 21.46 cm2.
3
a8 micrometres, 8 millimetres, 8 centimetres,
8 metres, 8 kilometres, 8 gigametres
b
4
aLearner’s own answers and explanations.
For example:
aWhen radius = 4, Area of
circle = π × 42 = 16π.
Marcus is correct. 1 tonne = 1000 kg.
Also 1 kg = 1000 g and
1 Mg = 1 000 000 g = 1000 kg = 1 t.
When radius = 4, side length of
square = 4 × 2 = 8 cm. Area of
square = 8 × 8 = 64.
Arun is incorrect. 1 litre = 1000 mL and
1 litre = 100 cL, so 1000 mL = 100 cL
→10 mL = 1 cL, not 100 mL = 1 cL
Shaded area = 64 − 16π = 16(4 − π) cm2.
b
c
d
i
25(4 − π) cm2
ii
9(4 − π) cm2
iii
36(4 − π) cm
iv
100(4 − π) cm
2
2
Learner’s own answers. For example:
The answer is always a number times the
bracket 4 − π. The number outside the
bracket is always the radius squared.
5
r2(4 − π)
Exercise 7.3
1
23
8 μm, 8 mm, 8 cm, 8 m, 8 km, 8 Gm
aA milligram is a very small measure of
mass. It is represented by the letters mg.
1 milligram = 0.001 grams which is the
same as 1 mg = 1 × 10−3 g.
You can also say that there are one
thousand milligrams in a gram or
1 milligram is one thousandth of a gram.
b A nanometre is a very small measure of
length. It is represented by the letters nm.
1 nanometre = 0.000 000 001 metres which
is the same as 1 nm = 1 × 10−9 m.
6
b
Learner’s own discussions.
c
Learner’s own answers and explanations.
d
Learner’s own discussions.
a2.5 Mm to m → 1 Mm = 1 000 000 m, so
2.5 Mm = 2.5 × 1 000 000 = 2 500 000 m
b
0.75 GL to L →1 GL = 1 000 000 000 L,
so 0.75 GL = 0.75 × 1 000 000 000
= 750 000 000 L
c
13.2 hg to g → 1 hg = 100 g, so 13.2 hg
= 13.2 × 100 = 1320 g
a364 cL to L → 100 cL = 1 L, so
364 cL = 364 ÷ 100 = 3.64 L
b
12 000 mg to g → 1000 mg = 1 g, so
12 000 mg = 12 000 ÷ 1000 = 12 g
c
620 000 μm to m → 1 000 000 μm = 1 m,
so 620 000 μm = 620 000 ÷ 1 000 000
= 0.62 m
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
7
From Earth to:
Distance in …
Unit 8 Getting started
Mars
78.34 Gm
1
Jupiter
628.7 Gm
Saturn
1.28 Tm
Uranus
2.724 Tm
2
Neptune
4.35 Tm
3
a
a
c
6
a
68
b
10
1
3
1
2
b
1
c
10
8
A and v, B and iv, C and i, D and iii, E and ii
4
a
9
aLearner’s own answers and explanations.
For example:
5
a
b
b
7
20
1
a
2
d
1
2
1
= 2 × = 2 × 0.25 = 0.5 which is a
c
9 460 000 000 000 000
3
d
6 × 9 460 000 000 000 000
= 56 760 000 000 000 000
= 5.676 × 1016
4
4
4
1
= 3 × = 3 × 0.25 = 0.75 which is a
4
terminating decimal
b
Learners own discussions.
1
= 0.2 which is a terminating decimal
5
2
1
5
D, B, C, A
= 2 × = 2 × 0.2 = 0.4 which is a
5
terminating decimal
b
2 147 483 648 bytes
4
c
10 880 photos
5
d
1864 films
11 Learner’s own answers and explanations. For
example: Magnar is incorrect. The fastest is
model B because 10 ns is quicker than 40 ns
and 60 ns.
2
a
39.27 cm
b
21.36 m
2
a
123 cm2
b
36.3 m2
3
49.1 cm2
4
170 cm
5
a5 nanograms, 5 micrograms, 5 milligrams,
5 grams, 5 kilograms, 5 tonnes
3
2
5 ng, 5 μg, 5 mg, 5 g, 5 kg, 5 t
4
5
a
b
Recurring decimal.
c
All recurring decimals.
.
2
ii
i
= 0.2
9
.
4
iv
iii
= 0.4
9
.
6
vi
v
= 0.6
9
.
8
vii = 0.8
9
.
.
3 1
6
2
= = 0.3 and = = 0.6
Check your progress
1
1
= 4 × = 4 × 0.2 = 0.8 which is a
terminating decimal
.
1
= 0.1
Reflection: Learner’s own answers.
b
7
15
1
= 0.25 which is a terminating decimal
4
terminating decimal
10 a
5
12
Exercise 8.1
299 792 458 × 60 × 60 × 24 × 365.25
= 9.460 730 473 × 1015
e
24
= 0.625 terminating
8
.
5
= 0.83 recurring
6
1
b 61
5
3
2
b
Sofia is correct.
300 000 000 × 60 × 60 × 24 × 365.25
= 9.467 28 × 1015, which rounds to
9.47×1015.
5
a
9
9
3
9
.
3 = 0.3
9
.
5
= 0.5
9
.
7
= 0.7
9
3
b
Learner’s own discussions. Their answers
.
are not different because 9 = 0.9 = 1.
a
1
= 0.125
8
b
Terminating decimal.
9
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
c
d
They are all terminating decimals.
i
2
= 0.25
8
ii
3
= 0.375
8
iii
4
= 0.5
8
iv
5
= 0.625
8
v
6
= 0.75
8
vi
7
= 0.875
8
aAlways true: 7 is odd and a prime number,
so all fractions with a denominator of
1
7 cannot be simplified. is a recurring
7
decimal, so all fractions with a
denominator of 7 are recurring.
b
Sometimes true: For example: 1 , the
6
denominator is a multiple of 2, and the
fraction is a recurring decimal. However,
it is not always true because they can
also be terminating decimals, e.g. 1 , the
4
denominator is a multiple of 2, and the
fraction is a terminating decimal.
c
Sometimes true: For example:
Learner’s own answers. The following
three fractions can be simplified.
ci
v
5
7
6
8
2
=
1
=
3
= 0.75
3
=
8
aNo,
6
= 0.25, iii
4
4
1
2
4
8
=
1
2
= 0.5 and
= 0.5 which is not a recurring
denominator is a multiple of 10, and
the fraction is a terminating decimal.
However, it is not always true because
they can also be recurring decimals
1
e.g. , the denominator is a multiple of
30
10, and the fraction is a recurring decimal.
decimal.
b
Yes.
c
Learner’s own explanations. For example:
6 is even, so it can be halved.
So 3 = 1 . However, 7 is odd and so it
6
2
cannot be halved, so there is not an
equivalent fraction such that ? = 1 .
d
7
15
6
c
2
2
2
1
1
etc.
Each decimal
=
0
.
0625
,
=
0
.
03125
,
24
25
can be divided by 2 to get the next
decimal in the sequence, so they will all be
terminating.
8
5
They are still recurring decimals. Learner’s
own explanations. For example: The
fractions that can be cancelled down still
have a denominator with a multiple of 3,
and once cancelled are not even.
d
aLearner’s own answers and explanations.
For example: Recurring decimals. All
the denominators are multiples of 7 and
they are all written in their simplest form
(apart from E).
b
Learner’s own answers and explanations.
For example: E is not written in its
simplest form, but when it is, it is
1
equivalent to which is recurring. So it
14
doesn’t change the answer to part a.
c
Learner’s own answers. For example: She
must add ‘when it is written in its simplest
form’ so her statement now is: Any
fraction which has a denominator that is
a multiple of 7, when it is written in its
simplest form, is a recurring decimal.
a
20 1
= recurring
60 3
b
36 3
= terminating
60 5
c
45 3
= terminating
4
60
Learner’s own explanations. For example:
3
1
B is now 3 = 1 , D is now = , E is now
12
4
6
2
3
1
= . These are all terminating decimals.
15 5
No. Learner’s own discussions.
Never true: A fraction with a denominator
which is a power of 2 is a terminating
decimal. 1 = 0.5, 12 = 0.25, 13 = 0.125,
aRecurring decimals. Learner’s own
explanations. For example: The
denominators are multiples of 3. The
numerators are all 1.
b
25
d
2
Learner’s own investigations and answers.
For example: If the denominator is
even, then there will be a fraction such
? 1
that = which will not be a recurring
? 2
decimal. If the denominator is odd and
the unit fraction is a recurring decimal,
then it’s possible that all the fractions with
the same denominator will be recurring
decimals as well. However, there are
1
exceptions such as: is recurring, but
15
3
1
= = 0.2 which is terminating.
1
, the
20
9
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
d
e
f
55 11
= recurring
60 12
8
2
1 = 1 recurring
60
15
21
7
3 = 3 terminating
60
20
b
7
10
Rewrite 10:
5
7
5×7
35
7
× =
= =
6 10 6 × 10 60 12
12
10 = 9
12
Subtraction:
9
Multiplication:
10 a
recurring
b
terminating
c
recurring
d
terminating
c
11 The fractions written in their simplest
form are:
5 ÷ +  
3
4
2
 3
12
7
5
− =9
12 12
12
2
2 2
2×2
4
× =
=
3 3
3×3
9
 2
 3 
2
Brackets:
3
4
20
= 5× =
4
3
3
=
Abi
21 1
=
168 8
Bim
28 1
=
168 6
Caz
32
4
=
168 21
Division: 5÷
Dave
35
5
=
168 24
Enid
40
5
=
168 21
Fin
42 1
=
168 4
Addition:
20 4 60 4 64
1
+ = + = =7
9
9
9
9
9
3
a, b Learner’s own decisions on how to sort
the friends into two groups.
2
For example:
Abi and Fin – the fractions they work are
terminating decimals.
3
Bim, Caz, Dave and Enid – the fractions
they work are recurring decimals.
a
2
c
2
3
1
4
d
3
3
4
aLearner’s own answers. For example:
7 + 3 − (6 − 3) = 10 − 3 = 7
1
12
OR
c
Learner’s own answers and explanations.
Abi, Bim and Fin – the fractions they
work are unit fractions.
d
Learner’s own discussions.
a
i
9 − (2 + 4) = 9 − 6 = 3
ii
3
3
40
b
i
8 + (2 − 1) = 8 + 1 = 9
ii
9
5
24
c
i
5 + 2 × 16 = 5 + 32 = 37
ii
42
4
9
d
i
16 − × = 16 − = 15
ii
15
11
24
a
Learner’s own answers.
b
Learner’s own answers. For example: It
might be easier to work with the whole
numbers and fractions separately and not
convert into improper fractions.
a
25 −  5 + 8
b
Learner’s own answer and explanation.
For example: Her estimate is too long as
the length of her third side is more than
the sum of the other two sides, which is
not possible in a triangle.
c
19
11 . Learner’s own answer and
4
Abi, Bim, Dave and Fin – the
denominators of the fractions they work
are even numbers.
Caz and Enid – the denominators of the
fractions they work are odd numbers.
5
etc.
Activity 8.1
Learner’s own answers.
Reflection: Learner’s own answers.
6
Exercise 8.2
a
b
7
OR
1
5
16
b
Caz, Dave and Enid – the fractions they
work are not unit fractions.
5 +  − 
2
3
3
5
1
2
Brackets:
Addition:
26
5
6
10 − ×
3 1
6
5
1
− = − =
5 2 10 10 10
2
1
20
3
23
5 + =5 + =5
3 10
30 30
30
1
 9
1
1
1
3
2
2
4
4
1
7
7
 or 25 − 5 − 8
15
9
15
45
explanation. For example: Yes, the third
side is less than the total of the other two
sides.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
7
8
1
4
3 kg
2
5
1
4
4
13
4
4
a
1
7
×5 =
b
65
b
m2
8
8
9
c
1
cm 2
8
c
d
3 2
m
11
1
3
1
2
1
1
1
2
1
2
1
2
1
4
4
5
1
1
1
1
1
2
2
2
2
4
1 ×1 = 1+ + +
1
= 2 . Marcus’s
4
method gets this answer.
1
Arun’s method only gets the 1 and the , it
4
1
doesn’t get the other two s.
2
b
General rule: change the mixed number
to an improper fraction. Square the
numerator, square the denominator.
Change the answer back to a mixed
number.
3
4
5
9
c
11 a
3
12 a
1
1
 1
1  1
1
 2  + 2 × 5 or 2 3 ×  2 3 + 5 2 
3
3
2
b
29
6
7
Learner’s own discussions.
