9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

Learner’s Book
answers
Unit 1 Getting started
1 a 144 b 9 8 a The square root of any integer between
c 125 d 4 16 and 25 is a possible answer.
b The square root of any integer between
2 a 512 b 128 144 and 169 is a possible answer.
3 a 15 7
b 15 3
9 a 14
4 a 4 and 3000 and 225 b 6
b All of them. 10 a i 1 ii 2 iii 3
5 106 b ( 5 + 1) × ( 5 − 1) = 4, and so on

Exercise 1.1 c ( N + 1) × ( N − 1) = N − 1
d Learner’s own answer.
1 a integer 3 b irrational
c irrational d integer 7 11 a No. It is not a repeating pattern.
b Learner’s own answer.
e irrational
5 Reflection:
2 a 1, 7 , −38 and − 2.25 are rational.
12
a i true ii true iii false
b 200 is the only irrational number. b No. It might be a repeating pattern or it
3 a integer b surd c surd might not.
d integer e integer f surd
Exercise 1.2
4 a irrational because 2 is irrational
1 a 3 × 105 b 3.2 × 105
b rational because it is equal to 4 = 2
c 3.28 × 105 d 3.2871 × 105
c irrational because 3 4 is irrational
2 a 6.3 × 107 b 4.88 × 108
d rational because it is equal to 3 8 = 2
c 3.04 × 106 d 5.2 × 1011
5 a Learner’s own answer. For example:
2 and − 2 . 3 a 5400 b 1 410 000

b Learner’s own answer. For example: c 23 370 000 000 d 87 250 000
2 and 2 − 2 4 Mercury 5.79 × 107 km; Mars 2.279 × 108;
6 a i 4 ii 6 Uranus 2.87 × 109

iii 10 iv 6 5 a Russia b Indonesia

b They are all positive integers. c The largest country is approximately 9
times larger than the smallest country.
c Learner’s own answer.
d Learner’s own answer. 6 a 7 × 10−6 b 8.12 × 10−4
c 6.691 × 10−5 d 2.05 × 10−7
7 a 7² = 49 and 8² = 64
b 4³ = 64 and 5³ = 125

1 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 1/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

7 a 0.0015 b 0.000 012 34 b

1 1
7 a 36 36
c 1 d 216
c 0.000 000 079 d 0.000 900 3
8 a 1 1 c d 1
b 1
8 a 30 b 9.11 × 10−25 kg 81 225 400
1 1
9 a z b y 9 a i 2 ii 4 iii 9
4 9

10 a 65 is not between 1 and 10. b i x=5 ii x = 10

b 6.5 × 105 10 a i 35 ii 39
c 4.83 × 107 iii 310 iv 36
11 a 1.5 × 10−2 b i 3 ii 3−1 iii 32
b 2.73 × 10−3 iv 3−2 v 3−3
c 5 × 10−8 c Learner’s own answers.
d Learner’s own answers.
12 a 6.1 × 10 6

b 6.17 × 105 11 a 56 b 52 c 5−2 d 5−6

c 1.75 × 10 5 12 a 6 −1
b 7 3

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 × 10−7 c x = −2 d x=5
14 a i 7 × 106 ii 3.4 × 107 14 a i 22 ii 43
iii 4.1 × 10−4 iv 1.37 × 10−3 iii 51 or 5 iv 23
b To multiply a number in standard form by b Learner’s own answers.
10, you add 1 to the index. c Learner’s own answers.
c To multiply a number in standard form
15 a 6−3 b 9−1
by 1000, you add 3 to the index. To divide
a number in standard form by 1000, you c 15−4 d 10−5
subtract 3 from the index.
16 a 25 b 87
Reflection: You can compare them easily. You c 5−6 d 122
can write the number without using a lot of zeros.
You can enter them in a calculator. 17 a 26 b 2−6 c 36
d 3−6 e 93 f 9−3
Exercise 1.3
1 a
1
b
1
c
1 Check your progress
4 8 81
1 1 1 1 a rational b irrational
d e f
216 10 000 32 c rational d irrational
2 3−3, 2−4 and 4−2 are equal, 5−1, 60 e rational
3 a 2−1 b 2−2 c 26 2 a rational because it is equal to 25 = 5
d 2−6 e 20 f 2−3 b irrational because it is 3 + 7 and 7 is
4 a 10 2
b 10 3
c 100 asurd

d 10−1 e 10−3 f 10−6 3 n=3

5 a 64−1 b 8−2 4 a 8.6 × 1010 b 6.45 × 10−6

c 4−3 d 2−6 5 C, D, A, B
6 a 3−4 or 9−2 or 81−1 1 1 1
6 a b c
49 81 128
b The three ways in part a.

