CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Workbook answers
11 aThe number is 7.142… and there is no
repeating pattern.
bLearner’s own answer. For example:
2 and 5 − 2 .
c Because the sum of two rational numbers
must be rational.
d No, because the product of two rational
numbers is rational.
12
B
3
5.5
√25
A
25
Exercise 1.1
1
Number
Rational
36

Irrational

48

64
3

100

2
a
27 , 500
3
a
integer
b
surd
c
surd
d
integer
e
integer
f
surd
a
irrational because 3 is irrational
i
20 + 2 = 6.4721…
b
rational because it is equal to 9 = 3
ii
20 − 2 = 2.4721…
c
rational because it is equal to 8 + 4 = 12
iii
16
d
irrational because it is 2 + an irrational
number
a
2.25
b
it is equal to 1.5
c
yes, it is equal to 4.5
d
yes, it is equal to 1.1
a
33 = 27 and 43 = 64
b
93 = 729 and 103 = 1000
c
1.12 = 1.21 and 1.22 = 1.44
4
5
6
7
b
−36, − 3 8
√25
13 a
b
5
519
She is correct. Substitute different values
to see that ( n + 2)( n − 2) = n − 4 seems
to be true.
Exercise 1.2
1
2
Learner’s own answers. For example:
3
a
2.6 × 106
b
9.2 × 108
c
4.62 × 105
d
2.08 × 107
a
5.5 × 104
b
5.5 × 107
c
6.4 × 108
d
4.06 × 108
a
53 000
b
53 800 000
c
711 000 000 000
d
133 100 000
a
5
b
a square root between 36 and 49
4
9.46 × 1012 km
c
2
5
a
3 × 10−5
b
6.66 × 10−7
8
a
12
c
5.05 × 10−5
d
4.8 × 10−10
9
aNo. All fractions are rational. In fact, the
repeating sequence is nine digits long.
6
a
b
c
d
0.0015
0.000 012 34
0.000 000 079
0.000 900 3
7
a
b
c
d
0.000 008
0.000 000 482
0.000 061
0.000 000 070 07
b
10 a
b
1
84
b
7
4
It is rational. It is 1 .
9
The answer is 8.
i
2 × 18 is a possible answer.
ii
3 × 27 is a possible answer.
iii
5 × 20 is a possible answer.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
8
4 × 10−7 m and 8 × 10−7 m
9
C, E, A, B, D
10 a
Exercise 2.1
1
b
22
12 a
106 or one million.
b
2.8 × 10
3 × 10
d
9.95 × 109
13 a
4.3 × 10−4
b
1.25 × 10−6
c
7 × 10−6
d
8 × 10−9
14 a
1.75 × 106
b
1.34 × 108
c
6.5 × 10−5
d
1.146 × 10−4
c
4.5 × 10
7
6
d
a
b
1
49
c
e
1
225
f
1
125
1
400
4−1
b
4−3
c
40
d
44
e
4−4
f
4−2
3
a
5−1 b
5−3 e
50
4
a
1
8
b
1
27
c
1
125
d
1 or 0.001
1000
a
122
b
12−1
c
12−3
d
123
a
5
3
b
4
c
8
−5
d
150 or 1
e
5
−12
7
a
73
c
76
8
a
125
b
5−7
c
3−4
d
251 or 25
a
6
b
−4
c
−2
d
4
10 a
−2
b
4
c
6
d
7
11 a
3
b 1 3
c
1
12 a
116 = 1 771 561
b
1
2
5
6
9
a
c
11−3 =
52 c 5−2 d
b
7−1
4
1
b
x ÷ 2 − 4 = 10 ÷ 2 − 4
= 5−4 =1
c
4 × x 2 = 4 × 10 2
= 4 × 100 = 400
d
3 × ( x + 2 ) = 3 × (10 + 2 )
= 3 × 12 = 36
9
2
A and iii, B and v, C and i, D and vi, E and ii,
F and iv
3
a
x + y = 6 + −2 = 6 − 2 = 4
b
x − y = 6 − −2 = 6 + 2 = 8
c
x 2 + y 2 = 62 + (−2)2 = 36 + 4 = 40
d
3x + y = 3 × 6 + −2 = 18 − 2 = 16
e
x + 4y = 6 + 4 × −2 = 6 − 8 = −2
f
3x + 4y = 3 × 6 + 4 × −2 = 18 − 8 = 10
a
2
b
−14
c
35
d
13
e
7
f
100
a
−4
b
5
c
−8
d
−26
e
94
f
−4
g
12
h
−11
Exercise 1.3
1
7
1
81
2 × x + 3 = 2 × 10 + 3
= 20 + 3 = 23
5.98 × 10 kg
11 aCopy and complete this sentence: 6.2 × 107
is 10 times larger than 6.2 × 106.
b
a
23
4
5
6
1
2
aIncorrect. He has worked out −12 and
not (−1)2.
Correct solution is
−4 × (−1)2 −3 × −4 = −4 + 12 = 8
−6
b
d
Incorrect. He has worked out that
(−4)3 = 64 and not −64.
Correct solution is ( −4 ) −
3
7−1
−4
2 × −1
= −64 −
−4
−2
= −64 − 2
= −66
7
Learner’s own values. For example:
a
x = 3 and y = 7, x = 4 and y = 44,
x = 5 and y = 105
4
9
b
x = −1 and y = −21, x = −2 and y = −28,
x = −3 and y = −47
112 = 121
c
x = 0 and y = −20, x = 1 and y = −19,
x = 2 and y = −12
1331
13 7
2
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
8
9
a
15
b
20
c
−20
d
11
e
8
f
−64
g
2
h
−7
i
8
j
2
k
−25
l
4
10 a
d
Let y = 2, so (2y)3 = (2 × 2)3 = 43 = 64 and
2y3 = 2 × 23 = 2 × 8 = 16
e
1
−1
n
f
64 ≠ 16, so (2y)3 ≠ 2y3.
g
3(n + 20)
h
3n
Let x = 4 and y = 2,
3x − 3y = 3 × 4 − 3 × 2 = 12 − 6 = 6 and
3(y − x) = 3(2 − 4) = 3 × −2 = −6
i
( 4n)2 − 3
j
6 3 n + 10
6 ≠ −6, so 3x − 3y ≠ 3(y − x).
k
 n
−9
 5 
a
6x
b
3x + 10
c
12x − 2
d
13x − 4
a
xy
b
y2
c
x3
d
16x2
a
g2 = 25, g(8 − g) = 15, 2g(3g − 11) = 40
b
80
c
g2 + g(8 − g) + 2g(3g − 11) =
g2 + 8g − g2 + 6g2 − 22g = 6g2 − 14g.
d
6g2 − 14g = 80
b
7
14 kg
4
5
Mass using
expression ①
10.5
13
15.5
18
20.5
Mass using
expression ②
10
12
14
16
18
Expression ②, 13.5 kg is closer to 14 kg
than 15.5 kg.
b
99
100
c2
100
5
− 3 × 5 × −3 − 5(5 − −3)
2
= 36 − 4 + 45 − 40
= 37
d +
8c
( c + d )2
= ( −3) +
3
+
9
2
8×5
( 5 + − 3 )2
2
2
= −27 + 10 + 25 + 29
= 37
c
2a + 16
ii
5a + 15
i
2a + 16 = 22
ii
5a + 15 = 30
i
2b + 2
ii
5b − 20
i
2b + 2 = 26
ii
5b − 20 = 40
i
4c − 16
ii
c2 − 8c
when c = 10,
d
Exercise 2.2
i
when b = 12,
2
3×5
−3
10
2n
when a = 3,
b
( ) − ( −4 − c )
+(
) − ( −4 − 5 )
3c
d
8
10 a
− 3cd − c(c − d )
= 4 × ( −3)2 −
3
18
n
1000
n
−5
4
3
3
12 4d 2 −
3
b
5
2n + 3
2
3
d
c
1
2
a
5
b
Age (A years)
1
c
n − 10
c
11 a
6
a
18 kg
d
b
6
40 ≠ 400, so 10x 2 ≠ (10x)2.
c
3
A and iii, B and vi, C and i, D and vii,