18
18
a
b
3
5
5 2
m
18
c
3
3
3
× 12 = × 12 = 3 × 3 = 9
4
1 4
5
7
4
5
d
3
8
27
× 28 =
× 45 =
× 72 =
5
17
4
15
3
18
7
4
× 45 =
4
3
5
20
3
3
7
9
2 10
× 45 =
7
2
= 13
2
52
× 13 =
× 8 = ×4=
36
16 ×
11
11
2
= 16 ×
24
3
=6
1
2
1
= 17
3
2
3
63
×9 =
2
24 3
= 2×
11
3
22
=
3
c
highest common factor
d
Learner’s own discussions.
a
84
b
140
c
1
2
2
10
21
8
39
6
35
1
8
d
22
a
= 31
1
2
=7
1
3
1
2
5
16
2
c
d
3
3
e
f
8
1
a
b
4
1
Lewis is correct, he travels 183 km which is
3
b
8
Exercise 8.3
1
39
=
27
×9 =
more than 180 km.
2
b
×8 =
13
× 39
2
aLearner’s own discussions. For example:
She cancelled using a common factor of
4, but she should have cancelled using the
highest common factor of 8.
b
2
2
4
5
3
9
× 36 =
28
× 39 =
10
not give the same answer. This can be
shown using a multiplication box.
1
5
6
10 aLearner’s own answer and explanation.
For example: They get different answers.
Marcus is correct. His method does
1
1
multiply 1 by 1 . Arun’s method
2
2
multiplies 1 by 1 and 1 by 1 , which does
×
4
3
× 36 =
4
30 65 95
3
+ = = 23
4
4
4
4
Addition:
3
8
Division: 6 ÷ 4 = 6 × 5 = 30
Multiplication: 3 × 5 =
9
a
Estimate
Accurate
a
1
3
1 ×3
2
5
1
1 ×4=6
2
5
2
5
b
2 ×3
1
4
2
3
2×3 = 7
1
2
8
1
4
c
1 ×3
1
8
1
6
1× 3 = 3
1
2
1
2
3
d
3 ×1
2
3
5
22
3 ×1 = 3
1
2
4
e
3 ×4
3
4
3
5
4 × 4 = 18
1
2
17
1
4
f
4 ×2
4
7
5
16
4 × 2 = 10
10
4
7
1
2
1
2
9
16
1
2
4
× 28 = 5 × 4 = 20
9
× 45 = 4 × 9 = 36
9
× 72 = 3 × 9 = 27
9
aLearner’s own working. For example:
1
2
8 × = 4 and 4 < 8, 4
5 × 3 = 1 and
9 10
6
1 2
1
× = 3 and 3 < 4 ,
2 3
2
1 5
<
6 9
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
iWhen you multiply any number by
an improper fraction, the answer will
always be greater than the original
number.
5
c
smaller, 2
c
bigger, 1 1
11 a
1
4
b
bigger, 8
4
5
2÷2=1
b 2 ÷1
1
4
2
3
2÷2=1
4 ÷5
1
8
1
6
4÷5=
4
5
99
124
d 2 ÷3
2
3
1
4
3÷3=1
32
39
e 5 ÷2
1
2
3
4
6÷3=2
2
4
5
2
3
5÷ 3 =1
2
3
1
4
5
1
4
10
11
1÷1=1
1
3
8
1
2
2
7
c
1
3
2
1
1
17
A , B 1 , C 9 , D 1, E 1 5 , F 2
12
3
16
33
9
3
b, c Learner’s own decisions on how to sort
the cards into two groups.
For example:
A, D and F are proper fractions; B, C and
E are improper fractions.
6
OR
f
4 ÷2
g
1 ÷
h
3
1
÷2
5
10
3÷
1
4
A, B, C, D and F have a denominator
which is a multiple of 3; E does not have a
denominator which is a multiple of 3, etc.
Exercise 8.4
a
b
3 7
5
21 ÷ = 21 × = 7 × 5 = 35
5
31
c
14 ÷ = 14 ×
d
8÷
2
9
7
9
= 7 × 9 = 63
21
7
8
4 2
11
= 8 × = 2 × 11 = 22
11
41
2
A and iii, B and i, C and iv, D and v, E and ii
3
a
8 4
8
7
2 × 7 14
5
÷ =
×
=
= =1
9 7
9
9 ×1
9
9
41
b
7
2 7 5
7 × 5 35
17
÷ = × =
= =1
5 9 2 9 × 2 18
18
9
2
2
2
6
3
6
14
÷ =
× 1
7 14
3
17
d
5 15
5 24
5 × 24
4
1
÷ = × =
= =1
6 24 6 15
3
3
1 6 × 15 3
=
1
4
28
2×2
=4
1×1
c
C
7
20
2
1
= 6 and 6 > 3, 1 ÷
1 5
2 8
2
1
= 2 and
2 3
4
3 5
1
3
= 3 and 3 >
4 8
6
4
iWhen you divide any number by an
improper fraction, the answer will
always be smaller than the original
number.
iiWhen you divide any number by a
mixed number, the answer will always
be smaller than the original number.
Reflection: Learner’s own answers.
1
1
2 >1 , ÷
b
4 4
7
16 ÷ = 16 × 1 = 4 × 7 = 28
7
4
1÷ 2 =
1
aLearner’s own working. For example:
B and E have an even number for the
denominator; A, C, D and F have an odd
number for the denominator.
OR
5
6
1
2
Learner’s own discussions.
10 a
Accurate
1 ÷1
a
iiWhen you multiply any number by a
mixed number, the answer will always
be greater than the original number.
Estimate
9
c
Learner’s own discussions.
a
bigger, 9 1
b
smaller, 4
c
smaller, 2
3
2
21
Learner’s own answer and explanation. For
example: His conjecture is not true. If you
divide a mixed number by a larger mixed
number the answer will be a proper fraction,
not a mixed number, e.g. 2 1 ÷ 3 1 = 10
a
c
e
14
15
1
1
7
11
27
2
4
b
6
2
7
1
1
9
1
1
11
d
f
13
4
20
1
9
9
, D , B 1 , A1
9
31
21
10
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Dae’s method – advantage: can work on
one step at a time and could easily do this
method mentally, disadvantage: method is
longer (which learners might not like).
10 aLearner’s answer and explanation.
For example: π ≈ 3, and
diameter = circumference ÷ π. 15 cm is
slightly more than the circumference,
15 ÷ 3 = 5, so the diameter will be just
under 5 cm.
1
7
14 ÷
b
11 a
1
2
3
22 99
7
9
1
= × = =4
7
7
22
2
2
1
b 1
2
c
1
b
Learner’s own answers and explanations.
c
Learner’s own answers and explanations.
iFor example: Dae’s method because
when 14 is multiplied by 2.5 it gives a
whole number.
1
3
1
2
12 92 km/h
2
iiFor example: Akeno’s method
because 15 cannot be divided by 2
exactly, so it is easier to use improper
fractions and to work out the answer
as a mixed number.
13   + 1 ÷ 5 is greater
2
 3
5
6
1
2
2
3
27
 1 4 34
= = and  2  + 1 5 ÷ 5 1 = 7 = 28
 2 −  ÷
 3
2 5
15
4
36
6
2 9
36
6
Exercise 8.5
1
a
0.28 × 52 ⇒ 0.28 =
a
2
1
2
1

1
 + 5.5 − 1 ⇒  + 5  = ( 6) = 36 ⇒ 36 − 1 = 35
2
2
2
3
3
⇒ 27 + 23 = 50
c
 3

 3 7
6 −  3 + 0.7 ⇒  3 +  = 4 ⇒ 62 = 36
 10

 10 10 
2
⇒ 36 − 4 = 32
2
a
3
a
c
3
3
 1

 1 1
 3 − 0.2 + 23 ⇒  3 −  = (3) = 27
5
5 5
b
48
c
49
12
1
4 100
3
2
6
13 3
13
, 4 − 4 = 60 ⇒
× 60 = 78
10
1 10
7
a
8
2 m2
9
aLearner’s own answers and explanations.
For example: Write the decimal as a
fraction, square the fraction then multiply
by the mixed number which has been
written as an improper fraction.
1
b
c
9
10
b
1
2
5
4
1.25 × 3 × 56 ⇒ ×
7
7 35
35
= ⇒
× 56 = 35 × 7 = 245
2
8
8
1
Learner’s own answers and explanations.
c
Learner’s own discussions.
d
0.82 × 7 = 4 ; example strategy:
0.8
c
2
3
11
4
4
2.75 × 18 ⇒ 2 × 18 =
11
2 4
9
× 18 =
2
= 49
a
5
aLearner’s own answers. For example:
Akeno’s method – advantage: shorter,
disadvantage: involves changing decimals
to improper fractions and cancelling before
multiplying (which learners might not like).
108
c
99
4
126
b
× 18 ⇒
41
4
For example: Fraction, because 3 and 9 are
both recurring decimals so it is easier to
write them as fractions.
10
5 15
15
= ⇒ × 40 = 15 × 10 = 150
2
4
4
1
b
× 25 = 7
1
7
2 30
7
30
,4 =
⇒
×
=3
10 7
7
10
71
1
2
1.5 × 2.5 × 40 ⇒ ×
29
28
3
2
7
1.3 × ( 4 − 4) ⇒ 1.3 =
b
100
, 52 = 25 ⇒
0.7 × 4 ⇒ 0.7 =
b
2
28
1
4
2
5
1
4
× 7 =  
2  5
2
8
1
=
2
×
15
2
3
16
15
×
25
21
5
8×3
5 ×1
24
=
5
4
=4
5
=
105
10 2 1 m
3
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
11 aLearner’s own answers and explanations.
For example: Write the decimal as a
fraction, square root the fraction then
complete the calculation using fractions.
b
c
d
2
Learner’s own answers and explanations.
For example: Fraction, because the
square roots are easier to work out if the
decimals are changed to fractions.
a
5, 7, 9, …
b
c
add 2
Pattern 4
d
Learner’s own discussions.
4.25 ×
7
2
1 = 5 ; example strategy:
9
3
7
9
1
4
16
9
4.25 × 1 = 4 ×
3
K=2
b
v=
c
v=
d
v=
2
3
term
5
7
9 11
2 × position number
2
4
6
2 × position number + 3
5
7
9 11
1
1
2
2
a
12 , 13, 13 , …, 17
b
0.5, 4.5, 8.5, …, 36.5
4
a
3n + 5
5
a
i
1 8
2×2
4
2
8
= 1 = 1 = , but v =1 ≠
1
3 9
9
2
2
2
4
4
4
2K
m
ii
2K
2 × 18
36
6
1
=
=
= = 1 and
m
25
25
5
5
2
1
1
25 36
 6
2
K = mv = × 25 ×   = × = 18
 5
2
2
2
25
Activity 8.5
b
i
x
6
11
y
6
8
x
−2 1
y
−15 2
y=
x
2
Learner’s own answers.
Exercise 9.1
Reflection: Learner’s own answers.
1
Check your progress
1
a
recurring
b
terminating
c
recurring
d
terminating
2
a
5
3
a
12
4
a
48
5
a
80
1
4
b
2
7
12
1
2
b
b
b
c
4
c
1
3
1
a
b
i
i
add 2
29
30
4
8
50
5
ii
subtract 0.3
ii
3
5 , 6
5
1
2
b
24 − 5n
18
25
12 15
1
2
1
2
8 11
1
2
1
2
35 52
ii
+3
1
2
y = 5(x − 1)
a
linear
b
linear
c
non-linear
d
non-linear
e
linear
f
non-linear
g
linear
h
non-linear
i
linear
Learner’s own explanations. For example:
The linear sequences go up/down by the same
amount each time. The non-linear sequences
do not go up/down by the same amount
each time.
4
15
3
4
Unit 9 Getting started
30
1
Position-to-term rule is:
term = 2 × position number + 3
1
17
4
×
3
1 4
17
=
3
2
=5
3
=
12 a
Position number
2
a
3.5, 4.2, 4.9, …
b
2, 5, 11, …
c
1
2
4 , 3 , 3, ...
3
3
d
40, 18, 7, …
e
1.25, 3.25, 7.25, …
f
1, 2 , 7, …
1
2
7.3, 7
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
3
A and iii, B and i, C and iv, D and ii
4
a
4, 5, 14, …
b
2, 7, 52, …
c
5, 9, 49, …
d
0, 9, 144, …
5
9
For example: The first term is 3 and when you
cube 3 you get 27. Then:
 If you subtract 24, you get a second term
which is also 3, so all the terms of the sequence
are 3 and so you don’t get a negative number.
a3, 3, 3, … All the terms of the sequence
are the same.
b
Learner’s own two sequences.
 If you subtract a number less than 24, the
second term is greater than 3, so all further
terms get bigger so you don’t get a negative
number – e.g. if you subtract 23, the sequence
will be 3, 4, 41, 68 898, …
For example: First term is 5, term-to-term
rule is square and subtract 20.
first term is 16, term-to-term rule is
subtract 12 and square.
c
Learner’s own answers.
For example: It is not possible if the
numbers are positive integers because if
you square then add or add then square,
you will have sequences where the terms
are getting bigger every time.