2 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 2/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

7 a 53 b 50 c 5−2 4 a Learner’s own answers.

8 a 6 5
b 12 −5 For example: Part a is incorrect as −32
should be written as (−3)2, which is 9 and
c 4−6 d 152 not −9; part b is incorrect as (−2)3 is −8
and not 8.
Unit 2 Getting started b Learner’s own answer.
x
1 +7
3 5 a x = 1 and y = 14, x = 2 and y = 11, x = 3 and
512 y=6
2 a 32 × 34 = 36 b = 53
59 b Learner’s own answer. For example:
(72 ) = 710
5
c x = −4 and y = −1, x = −5 and y = −10,
x = −6 and y = −21
3 a x2 + 2x b 12y2 − 21yw
c Learner’s own answer. For example:
4 a 4(x + 3) b 2x(2x + 7) x = −1 and y = 14, x = −2 and y = 11,
x = −3 and y = 6 or x = 4 and y = −1,
17
5 a or 1 5 b
6
or 11 x = 5 and y = −10, x = 6 and y = −21
12 12 5 5
6 a 4( m + 2 p ) = 4( 2 + 2 × −4)
F
6 a F = 25 b a= = 4( 2 − 8)
m
c a=6 = 4 × −6
= −24
Exercise 2.1 b p 3 − 3mp = ( −4) 3 − 3 × 2 × −4
1 a x − 2y = 3 − 2 × 5 = −64 + 24
= 3 − 10 = −40
= −7  p
5 5
 − 4
 m  + ( p ) =   + ( −4 )
3 3
c
2
b x + xy = 3 + 3× 5
3 3
5
= 27 + 15 = ( − 2) − 64
= 42 = −32 − 64
= −96
10x 10 × 3
= (5 ) −
2
c y2 − 7 a 21 b 36 c 16
y 5
30 d 64 e 68 f −18
= 25 −
5 g 14 h −25 i −7
= 25 − 6 j 82
= 19
2 a 9 b 4 c 9
Activity 2.1
Learner’s own answer.
d 8 e 8 f 30
g 5 h 47 i −30 8 Learner’s own counter-examples.
j −4 a For example: When x = 2,
3x2 = 3 × 22 = 3 × 4 = 12, and
3 a Learner’s own answers. For example: (3x)2 = (3 × 2)2 = 62 = 36, and 12 ≠ 36
i a = 3, b = 10, c = 12, d = 2 b For example: When y = 2, (−y)4 = (−2)4 = 16
ii a = −3, b = −10, c = −12, d = −2 and −y4 = −24 = −16, and 16 ≠ −16
iii a = 3, b = 4, c = −36, d = 3 c For example: When x = 3 and y = 4,
2(x + y) = 2(3 + 4) = 2 × 7 = 14 and
b Learner’s own answers.
2x + y = 2 × 3 + 4 = 10, and 14 ≠ 10
c Learner’s own answers.
9 a 26
b 49

3 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 3/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