E and viii, F and ii, G and iv, H and v
Let x = 2, so 10x 2 = 10 × 22 = 10 × 4 = 40
and (10x)2 = (10 × 2)2 = 202 = 400
b
a
5
Learner’s own counter-examples. For example:
a
4
i
4c − 16 = 24
ii
c2 − 8c = 20
i
2d 2 + 14d
ii
7d 3
ii
7d 3 = 875
when d = 5,
a
6
b
12
c
x+2
d
z+2
a
2
b
5
c
y−3
d
z−3
a
10
b
20
c
5a
d
5b
i
11 a
2d 2 + 14d = 120
i2(a + 3) + 2(3a + 1) = 8a + 8,
4(2a + 2) = 8a + 8
ii3(a + 3) + 3(3a + 1) = 12a + 12,
6(2a + 2) = 12a + 12
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
iii5(a + 3) + 5(3a + 1) = 20a + 20,
10(2a + 2) = 20a + 20
b
c
n black rods + n striped rods = 2n white rods
(or similar explanation given in words)
i4(a + 3) + 2(2a + 2) = 8a + 16,
8(a + 2) = 8a + 16
ii6(a + 3) + 3(2a + 2) = 12a + 24,
12(a + 2) = 12a + 24
iii8(a + 3) + 4(2a + 2) = 16a + 32,
16(a + 2) = 16a + 32
d
12 a
2n black rods + n white rods = 4n grey rods
(or similar explanation given in words)
i
b
$10
d
16 + 10d
13 a
ii
$26
c
$46
$16
a2
42
When a = 4, + 3a = + 3 × 4 = 20 and
2
2
when b = 5, 2b (b2 − 4b − 3) =
2 × 5(52 − 4 × 5 − 3) = 10(25 − 20 − 3) = 20
15 aSide length = 3 27 = 3 cm, cube
has 12 edges, so total length of
edges = 12 × 3 = 36 cm
b
80
c
i
2a + 12a
1
A and ii, B and iv, C and i, D and iii
2
A and iii, B and iv, C and ii, D and i
3
a
True b
False y 5 × y 4 = y 9
c
True
d
False y 9 ÷ y 3 = y 6
a
g8
b
h30
c
i 21
d
j 20
a
8x2
b
16x3
c
4y 4
d
11y 6
a
a7
b
b10
c
c8
d
d4
e
e4
f
f7
g
g32
h
y14
i
i 72
j
13j 2
k
k3
l
−3l5
a
6a4
b
16b7
c
36c12
d
10e11
e
8g8
f
3h6
g
5x8
h
5x4
8
a
B
9
aWhen the terms are simplified, one group
has x6 terms and one group has x9 terms.
4
5
6
ii
8b − 32b − 24b
7
2
3
2
d
When a = 4, 2a + 12a = 80 and when b = 5,
8b3 − 32b2 − 24b = 80
e
Yes. Learner’s own explanations.
2
For example: When a is a positive integer,
f
g
i
−10
iii
−18
ii
−16
No, because the perimeter cannot be a
negative number.
For a < −6 the perimeter is positive, so is a
valid measurement.
14 a2(4x2 + 3x) + 2(2x2 − 5x) =
8x2 + 6x + 4x2 − 10x = 12x2 − 4x
4
b
12x2 − 4x = 4x(3x − 1)
c
Arun is incorrect. When x = 3,
perimeter = 96 and when x = −3
perimeter = 120.
b
A
c
A
d
D
x6 terms: 3x3 × 2x3, 9x9 ÷ 3x3, 2x × 3x5
a2
a is positive, so is positive. Also 3a
2
2
is positive. When you add two positive
numbers, you will get a positive answer, so
the perimeter of the rectangle will always
be positive.
12 3 x
Exercise 2.3
As the side lengths are both 20, it must be
a square.
b
c
48 cm
x9 terms: x6 × 3x3, 12x12 ÷ 4x3, 6x6 × x3
b
10 a
9x12 ÷ x9 = 9x3: this is the only card, which
when simplified, has an x3 term; all others
have x6 terms or x9 terms.
Zara is correct. (2x3)2 = 22 × x3×2 = 4x6
b
i
9x14 ii
11 a
C
b
12 a
c
e
13 a
c
m−8 =
A
64y27 iii
32z15
c
B
4−4 = 4
1
4
b
5−3 = 3
8−5 = 5
1
8
d
x−4 = 4
y−7 = 7
1
y
f
z−1 = 1 =
x−3 = 3
1
x
b
y−4 = 4
1
m8
d
n−5 = 5
d
D
1
5
1
x
1
z
1
z
1
y
1
n
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
14 a
A and v, B and iii, C and i, D and vii,
E and ii, F and iv
b
Any expression that simplifies to give
d
5
.
2 y7
For example: 10y3 ÷ 4y10
15
2 n × 3n
( 2 n 2 )3
2
16 Yes,
5
=
4 4
3x × x 2 × 3x
6n
3n
=
8n6
4
a
=
9
3
a
81x16
= 9x 4
9x12
23 × 34
b
a
+9
+9x
+27
(x + 5)(x − 3)
×
x
+5
x
2
x
+5x
−3
−3x
−15
(x + 6)(x − 2)
20
3
×
x
+6
90
x
2
x
+6x
4
80
12
−2
−2x
−12
x2 + 6x − 2x − 12 = x2 + 4x − 12
18 × 42
c
(x − 7)(x + 4)
×
10
8
×
x
−7
40
400
320
x
2
x
−7x
2
20
16
+4
+4x
−28
x2 − 7x + 4x − 28 = x2 − 3x − 28
d
(x + 2)(x + 3)
(x − 8)(x + 2)
+2
×
x
−8
x
2
x
+2x
x
x2
−8x
+3
+3x
+6
+2
+2x
−16
×
x
(x + 1)(x + 4)
x2 − 8x + 2x − 16 = x2 − 6x − 16
4
a
(x − 1)(x − 3)
×
x
+1
×
x
−1
x
2
x
+x
x
x2
−x
+4
+4x
+4
−3
−3x
+3
(x + 5)(x + 6)
x2 − x − 3x + 3 = x2 − 4x + 3
b
(x – 4)(x – 8)
×
x
+5
×
x
−4
x
2
x
+5x
x
x2
−4x
+6
+6x
+30
−8
−8x
+32
x2 + 5x + 6x + 30 = x2 + 11x + 30
5
+3x
600
x2 + x + 4x + 4 = x2 + 5x + 4
c
x
×
x2 + 2x + 3x + 6 = x2 + 5x + 6
b
x
30
400 + 320 + 20 + 16 = 756
2
+3
2
x2 + 5x − 3x − 15 = x2 + 2x − 15
600 + 90 + 80 + 12 = 782
b
x
x2 + 3x + 9x + 27 = x2 + 12x + 27
Exercise 2.4
1
×
7
6 x 2 × 3x 6 × 2 x 9 36 x17
= 13 = 9x 4 and
4 x13
4x
(3x )
(x + 3)(x + 9)
x2 − 4x − 8x + 32 = x2 − 12x + 32
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
Learner’s own answer.
6
a
x2 + 7x + 10
b
x2 + 2x − 8
c
x2 − 3x − 18
d
x2 − 6x + 9
e
x2 + 15x + 50
f
x2 − 13x + 40
g
x2 + 5x − 50
h
x2 − 3x − 40
7
a
B b
C
d
8
1(x + 4)(x + 3) = x + 7x + 12 Rohan had the
final term incorrect – he added 4 and 3 to
get 7, not multiplied 4 by 3 to get 12.
c
A
(x + 8)2 and (x + 7)(x + 9) giving x2 + 16x + 64
and x2 + 16x + 63.
There is still a difference of 1.
15 a(2x + 1)(3x + 2) = 6x2 + 4x + 3x + 2 =
6x2 + 7x + 2
b
C
12x2 + 19x + 5
ii
8y2 − 14y − 15
2
Exercise 2.5
a
1 1 2
+ =
3 3 3
b
1 2 3
+ =
5 5 5
c
2 3 5
+ =
7 7 7
d
1 3 4 1
+ = =
8 8 8 2
e
1 2 3 1
+ = =
2 9 9 3
f
3
3
6 3
+ = =
10 10 10 5
a
x x 2x
+ =
3 3
3
b
x 2 x 3x