However, if you use fractions, it is possible
1
2
– e.g. first term is , term-to-term rule is
1
1
4
‘square and add ’, or first term is 9 ,
term-to-term rule is ‘add
2
and square’.
9
It is also possible if you add negative
numbers – e.g. first term is 2, term-to-term
rule is ‘square and add −2’, or first term is
9, term-to-term rule is ‘add −6 and square’.
6
7
8
d
Learner’s own discussions.
a
2, 3 , 4 , 5 , 7 , 8 , 9
b
90, 84 , 79 , 74 , 69, 63 , 58
c
−4, −3.7, −3.4, −3.1, −2.8, −2.5, −2.2
d
31, 24.8, 18.6, 12.4, 6.2, 0, −6.2
a
C
b
The fifth term, which is 126 382 570
(fourth term = 11 242 which is less than
one million)
4
7
2
7
3
4
6
7
1
7
1
2
3
7
1
4
a
3, 4, 6, 9, …
b
6, 8, 12, 18, …
c
20, 19, 16, 11, …
d
100, 90, 75, 55, …
 If you subtract a number greater than
24, the second term is smaller than 3, so all
further terms get smaller so you do get a
negative number – e.g if you subtract 25, the
sequence will be 3, 2, −17, −4938, …
10 a
4, 8, 216, …
c
2, 4, 244, …
1
2
11 Tania’s method is incorrect. Learner’s own
explanation. For example: She needs to reverse
the term-to-term rule to find the previous
terms in the sequence, not just halve the 6th
term to get the 3rd term.
Correct answer is: 5th term = 486 ÷ 3 = 162,
4th term = 162 ÷ 3 = 54, 3rd term = 54 ÷ 3 = 18.
13 3
Reflection: Learner’s own answers.
Exercise 9.2
1
a1st term = 4 × 1 − 5 = −1
2nd term = 4 × 2 − 5 = 3
3rd term = 4 × 3 − 5 = 7
4th term = 4 × 4 − 5 = 11
b
1st term = 12 + 1 = 2
2nd term = 22 + 1 = 5
3rd term = 32 + 1 = 10
4th term = 42 + 1 = 17
c
Activity 9.1
Learner’s own questions and discussions.
d
31
b −6, −8, −64, …
12 4th term = (11.5 − 6) × 2 = 11,
3rd term = (11 − 6) × 2 = 10,
2nd term = (10 − 6) × 2 = 8
5
7
3
4
Zara is correct. Learner’s own explanation.
2
1
3
3
3rd term = 3 = 1
2nd term = 3
1st term = 13 = 1
2nd term = 23 = 8
3rd term = 33 = 27
4th term = 43 = 64
1st term =
4
1
4th term = 3 = 1 3
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
2
a
7, 11, 15, …, 43
c
3 , 4, 4 , …, 8
e
0, 3, 8, …, 99
1
1
2
2
b
−3, −1, 1, …, 15
d
1 2 3
, , , …, 2
5 5 5
8
9
3
A and iv, B and iii, C and i, D and ii
4
a
Learner’s own answer and reason.
b
Card A has the greater value.
A: 8th term = 82 − 14 = 50,
Learner’s own discussions.
a
A 12 = 4 , B
14 7
9
3
= ,C =
18 9
12
4
b
C 3, B 7, A
4
5
d
Learner’s own answer.
a
7, 8, 11, 16, 23, 32, …
b
7, 8, 11, 16, 23, 32, …
+2
c
d
+2
+2
+4
+2
ii
+2
+5
+2
+6
+1
iii
+6
+8
+2
+7
+1
+1
+2 +6
+4
+4
+1
+4
+ 18
32
1
1
1
1
2
2
2
1
1
2
2
2
4
1
9 − n
b
20.2 − 0.2n
c
−1 − n
d
−3.5 − 1.5n
2
Exercise 9.3
1
a
i
+4
i
quadratic
ii
linear
iii
neither
iv
linear
v
neither
vi
quadratic
f
Learner’s own discussions.
a
n +3
b
n + 10
c
n2 − 1
d
n2 − 9
2
1
12 a
+9
+ 10 + 14
2
when n = 3, 4 − × 3 = 3, etc.
3, 5, 11, 21, 35, 53, …
1
2
when n = 2, 4 − × 2 = 3 ,
+2
+8
1
when n = 1, 4 − × 1 = 4 ,
ii
In each sequence the second
differences are all the same.
7
iYes, when n = 13, 132 − 76 = 93, so 93 is
the 13th term.
11 Marcus is correct. Learner’s own explanations.
For example:
+ 10
2, 7, 13, 20, 28, 37, …
6
9
5, 7, 11, 17, 25, 35, …
e
4
No, when n = 16, 163 = 4096, when
n = 17, 173 = 4913 and 4896 lies
between 4096 and 4913, so cannot be
in the sequence.
+9
The second differences are all the
same (+2).
i
5
OR
+7
+2
15
n
6
iiNo, 3 4896 = 16.98 …, so not a whole
number.
c
+5
ii
10 a, b, c Learner’s own answers.
4
+3
iii
b
5
+1
n
8
i
B: 20th term = × 20 + 33 = 49
5
n
7
a
b
i
x
0
1
2
3
y
0
1
4
9
x
0
1
2
y
0
1
8
0 1 2 3 4 5 6 7 8 9 10
x
y
0 1 2 3 4 5 6 7 8 9 10
2
Learner’s own explanation. For example: When
you square a number you get a positive answer
and then once you add 5 you know that all
the terms in the sequence will be positive.
You cannot have a first term of −1 as this is a
negative number not a positive number, so it
cannot be in the sequence.
ii
0 1 2 3 4 5 6 7 8 9 10
x
y
0 1 2 3 4 5 6 7 8 9 10
c
i
y = x2
ii
y = x3
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
2
a
i
ii
3
b
i
a
i
ii
iii
4
x
2
5
y
7
28 84 124
x
1
3
y
−2 24 122 997
5
10
ii
y=x +3
2
x
−3
1
3
1
2
y
18
2
9
1
2
x
−5
1
4
1
2
y
100
1
4
1
0
3
8
125
Learner’s own discussions.
c
i
y = 2x2
ii
y = (2x)2
iii
y = (x + 2)3
d
Learner’s discussions.
a
i
x
y
iii
b
−2
1
2
12
1
3
3
4
1
4
1
2
1 33
64
5
8
y = (x + 5)2
ii
y = 3x2
iii
y=x +
3
7
a
4
x
−4 −3
3
4
y
16
9
16
i
9
i
20
80
500
i
i
x
2
4
5
y
8
32
50 288
x
7 10
11
13
y
16 49
64
100
12
y = 2x2
ii
y = (x − 3)2
ii
x=± y
ii
x= 3 y
ii
x = ±2 y
ii
x = ± y −3
Activity 9.3
Learner’s own answers.
8
a
b
2
×4
y
c
d
33
1
ii
1
2
4
9
iiiLearner’s own discussions. For
example: You could say that either all
the x-values are positive or that all the
x-values are negative.
a
x
1
4
5
b
i
y
iiLearner’s own answer. For example:
There are two possibilities for x for
each y-value.
15
1
3
y −7 2
1
1 or −1 2 or −2 4 or −4 10 or −10
y
1 400
x −2
1
2
y=x −3
x
−8
9
1
3
3
b
−4
y
a
1
4
iiiLearner’s own discussions. For
example: Yes, when you square +x
and −x, you get x2.
y
−8 −4
6
x
iiLearner’s own answer. For example:
x = −4 and 4 have the same y-value.
x = −3 and 3 have the same y-value.
x
x
b
11
b
ii
5
9
i
y = x2
iii
Learner’s own check.
i
y = x3
iii
Learner’s own check.
i
y =  
iii
Learner’s own check.
i
y = x2 + 3
iii
Learner’s own check.
x
 2
2
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
e
f
9
i
y = (x − 4)2
iii
Learner’s own check.
i
y = (2x)3
iii
Learner’s own check.
ii
x=± y+4
ii
x=
3
4
y
2
The number 178 is not a term in this sequence,
because when you solve the equation
n2 + 32 = 178 to find the value of n you do not
get a whole number.
n 2 + 32 = 178
n 2 = 178 − 32
A and iii, B and i, C and v, D and vi,
E and iv, F and ii
n 2 = 146
10 They are both correct. Learner’s own
explanations. For example: The x-values match
the y-values for both function equations and
y = (2x)2 = 2x × 2x = 4x2.
n = 146 = 12.08 ...
5
a
i
11
2
x
×8
y
x
−2
4
5
y
2
8
12
ii
x
1
4
y
1
2
1
2
3 or –3
2
72
b
y = 8x
2
12 Arun is incorrect. Learner’s own explanations.
For example:
He is correct for the function y = 2x4 because
any positive or negative number to the power
of four gives a positive answer. This is then
multiplied by two to still give a positive
answer.
1
He is incorrect for the function y = x
because when a negative number 2
is cubed, the answer will be negative. When
this is multiplied by 1 , the answer will still be
2
negative.
34
a
3, 4, 11, 116, …
b
−3, 1, 9, 121, …
c
5, 6, 9, 14, …
d
40, 38, 34, 28, …
2
a
1
3
a
n2
3
2
2
b
n2 − 2
49
y=
2
1
4
x2
2
1
2
1
4
81
144
ii
y = (x + 8)2
a
$155
2
a
x
−2 −1
0
1
2
3
y
−5 −3
−1
1
3
5
3
b
c = 20d + 35
b
Learner’s own graph; A straight line
through (0, −1), (0.5, 0) and (3, 5).
c
2
d
−1
a
40 °C
b
20 °C
c
At the start
Exercise 10.1
1
b
, 1, , ..., 5
y
1
40
1
Reflection: Learner’s own answers.
1
−15 − 8
1
2
Unit 10 Getting started
3
Check your progress
i
x
9
8, 11, 16, …, 107
c
n
9
a
$31
b
The number of days multiplied by 3 plus
10 for the fixed charge.
2
a
3 days
b
t = 10n +15
3
a
27 kg
b
b = 2g − 3
4
a
s + f = 50
b
s + f = 52
c
s + f = 60
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
6
7
8
9
6
a
6 × 5 + 6 × 10 = 90
b
5f + 10t = 90
c
8
a
12 × 6 + 12 × 4 = 72 + 48 = 120
b
6l + 4s = 120
c
s = 2l
a
The total value is 80 cents.
b
4
a
3x + 2y = 50
b
If 3x = 21 then 2y = 19 and that is
impossible if y is a whole number.
b
7
8
a–d Learner’s own answers.
10 a
32
b
3r + 4q = 100
c
10
d
32
e
No. Each pair would have a total of 7
edges and 7 is not a factor of 100.
f
q = 3r − 5
Reflection: Two possible ways are x = 20 − 2y
and 2y = 20 − x.
Exercise 10.2
1
2
3
4
5
a
35
x
−1
0
1
2
3
y
5 15
25
35
45
b
When x = 5, then y = 10 × 5 + 15 = 65
a
x
−10
0
10
20
30
y
−30
−10
10
30
50
b
At (0, −10)
c
When x = 23, then y = 2 × 23 − 10 = 36, so
(23, 36) is on the graph.
a
x
0
1
3
5
6
y
20
16
8
0
−4
b
At (0, 20) and (5, 0)
a
x
0
10
20
30
40
y
12
8
4
0
−4
b
2 × 15 + 5 × 6 = 60
a
x
−2
0
2
4
6
y
10
6
10
22
42
b
a
When x = 5, then y = 52 + 6 = 31
9
x
0
2
6
10
16
y
8
7
5
3
0
ii
(0, 8)
i
a
(16, 0)
x
0
1
2
4
6
y
9
7.5
6
3
0
b
i
ii
c
Learner’s own graph. A straight line
through (6, 0) and (0, 9).
a
x
15
10
5
0
y
0
1
2
3
(6, 0)
(0, 9)
b
Learner’s own graph. A straight line
through (15, 0) and (0, 3).
a
x
0
2
4
6
8
10
y
10
8
6
4
2
0
b
Learner’s own graph. A straight line
through 10 on each axis.
c
Learner’s own graph. A straight line
through 7 on each axis.
d
x
−1
0
1
2
3
4
5
y
5
4
3
2
1
0
−1
e
Learner’s own graph. A straight line
through 4 on each axis.
f
A straight line through c on each axis.
g
Learner’s own graph. A line parallel to the
others through the origin.
10 aLearner’s own graph. A straight line
through 12 on each axis.
b
x
0
1
2
3
4
5
6
y
12
10
8
6
4
2
0
c
Learner’s own graph. A straight line
through 12 on the y-axis and 6 on the
x-axis.
d
Learner’s own graph. A straight line
through 12 on the y-axis and 4 on the
x-axis.
e
Learner’s own graph. A straight line
through 12 on the y-axis and 3 on the
x-axis.
f
A straight line through 12 on the y-axis
12
and on the x-axis.
k
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
11 aLearner’s own graph. A straight line
through (14, 0) and (0, 7).
b
1
A straight line through (n, 0) and (0, n).