2
10 5a2 − 9(b − a ) + + 7ab = 5 × ( − 2)2 − 9( − 1 − − 2) + c Length of rectangle = x + 5 = 3 + 5 = 8
b5
Width of rectangle = 2x = 2 × 3 = 6
2
+ 7 × −2× −1
( −1 )
5 Perimeter = 2 × length + 2 × width =
2 × 8 + 2 × 6 = 28
2
= 5 × 4 − 9 ×1 + + 14 Area = length × width = 8 × 6 = 48
−1
= 20 − 9 − 2 + 14 d Perimeter = 6x + 10 = 6 × 3 + 10 = 28
= 23 Area = 2x2 + 10x = 2 × 32 + 10 × 3 =
18 + 30 = 48
−5a 3 4 9 − 5 × −2 3
− 6a − ( ab) + = − 6( −2) − e Learner’s own answer.
b b2 − a3 −1
9 5 a i P = 2x + 10
( − 2 × − 1) 4 + 2 3
( − 1) − (− 2 ) ii A = 3x + 6
10 4 9 iii When x = 4, P = 18 and A = 18
= − 6 × − 8 − (2 ) +
−1 1+8
b i P = 2y − 4
9
= −10 + 48 − 16 + ii A = 4y − 24
9
= 22 + 1 iii When y = 10, P = 16 and A = 16
= 23 c i P = 4n + 8
ii A = n2 + 4n
Reflection: Learner’s own answers.
iii When n = 6, P = 32 and A = 60
Exercise 2.2 d i P = 2p2 + 8p
ii A = 4p3
1 a n+5 b 5n − 5
iii When p = 2, P = 24 and A = 32
n
c +5 d 5(n + 5)
5 6 a i 2 red + 2 yellow = 4 green;
n−5 both = 8x + 4
e f 5−n
5 ii 3 red + 3 yellow = 6 green;
2 a 7x b 20 − x both = 12x + 6
x iii 4 red + 4 yellow = 8 green;
c 2x + 9 d −4 both = 16x + 8
6
100 b n red + n yellow = 2n green (or similar
e x2 f explanation given in words)
x
g 5(x − 7) h x c i 6 red + 2 yellow = 12 blue;
3 both = 12x + 12
i x3 j x
ii 9 red + 3 yellow = 18 blue;
k (3x)2 + 7 or 9x2 + 7
both = 18x + 18
l (2x)3 − 100 or 8x3 − 100
iii 12 red + 4 yellow = 24 blue;
3 a i 2x + 2y ii xy both = 24x + 24
b i 6x + 2y ii 3xy d 3n red + n yellow = 6n blue (or similar
explanation given in words)
c i 6x + 4y ii 6xy
e Learner’s own answer.
d i 4x ii x2
e i 8x ii 4x2 7 a (3w)2 = 36, 2v(3v − 2w) = 30, 5w(w + v) = 50
f i 2
2x + 4x ii 2x 3 b 116
c (3w)2 + 2v(3v – 2w) + 5w(w + v) =
4 a Perimeter = 2(x + 5) + 2(2x) =
9w2 + 6v2 − 4vw + 5w2 + 5vw =
2x + 10 + 4x = 6x + 10
14w2 + vw + 6v2
b Learner’s own answer.
d 116

4 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 4/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

8 a 3a2 − 7b = 61, 8b – 3a = 31, a2 + 6b = 37, 2 a m14 b n12 c p7

4(a + 3b) = 4 d q5 e r3 f t5
b 133 g x21 h y10 i z12
c 3a2 − 7b + 8b − 3a + a2 + 6b + 4(a + 3b) = j 5t7 k 5g2 l −h9
4a2 + 7b − 3a + 4a + 12b = 4a2 + a + 19b
3 a Sofia is correct. x2 ÷ x2 = x2−2 = x0 = 1
d 133
e 11 b Learner’s own answer.

f Not valid because although the perimeter c x2 ÷ x2 = 1

is positive, three of the side lengths are d All the answers are 1. Learner’s own
negative, which is not possible. explanations. For example:
9 a 2(3x2 + 4) + 2(5 − x2) or When simplified, all the expressions have
3x2 + 4 + 3x2 + 4 + 5 − x2 + 5 − x2 an index of 0, and anything to the power
of 0 = 1.
b 2(3x2 + 4) + 2(5 − x2) =
6x2 + 8 + 10 − 2x2 = 4x2 + 18 = 2(2x2 + 9) or Any expression divided by itself, always
gives an answer of 1.
or 3x2 + 4 + 3x2 + 4 + 5 − x2 + 5 − x2 =
4x2 + 18 = 2(2x2 + 9) 4 a 6x5 b 12y9 c 30z7
c Arun is correct. Learner’s own d 4m7 e 4n13 f 8p3
explanation.
5 a Learner’s own answer.
For example: The variable x only appears
b Learner’s own answer.
in the expression for the perimeter when
it is squared. When you square 2 and −2 c Learner’s own answer.
you get the same answer. Sasha’s method would be easiest to use to
or: 2(2(−2)2 + 9) = 2(2 × 4 + 9) = simplify these expressions:
2(8 + 9) = 34 2 5 2
4x5 ÷ 6x3 = 3 4x3 = 2 x ,
and 2(2(2)2 + 9) = 2(2 × 4 + 9) = 6x 3
2(8 + 9) = 34 3
12 y7 3y
12y ÷ 8y =7 6
2
= and
10 a Side length = 25 = 5 cm, 8 y6 2