+ =
5
5
5
c
2 y 3y 5 y
+ =
7
7
7
d
y 3y 4 y y
+ = =
8
8
8
2
e
m 2 m 3m m
+ = =
9
9
9
3
f
3n 3n 6 n 3n
+ = =
10 10 10
5
a
1 3 2 3 5
+ = + =
4 8 8 8 8
b
1 2 3 2 5
+ = + =
3 9 9 9 9
a2 − 16
c
2 1 4 1 3 1
− = − = =
3 6 6 6 6 2
b
There is no term in a, and the number
term is a square number.
d
11 1 11 2
9 3
− = − = =
12 6 12 12 12 4
c
a2 − 64
4
a
5x
8
b
5y
9
c
p
2
d
3b
4
d
a2 − b2
11 ( x + 4 )( x – 3) + x (5 – x ) = x 2 − 3x + 4 x − 12 + 5x − x 2 5
a
x
2
b
4x
5
c
12
x
d
6x
7
e
5
4x
f
y
6
g
2y
9
h
y
18
i
5
16y
j
17
24y
2(x + 5)(x − 9) = x2 − 4x − 45 Rohan
simplified 5x − 9x to be 4x not −4x.
1
3(x − 3)(x − 2) = x2 − 5x + 6 Rohan had the
final term incorrect – he multiplied −3 by
−2 to get −6, and it should be +6.
9
a
b
c
10 a
i
a2 + 4a + 4
ii
a2 − 4a + 4
iii
b2 + 8b + 16
iv
b2 − 8b + 16
v
c2 + 2c + 1
vi
c2 − 2c + 1
2
Learner’s own answer. For example:
The first and last terms are the same, the
middle terms have different signs.
(x + y)2 = x2 + 2xy + y2 so
(x − y)2 = x2 − 2xy + y2
i
a2 − 1
iii
a − 81
3
ii
2
= 6 x − 12
= 6 ( x – 2)
12 a
b
13 a
b
i
x2 + 12x + 36
ii
x2 + 12x + 35
Learner’s own answer. For example: There
is a difference of 1.
i
x2 + 14x + 49
ii
x2 + 14x + 48
Learner’s own answer. For example: There
is a difference of 1.
14 Learner’s own answer. For example:
(x + 5)2 and (x + 4)(x + 6) giving x2 + 10x + 25
and x2 + 10x + 24
6
i
6
1
4
b
7
x
and
4
1
x
B, C both equal x or .
2
2
1
x
E, which equals x or .
3
3
aA, D, F all equal x or
a
x+ y
2
b
2x + y
6
c
9x + y
12
d
15x − y
18
e
7x − 8 y
12
f
21a + 4b
28
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
8
g
10a + 15b
18
i
8ab − 45
36b
a
17
c
17 ≠ 32.
h
ab − 35
7b
4
b
b
5
32
Learner’s own explanation. For example: She
has just crossed the 2s off and not cancelled
properly.
9
8x + 2 2( 4 x + 1) 2 ( 4 x + 1)
=
=
= 4x + 1
1
2
2
2
a
2x + 1
b
5x + 1
c
3x − 4
d
3x − 4
c
1
11 a
b
i
2x + 4 and 2(x + 2)
ii
3x + 9 and 3(x + 3)
iii
6x − 9 and 3(2x − 3)
iv
4 − 6x and 2(2 − 3x)
12 a
2x + 3
2
b
c
2x − 3
4
d
y+x
x+ y
or xy
xy
b
d +c
c+d
or
cd
cd
c
y−x
xy
d
2b + a
a + 2b
or
ab
ab
e
5n − 2 m
mn
f
3h − 4 g
gh
13 a
Exercise 2.6
ii
45
A=b×h
swap sides:
A
h
b × h = A
reverse the ×: b =
F = bg
F=b×g
b×g=F
reverse the ×: b = g
T = mb
T=m×b
m×b=T
reverse the ×: b =
swap sides:
F
swap sides:
T
m
i
b
S = , S = 20
a
Polly’s age: d + 3, Max’s age: d − 2
b
T = 3d + 1
c
T = 25
d
d=
T −1
3
e
d = 11
a
F = 25
b
F = 54
c
I = 40
d
e=5
e
a=7
a
50%
b
8%
c
110%
10 a
450 m
b
1303 m
c
1078 m
d
1615 m
11 a
A
b
B
c
A
d
C
12 a
n=
p+8
3
b
n = 7(q − k)
n = 2pw − r
d
n=
9
c
ii
D = 150
D
T
c
D = 180
D
S
T = , T = 5.5
hr 2 + 2
5
1
A and v, B and iv, C and ii, D and iii,
E and vi, F and i
13 Arun is correct. 20 °C = 68 °F and 68 °F > 65 °F.
2
a
14 F = 120. Learner’s own explanation and
working. For example:
3
i
24
ii
48
iii
72
iv
24d
Use the formula a =
b
H = 24d = 24 × d = 24 × 10 = 240
a
i
7
ii
14
iii
21
iv
7w
b
7
19
a
6
8
2x + 3
5
5 − 7x
2
i
M = b − kn swap sides: b − kn = M
reverse the −: b = M + kn
7
2
8x + 24 8 ( x + 3)
=
= 2(x + 3)
1
4
4
7w + d
25
e
1
8x + 24 4 ( 2 x + 6 )
=
= 2x + 6 and
1
4
4
iii
ii
X = b + rt
swap sides: b + rt = X
reverse the +: b = X − rt
1
x − 2 + 4x + 3 = 5x + 1
9
d
10 Evan is correct.
7 x − 14 8x + 6 7 ( x − 2 ) 2 ( 4 x + 3)
+
=
+
=
1
1
7
2
7
2
i
a
A = bh
b
1
d
a
= 7w, 56 days
So a =
v−u
to find the value of a.
t
v − u 32 − 12
=
= 4.
t
5
Then use the formula F = ma to work out the
value of F. So F = 30 × 4 = 120.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
15 a
r=
2A
π
b
4.8 cm
16 a
bh
A = a2 +
2
b
A = 61
bh
2
d
a = 12
c
17 a
a= A−
6
side length of the larger cube = 2x
c
V
9
b
V = 9x3
d
Learner’s explanation and working.
Example:
Used the formula x = 3
x= 3
a
3.4
b
3.4
c
0.034
d
0.034
e
0.034
f
0.034
g
34
h
3.4
i
3400
j
30 400
k
30
l
340
7
POWERS OF TEN – EASY!
8
a
V
to work out the
9
i
5000
ii
500
iii
50
iv
5
v
0.5
vi
0.05
b
larger
a
i
0.099
ii
0.99
Side length of larger cube is 2 × 4 = 8 cm
iii
9.9
iv
99
Area of one face of larger cube = 8 × 8 =
64 cm2
v
990
vi
9900
576
value of x. x = 3
= 4 cm
9
9
Surface area of larger cube = 6 × 64 =
384 cm2
b
smaller
10 a
0.004 × 103
Exercise 3.1
1
2
a
3.4 × 102 = 3.4 × 100 = 340
b
4.8 × 103 = 4.8 × 1000 = 4800
c
12.5 × 101 = 12.5 × 10 = 125
d
5 × 105 = 5 × 100 000 = 500 000
e
14 × 103 = 14 × 1000 = 14 000
A and ii, B and v, C and iv, D and i, E and iii
4
a
3.4 × 10–2 = 3.4 ÷ 100 = 0.034
b
8 × 10–3 = 8 ÷ 1000 = 0.008
c
15 × 10–4 = 15 ÷ 10 000 = 0.0015
d
12 × 10–1 = 12 ÷ 10 = 1.2
a
2800
b
c
280
e
0.04 ÷ 10−2
=4
A and ii, B and vi, C and iv, D and i, E and iii,
F and v
3
5
8
4 × 100
400 ÷ 102
b
0.4 × 101
40 ÷ 101
67 ÷ 102
670 × 10−3
670 ÷ 103
6.7 ÷ 101
= 0.67
67 × 10−2
6.7 × 10−1
11 a
45: A, D, H 4.5: B, E, J 0.45: C, G, I
b
0.045: F is spare. Learner’s own answers.
For example: 45 ×10−3, 4.5 ×10−2, etc.
12 a