2
12 a
b
x
−3
−2
−1
0
1
2
3
y
9
4
1
0
1
4
9
Learner’s own graph. A parabola with the
base at the origin.
c
x
−3
−2
−1
0
1
2
3
y
11
6
3
2
3
6
11
6
d
Learner’s own graph. A parabola with the
base at (0, 2).
e
x
−3
−2
−1
0
1
2
3
y
5
0
−3
−4
−3
0
5
f
Learner’s own graph. A parabola with the
base at (0, −4).
g
A curve with the y-axis as a line of
symmetry and the lowest point at (0, c).
(Learners are not expected to know the
word parabola.)
7
x
0
2
4
6
8
10
y
5
4
3
2
1
0
b
Learner’s own graph. A straight line
through (10, 0) and (0, 5).
c
y = 5− x
d
gradient − and y-intercept 5
e
Learner’s own check.
a
y = 15 − 3x
b
gradient −3 and y-intercept 15
c
x
0
5
2
4
y
15
0
9
3
1
2
1
2
d
Learner’s own graph. A straight line
through (0, 15) and (5, 0).
e
Learner’s own check.
a
y = 6− x
b
c
gradient − and y-intercept 6
3
4
3
4
x
0
8
4
13 A and iii, B and iv, C and i, D and ii
y
6
0
3
14 a
Learner’s own (correct) values in the last
column.
x −5 −4 −3 −2 −1
y
16
7
0
1
2 3
4
0 −5 −8 −9 −8 −5 0
1
2
3
4
5
7 16
d
Learner’s own graph. A straight line
through (0, 6) and (8, 0).
e
Learner’s own check.
a
i
y = 18 − 2x
b
Learner’s own graph. A parabola with the
bottom at (0, −9).
c
i
(−10, 91)
iii
(20, 391)
ii
y = 9− x
iv
(−3, 0) or (3, 0)
iii
y = 9 − 2x
v
(6, 27) or (−6, 27)
iv
y = 3− x
ii
(8, 55)
Exercise 10.3
36
a
8
b
a
gradient 4 and y-intercept −6
b
gradient 6 and y-intercept 4
c
gradient −6 and y-intercept 4
a
gradient 0.5 and y-intercept 3
b
gradient −1 and y-intercept 8
c
gradient and y-intercept 0
a
3
b
1
c
1
3
a
1
−
2
b
−1
c
−4
1
4
c
1
2
1
2
Line
Gradient y-intercept
2x + y = 18
−2
18
x + 2y = 18
−
1
2
9
4x + 2y = 18
−2
9
3x + 6y = 18
−
1
2
3
The gradient is − a and the y-intercept is
b
18
.
b
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
9
a
y = 4x + 8
b
straight line, gradient 4, y-intercept 8,
passes through (0, 8) and (−2, 0)
10 a
b
i
i
7
3 ii
1
iii
2
3 ii
1
−
2
iii
y=
1
2
x+3
1
y = − x+3
8
2
Exercise 10.4
1
2
4
5
6
24 dollars
c
1.6
d
1.6 dollars
e
y = 1.6x
f
152 dollars
g
62.5 kg
aLearner’s own graph. A straight line from
(0, 20) going through (30, 32).
b
24 °C
250 m
b
16 s
c
0.4
c
12.5 m/s
d
d =12.5t
d
y = 0.4t + 20
e
625 m
e
44 °C
a
400
f
200 seconds
c
i
g
Learner’s own answers.
b
20
8
9
aLearner’s own graph. A straight line from
(0, 100) going through (8, 72).
d
y = 8x
b
79 litres
e
920 HK dollars
c
3.5 litres/hour
a
28 dollars
b
15
d
y = 100 − 3.5h
c
1.4 dollars
d
y = 1.4x
10 a
e
64.12 dollars
f
36.5 litres
11 aThe y-intercept is 24.
a
1m
b
Weeks
0
Height (m)
1 1.2 1.4 1.6 1.8 2
1
2
3
4
0.2 m
d
y = 0.2t + 1
e
3.2 m
a
1500 m
b
750 m
c
50 m/minute
d
y = 1500 − 50x
e
350 m
f
30 minutes
b
45
18
27
i
0.9
ii
1 dollar buys 0.9 euros
d
y = 0.9x
e
252
f
170
b
12 a
aLearner’s own graph. A straight line from
(0, 0) through (50, 45).
b
Dollars
50 20 30 15
13.5
32 − 24
10 − 0
33 000
= 0.8, so the equation of
the line is p = 0.8t + 24.
5
c
c
b
4800
Gradient =
Euros
37
b
a
iiThere are 8 HK dollars to 1 US
dollar.
3
aLearner’s own graph. A straight line from
the origin through (25, 40).
36 = 0.8t + 24 so t =
years.
36 − 24
0.8
= 15; it takes 15
8 m/s
Marcus. Arun’s speed is 5 m/s.
13 The rate for A is 2 cm/minute and the rate for
B is 5 cm/minute.
14 a
b
c
15 a
120
= 12 m/s
10
280 − 120
= 16 m/s
10
400 − 280
5
= 24 m/s
Decreasing at a rate of 2 litres/hour.
b
y = 18 − 2t
c
9 hours
16 Learner’s own answers.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Check your progress
3
a
$125
b
$200
1
$18
b
$42
2
3
4
a
5x + 10y = 100
4
a
b
10y = 100 − 5x, then divide both sides
by 10.
5
aSand: 2 parts = 15 kg,
1 part = 15 ÷ 2 = 7.5 kg
c
−
a
Cement: 1 part = 7.5 kg
1
2
Gravel: 4 parts = 4 × 7.5 = 30 kg
x
0
1
2
3
4
5
y
15
12
9
6
3
0
b
Learner’s own graph. A straight line
through (0, 15) and (5, 0).
c
−3
b
Total = 15 + 7.5 + 30 = 52.5 kg
6
a
24 and 42
7
a
Learner’s own answers.
b
Learner’s own answers.
c
Learner’s own discussions.
750 mL
b
120
aLearner’s own graph. The usual parabola
shape with the bottom at (0, 5).
8
a
b
5 and −5
9
1. Difference in number of parts = 4 − 1 = 3
a
4.5 m
b
0.3 m/year
2. 3 parts = 39 g
c
y = 0.3x + 3
d
5.7 m
3. 1 part = 39 ÷ 3 = 13 g
b
1.5 L
4. 4 parts = 13 × 4 = 52 g
Unit 11 Getting started
5. Total mass = 13 + 52 = 65 g
1
a
20 : 1
b
1:4
c
1:5
2
a
90
b
108
c
72
3
a
4
7
4
a
Sky blue: , Ocean blue:
b
Sky blue is lighter. Learner’s own method.
For example:
b
3
4
32
5
7
10 a
b
There is more white in sky blue, so this
shade is lighter.
5
$6.80
b
aCherries: 2 parts = 80 g, 1 part = 80 ÷ 2 = 40 g
Sultanas: 5 parts = 5 × 40 = 200 g
b
2
38
Total = 80 + 200 = 280 g
aStrawberries: 2 parts = 400 g,
1 part = 400 ÷ 2 = 200 g
b
i
Learner’s own answers.
iiThere are two possible solutions.
Either the first number is 6 or the
second number is 6.
iii6 : 9 → dividing both numbers by
3 gives 2 : 3
4 : 6 → dividing both numbers by
2 gives 2 : 3
c
Learner’s own discussions.
12 0.18 or 1.28; Check: 0.48 : 0.18 = 8 : 3 or
1.28 : 0.48 = 8 : 3
Exercise 11.1
1
Moira gets $21 and Non gets $49.
11 aThere are two possible solutions. The
numbers are either 6 and 9 or 4 and 6.
Sky blue 1 : 3 = 2 : 6 = 2 parts blue and 6
parts white
Ocean blue 2 : 5 = 2 parts blue and 5 parts
white
$70
13 440 g of oats, 220 g of butter and 110 g of
syrup. Learner’s own method. For example:
Butter: 250 ÷ 2 = 125 g per part, Oats:
440 ÷ 4 = 110 g per part.
Use 110 g per part as smallest amount.
Syrup: 1 × 110 g = 110 g, Butter:
2 × 110 g = 220 g, Oats: 4 × 110 g = 440 g
Raspberries: 1 part = 200 g
14 12 g
Total = 400 + 200 = 600 g
15 3 : 4 : 5
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
16 Learner’s own working. For example:
7
When working out the number of members of
staff the number must be rounded up to make
sure there are enough members of staff.
Age of
children
Child :
staff
ratios
Number Number of
of
members of
children staff
up to 18
months
3:1
10
10 ÷ 3 = 3.3… = 4
18 months
up to 3
years
4:1
18
18 ÷ 4 = 4.5 = 5
3 years up
to 5 years
8:1
15
15 ÷ 8 = 1.875 = 2
5 years up
to 7 years
14 : 1
24
24 ÷ 14 = 1.7…= 2
1
6
10
8
a–d Learner’s own answers and discussions.
9
2 hours 24 minutes
5
10 aLearner’s own answers. Marcus is correct
because the length of the ride is 4 minutes
and it doesn’t matter how many people
are on the roller coaster.
b
Learner’s own discussions.
Learner’s own questions and answers.
Height of bounce (cm)
direct proportion
b
neither
c
inverse proportion
d
direct proportion
e
neither
f
inverse proportion
g
neither
a
$7
b
c
$1.75
3
a
50 g
4
a
5
2
Cost per
300 100 600 1200 200 120 240
person (€)
Learner’s own answers.
a
2
12
11 aYes. Learner’s own explanations. For
example: The height of bounce is 0.8 × the
height it is dropped from.
Exercise 11.2
1
4
Activity 11.2
Total number of members of staff
needed = 4 + 5 + 2 + 2 = 13
Reflection:
Number
of people
b
96 cm
c
i
250
200
150
100
50
0
Height of ball before and after bounce
0
50
100 150 200 250
Height when dropped (cm)
ii
They are in a straight line.
$17.50
iii
Yes
d
$8.75
iv
225 cm
150 g
c
4 horses = 2 days
÷4
1 horse = 8 days
×4
b 4 horses = 2 days
×2
8 horses = 1 day
÷2
b
1.875 L
a normal speed = 36 seconds
÷2
1
speed = 72 seconds
×2
b normal speed = 36 seconds
×3
3 × speed = 12 seconds
÷3
300
12 aDirect proportion. Learner’s own
explanation. For example: The
mass : length increase ratio is the same as
5 g : 3 mm for all pairs of values
2
6
39
a
20 minutes
b
30 km/h
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Length of increase (mm)
b
5
a
0.084
6
a
0.85
7
a
0.4
b
0.52
c
0.6
d
0.48
a
1
8
b
c
d
e
2 1
=
4
8
3
8
2 1
=
8
4
5
8
9
a
0.45
b
0.7
Use your graph to work out
10
4
1
2
P(A) = ; P(B) = ; P(C) =
7
7
7
i
27 mm
11 a
0.2
b
0.95
c
0.4
ii
33 g −34 g (accurate answer is 33 g)
12 a
0.1
b
0.09
c
0.19
True. Learner’s own explanation. For
example: Because one set of values is a
multiple of the other, so the gradient of
the line is constant.
13 a
1
12
Length of increase of string when different
masses are added
30
25
8
20
15
10
25
20
c
d
30
35
40
Mass (g)
45
50
1
3
Check your progress
b
If A happens, the number is 2, 3 or 4 and
1
then P(1 or 2) = 3 ; if A does not happen, the
1
number is 1, 5 or 6 and then P(1 or 2) = 3 ; as
these are the same, the events are independent.
3
No. If the first two spins are tails then the
1
probability that all three are = P(Y) = 2 . If
the first two spins are not both tails then Y is
impossible and P(Y) = 0.
4
They are independent. The coin is fair and so
1
the probability is always . The coin has no
2
memory of the previous throws!
5
Fog will decrease the probability that the flight
will leave on time because the flight could be
cancelled.
6
aIf R happens, the number is 1, 2, 3, 4, 5
3 1
or 6 and P(even) = = . If R does not
6 2
happen, the number is 1, 2, 3 or 4 and
2 1
P(even) = = . The probabilities are the
4 2
same and so the events are independent.
3
Sugar = 50 g, Butter = 100 g, Flour = 400 g
4
a
$6
b
$18
c
$4.50
5
a
12 days
b
3 days
c
6 people
Unit 12 Getting started
H1 H2 H3 H4 H5 H6
T4 T5 T6
1
12
b
i
3
a
4
a
13
= 0.26
50
4
b
5
3
25
ii
1
3
=
12
4
b
1
or 0.2
5
5
d
8
c
3
32
Exercise 12.1
40
1
25%
2
a
1
6
3
a
4
2
=
10 5
b
3
10
c
7
10
d
4
2
=
6
3
3
10
4
a
0.3
b
0.45
c
0.7
d
0.25
b
Learner’s own answers. For example: The
smallest possible numbers are black 3,
white 8, yellow 1. Or learners could have
any multiples of these.