Perimeter = 4 × 5 = 20 cm 6z 9
z 5
6z9 ÷ 36z4 = 6 = .
b Side length = 49 = 7 cm, 36 z4 6
Perimeter = 4 × 7 = 28 cm 6 a 3q4 b 3r4 c 3t6
c Perimeter = 4 × x or 4 x d 2u5 e 2v4 f 5w
11 a Volume = x3 7 a D 1 x3 b A 2 y6
2 5
b Side length = 3 y 5 1
c C k d B3
3 3
Exercise 2.3 8 a Arun is correct. Learner’s own
1 a x 4 × x5 = x 4 + 5 b y2 × y4 = y2 +4 explanation. For example:
= x9 = y6 (3x2)3 = 33 × (x2)3 = 27 × x6 = 27x6
or (3x2)3 = 3x2 × 3x2 × 3x2 =
c u8 ÷ u6 = u8 −6 d w5 ÷ w = w5 −1 3 × 3 × 3 × x2 × x2 × x2 = 27 × x6 = 27x6
= u2 = w4 or (3x2)3 means everything inside the
bracket must be cubed. That means the 3
(g ) (h )
12
3 2
e =g 3× 2
f
5
=h 5 ×12
must be cubed as well as the x2.
6 60
=g =h b i 16x10 ii 125y12
g 3
5m + 3m = 8m 3 3
h 2
8n − n = 7n2 2 iii 16z28

5 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 5/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

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. e p2 − 11p + 30 f n2 − 30n + 200
Forexample: 5 a The plus at the end would change
3x2 × 12x10 to a minus and the 9 changes to a 1.
x2 + 1x − 20
4x8 × 9x4
b The plus at the end would change to
36x14 ÷ x2 a minus and the 9 changes to a −1.
72x20 ÷ 2x8 x2 − 1x − 20

(6x6)2 c The plus in the middle would change to a

minus. x2 − 9x + 20
36(x3)4
d i (x + A)(x + B) = x2 + Cx + D
c Learner’s own answers. ii (x + A)(x − B) = x2 + Cx − D

9 a q−3 =
1
b r−2 = 12 iii (x − A)(x + B) = x2 − Cx − D
q3 r
iv (x − A)(x − B) = x2 − Cx + D
c t−5 = 15 d v−1 = 1 6 a 2
C w + 12w + 27 b A x2 + 2x − 35
t v

10 a A and iii, B and iv, C and i, D and vii, c B y2 − 2y − 48 d A z2 − 9z + 20

Eand vi, F and v. 7 a (x + 2)2 = (x + 2)(x + 2)
b Learner’s own answer. Any expression = x2 + 2x + 2x + 4
1
that simplifies to give . = x2 + 4x + 4
6y 7
5 y2 b (x − 3)2 = (x − 3)(x − 3)
For example:
30 y9
= x2 − 3x − 3x + 9
Reflection: Learner’s own answers.
= x2 − 6x + 9
Exercise 2.4 8 a i y2 + 10y + 25
ii z2 + 2z + 1
1 a (x + 4)(x + 1) = x2 + 1x + 4x + 4
iii m2 + 16m + 64
= x2 + 5x + 4
iv a2 − 4a + 4
b (x − 3)(x + 6) 2
= x + 6x − 3x − 18 v p2 − 8p + 16
= x2 + 3x − 18 vi n2 − 18n + 81
b (x + a)2 = x2 + 2ax + a2
c (x + 2)(x − 8) = x2 − 8x + 2x − 16
9 a (x + 3)(x − 3) = x2 + 3x − 3x − 9 = x2 − 9
= x 2 − 6x − 16
b i x2 − 4
d (x − 4)(x − 1) = x2 − x − 4x + 4 ii x2 − 25
= x2 − 5x + 4 iii x2 – 49
2 a x2 + 10x + 21 b x2 + 11x + 10 c There is no term in x, and the number
term is a square number.
c x2 + 2x − 15 d x2 + 4x − 32
d x2 − 100
e x2 − 9x + 14 f x2 − 14x + 24
e x2 − a2
3 a Learner’s own answers and explanations.
b Learner’s own answers and explanations. Activity 2.4
c Learner’s own answer. a ① 33 × 29 = 957, ② 28 × 34 = 952,
③ 957 − 952 = 5