b

c

28 000
d

e

f

d
2880
13 a
270 b
280 000
f
0.2
g
28
h
0.2
14 a
B
i
0.028
j
0.28
Exercise 3.2
k
0.028
l
28.8
1
b
0.0048 c
125 000
c
d
A
C
B
a
4 × 0.3
4 × 3 = 12
so 4 × 0.3 = 1.2
b
7 × 0.4
7 × 4 = 28
so 7 × 0.4 = 2.8
c
9 × −0.1 9 × −1 = −9
9 × −0.1 = −0.9
so
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
d
−15 × 0.2 − 15 × 2 = −30 so
−15 × 0.2 = −3
e
8 × 0.02
f
−5 × −0.04 − 5 × −4 = 20 so
−5 × −0.04 = 0.2
g
11 × 0.07 11 × 7 = 77
11 × 0.07 = 0.77
8 × 2 = 16
so 8 × 0.02 = 0.16
so
2 a6 ÷ 0.3
0.3 × 10 = 3
3
9
Denominator should be: 5 × 0.1 = 0.5, not 50.
Answer = 1.
10 a
20
b
30
c
500
d
0.2
11 a
6 × 10 = 60
60 ÷ 3 = 20
b
8 ÷ 0.2
0.2 × 10 = 2
8 × 10 = 80
80 ÷ 2 = 40
c
−9 ÷ 0.1
0.1 × 10 = 1
−9 × 10 = −90
−90 ÷ 1 = −90
d
12 ÷ 0.4
0.4 × 10 = 4
12 × 10 = 120
120 ÷ 4 = 30
e
6 ÷ −0.02
−0.02 × 100 = −2
6 × 100 = 600
600 ÷ −2 = −300
f
8 ÷ 0.04
0.04 × 100 = 4
8 × 100 = 800
800 ÷ 4 = 200
g
−16 ÷ −0.08
−0.08 × 100 = −8
−16 × 100 = −1600
−1600 ÷ −8 = 200
a
1.2
b
3.6
d
−8.1
e
3.3
f
−0.24
g
0.28
h
0.45
i
1.4
j
−5.55
i
1.1
ii
2.2
iii
3.3
iv
4.4
v
5.5
vi
6.6
b
i
larger
ii
smaller
c
i
80
ii
40
iii
20
iv
16
v
10
i
larger
ii
larger
d
12 a
158.4
b
158.4
c
0.015 84
d
352
e
0.352
f
3.52
13 a
2.6
c
Hassan is incorrect. Numerator should be:
2.5 × 0.2 = 0.5, not 5.
Estimate: 6 × 40 = 240
Accurate: 271.377
b
Estimate: 200 ÷ 0.4 = 500 Accurate: 495
c
Estimate:
80 × 5
0.2
=
400
0.2
= 2000
Accurate: 2400
14 a
5.4 m2
b
7.2 m2
c
0.48 m2
d
0.124 m2
True
b
True
15 4 m
4
A, C, E, I (0.024); D, G, J, L (0.24);
B, F, H, K (2.4)
5
a
20
b
40
17 a
c
30
d
−40
c
False, 0.0025
d
False, 0.3
e
200
f
−250
e
True
f
True
g
300
h
3000
i
200
j
−400
6
a
B
c
C
7
a
0.12
b
1.35
c
0.072
d
0.15
e
0.055
f
30
$200 increased by 20%
100% + 20% = 120%
multiplier is 1.2
g
9
h
5
$200 × 1.2 = $240
i
7
j
40
a
True
b
True
c
False
d
True
8
b
B
16 0.35 m
Exercise 3.3
d
B
1
a$300 increased by 15%
100% + 15% = 115% multiplier is 1.15
$300 × 1.15 = $345
b
c
$400 increased by 32%
100% + 32% = 132%
multiplier is 1.32
$400 × 1.32 = $528
2
a$300 decreased by 15%
100% − 15% = 85% multiplier is 0.85
$300 × 0.85 = $255
9
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
$200 decreased by 20%
100% − 20% = 80%
multiplier is 0.8
c
The value of the scooter after 12 years.
d
1800 × (0.88)4 = $1079.45 and
1800 × (0.88)5 = $949.92
e
1800 × (0.88)n
$200 × 0.8 = $160
c
$400 decreased by 32%
100% − 32% = 68%
multiplier is 0.68
$400 × 0.68 = $272
3
A and vi, B and iii, C and i, D and iv, E and ii,
F and v
4
a$800 increased by 10%, then increased
by 20%.
800 × 1.1 = 880
b
15% decrease then 12% increase →
multiplier = 0.952 → $800 × 0.952 = $761.60
c
45% increase then 24% increase →
multiplier = 1.798 → $400 × 1.798 = $719.20
720 × 0.8 = $576
→
14 a
a
$2400
1.5 × 0.4 = 0.6 → 50% increase and 60%
decrease
1.2 × 0.5 = 0.6 → 20% increase and 50%
decrease
840 × 0.85 = $714
b
1.25 × 1.2 = 1.5 → 25% increase and 20%
increase
i
b
=
c
i
=
ii
=
a
i
85.8
ii
362.5
b
i
891
ii
48.72
7
a
1.071
8
a
i
33.6
ii
120
a115, 116, 117, 118, 119, 120, 121, 122, 123,
124
b
i
127.5
ii
76.95
b
115
a
0.63
c
124
a
65, 66, 67, 68, 69, 70, 71, 72, 73, 74
b
65
c
74
9
10
ii
b
15 Learner’s own answers. For example:
1000 × 0.6 = $600
→
1.1016
a
6
198
b
198
$1.29
b
Calculation:
Amount:
1
4000 × 1.05
$4200.00
b
2 × 0.75 = 1.5 → 100% increase and 25%
decrease
16 500 000
Exercise 3.4
1
$529.20
End of year:
11 a
10
b
$800 increased by 5%, then decreased
by 15%.
800 × 1.05 = 840
5
→
E and 0.54
13 a60% increase then 45% decrease →
multiplier = 0.88 → $600 × 0.88 = $528
880 × 1.2 = $1056
$800 increased by 25%, then decreased
by 40%.
800 × 1.25 = 1000
d
b
$800 decreased by 10%, then decreased