2
24, 30 and 42 b
T3
0.81
P(S) is always whether the first spin is a head
2
or a tail.
a
T1 T2
d
1
2
a
0.05
114
750 g
2
c
0.7
Exercise 12.2
a
0.85
b
0.916
1050 g or 1.05 kg
1
1
b
b
3
1
=
2
6
c
1
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
1
If B happens, the number is 1, 2, 3 or 4 and P(2) = 4 . If B does not happen, the number is 1, 2, 3, 4,
1
5 or 6 and P(2) = 6 . The probabilities are not the same, so the events are not independent.
7
aThey are independent. If the first ball is replaced then the situation is exactly the same both times.
b
8
aLearner’s own explanation. For example: Arun and Sofia are not friends and do not travel together
and there are no external factors such as weather or traffic.
b
9
They are not independent. If the first ball is black, the probability that the second ball is black is
smaller than if the first ball is white.
Learner’s own explanation. For example: Arun and Sofia are brother and sister and travel to
school together.
If X happens then one of the cards must be A, C or D. Of these, 2 out of 3 are in the word CODE, so
2
the probability of Y is . If X does not happen the card must be B or E. Then 1 out of 2 is in the word
3
1
CODE, so the probability is . These probabilities are different, so the events are not independent.
2
Exercise 12.3
1
a
1
4
b
1
4
c
1
4
2
a
1
36
b
1
12
c
1
12
3
a
1
6
b
1
9
c
25
36
4
a
0.09
b
0.49
c
0.21
d
0.21
5
a
0.48
b
0.32
c
0.12
d
0.08
6
a
i
ii
0.085
iii
0.135
iv
0.765
b
Learner’s own explanation. For example: They are mutually exclusive and one of them must
happen.
7
0.015
First
a
1
9
8
9
41
Second
Outcome
1
9
5
5, 5
1
9
1
× 19 = 81
8
9
not 5
5, not 5
1
9
8
× 89 = 81
1
9
5
not 5, 5
8
9
8
× 19 = 81
8
9
not 5
not 5, not 5
8
9
× 89 = 64
81
5
not 5
1
81
b
i
c
Not getting a 5 either time.
ii
64
81
iii
8
81
iv
8
81
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
8
a
First
0.6
Second
Outcome
red
red, red
0.3 × 0.6 = 0.18
not red
red, not red
0.3 × 0.4 = 0.12
red
not red, red
0.7 × 0.6 = 0.42
not red
not red, not red
0.7 × 0.4 = 0.28
red
0.3
0.4
0.7
0.6
not red
0.4
9
b
i
ii
c
Learner’s own explanation. For example: They are mutually exclusive and one of them must
happen.
0.18
a
iii
0.28
Blackbird
iv
0.12
Robin
Outcome
0.8
Yes
Yes, Yes
0.9 × 0.8 = 0.72
0.9
0.2
No
Yes, No
0.9 × 0.2 = 0.18
0.1
0.8
Yes
No, Yes
0.1 × 0.8 = 0.08
No
No, No
0.1 × 0.2 = 0.02
0.42
Yes
No
0.2
b
i
c
0.98
10 a
1
3
42
0.02
First
2
3
b
ii
0.72
i
1
6
Second
Outcome
1
4
Blue
Blue, Blue
2
3
2
× 14 = 12
= 16
3
4
Yellow
Blue, Yellow
2
3
6
× 34 = 12
= 12
1
4
Blue
Yellow, Blue
1
3
1
× 14 = 12
3
4
Yellow
Yellow, Yellow
1
3
3
× 34 = 12
= 14
Blue
Yellow
ii
1
4
iii
1
2
iv
3
4
v
5
6
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
11 aLearner’s own diagram. For example: The best way to do this is with a tree diagram.
First
Second
Outcome
0.9
Yes
Yes, Yes
0.4 × 0.9 = 0.36
No
Yes, No
0.4 × 0.1 = 0.04
Yes
No, Yes
0.6 × 0.9 = 0.54
No
No, No
0.6 × 0.1 = 0.06
Yes
0.4
0.1
0.9
0.6
No
0.1
b
Miss the first time, and get a basket the second time.
c
0.94
Exercise 12.4
1
a
3
or 0.12
25
2
a
Red 0.39; white 0.27; blue 0.34
b
The probability of each colour is 0.333. Blue is closest to this, white is furthest from this.
a
Rolls
10
20
30
40
50
60
70
80
90
100
Total frequency
2
4
5
8
9
10
11
16
17
18
Relative frequency
0.2
0.2 0.167 0.2
3
4
5
6
43
b
7
or 0.28
25
c
1
or 0.2
5
0.18 0.167 0.157 0.2 0.189 0.18
b
Learner’s own graph. Check that the relative frequency values from the table in part b have been
plotted correctly.
c
Line through 0.167 on vertical axis.
a
Flips
20
40
60
80
100
Frequency of heads
8
19
30
38
44
Relative frequency
0.4
0.475
0.5
0.475
0.44
b
Learner’s own graph. Check that the relative frequency values from the table in part a have been
plotted correctly.
c
The probability is 0.5. The relative frequency values are close to this. The values are below or equal
to this.
a
Learner’s own table. Check that they have calculated the relative frequencies correctly.
b
Learner’s own graph. Check that the relative frequency values from the table in part a have been
plotted correctly.
c
Learner’s own estimate. The probability is 0.583 and the estimate could be close to this.
d
a
Learner’s own discussions.
Draws
20
40
60
80
100
120
140
160
180
200
Frequency
10
14
27
36
42
50
55
62
70
79
Relative frequency
0.5
0.35 0.45 0.45 0.42 0.417 0.393 0.388 0.389
0.395
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
c
7
Learner’s own graph. Check that the
relative frequency values from the table in
part a have been plotted correctly.
2
aIf X happens then the number is 2, 4 or 6
1
and P(Y) = . If X does not happen then
3
Learner’s own estimate. For example: 8
black and 12 white.
b
If X happens the numbers are 2, 4 or 6
1
then P(Z) = . If X does not happen
3
the numbers are 1, 3 or 5 then
2
P(Z) = . Different probabilities so they
3
are not independent.
3
a
0.36
b
0.16
4
a
0.2
b
0.22
c
The probability is 0.2. The relative
frequencies are the same or similar.
a
Digits
20
40
60
80
100
Frequency
of 0
2
5
7
7
8
Relative
frequency
0.1 0.125 0.117
0.088
0.08
b
Learner’s own graph. Check that the
relative frequency values from the table in
part a have been plotted correctly.
c
20
40
60
80
100
Frequency
of 0
2
6
8
9
15
Relative
frequency
0.1
0.15 0.133 0.113 0.15
d
Unit 13 Getting started
1
Digits
1
3
the number is 1, 3 or 5 and again P(Y) = .
a
N
b
N
B
A155°
60°
Learner’s own graph. Check that the
relative frequency values from the table in
part c have been plotted correctly.
A
B
e
Digits
100
200
300
400
500
Frequency
of 0
11
27
40
52
60
Relative
frequency
0.11
0.135 0.133
0.13
0.12
8
f
Learner’s own graph. Check that the
relative frequency values from the table in
part e have been plotted correctly.
g
The probability is 0.1. The probabilities
vary around this value. Sofia has the
closest final value. You might expect her
final value to be close because she has the
largest sample size.
Learner’s own answers and experiments.
N
c
N
d
B
A
A 220°
305°
B
2
a
16 km
b
30 cm
3
a
(8, 8)
b
(5, 8)
4
a
 −5
 −6
b
 5
 6
Check your progress
1
44
Learner’s own answers. There are many
possible answers. For example:
a
Roll a 2 and roll an odd number.
b
Roll a 2 and roll an even number.
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
a
0
0
a
N
y=4
140°
230°
1 2 3 4 5 6 7 8 9 10 x
5 cm
7 cm
y
6
5
4
3
2
1
b
6
2
y
6
5
4
3
2
1
J
R
b
1 2 3 4 5 6 7 8 9 10 x
x=6
3
Rotation, 90 ° clockwise, centre (−1, 0).
7
Learner’s own measurement. Answer in
range 85 m–88 m.
Yes they could meet. Learner’s own answers
and discussions. Learner’s own explanation.
For example: In a sketch of the situation,
the two lines cross, showing the point where
the yacht and the speedboat could meet. You
don’t know if the yacht and the speedboat will
meet because you don’t know their speeds, but
if they do meet it will be at this point.
4
N
A
152°
Exercise 13.1
1
Distance on scale drawing = 800 ÷ 100 = 8 cm
8 cm
N
Ship
8 cm
42°
50°
B
5
45
aTeshi’s sketch is incorrect. He has drawn
Yue south of Jun instead of Jun south
of Yue.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
Yue
N
b
Learner’s own measurement. Answer in
range 12 km–13 km (Accurate answer is
12.4 km to 3 s.f.)
137°
c
Learner’s own measurement. Answer in
range 140 °–145 ° (Accurate answer is 143 °
to 3 s.f.)
4.1 cm = 8.2 km
Activity 13.1
Learner’s own question and discussions.
(8 km)
4 cm
8
a, b i, c i, d i
N
70°
(6 km)
3 cm
Jun
6
a
Farm
house
N
N
Q
café
N
P
5 cm
(100 km)
120°
30°
Shop
4 cm
(80 km)
7
b
Learner’s own measurement. Answer in
range 125 km–130 km (Accurate answer is
128 km to 3 s.f.)
c
Learner’s own measurement. Answer in
range 246 °–252 ° (Accurate answer is 249 °
to 3 s.f.)
d
Learner’s own discussions.
9
N
a
b
iiLearner’s own measurements. In the
range 14 km–14.5 km and 275 °–280 °.
c
iiLearner’s own measurements. In the
range 6.5 km–7 km, 140 °–145 °.
d
iiLearner’s own measurements. In the
ranges: Distance from P = 11.5 km–
12 km, Distance from Q = 1.2 km–
1.6 km.
a
Ship
N
N
8 cm (16 km)
275°
N
45°
46
6 cm
(12 km)
P
7.5 cm (75 km)
L
b
Learner’s own measurements. In the range
46 km–47 km.
c
Learner’s own measurements. In the range
53 km–54 km.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
10 a
b
i
120 ° ± 2 °
iii
351 ° ± 2 °
ii
c
20 km
247 ° ± 2 °
7
R (12, 15)
8
aLearner’s own explanation. For example:
The point (1, 2) is not on the line. It is two
units to the left of where the line starts at
point A.
$1120
Reflection: Learner’s own answers.
Exercise 13.2
1
2
3
Learner’s own diagram. Check that all of the
points are plotted and labelled correctly.
5
6
c
The coordinates of C are (3 + 1, 2 + 2)
= (4, 4). Learner’s own check.
A (0, 2) and B (3, 2)
b
(1, 2)
d
i
c
(2, 2)
e
Learner’s own discussions.
d
C (4, 0) and D (4, 8)
e
(4, 2)
f
(4, 6)
9
ii
1:5
1:4
2
3
Difference in x-coordinates = 9 − 3 = 6, × 6 = 4
2
3
Difference in y-coordinates = 13 − 4 = 9, × 9 = 6
A and v, B and iii, C and vi, D and ii,
E and iv, F and i
H = F(3, 4) + (4, 6) = (3 + 4, 4 + 6) = (7, 10)
10 a
b
a–d Learner’s own answers.
Learner’s own answer. For example:
Chesa’s method will work as she takes
into account the position of S. When S
moves she will add her distances on to the
coordinates of S. Tefo’s method will not
work as he is just finding the fraction of
the coordinates of T. When S moves this
will give the incorrect answer.
f
Learner’s own discussions.
a
B (4, 3)
b
A (12, 9)
c
C (2, 3)
d
B (8, 12)
a
B (4, 6)
b
C (6, 9)
c
J (2 × 10, 3 × 10) = (20, 30)
d
P (2 × 16, 3 × 16) = (32, 48)
e
(2 × 20, 3 × 20) = (40, 60)
f
Coordinates of the point labelled with the
nth letter are (2n, 3n).
L (10, 11)
Learner’s own check using a diagram.
11 a, b Learner’s own diagram. Check that the
points and diagonals are drawn accurately.
c
 1 1
E 3 ,3 
 2 2
d
 2 + 5 5 + 2   7 7  1 1 
,
 =  ,  = 3 , 3 
 2
2   2 2  2 2 
e
Difference in x-coordinates of
5
1
×4=2
AC = 5 − 1 = 4
8
aYes. Learner’s own explanation. For
example: E is at (4 × 3, 4 × 7) = (12, 28).
b
No. Learner’s own explanation. For
1
example: OD lies of the distance OE
3 4
and so DE lies of the distance OE. This
4
1 3
means the ratio OD : DE is : = 1 : 3 and
4 4
not 1 : 4.