6 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 6/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

b ① 16 × 12 = 192, ② 11 × 17 = 187, 5 a
1 2
+ = + =
3 2 5
③ 192 − 187 = 5 2 6 6 6 6
1+ 2 3
c The answer is always 5. b = = 11
2 2 2
d n n+1 5 1
c ≠1
6 2
n+5 n+6
d She cannot cancel the 3 with the 6,
because the expression is 3x + y, all
e ① (n + 5)(n + 1) = n2 + 6n + 5,
divided by 6, not just 3x divided by 6.
② n(n + 6) = n2 + 6n, x y 3x y 3x + y
+ = + =
③ n2 + 6n + 5 − (n2 + 6n) = 2 6 6 6 6
n2 + 6n + 5 − n2 − 6n = 5 e Learner’s own answer.
The answer is always 5. f i correct
Learner’s own answer. ii incorrect. Learners should show that
4x − y
Exercise 2.5 the correct answer is
10
iii correct
2x 4x
1 a b iv incorrect. Learners should show that
5 7
9x − 8
8 the correct answer is
c d x 20
x
a+b 5a + 9b
2x 4 6 a i ii
e f 5 12
5 x
2a + 9 ab + 12
iii 15
iv 4b
2 y 3y 4 y y y
2 a + = +3 =7
5 10 10 10 10 3ab + 40 8ab + 27
v vi
2 1 10 1 9 10 b 18b
b − = − =
5y 25y 25y 25y 25 y b Learner’s own checks.
3y 3y
c 4
d 8
Activity 2.5
11
Learner’s own answers.
3y
e 9y
f
14 6× 3+ 2 +
7 a = 18 2 = 20 = 10
2 2 2
a a 5 a 2a b b b b
3 a + = + b + =3 + 4 b 3 × 3 + 1 = 9 + 1 = 10
2 5 10 10 4 3 12 12
a+ a 3b + 4 b
=5 2 = c 10 = 10
10 12
a d Learner’s own explanation. For example:
=7 =7
b
10 12 He factorises the bracket to give
2 × bracket, which is then divided by 2.
c 5
+ 2 = 25 + 14 d 5d − 3d = 25 d − 18d The × 2 and ÷ 2 cancel each other out,
7c 5c 35c 35c 6 5 30 30
25 + 14
leaving just the bracket.
= 25 − 18
d d
=
35 c 30 e When x = 3, 6 × 3 + 1 = 18 + 1 = 19, 19 ≠ 10,
39 7d
= = so the answer is wrong.
35c 30
Learner’s own explanation. For example:
7e e e
e − 2 = 21 − 16e f 9
− 3 = 18 − 15 The expression shows that 6x + 2 must all
8 3 24 24 10f 4f 20f 20 f be divided by 2.
21e −16 e
=
24
= 18 −15 Arun has only divided the 2 in the
20 f
=
5e numerator by 2, and not the 6x by 2 as well.
24 = 3
20 f f Learner’s own answer.
4 a A, D, F b B, C, E
8 a 2x + 1 b x+2
x
c G; the answer is c 2x − 3 d 2x − 5
3

7 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 7/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

6x − 4 20 x + 25 2(3x − 2 ) 5(4 x + 5 ) 4 a Ben’s age is x + 2, Alice’s age is x − 6