by 20%.
800 × 0.9 = 720
c
→
12 a
A and ii, B and v, C and i, D and iv,
F and iii
2
4000 × (1.05)
2
$4410.00
3
3
4000 × (1.05)
$4630.50
4
4000 × (1.05)4
$4862.03
5
4000 × (1.05)5
$5105.13
2
3
4
a24.5, 24.6, 24.7, 24.8, 24.9, 25.0, 25.1,
25.2, 25.3, 25.4
b
24.5
c
25.4
a
7.5, 7.6, 7.7, 7.8, 7.9, 8.0, 8.1, 8.2, 8.3, 8.4
i
1800 × 0.88
b
7.5
ii
1800 × (0.88)2
c
8.4
iii
1800 × (0.88)3
5
a
2.5
b
3.5
6
a
85
b
95
The value of the scooter after 7 years.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
7
8
9
a
6.5 ⩽ x < 7.5
Exercise 4.1
b
27.5 ⩽ x < 28.5
c
134.5 ⩽ x < 135.5
1
d
558.5 ⩽ x < 559.5
a
45 ⩽ x < 55
b
415 ⩽ x < 425
c
3735 ⩽ x < 3745
d
5205 ⩽ x < 5215
a
750 ⩽ x < 850
b
1150 ⩽ x < 1250
c
6650 ⩽ x < 6750
d
9050 ⩽ x < 9150
a
y
2
b
19.5 m2
2
c
18.5 m2 ⩽ x < 19.5 m2
y
b
2
y
ii
55
12 A, ii and c; B, ii and a; C, i and e; D, iii and b;
E, i and f; F, iii and d
13 a
b
14 a
b
2
a
495 g ⩽ x < 505 g
i
2 × 495 g = 990 g
ii
2 × 505 g = 1010 g
i
145 cm
ii
155 cm
3y
iii
145 cm ⩽ x < 155 cm
4
18 = 9 x
−2
18
x = −2
9
3y
4
ii
6.25 litres or 6250 mL
iii5.75 litres ⩽ x < 6.25 litres or
5750 mL ⩽ x < 6250 mL
+1 = 7
d
=x
3(y + 5) = 2(20 − y)
3 y + 15 = 40 − 2 y
3 y + 2 y = 40 − 15
5 y = 25
= 7 −1
3y
=6
4
3y = 6 × 4
3 y = 24
Carlos has worked out the correct
answer as all pieces of wood can vary
between 145 cm and 155 cm, so you
must multiply the upper and lower
bounds by 3.
5.75 litres or 5750 mL
6(3 − x) = 3x
18 − 6 x = 3 x
18 = 3 x + 6 x
x=2
c
i
b
3
4
x=
y=
y=
25
5
y=5
24
3
y=8
3
a
30
x
4
=5
63
b
y +1
=9
30 = 5 x
63 = 9( y + 1)
30
63
5
=x
9
= y +1
x=6
7 = y +1
7 −1 = y
y=6
a
g = 12
b
g = −10
c
p=7
d
g=7
e
y=5
f
y = 12
g
x = −3
h
x = −2
iii1.15 litres ⩽ x < 1.25 litres or
1150 mL ⩽ x < 1250 mL
b
5 − 2x = 9
−2 x = 9 − 5
−2 x = 4
iii
1.25 litres or 1250 mL
6
y = 4×2
y = 8
505 g
ii
x=2
d 5y + 3 = 9 + 2y
5y − 2y = 9 − 3
3y = 6
y=2
ii
1.15 litres or 1150 mL
12
y=
495 g
i
24
x=
=4
i
Pepe is incorrect as he has multiplied the
rounded number by three then worked
out +/− 5 cm from that answer instead of
+/− 15 cm from that answer (as there are
three pieces of wood).
15 a
= 1+ 3
2
65
55 ⩽ x < 65
12 x = 32 − 8
12 x = 24
2
−3=1
18.5 m
i
4(3x + 2) = 32
12 x + 8 = 32
x=8
c
b
16
x=
10 a
11 a
11
2x − 6 = 10
2 x = 10 + 6
2 x = 16
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
6
7
a
5x + 15 = 10x − 20 → x = 7
b
x + 3 = 2x − 4 → x = 7
b
y = 14
c
Learner’s own answers.
c
48
a
8x − 32 + 20 − 4x = 0 → 4x − 12 = 0 → x = 3
15 a
b
2(x − 4) + 5 − x = 0 → 2x − 8 + 5 − x = 0 →
x−3=0 → x=3
b
c
Learner’s own answers.
14 a
a5(23 + 4) = 5 × 27 = 135 and
2(30 − 23) = 2 × 7 = 14, 135 ≠ 14
b
16 a
Line 1: he added 5 and 4 instead of
multiplying 5 and 4.
Line 2: he subtracted 2x instead of adding
2x and added 9 instead of subtracting 9.
c
b
b=
c
c=2
d
3
d =4
5
b
n = 28
c
28 and 62
11 a
b
1
C x=
5
c
x = 32
s = 19
c
43 cm
y 1 5 9
y 4 6 8
1
D x=
5
c
12
= −3
y=x+4 x 0 2 4
s + 2s + 2s + 5 = 100 → 5s + 5 = 100
b
4
y = 2x + 1 x 0 2 4
x + 50 and 2x + 80
2x + 80 = 144
−12
2 y = 6x + 3
= 6 × −3 + 3
= −18 + 3
= −15
4 x = −3 and y = −15
b
A x = 6480 B x = 5
E x=5
b
13 a
6x + 3 = 2x − 9
6 x − 2 x = −9 − 3
4 x = −12
20
B and E
x=2
3 y = 2x − 9
= 2 × −3 − 9
= −6 − 9
= −15
a
3
5(x − 8) = 2(x + 10)
b
1
B and E give the correct answer of five
grandchildren.
12 a
= 60
9−x
x=
n + 2(n + 3) = 90 → 3n + 6 = 90
b
7 cm and 15 cm
420
2 y = 2x − 1
= 2 × 6 −1
= 12 − 1
= 11
3 Check values are correct. y = x + 5
= 6+5
= 11
4 x = 6 and y = 11
1
4
a
10 a
ii
2
Learner’s checks.
9
3 cm
i3(a − 2) + 3(a − 2) + a + a = 44 or
2a + 6(a − 2) = 44 or a + 3(a − 2) = 22
or 4a − 6 = 22
1 2x − 1 = x + 5
2 x − x = 5 + 1 x=6
5
7
a
ii
3(a − 2) = a
1
5
25
4
4
 5