47
Learner’s own explanation. For example:
She needs to add (1, 2) on to the
coordinates of A (3, 2).
a
e
4
b
2
Difference in y-coordinates of
5
1
×4=2
AC = 5 − 1 = 4
8
2
1
1
 1 1 
E = A (1, 1) +  2 , 2  = 1 + 2 , 1 + 2 
 2 2 
2
2
 1 1
= 3 , 3 
 2 2
12 J (13, 13). Learner’s own working. For
example:
Difference in x-coordinates is 17 − 5 = 12
Difference in y-coordinates is 19 − 1 = 18
There are six points after F, so the
x-coordinates increase by 12 ÷ 6 = 2 for each
point, and the y-coordinates increase by
18 ÷ 6 = 3 for each point.
Points:
F
G
H
I
J
K
L
x-coordinates
5
7
9 11 13 15 17
y-coordinates
1
4
7 10 13 16 19
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 13.3
5
1
a and iii, b and i, c and ii
2
aLearner’s own diagram. The image should
have vertices (2, 0), (4, 0), (4, 1) and (3, 1).
3
4
b
Learner’s own diagram. The image should
have vertices (3, 0), (4, 0), (4, 1) and (3, 2).
c
Congruent. Learner’s own explanation.
For example: In both parts the object and
the image are identical in shape and size.
aLearner’s own diagram. The image should
have vertices (2, 1), (4, 3) and (1, 3).
b
Learner’s own diagram. The image should
have vertices (−2, 0), (−5, 0) and (−3, 2).
c
Learner’s own diagram. The image should
have vertices (−2, 1), (−2, 4) and (0, 3).
d
Learner’s own diagram. The image should
have vertices (2, −1), (4, −3) and (2, −4).
a
iLearner’s own diagram. The image
should have vertices (−1, −3), (1, −3),
(0, −2) and (0, −3).
6
7
8
a
Reflection in the y-axis.
b
Reflection in the x-axis.
c
Reflection in the line y = 1.
d
Reflection in the line x = 1.
a
 2
 −1 
b
 5
 1
c
 −6
 −4
d
 5
 −1
e
 −4 
 4 
f
 1
 3
a
90 ° clockwise, centre (3, 3)
b
90 ° anticlockwise, centre (3, 0)
c
180 °, centre (3, 0)
d
90 ° clockwise, centre (−1, 0)
e
90 ° anticlockwise, centre (−1, −1)
a
iRotation 180 °, centre (−2, 1) OR
reflection in the line y = 1 OR
 0
 
translation  −3 .
 2
OR rotation 180 °,
 −4
iiTranslation 
iiLearner’s own diagram. The image
should have vertices (−3, 5), (−5, 5),
(−4, 4) and (−4, 5).
b
iLearner’s own diagram. The image
should have vertices (3, −3), (5, −2),
(5, −1) and (4, −1).
centre (2.5, 3)
iiiReflection in the line x = 4.5 OR
rotation 180 °, centre (4.5, 1) OR
 2
translation  0 .
b
Learner’s own discussions. For example:
Yes, for all of them there is more than one
transformation. Because each object and
image are in the same orientation, they
can all just be translated from one shape
to the other shape. The shapes can all also
be rotated 180 °. For the two pairs where
the translation is either horizontal only or
vertical only, it is also possible to reflect
the shapes in a vertical or horizontal
mirror line.
c
iFor example: Rotation 180 °, centre
iiLearner’s own diagram. The image
should have vertices (−1, −3), (1, −2),
(1, −1) and (0, −1).
c
iLearner’s own diagram. The image
should have vertices (−2, 2), (−2, 4),
(−3, 3), (−4, 3) and (−4, 2).
iiLearner’s own diagram. The image
should have vertices (−2, −4), (−2, −6),
(−4, −6), (−4, −5) and (−3, −5).
d
iThe positions of the shapes are
different, even though the elements of
the transformations are the same.
iiYes. Learner’s own explanation.
For example: A different order
often results in a different finishing
position.
iiiLearner’s own discussions.
ivLearner’s own transformations. For
example: Reflection in line y = −2,
then reflection in line x = 3.
v
48
 1
 
(3, 5) followed by a translation  −4 .
iiFor example: Rotation 90 °
anticlockwise, centre (1, 4) followed
 −2
by a translation   .
 −5
iiiFor example: Rotation 90 °
anticlockwise, centre (3, 0) followed
 −4
by a translation  2 .
Learner’s own checks.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
d
9
Learner’s own discussions. For example:
Yes, for all of them there is more than one
combined transformation. Each object can
be rotated about any point to get it in the
same orientation as the image, and then
you can use a translation to move it into the
correct position. In part i, you could also
use a reflection in any vertical or horizontal
line and then you can use a translation to
move it into the correct position.
aThey are both correct. When you start
with triangle G and follow Sofia’s
instructions, the final image is triangle
H. When you start with triangle G and
follow Zara’s instructions, the final image
is triangle H.
b
Exercise 13.4
1
2
a
For example: Reflection in the line x = 3
 −6
then translation  −2 .
 −8
For example: Translation  −2 then
reflection in the line x = −4.
c
10 a
There are an infinite number of
combined transformations. Learner’s
own explanation. For example: G can be
reflected in any line x = ‘a number’ then
translated to H.
b
iLearner’s own diagram. Shape B with
vertices (6, 4), (8, 5), (8, 2) and (6, 2).
Shape C with vertices (2, 5), (4, 6),
(4, 8) and (2, 8).
ii
b
Scale factor 2
Reflection in the line y = 5.
Scale factor 3
iLearner’s own diagram. Shape D with
vertices (5, 8), (8, 8), (8, 10) and
(6, 10). Shape E with vertices (2, 5),
(5, 5), (5, 7) and (3, 7).
c
iiRotation 90 ° anticlockwise,
centre (2, 5).
Activity 13.3
Learner’s own answers and discussions.
11 a
A to B
b
A to C
c
B to D
d
C to E
Scale factor 4
Reflection:
49
a
It is the same shape and size.
b
•
•
•
corresponding lengths are equal
corresponding angles are equal
the object and the image are congruent
3
aLearner’s own explanation. For example:
She hasn’t enlarged the shape correctly
from the centre of enlargement. She has
incorrectly used the centre as one of the
vertices of the triangle.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
7
Perimeter = 60 cm, Area = 150 cm2
8
Shape G is an enlargement of shape F, scale
factor 3 and centre of enlargement at (1, 2).
9
a
Enlargement scale factor 2, centre (−5, 2).
b
Enlargement scale factor 4, centre (−6, −2).
10 Enlargement scale factor 3, centre (4, −5).
11 Learner’s own answers and justification. For
example: Arun is incorrect. When one shape is
an enlargement of another, and the centre of
enlargement is inside the shapes, you can use
ray lines to find the centre of enlargement.
Activity 13.4
Learner’s own enlargements and discussions.
12 Enlargement scale factor 3, centre (6, 5).
4
13 Enlargement scale factor 2, centre (4, 4).
5
aLearner’s own diagram. Check that the
shape has been enlarged correctly. Vertices
of the image should be at (1, 7), (5, 7),
(5, 3) and (1, 3).
b
Learner’s own diagram. Check that the
shape has been enlarged correctly. Vertices
of the image should be at (2, 6), (8, 6),
(8, 0) and (2, 0).
c
Learner’s own diagram. Check that the
shape has been enlarged correctly. Vertices
of the image should be at (1, 9), (9, 9),
(9, 1) and (1, 1).
a
Check your progress
1
N
9 cm
(90 km)
iPerimeters: A = 8 cm, B = 16 cm,
C = 24 cm and D = 32 cm
Scale
Ratio Ratio
Ratio of
Squares factor of
of
of peri­
areas
enlargement lengths meters
A:B
2
1:2
1:2
1:4=1:2
A:C
3
1:3
1:3
1 : 9 = 1 : 32
A:D
4
1:4
1:4
1 : 16 = 1 : 42
2
c
ratio of lengths = ratio of perimeters.
d
ratio of lengths squared = ratio of areas.
e
Yes. Yes.
f
Learner’s own discussions.
Perimeter of R = 14 cm → Perimeter of
T = 14 × 3 = 42 cm
Area of R = 10 cm2
12 cm
(120 km)
50°
b
Answer in range 148 km–152 km
(accurate answer 150 km).
c
Answer in the range 264 °–270 °
(accurate answers 267 ° to 3 s.f.)
2
a
(5, 3)
3
L (4, 10)
4
a
b
50
N
140°
iiAreas: A = 4 cm2, B = 16 cm2,
C = 36 cm2 and D = 64 cm2
6
a
b
(6, 10)
iLearner’s own diagram. The vertices
of triangle B should be at (3, 3), (5, 3)
and (4, 4).
iiLearner’s own diagram. The vertices
of triangle C should be at (3, 3), (4, 2)
and (4, 4).
b
i
Rotation of 180 °, centre (3, 4).
iiRotation 90 ° anticlockwise,
centre (2, 3).
→ Area of
T = 10 × 32 = 90 cm2
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
6
7
4
a
Yes. The cross-section is a circle.
b
Area of circle × height
c
V = πr2h
d
Learner’s own discussions.
5
Learner’s own explanation. For example: The
radius and height are in different units. She
needs to change the 5 mm to cm or change the
2 cm to mm before she works out the volume.
Volume = 1570 mm3 (3 s.f.) or 1.57 cm3 (3 s.f.)
6
a
942.5 cm3
b
353.4 cm3
c
17 592.9 mm3
Scale factor 3, centre of enlargement
at (10, 4).
Perimeter = 54 cm and area = 180 cm2.
Learner’s own answers. For example:
Activity 14.1
Unit 14 Getting started
Learner’s own cylinders, answers and discussions.
1
25.13 cm
7
2
a
27 mm2
3
a
120 cm3
4
a
480 cm3
b
5
Radius
of circle
Area of
circle
Height
of
cylinder
Volume
of
cylinder
a
2.5 m
19.63 m2
4.2 m
82.47 m3
Learner’s own diagram. Any correct net.
b
6 cm
113.10 cm2 4.48 cm
507 cm3
c
528 cm2
c
a
1
b
b
b
c
21 cm2
c
2
78.5 m2
158 cm2
6
d
0
Exercise 14.1
1
a
120 cm3
b
130 cm3
c
134.4 cm3
2
3
20 m
2.5 m
50 m3
d
4.56 mm
65.25 mm2
16 mm
1044 mm3
8
a
9
Learner’s own methods and answers. For
example:
5.5 cm
b
4.2 cm
c
2.1 cm
Volume of cylinder: V = πr2h = π × 62 ×18
= 2035.75 cm3 (2 d.p.)
Area of
cross-section
Length of
prism
Volume
of prism
a
12 cm2
10 cm
120 cm3
b
24 cm2
8.5 cm
204 cm3
c
18.5 m2
6.2 m
114.7 m3
aLearner’s own explanation. For example:
Yusaf hasn’t used the correct crosssection. Instead of using the trapezium
as the cross-section, he has used the side
rectangle (which is not the cross-section
of the prism).
b
2.52 m
2
Area of trapezium =
1
2
Volume of cube: V = 83 = 512 cm3
Volume of water: 1.5 litres = 1500 mL = 1500 cm3
Volume of cube + 1.5 litres = 512 + 1500
= 2012 cm3
The total volume of the cube and water is less
than the volume of the cylinder, so the water
will not come over the top of the cylinder.
2012 cm3 < 2035.75 cm3
Reflection: Learner’s own explanations.
(8 + 14) × 4 = 44 cm2
Volume of prism = 44 × 20 = 880 cm3
51
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 14.2
1
e
Add 1 to the number in front of the
r, then double it. This gives you the
number in front of the πr2. So, 19 + 1 = 20,
20 × 2 = 40, so SA = 40πr2.
f
Learner’s own discussions.
Answer using rounded intermediate values:
Area of circle = π r 2
= π × 52
= 78.54 cm 2 (2 d.p.)
Circumference of circle = π d
= π × 10
= 31.42 cm (2 d.p.)
5
226 cm2 (3 s.f.)
6
Learner’s own methods and answers. For
example:
a
Area of rectangle = 31.42 × 12
ii
= 377.04 cm 2 (2 d.p.)
Total area = 2 × 78.54 + 377.04
= 534 cm 2 (3 s.f.)
Answer using accurate intermediate values:
Area of circle = π r
7
2
=π ×5
Total area = 2 × 78.5398... + 376.9911...
= 534 cm 2 (3 s.f.)