9 + = + =
2 5 2 5
b T = 3x − 4 c T = 53
3x − 2 + 4x + 5 = 7x + 3
T+4
d x= e x = 22
10 a 2(x + 3) = 2 × x + 2 × 3 = 2x + 6 3

b Learner’s own choice and explanation. 5 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 6 a 20% b 60%
iv 3(1 − 3x) or 3 − 9x c 125%
Reflection: Learner’s own answers. 7 a 65 kg b 49.1 kg (1 d.p.)
c 95.9 kg (1 d.p.) d 57.3 kg (1 d.p.)
Exercise 2.6
y− z
1 a S = 60M b S = 900 8 a i B x= 2

2( y + 3 h )
c M= S d M = 22.5 ii C x=
60 5

2 a i F = 60 ii F = −78 iii A x = 7k(y − 6)

F iv C x = 3ny + m
b m= , m = 12
a
v A x =w − y
F 7
c a = , a = −1.75
m b Learner’s own answer.
3 a m−9
9 a t= 7 b t = 5(k + m)
3D Shape Number Number Number 9q + w
of faces of of c t = pv − h d t=
5
vertices edges
10 a A = a2 + bc
Cube 6 8 12
b A = 49.5
Cuboid 6 8 12
c A = a2 + bc, A − bc = a2, a = A − bc
Triangular
5 6 9 d a=8
prism
A
Triangular- 11 a 78.5 cm b r=
π
based 4 4 6
c 6.25 cm
pyramid
Square-based 12 a l = 3V b 2 cm
5 5 8
pyramid 13 Sasha is correct as 30 °C = 86 °F and
b E = F + V − 2, or any equivalent version 86 °F > 82 °F (or 82 °F = 27.8 °C and
27.8 °C < 30 °C).
c V=E−F+2
14 a She is not underweight as her BMI is
i V=6 ii V=7
20.05, which is greater than 18.5.
d c i is a pentagonal-based pyramid and
b 3.7 kg
c ii is a hexagonal-based pyramid
e F = E − V + 2, F = 0, it is not possible Check your progress
to have a shape with five edges and
sevenvertices. 1 a 39 b 161
f Learner’s own answer. c 12
2 perimeter = 16x + 8,
area = 5x(3x + 4) = 15x2 + 20x

8 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 8/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

3 a x5 b q6 c h10 c 320 ÷ 101 = 320 ÷ 10 = 32

d 15m9 e 2u2 f 3p2 d 320 ÷ 100 = 320 ÷ 1 = 320
4 a x2 + 7x + 10 b x2 + x − 12 6 a 2.7 b 0.45
c x2 − 3x − 54 d x2 – 14x + 40 c 0.36 d 0.017
e x2 − 64 f x2 − 12x + 36 e 0.08 f 0.0248
2x 2y g 9 h 0.0025
5 a 3
b 15
7 a Learner’s own answer.
12x − y
c d 3x − 5 b i 6.8 ÷ 10−3 = 6800
20
z=x − y ,
2
6 a x = 31 b z=6 ii 0.07 ÷ 10−4 = 700
5
c y = ± x − 5z , y = ±6 c Learner’s own answer.
d Learner’s own answer. For example: An
Unit 3 Getting started alternative method is to realise that ÷ by
10−x and × by 10x are the same. So, in this
1 a 8 b 32.5 c 6 d 0.85 case 2.6 ÷ 10−2 = 2.6 × 102
e 90 f 625 g 700 h 32 e Learner’s own answer.

2 B 8 a 3.2 ÷ 103 = 3.2 ÷ 1000 = 0.0032

3 a 15.4 b 640 b 3.2 ÷ 102 = 3.2 ÷ 100 = 0.032

c 3.2 ÷ 101 = 3.2 ÷ 10 = 0.32
4 a $345 b $240
d 3.2 ÷ 100 = 3.2 ÷ 1 = 3.2
5 63.6 cm2 (3 s.f.)
e 3.2 ÷ 10−1 = 3.2 × 10 = 32

Exercise 3.1 f 3.2 ÷ 10−2 = 3.2 × 100 = 320

g 3.2 ÷ 10−3 = 3.2 × 1000 = 3200
1 a, D and ii; b, A and v; c, E and iv; d, C and i;
e, B and iii h 3.2 ÷ 10−4 = 3.2 × 10 000 = 32 000