5  5 + 4 = 5 × 9 = 45 + = 45 + 3 = 48
 7

7
7
7
7
8
i
Exercise 4.2
x=5 ,
5
2
4

and 2  30 − 5  = 2 × 24 = 48


7
7
7
a = 21
y + 3y + y − 2 + 4(y − 2) = 116
y
10
9
8
7
6
5
4
3
2
1
0
y = 2x + 1
y=x+4
0 1 2 3 4x
(3, 7); x = 3, y = 7
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
d
1 2x + 1 = x + 4
2x − x = 4 − 1
x=3
2 y = 2x + 1
= 2 × 3 +1
= 6 +1
=7
3 y = x+4
= 3+ 4
=7
4
9
4 x = 3 and y = 7
a
x = 18, y = 2
b
x = 9, y = 3
c
x = 9, y = 6
d
x = 12, y = 14
10 a
x = 10, y = 20
b
x = 3, y = 24
c
x = 14, y = −9
d
x = −2, y = 4
11 a
b
e
Learner’s own answer.
a
x + y = 10 and x − y = 4
12 a
b
1
x + y = 10
2 7 + y = 10
+
x − y= 4
y = 10 − 7
2x + 0y = 14
=3
2x = 14, x =
x + y = 37.74, x − y = 9.24
$23.49 and $14.25
14 a = 9, b = 3, c = 4, d = 10, e = 5, f = 11
14
=7
2
4 x = 7 and y = 3
1
x + 5y = 28
2 x + 5 × 5 = 28
−
x + 3y = 18
x = 28 − 25
0x + 2y = 10
= 3
1
2
mean =
b
range = 11 − 3 = 8
x − 2y = 6
4x = 40, x =
b
x ⩾ −6
c
x<0
d
x ⩽ 10
e
−8 ⩽ x < 0
f
−3 < x ⩽ 3
0
1
a
13
x = 6, y = 18
6
x = 2, y = 5
7
x = 6, y = −3
8
a
i, ii x = 2, y = 5
b
Learner’s own check.
c
Learner’s own answers.
–1
–4 –3 –2 –1 0
d
5
10 15 20
0
1
e
4
2
f
3
4
2
3
4
5
6
7
–5 – 4 –3 –2 –1 0 1 2 3 4 5 6
3 10 − 2 × 2 = 6
5
–2
0 0.5 1 1.5 2 2.5 3 3.5
2y = 4, y = = 2
4 x = 10 and y = 2
=7
x>2
2y = 34 − 30
40
= 10
4
6
c
2 3 × 10 + 2y = 34
4x + 0y = 40
42
a
3x + 2y = 34 and x − 2y = 6
+
=
b
4 x = 3 and y = 5
3x + 2y = 34
6
–3
3 3 + 3 × 5 = 18
1
9 + 3 + 4 + 10 + 5 + 11
a
Exercise 4.3
x + 5y = 28 and x + 3y = 18
10
2y = 10, y = = 5
2
c
cost of a cake, x = $1.50 and the cost of a
coffee, y = $2
13 x = 13, y = 8, so 2x + 3y = 50
3 7−3=4
b
2x + 3y = 9, 2x + y = 5
a
9
b
−6
c
−3, −2, −1, 0, 1, 2
a
Could be true.
b
Could be true.
c
Must be true.
d
Cannot be true.
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
a
b
6x > 18
2x − 3 < 19
2 x < 19 + 3
2 x < 22
18
x>
6
x>3
c
x<
5x + 1 ⩽ −9
5x ⩽ −9 − 1
5x ⩽ −10
x⩽
−10
5
3x > 4x + 12 b
3 x − 4 x > 12
− x > 12
22
−x
2
−1
x < 11
3(x − 4) ⩾ 9
3x − 12 ⩾ 9
3x ⩾ 9 + 12
3x ⩾ 21
d
11 a
c
7
a
x ⩾ 0.5
b
x<3
c
x ⩽ 13
d
x < 6.5
a
0
1
2
3
4
0
1
2
3
4
10
11
12
13
14
4
5
6
7
8
c
d
8
a
b
3(y − 4) + 7y ⩾ 8y − 5
3y − 12 + 7y ⩾ 8y − 5
10y − 8y ⩾ − 5 + 12
2y ⩾ 7
y ⩾ 3.5
i
y = 33(3 − 4) + 7×3 ⩾ 8 × 3 − 5;
3 × −1 + 21 ⩾ 24 − 5;
18 ⩾ 19 false
ii
y = 3.53(3.5 − 4) + 7 × 3.5 ⩾ 8 × 3.5 − 5;
3 × −0.5 + 24.5 ⩾ 28 − 5;
23 ⩾ 23 true
iii
y = 43(4 − 4) + 7 × 4 ⩾ 8 × 4 − 5;
3 × 0 + 28 ⩾ 32 − 5;
28 ⩾ 27 true
9
a
x ⩽ 10
b
x>4
c
x⩾2
d
x < 20
Learner’s own checks.
10 a
14
−2
>
−14
−2
x>7
−5x −18
⩾
−5
−5
x⩾3
12 a
3
5
x + 2x + x + 30 < 360 or 4x + 30 < 360
b
x < 82.5 °
c
No, x cannot be 90 ° because it has to be
less than 82.5 °.
13 a
A + A + 5 + 2(A + 5) < 100 → 4A + 15 < 100
5
b
−2 x
−1
6 − 5x ⩽ −12
−5x ⩽ −12 − 6
−5x ⩽ −18
x⩾7
6
12
x < −12
21
x⩾
3
x ⩽ −2
<
3x − 3 < 5x − 17
3 x − 5 x < −17 + 3
−2 x < −14
b
A < 21.25
c
No, because A < 21.25, so 2(A + 5) < 52.5.
14 a
x + 2x + 3(x − 10) < 360 → 6x − 30 < 360
b
x < 65
c
Yes. 2x = 3(x − 10) → x = 30 and this is in
the solution set.
15 a
2z + 9 > 13
b
3(z − 4) > −6
c
4 + 2z > 8
d
5(3z − 2) > 20
16 a
0<x⩽7
5
6
0
1
2
b
2 ⩽ y ⩽ 10
0
c
2
4
7
8
4
6
8
10
12
0
1
2
3
4
2
3
−1 < n < 3
–2
d
3
–1
0 < m < 4.5
0
1
4
5
5x − 14 > 2x + 1
b
x>5
c
Learner’s own checks.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 5.1
ii
1
a = 80, b = 80, c = 55, d = 125
2
B and C
3
B = 48, C = 66, A = 180 − (48 + 66) = 66
so A and C are equal.
4
a = 100, b = 105, c = 25
5
a = 49, b = 49, c = 48
6
x = 30
7
a, bLearner’s own sketch showing the exterior
angles 135 °, 120 °, 105 °
8
x = 95, y = 39, z = 124
Exercise 5.3
9
x = 50, y = 30, z = 80
1
a = 123; b = 109
2
a
3
Two exterior angles are 74 ° and two are 106 °.
4
60
5
72 °
6
a
7
20, 30 and 40 are factors of 360. 50 is not a
factor of 360.
8
a
9
12
10 a
b
11 a
b
Using the exterior angle property,
A = 122 − 59 = 63 and E = 122 − 63 = 59.
The third angle is 180 − 122 = 58, so both
triangles have angles of the same size.
13 It is correct. Substitute values of n. To show it
algebraically requires factorising.
Exercise 5.2
b
1
a
2
aLearner’s own answer; divide pentagon
into three triangles.
c
1080 °
1800 °
b
60 °
c
All the angles are 108 °. The fifth angle is
also 108 degrees. It is a regular polygon if
all the sides are the same length but this
may not be the case.
3
a
1260 °
4
a
7 sides
b
The sum must be a multiple of 180.
b
140 °
5
One of the angles marked is not inside the
hexagon. The angle there is 360 − 90 = 270 °.
6
84
7
a
8
1800 ÷ 180 = 10 so that is 12 sides. However,
180 is not a factor of 2000.
9
a
Angles are 135 ° + 135 ° + 90 ° = 360 °
b
i2 × 120 ° + 2 × 60 ° = 360 ° OR
120 ° + 4 × 60 ° = 360 °
900 °
Third angle = 360 − 108 − 108 = 144 °
12 36 °
Angles B and D are not equal.
360 °
1440 °
11 30 °
10 Angles of the quadrilateral are 118 °, 127 °, 75 °;
a = 360 − 320 = 40 °
15
There are three ways:
b
50 °
b
90 °
6
b
10 °
8
c
b
120 °
9
d
10
36
10 36
11 12 sides; the interior angle is 150 °, the exterior
angle is 30 °, 360 ÷ 30 = 12.
12 55
13 142 °
Exercise 5.4
All questions except questions 7, 9 and 10 have the
answer included for self-assessment.
1–6 Learner’s own diagrams and checks.
7
Learner’s own pattern. Assess by looking.
8
Learner’s own diagrams and checks.