52
Learner’s own discussions.
c
408 cm2
a
SA = 660 cm2
b
SA = 1188 mm2
c
SA = 23.3 m2
a, b Learner’s own shapes. For example:
A cuboid with length 10 cm, width
10 cm and height 8 cm (V = 800 cm3,
SA = 520 cm2); A triangular prism of
length 33 cm with a right-angled
cross-section with base length 6 cm, height
8 cm and hypotenuse 10 cm (V = 792 cm3,
SA = 840 cm2); A cylinder with height
16 cm and cross-section radius 4 cm
(V = 804 cm3, SA = 503 cm2).
= 376.9911... cm 2
4
b
Activity 14.2
Area of rectangle = 31.4159... × 12
3
Pythagoras’ theorem
2
= 78.5398 ... cm 2
Circumference of circle = π d
= π × 10
= 31.4159 ... cm
2
iThe hypotenuse of the triangular
cross-section.
a
SA = 477.5 cm2
c
Learner’s own answers and explanations.
b
SA = 322.0 cm2
d
Learner’s own discussions.
c
SA = 4272.6 mm2
The pyramid has a greater surface area than
the cylinder. 132 cm2 > 125.66 cm2.
1

Pyramid: SA = 4 ×  × 6 × 8 + 6 × 6 = 132 cm 2
2

Cylinder: SA = π × 22 × 2 + π × 4 × 8 = 125.66 cm2
Learner’s own methods and answers.
For example:
8
754 cm2
9
15 labels is the maximum using Method 1
below.
Method 1:
120 ÷ 23.6 = 5 whole lengths
35 ÷ 10 = 3 whole lengths
Number of labels = 5 × 3 = 15
a
SA = πr2 + πr2 + 2πrh
b
SA = πr2 + πr2 + 2πrh = 2πr2 + 2πrh =
2πr(r + h)
c
SA = 2πr(r + h) = 2πr(r + 2r) = 2πr × 3r =
6πr2
35 ÷ 23.6 = 1 whole length
d
i
SA = 8πr2
Number of labels = 12 × 1 = 12
iii
SA = 12πr2
ii
SA = 10πr2
Method 2:
120 ÷ 10 = 12 whole lengths
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 14.3
6
1
a, b and c Learner’s own drawings. Check that
the planes of symmetry are drawn correctly.
Shapes a and b have vertical planes of
symmetry. Shape c has a horizontal plane of
symmetry.
2D
regular
polygon
Triangle
3
Triangular
4
a, b Learner’s own drawings. Check that the
planes of symmetry are drawn correctly. Shape
a has one vertical and one horizontal plane of
symmetry. Shape b has two vertical and one
horizontal plane of symmetry.
Square
4
Square
5
Pentagon
5
Pentagonal
6
Hexagon
6
Hexagonal
7
Octagon
8
Octagonal
9
a, b Learner’s own drawings. Check that the
plane of symmetry is drawn correctly. The
plane of symmetry should be vertical.
b
2
3
c
4
The plane of symmetry is a vertical plane
of symmetry.
a, b Learner’s own lines of symmetry. Any of
these:
c
A cube has a total of nine planes of
symmetry.
d
Learner’s own justification. All nine
diagrams shown in the answer to part b.
e
Learner’s own discussions.
aThere are two vertical and one horizontal
planes of symmetry.
b
Number
of lines of 3D prism
symmetry
Number
of planes
of
symmetry
Learner’s own answers and explanations.
For example:
Number of planes of symmetry = number
of lines of symmetry + 1. This happens
because the planes of symmetry can be
drawn, the length of the prism, in the
same place as the lines of symmetry on the
cross-section of the prism. There is then
the extra plane of symmetry that divides
the prism halfway along its length.
7
5
a
c
i
d
Learner’s own discussions.
11
ii
13
a, bLearner’s own diagram. Check that the
plane of symmetry passes through the
circular ends of the cylinder, dividing the
circular cross-section into two identical
semi-circles.
c
Learner’s own diagram. Check that the
plane of symmetry passes halfway along
the height, splitting the cylinder into two
identical cylinders.
d
Learner’s own answers and explanations.
For example:
It has an infinite number of planes of
symmetry. A circle has an infinite number
of lines of symmetry, so this is the same
in 3D for the cylinder. When the cylinder
is placed upright there is always one
horizontal plane of symmetry, but an
infinite number of vertical ones.
Reflection: Learner’s own answers.
53
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Check your progress
3
a
Mean Median Mode Range
1
120 cm3
2
14 m2
History
12.9
13
16
7
3
452 cm3
Chemistry
14
16
18
15
4
The square-based pyramid has the greater
surface area.
b
The Chemistry group has better marks on
average, because the mean, median and
mode are all greater than for the History
group.
c
The History group has more consistent
marks because the range is lower.
Pyramid: SA = 340 cm2,
Cylinder: SA = 320.44 cm2, 340 > 320.44
5
aThe shape has two vertical, one horizontal
and two diagonal planes of symmetry.
b
Learner’s own diagrams showing the five
planes of symmetry correctly as described
in the answer to part a.
1
a
Height, h (cm)
Frequency Midpoint
7
145
Unit 15 Getting started
140 ⩽ h < 150
150 ⩽ h < 160
13
155
1
160 ⩽ h < 170
6
165
170 ⩽ h < 180
2
175
a
b
c
2
Age, a (years)
10 < a ⩽ 15
Frequency
3
15 < a ⩽ 20
6
20 < a ⩽ 25
7
25 < a ⩽ 30
4
Learner’s own diagram. Frequency
diagram showing the data in part a. Make
sure the axes are labelled correctly and
that a sensible scale is used. Make sure the
bars are the correct width and height.
2
b
Learner’s own diagram. Frequency
polygon with points (145, 7), (155, 13),
(165, 6) and (175, 2) joined with straight
lines. Make sure that the axes are labelled
correctly and that a sensible scale is used.
a
Mass, m (kg)
11
a
Class 9P test results
40 ⩽ m < 50
Frequency
4
Midpoint
45
50 ⩽ m < 60
12
55
60 ⩽ m < 70
8
65
b
Learner’s own diagram. Frequency
polygon with points (45, 4), (55, 12) and
(65, 8) joined with straight lines. Make
sure that the axes are labelled correctly
and that a sensible scale is used.
0
3 8 9
1
2 4 6 7 8 9
2
2 3 4 4 6 7 8
3
0 1 6 8 9 9 9
c
24
4
0 0
d
2
3
e
Arun is incorrect. Learner’s own
explanation. For example: You do not
know how heavy the heaviest student is.
You only know that their mass is in the
interval 60 kg ⩽ m < 70 kg.
Key: 0 3 means 03 marks
54
Exercise 15.1
b
32%
c
1
5
d
14
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
3
aLearner’s own frequency table. For
example:
b
c
4
Frequency
10 ⩽ a < 25
6
25 ⩽ a < 40
9
40 ⩽ a < 55
7
55 ⩽ a < 70
4
70 ⩽ a < 85
2
Learner’s own diagram. Frequency
polygon with points (17.5, 6), (32.5, 9),
(47.5, 7), (62.5, 4) and (77.5, 2) joined
with straight lines. Make sure that the
axes are labelled correctly and that a
sensible scale is used.
6
b
a
Learner’s own diagram. Two frequency
polygons drawn on one grid. Oaklands
points (5, 25), (15, 10), (25, 12) and (35, 3)
joined with straight lines. Birchfields
points (5, 8), (15, 14), (25, 17) and (35, 11)
joined with straight lines. Make sure that
the axes are labelled correctly and that a
sensible scale is used.
d
Learner’s own comments. For example:
Over three times as many people waited
less than 10 minutes in Oaklands surgery
compared to Birchfields surgery. More
people waited over 10 minutes in Birchfields
surgery compared to Oaklands surgery.
Time, t (minutes)
Frequency
10 ⩽ a < 20
4
20 ⩽ a < 30
8
a, bLearner’s own comments. For example:
Using Sofia’s method you don’t need
to work out the midpoints. When you
have drawn the bars it is easy to join the
midpoint of each bar with straight lines.
Her method will take longer though, as
you have to draw all the bars first. Using
Zara’s method is quicker as you don’t have
to draw all the bars, but you do need to
work out the midpoints, and if you make
a mistake with one of the midpoints you
might not notice when you plot the point.
30 ⩽ a <40
9
c
40 ⩽ a < 50
3
Learner’s own discussions.
Learner’s own frequency tables and polygons.
For example:
a
5
Age, a (years)
c
7
Learner’s own diagram. Frequency
polygon with points (15, 4), (25, 8),
(35, 9) and (45, 3) joined with straight
lines. Make sure that the axes are labelled
correctly and that a sensible scale is used.
aLearner’s own diagram. Frequency
polygon with points (5, 2), (15, 4), (25, 8)
and (35, 6) joined with straight lines.
Make sure that the axes are labelled
correctly and that a sensible scale is used.
b
50 at each surgery.
Oaklands Surgery
Time, t (minutes) Frequency Midpoint
0 ⩽ t < 10
25
5
10 ⩽ t < 20
10
15
20 ⩽ t < 30
12
25
30 ⩽ t < 40
3
35
0 ⩽ t < 10
8
5
10 ⩽ t < 20
14
15
b
20 ⩽ t < 30
17
25
30 ⩽ t < 40
11
35
Birchfields Surgery
55
Learner’s own comments. For example:
The plants that were grown in the
greenhouse grew higher than the plants
that were grown outdoors. 14 of the
plants grown in the greenhouse were over
20 cm tall, whereas only six of the plants
grown outdoors were over 20 cm tall.
aLearner’s own diagram. Two frequency
polygons drawn on one grid. Boys’ points
(2, 5), (6, 10), (10, 15), (14, 7) and (18, 3)
joined with straight lines. Girls’ points
(2, 7), (6, 8), (10, 12), (14, 18) and (18, 5)
joined with straight lines. Make sure that
the axes are labelled correctly and that a
sensible scale is used.
Time, t (minutes) Frequency Midpoint
8
Learner’s own discussions.
Learner’s own comments. For example:
More girls spend between 0 and 4 and
between 12 and 20 hours doing homework
each week. More boys spend between 4
and 12 hours doing homework each week.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
9
Exercise 15.2
c
40 boys and 50 girls
d
Learner’s own comments. For example:
No, as there were 10 more girls than boys
surveyed. There should have been the
same number of boys and girls in order to
make a fair comparison.
aLearner’s own diagram. Frequency
polygon with points (200, 5), (220, 8),
(240, 11), (260, 7), (280, 5) and (300, 4)
joined with straight lines. Make sure that
the axes are labelled correctly and that a
sensible scale is used.
b
i
Length, l (cm)
Frequency
190 ⩽ l < 230
13
230 ⩽ l < 270
18
270 ⩽ l < 310
9
iiLearner’s own diagram. Frequency
polygon with points (210, 13),
(250, 18) and (290, 9) joined with
straight lines. Make sure that the
axes are labelled correctly and that a
sensible scale is used.
c
d
2
Learner’s own answers and explanations.
For example: The first frequency polygon
gives you better information because there
are more groups so it shows you more
information on the lengths of the turtles.
The second frequency polygon only has
three groups so less information can be
taken from the graph.
i
12
iiNo, Arun cannot fill in the correct
frequencies in his table. Learner’s own
explanation. For example: From the
first table Arun knows that there are
five turtles between 190 and 210 cm.
But this does not tell him how many
turtles there are between 190 and
200 cm and how many turtles there
are between 200 and 210 cm, so it
is impossible for him to complete
his table. He would have to find the
original data, before it was grouped,
in order to use the groups he wants to.
56
1
3
aLearner’s own scatter graph. Horizontal
axis showing ‘Hours doing homework’
and vertical axis showing ‘Hours watching
TV’. Points (14, 7), (11, 12), (19, 4),
(6, 15), (10, 11), (3, 18), (9, 15), (4, 17),
(12, 8), (8, 14), (6, 16), (15, 7), (18, 5),
(7, 16) and (12, 10) plotted. Make sure
that the axes are labelled correctly and
that a sensible scale is used.
b
Negative correlation. The more time the
student spends doing homework, the less
time they spend watching TV.
c
Student’s line of best fit. Strong
negative correlation.
d
Correct answer from learner’s line of
best fit. Answer should be within range
16–17.
a
Learner’s own answer and explanation.
b
Learner’s own scatter graph. Horizontal
axis labelled ‘Maximum daytime
temperature’ and shown from 25 to 35.
Vertical axis labelled ‘Number of cold
drinks sold’ and shown from 20 to 40.