2 a 3.2 × 103 = 3.2 × 1000 = 3200 9 a Yes. Learner’s own explanation.

b 3.2 × 102 = 3.2 × 100 = 320 b i greater ii the same

c 3.2 × 101 = 3.2 × 10 = 32 iii smaller
d 3.2 × 100 = 3.2 × 1 = 3.2 10 a 2.5 b 47 600
e 3.2 × 10−1 = 3.2 ÷ 10 = 0.32 c 70 d 8.5
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 12 a i 400 ii 40
h 3.2 × 10−4 = 3.2 ÷ 10 000 = 0.000 32 iii 4 iv 0.4
3 a Yes. Learner’s own explanation. v 0.04 vi 0.004
b i smaller ii the same b Smaller
iii greater c Smaller
4 a 1300 b 7800 c 240 d i 0.12 ii 1.2
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 e Larger
5 a 320 ÷ 103 = 320 ÷ 1000 = 0.32 f Larger

b 320 ÷ 102 = 320 ÷ 100 = 3.2 g Learner’s own answer.

9 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 9/61
9/18/23, 7:45 PM Ls maths 9 2ed tr learner book answers

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

13 a c i 60 ii 30 iii 20
iv 15 v 12 vi 10
0.8 × 101 8÷ 100 d i Smaller ii Larger
e Learner’s own answer.
80 × 10–1 =8 0.08 ÷ 10–2
8 a False b True
c False d True
0.008 × 103 800 ÷ 102
9 He has made a mistake. The denominator is
0.12, not 1.2; he wrote the answer with only
b
one decimal place. Answer = 50.
32 ÷ 102 0.32 × 100 10 a 200 b 120
c 300 d 40
3.2 ÷ 101 = 0.32 320 ÷ 103 11 a A and iv, B and v, C and vi, D and vii,
E and iii, F and i
32 × 10–2 3.2 × 10–1 b Learner’s own answer. Any question that
gives an answer of 0.024. For example:
0.03 × 400 × 0.002
Activity 3.1
c Learner’s own answer.
Learner’s own answers.
12 Learner’s own answers and discussions.
Reflection: Learner’s own answers.
For example: 28 × 0.057 = 1.596,
Exercise 3.2 2.8 × 0.57 = 1.596, 28 × 5.7 = 159.6,
2.8 × 5.7 = 15.96
1 a 1.6 b −5.6 c −5.4
15.96 ÷ 0.57 = 28, 159.6 ÷ 0.57 = 280,
d 6 e 0.3 f −0.66 15.96 ÷ 28 = 0.57, 15.96 ÷ 280 = 0.057
g 3.6 h −0.44 13 a 123 × 57 = 7011
2 a 0.08 × 0.2 8 × 2 = 16 b i 701.1 ii 701.1 iii 70.11
8 × 0.2 = 1.6 0.08 × 0.2 = 0.016 iv 7.011 v 7.011 vi 0.070 11
b 0.4 × 0.007 4 × 7 = 28
14 a Learner’s own answer.
4 × 0.007 = 0.028 0.4 × 0.007 = 0.0028
b Learner’s own answer.
3 C, D, I, K (0.015); A, F, H, J (0.15); c i Estimate: 4 × 30 = 120
B, G, L (1.5); E (15) Accurate: 119.625
4 a 20 b −50 ii Estimate: 10 ÷ 0.2 = 50
c −30 d 600 Accurate: 62

e 40 f −400 iii Estimate: 60 × 4 = 24 000

0.01
g 200 h −300 Accurate: 19 200
0.81 ×100
5 a 0.09 × 100
= 81
9
=9 15 a 0.2 ÷ 0.4 = 0.5 m
6.4 × 1000 b 0.45 m
b 0.004 × 1000
= 6400
4
= 1600
c Learner’s own answer.
6 a D b B c C d D
Exercise 3.3
7 a i 0.8 ii 2.4 iii 4
1 a 200 × 1.1 = $220 220 × 1.15 = $253
iv 5.6 v 7.2 vi 8.8
b 200 × 0.9 = $180 180 × 0.85 = $153
b i Larger ii Smaller
c 200 × 1.2 = $240 240 × 0.95 = $228

10 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021

about:blank 10/61