9
Learner’s own pattern. Assess by looking.
10 a
b
Learner’s own diagram.
10 cm
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 5.5
e
1
a = 10 cm; b = 13 cm; c = 34 cm
2
a = 8.6 cm; b = 14.4 cm; c = 16.7 cm
aLearner’s own answer. For example: Time
of day, size of car, reason for travel, day
of the week.
3
a = 12 cm; b = 7.5 cm; c = 14 cm
b
4
a = 35.3 cm; b = 17.9 cm; c = 16.2 cm
5
a
10.1 cm
b
Learner’s own diagram.
Learner’s own questions. For example:
Do larger cars have more passengers? Are
there more passengers in cars early in the
morning? Are cars likely to have more
passengers at the weekend?
a
i
4.2 cm
ii
7.1 cm
c
iii
11.3 cm
iv
14.1 cm
Learner’s own predictions. For example:
During the rush hour cars are more likely
to have only one passenger. Cars on
Sundays will have more passengers than
cars on Mondays.
d
Learner’s own answer. For example:
Observing cars at different times of day or
different days of the week.
e
Learner’s own answer and explanation.
6
b
Learner’s own checks. Check with answers
in part a and try other side lengths.
7
3.7 m OR 3.71 m
8
14.7 m OR 14.72 m
9
a
10 a
2
17.3 cm
b
173 cm²
12 cm
b
15 cm
3
11 The calculator answer is 36.37… so 36.3 cm or
36.4 cm are acceptable answers.
12 a
b
24 cm
720 cm²
There are alternative answers to many questions in
this unit.
1
aLearner’s own answer. For example:
Gender, other interests, availability of
equipment.
b
c
d
aLearner’s own questions. For example:
Are young people faster using a keyboard
than older people? Is there a difference
between the speed of young people and
old people writing on paper? Is there
a difference between boys’ speed and
girls’ speed?
b
Exercise 6.1
16
Learner’s own answer and explanation.
Learner’s own questions. For example:
Do girls spend the same amount of time
playing computer games as boys? Do
young children play on computer games
less than older children? Does playing
computer games affect the time spent
doing homework?
Learner’s own predictions. For example:
Girls spend less time playing computer
games than boys. Learners who play
sports spend less time playing computer
games than learners who don’t play
sports.
Learner’s own predictions. For example:
Girls can type more quickly than boys.
Older people can write more quickly on
paper than younger people.
c, d, e Learner’s own answers. These will
depend on the predictions in part b.
Exercise 6.2
1
a
34
2
a
Teacher might not choose at random.
b
Learner’s own answer. For example:
Using random numbers or names from a
hat or particular positions in the register.
3
b
26
aPeople might be more likely to phone if
they have a complaint.
b
Learner’s own answer.
Learner’s own answer. For example: Use
random numbers or names from a hat or
a number of learners from different year
groups.
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
4
a Using
social
media
Advantage
Disadvantage
Easy to do
Some
people do
not use
social media
b Sending Can select
letters to
who to ask
people
c Asking
people in
the street
5
Age
Can choose a Can be
representative expensive
sample
and take a
lot of time
c
31
2
diameter, d = 2 × 4 = 8 cm
C = π × d = π × 8 = 25.14 cm
3
C = π × d = π × 12 = 37.704 cm
1
of the circumference = 37.704 ÷ 2 = 18.852 cm
2
Perimeter = 12 + 18.852 = 30.85 cm
4
a
b
42
9
Learner’s own answer. Any method should
take account of the fact that parents might
have more than one child in the school and
you do not want to choose any parent twice.
Exercise 7.1
d = 6 cm b
= 50.24 cm 2
r = 1 cm
A = π × r 2 = π × 12
= π ×1
c
= 3.14 cm 2
r = 6 cm
A = π × r 2 = π × 62
= π × 36
Sample size.
Q1: How many portions of fruit or
vegetables did you eat yesterday? Q2: How
often do you eat meat in the main meal
of the day? Q3: Why do you think people
are overweight? This could be a multiplechoice question.
r = 4 cm
A = π × r 2 = π × 42
= π × 16
Learner’s own questions. For example:
How were the adults chosen? What age
were the adults? What questions were the
adults asked? How did the adults measure
their energy levels?
8
a
radius, r = 4 cm
12
aQ1: People might say yes because they
think they should. Q2: This question will
encourage people to say no. Q3: This
question asks people to say something
they might feel uncomfortable about
because it is being rude about people.
b
1
= 28.26 cm
aNo. Adults from a small sample said that
vitamins gave them more energy but that
is not the same thing as proving that
they work.
b
7
7
d = 9 cm
C = π×d=π×9
Under 18 18 to 55 Over 55
Sample
6
People
might not
reply
c
5
= 113.04 cm 2
diameter, d = 6 cm
radius, r = 6 ÷ 2 = 3 cm
A = π × r2 = π × 32
= π × 9 = 28.28 cm2
6
A = π × r2 = π × 52
= π × 25 = 78.55 cm2
Area of semicircle = 78.55 ÷ 2 = 39.3 cm2
7
8
a
12.6 cm2
b
44.2 m2
c
616 cm2
d
8.04 m2
a
Learner’s own answers and explanations.
For example: Dipti has the incorrect
answer. She has not halved the diameter
to get the radius.
For example: Gabir has the incorrect
answer. He has used the formula for the
circumference not for the area.
d = 5 cm
C = π × d = π × 6 C = π × d = π × 5
= 18.84 cm
17
= 15.7 cm
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
c
Area = πr2
d = 2.4, so r = 2.4 ÷ 2 = 1.2
9
r2 = 1.22 = 1.44