Points (28, 25), (26, 22), (30, 26), (31, 28),
(34, 29), (32, 27), (27, 24), (25, 23), (26, 24),
(28, 27), (29, 26), (30, 29), (33, 31) and
(27, 23) plotted.
c
Positive correlation. The higher the
temperature, the more cold drinks were
sold.
d
Learner’s own answer.
e
Learner’s own line of best fit.
f
Learner’s own comments. For example: It is
not possible to predict from a line of best
fit a value higher or lower than the data
given, as there are no data to show that
the correlation is the same after or before
these points. With a temp of 44 °C the store
might not sell many drinks as people might
not go outside in that temperature.
g
Learner’s own discussions.
a
Learner’s own answer and explanation.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
4
5
6
Learner’s own scatter graph. Horizontal
axis labelled ‘History result’ and shown
from 0 to 100. Vertical axis labelled ‘Music
result’ and shown from 0 to 100. Points
(12, 25), (15, 64), (22, 18), (25, 42), (32, 65),
(36, 23), (45, 48), (52, 24), (58, 60), (68, 45),
(75, 68), (77, 55), (80, 42), (82, 32) and
(85,76) plotted.
c
No correlation. Getting a good result in
one subject does not mean a student will
get a good, or bad, result in the other
subject.
d
Learner’s own answer and explanation.
a
Strong positive correlation.
b
6 km in 16 minutes. Learner’s own
explanation. For example: It should have
taken less time, so the taxi might have
been delayed in traffic.
b
Weak negative correlation.
c
Learner’s own line of best fit, and correct
answer from their line, for number of fish
when the temperature is 27 °C. Answers
should be within range 74–78.
d
It is not a good idea to use the line of
best fit to predict the number of fish in
the Red Sea when the temperature of the
sea is 30 °C, 35 °C or even higher, because
you do not know what happens beyond
the data you are given. There may be no
fish at 30 °C and the number cannot keep
dropping after that.
e
Learner’s own answers.
Reflection: Learner’s own answers.
7
a
Learner’s own answers.
b
Learner’s own answers. For example: Try
to get an equal number of points on either
side of the line (not always possible). The
line can go through some of the points.
Make lines long enough to go through all
the data, don’t make the lines too short.
Work out the mean of the data and make
the line go through this point.
aLearner’s own explanation. For example:
It is a coincidence that the graph shows a
positive correlation. In a school the older
learners might have longer feet, and they
might be better at maths as they have been
in school longer than the younger students,
but they might not. Also, when your feet
stop growing, it doesn’t mean that you are
going to stop getting better at maths. Your
ability in maths does not depend on the
length of your foot. Your ability in maths
depends on how hard you work.
c
Learner’s own discussions.
b
d
It is not a good idea to use the line of
best fit to make predictions outside the
range of the data, because you do not
know what happens beyond the data you
are given. It could be that after a body
length of 60 cm, a bird’s wingspan hardly
changes in length.
Learner’s own discussions.
a
150
Number of fish at different points in the Red Sea
Number of fish
140
130
120
110
100
90
80
70
57
18
20
22
24
26
Temperature (ºC)
28
30
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 15.3
1
Shop A
a
Shop B
8 9
0
8
9 0 7 7 4 6 2 8
1
9 9
1 4 3 9
2
9 5 4 7 7
1 0
3
6 3 2 5 6 0 1 2
Key: 9 0 means 9 years old
Key: 3 6 means 36 years old
Shop A
b
Shop B
9 8
0
8
9 8 7 7 6 4 2 0
1
9 9
9 4 3 1
2
4 5 7 7 9
1 0
3
0 1 2 2 3 4 6 6
Key: 9 0 means 9 years old
c
Learner’s own checks.
d
i
Shop A
ii
Shop B
e
2
a
Key: 3 6 means 36 years old
Learner’s own answers. For example: Shop A sells clothes for younger people and shop B sells
clothes for older people.
Beach car park
City car park
3
0 4 9
7 6 6 6 5
4
2 5 5 5 7
9 7 7 6
5
4 6 9
2 2 1 0 0
6
8 8 9
Key: For the Beach car park, 5 4 means 45 ice-creams
For the City car park, 3 0 means 30 ice-creams
b
58
i Mode ii Median iii Range
Beach
car park
46
57
17
City car
park
45
46
39
c
Learner’s own answers. For example: On average the vendor had better sales at the Beach car park.
Their median was higher. This shows that 50% of their daily sales were 57 ice-creams or more,
compared to only 46 for the City car park. Their mode was also higher. The range was smaller,
showing that their sales were more consistent. However, it was at the City car park where they had
their highest daily sale of 69 ice-creams.
d
Learner’s own answers. For example: No. The vendor’s sales were better at the Beach car park as
they had a higher median and mode and sales were more consistent.
e
Learner’s own discussions.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
3
a
Boys’
times
17.4 s
16.3 s
2.9 s
16.56 s
Girls’
times
16.8 s
17.5 s
4s
17.72 s
b
4
Learner’s own answers. For example: No,
as the girls’ mean and median are both
slower. This shows that on average the
boys are faster.
a
A ,B2
b
A 25%, B 0%
c
The variation is the same for A and B.
They both have a range of 31 g.
.
A: mean = 408.83
. g, median = 409 g
B: mean = 395.6 g, median = 395 g
e
5
Learner’s own answers. For example: On
average the boys ran faster than the girls,
as their mean and median were lower.
The girls had the fastest modal time, but
they had a larger range showing that their
times were more varied than the boys.
c
d
1
4
3
c
Learner’s own answers. For example:
Neither website appears to be better.
Website A was more consistent. Website
B was only slightly better on average than
Website A.
d
Learner’s own discussions.
Exercise 15.4
1
a
a
Website A
Website B
4 3 0 0 13
ii
150 cm ⩽ h < 160 cm
Learner’s own explanation. For example:
You can only give the modal class and
class where the median lies, because the
data is grouped and you don’t know the
individual values.
c
40 cm
Midpoint Frequency
4 6 8
5 5 5 6 6 8
9 8 5 3 3 2 2 15
4 5 6 7 7 8
6 7 8 9
Key: For Website A, 0 13 means 130 hits
For Website B, 12 8 means 128 hits
b
150 cm ⩽ h < 160 cm
d
8 9
8 7 6 5 5 5 2 1 14
1 0 16
i
b
Learner’s own answers. For example:
Location A because on average the mass
of the hedgehogs is greater.
12
59
Learner’s own answers. For example:
Website A and Website B both had the
same mode and almost the same median.
The median for Website B was only one
more than Website A, so this average is
almost the same. The mean was also very
similar with only a difference of 2.8 hits
per day. So, on average Website B had
slightly more hits than Website A. Website
B’s range is a lot higher than Website A,
showing that the number of hits it had per
day varied a lot more.
i Mode ii Median iii Range iv Mean
145
7
145 × 7 = 1015
155
13
155 × 13 = 2015
165
6
165 × 6 = 990
175
2
175 × 2 = 350
Totals:
28
4370
Estimate of mean =
145
147
31
147.1
Website
B
145
148
41
149.9
4370
28
2
a
Mode Median Range Mean
Website
A
Midpoint × frequency
b
c
i
50 kg ⩽ m < 60 kg
ii
i
50 kg ⩽ m < 60 kg
.
56.6 kg or 57 kg
ii
30 kg
= 156 cm
Learner’s own explanation. For example:
Answers are estimates because the data
is grouped and you do not know the
individual values.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
3
4
aLearner’s own answers and explanations.
For example: Using the midpoint would
be best. If you use the smallest value in
each class the estimate of the mean will be
too low, because not all the values will be
the smallest value. If you use the highest
value the estimate of the mean will be too
high, because not all the values will be the
highest value.
b
Learner’s own discussions.
a
40 at The Heath and 50 at Moorlands.
Table B
Score
Tally
Frequency
2–5
llll l
6
6–9
llll lll
8
10–13
llll l
6
Modal
class
interval
Class
interval
where the
median lies
Estimate
of mean
Table A
8–10
8–10
7.2
Table B
6–9
6–9
7.5
e
b
Hospital
Modal
class
interval
Class
interval
where the
median lies
The
Heath
10 ⩽ t < 20
minutes
10 ⩽ t < 20
minutes
17.25
minutes
Moorlands 0 ⩽ t < 10
minutes
20 ⩽ t < 30
minutes
19.4
minutes
c
d
5
Estimate
of mean
f
iiLearner’s own answers. For example:
The accurate median lies in both the
class intervals containing the median.
Learner’s own answers. For example:
The modal class interval is lower for
Moorlands than The Heath, but the class
interval containing the median is lower
for The Heath than Moorlands. The mean
is just over 2 minutes less waiting time in
The Heath than Moorlands.
Learner’s own answers. For example: The
Heath, because the mean is lower and the
median is lower. Even though the modal
group is lower at Moorlands, on average I
think waiting times will be less at
The Heath.
iiiLearner’s own answers. For example:
The accurate modal value is 3, but
this isn’t reflected at all in either of
the modal class intervals, which are
totally different.
g
Learner’s own discussions.
Activity 15.4
a
1
b
36
a
2
c–i Learner’s own data, tables, answers and
discussions.
b
13
6
c
Mean = 7.15, Median = 8, Mode = 3
d
a
b
Table A
i
750 g ⩽ m < 800 g
ii
750 g ⩽ m < 800 g
i
798 g
ii
Mass, m (g)
c
400 g
Score
Tally
Frequency
2–4
llll l
6
600 ⩽ m < 700
Frequency
7
5–7
lll
3
700 ⩽ m < 800
19
8–10
llll lll
8
800 ⩽ m < 900
18
11–13
lll
3
900 ⩽ m < 1000
6
d
e
60
iLearner’s own answers. For example:
When there are more groups, the
estimate of the mean is closer to the
accurate mean.
i
700 g ⩽ m < 800 g
ii
700 g ⩽ m < 800 g
i
796 g
ii
400 g
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
f
iLearner’s own answers and
explanations. For example: I think the
answers in parts a and b are the more
accurate answers because the groups
are smaller in size so the individual
values are more likely to be nearer the
midpoints in the smaller groups than
in the bigger groups. The range is the
same for both sets of answers because
the smallest and greatest possible
values are the same.
work to Shoprite than Kabayan, whereas
nearly four times as many employees
took between 30 and 45 minutes to
travel to Kabayan than Shoprite. Only
five employees (8%) from Shoprite took
longer than 45 minutes to travel to work,
compared with nine employees (15%)
from Kabayan.
2
iiLearner’s own answers and
explanations. For example: The
answers in parts d and e were quicker
to work out because there were fewer
groups, so there were fewer calculations
to do for the median and mean.
Check your progress
1
a
aLearner’s own scatter graph. Horizontal
axis labelled ‘Age (years)’ and shown
from 0 to 16. Vertical axis labelled
‘Value ($)’ and shown from 0 to 16 000.
Points (8, 8500), (10, 6000), (2, 13 500),
(3, 12 500), (15, 3500), (1, 15 000),
(12, 4000), (5, 10 000), (9, 6500) and
(4, 12 000) plotted.
b
Negative correlation.
c
Learner’s own line of best fit and correct
estimate of the value of a car that is six
years old. Answer should be within range
9600–10 400.
a
i–iv
60
b
3
Kabayan Supermarket
Time, t (minutes)
Frequency
Midpoint
0 ⩽ t < 15
5
7.5
15 ⩽ t < 30
8
22.5
30 ⩽ t < 45
38
37.5
45 ⩽ t < 60
9
52.5
i Mode ii Median
Boys’
times
67 s
69 s
32 s
69.1 s
Girls’
times
56 s
63 s
32 s
64.5 s
b
Learner’s own answers. For example: The
range is the same for the boys and the
girls so they are both as varied as each
other. The median and the mean for the
boys and girls are all over 60 seconds. The
boys’ mean and median are higher than
the girls’. The girls’ mean and median
are closer to 60 seconds. The girls’ mode
is only 4 seconds under 60 seconds,
whereas the boys’ mode is 7 seconds over
60 seconds.
c
Learner’s own answers. For example:
No, the boys’ median is higher, but is
further away from 60 seconds, as is their
mean, so the boys are worse at estimating
60 seconds.
a
i
Shoprite Supermarket
Time, t (minutes)
Frequency
Midpoint
0 ⩽ t < 15
32
7.5
15 ⩽ t < 30
13
22.5
30 ⩽ t < 45
10
37.5
45 ⩽ t < 60
5
52.5
c
d
61
Learner’s own diagram. Two frequency
polygons drawn on one grid. Kabayan
Supermarket points (7.5, 5), (22.5, 8),
(37.5, 38) and (52.5, 9) joined with
straight lines. Shoprite Supermarket
points (7.5, 32), (22.5, 13), (37.5, 10) and
(52.5, 5) joined with straight lines. Make
sure that each line is labelled clearly. Make
sure that the axes are labelled correctly
and that a sensible scale is used.
Learner’s own answers. For example:
More than six times as many employees
took less than 15 minutes to travel to
iii Range iv Mean
4
6 ⩽ t < 8 hours
6 ⩽ t < 8 hours
.
i7.26 hours or 7 hours 16 minutes or
7.3 hours
ii
b
ii
6 hours
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021