Circle A :
Circle B
Radius Circumference Area
(cm)
(cm)
(cm2)
Area = π × 1.44 = 4.5238…
Ratio
4:8
8π : 16π
16π : 64π
Area = 4.52 cm2 (3 s.f.)
Ratio in its
simplest
form
1:2
1:2
1:4
a
i
A = 22.9 cm2 ii
C = 17.0 cm
b
i
A = 1590.4 mm2 ii
C = 141.4 mm
10 a
i
A = 113.5 cm
b
i
c
d
ii
P = 43.7 cm
A = 904.8 mm
ii
P = 123.4 mm
i
A = 402.1 cm
ii
P = 82.3 cm
i
A = 88.4 m2 ii
P = 38.6 m
2 2
2
11 Sofia is correct. Learner’s own explanation
and working. For example:
d
Learner’s own answers. For example: The
ratios of the radius and circumference are
the same.
e
Learner’s own answers. For example: The
ratio of the areas is the square of the ratio
of the radius.
f
i1 : 3 (the ratios of the radius and
circumference are the same)
ii12 : 32 = 1 : 9 (the ratio of the areas is
the square of the ratio of the radii)
Area of semicircle
1
1
2
2
= × π × r2 =
× π × 22 = 6.28 m2
Area of quarter-circle =
1
4
× π × r2 =
1
4
12.57 m > 6.28 m
2
Perimeter of semicircle =
×π×d +d =
1
2
1
2
12 Marcus is incorrect. Learner’s own
explanation and working. For example:
2
Perimeter of semicircle =
× π × 42 = 12.57 m2
2
1
19 Zara is incorrect.
4
3
4
1
a
b
× π × 6 + 3 + 3 = 20.14 m
Area = base × height
=6×4
Area =
1
2
=
1
2
13 a, b A and v, B and ii, C and i, D and vi,
E and iv, F and iii
c
Area =
1
2
15 7.59 m = 759 cm
=
Circumference = 6π cm
b
Area = 49π m2
Circumference = 14π m
c
Area = 100π mm2 Circumference = 20π mm
18 aArea = πr2 = π × 42 = 16π cm2
Circumference = πd = π × 8 = 8π cm
b
18
1
2
2
Area = 9π cm2
Area = πr2 = π × 82 = 64π cm2
Circumference = πd = π × 16 = 16π cm
× base × height
×6×4
= 12 cm 2
14 1.4 cm = 14 mm
17 a
× π × 6 + 6 = 3π + 6 m
= 24 cm 2
20.57 m > 20.14 m
16 27 cm
2
Exercise 7.2
× π × 8 + 8 = 20.57 m
×π×d +r+r =
1
not 6π + 6 cm
Perimeter of three-quarter circle =
3
×π×d +d =
× π × radius 2
× π × 32
= 14.14 cm 2
2
a
Area = rectangle + triangle
= 24 + 12
= 36 cm2
b
Area = rectangle + semicircle
= 24 + 14.14
= 38.14 cm2
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
c
3
= 14.14 + 12
= 26.14 cm
a
The two semicircles make one circle, so:
Area = semicircle + triangle
Area of circle = πr2 = π × 5.52 = 95.033…
2
Total area = 110 + 95.033… = 205 cm2 (3 s.f.)
Area A = l × w = 8 × 10 = 80
9
Area B = l × w = 12 × 1 = 12
b
Total area = 80 + 12 = 92 cm2
Area of large semicircle
Area A = l × w = 6 × 6 = 36
= πr 2 =
1
1
Area B = × b × h = × 6 × 4 = 12
2
2
Total area = 36 + 12 = 48 mm2
c
1
1
2
2
× π × 52 = 39.27
Total area = 30 + 39.27 = 69.27 cm2
1
2
1
2
Area triangle = × b × h = × 2 × 6 = 6
Area circle = πr2 = π × 42 = 50.26
Shaded area = 50.26 − 6 = 44.26 cm2
4
a
5 cm and 3 cm
b
Area A = base × height
= 3 × 7
= 21 cm2
7 cm
ii
135 cm2
b
i
3 cm, 6 cm
ii
90 cm2
6
a
104 cm2
7
a
i
Estimate = 42 cm2
ii
A = 39.44 cm2
i
Estimate = 49.5 m2
ii
A = 47.7 m2
i
Estimate = 108 cm2
ii
A = 120.7 cm2
i
Estimate = 3600 mm2
ii
A = 4156.3 mm2
d
8
Area of small semicircle
1
1
2
2
× π × 3.42 = 18.16
1
1
2
2
Area of triangle = bh =
× 6.8 × 9.2 = 31.28
Total area = 33.24 + 18.16 + 31.28 = 82.68
= 83 cm2 (2 s.f.)
10 a
60 m2
b
54.54 cm2
c
59.69 m2
Shape B, shaded area = π × 42 − 5.662 = 18.23 cm2
i
c
× π × 4.62 = 33.24
Shape A, shaded area = 82 − π × 42 = 13.73 cm2
a
b
2
12 Learner’s own answers and explanations. For
example: Arun is incorrect. The shaded area
in Shape A is less than, not greater than, the
shaded area in Shape B.
Total area = Area A + Area B
= 21 + 15 = 36 cm2
b
1
2
11 338 cm2
Area B = base × height
= 5 × 3
= 15 cm2
5
1
= πr 2 =
Area A = l × w = 10 × 3 = 30
Area B = π r 2 =
d
Chatri is not correct as the area of this
compound shape is 83 cm2 not 82 cm2 (2 s.f.)
13 aWhen radius = 2 cm,
height of rectangle = 4 cm and
length of rectangle = 3 × 4 = 12 cm.
152.55 cm2
Seb’s method is incorrect. Learner’s own
explanation and working. For example: When
he works out the area of the circle he doesn’t
use the correct radius. He actually uses the
diameter of 11 cm rather than the radius of
5.5 cm. The answer should be:
Shaded area = 4 × 12 − π × 22 = 48 − 4π
= 4(12 − π) cm2.
b
i
12 − π cm2
ii
9(12 − π) cm2
iii
25(12 − π) cm2
iv
100(12 − π) cm2
c
Learner’s own answer and explanation.
For example: The number outside the
bracket is the radius squared, and inside
the bracket is always (12 − π).
d
r2(12 − π) cm2
14 a
400 m
c
46.56 m
e
14 669 m
2
b
10 465 m2
d
461 m
f
4204 m2
15 8.8 cm
Area of rectangle = 10 × 11 = 110
19
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 7.3
1
2
3
a
1 hectometre = 100 metres
b
1 kilogram = 1000 grams
c
1 megatonne = 1 000 000 tonnes
a1 km = 1000 m, so 17.2 km = 17.2 × 1000
= 17 200 m
d
1 gigalitre = 1 000 000 000 litres
b
1 hL = 100 L, so 0.9 hL = 0.9 × 100 = 90 L
a1 centimetre = 0.01 metres OR
1 metre = 100 centimetres
c
1 Gg = 1 000 000 000 g, so 1.5 Gg
= 1.5 × 1 000 000 000 = 1 500 000 000 g
1 microlitre = 0.000 001 litres OR
1 litre = 1 000 000 microlitres
b
1000 mL = 1 L, so
43 000 mL = 43 000 ÷ 1000 = 43L
d
1 nanometre = 0.000 000 001 metres OR
1 metre = 1 000 000 000 nanometres
c
1 000 000 μg = 1 g, so
900 000 μg = 900 000 ÷ 1 000 000 = 0.9 g
a3 nanolitres, 3 millilitres, 3 centilitres, 3
litres, 3 teralitres
9
3 nL, 3 mL, 3 cL, 3 L, 3 TL
a9 micrograms, 9 milligrams, 9 grams,
9 kilograms, 9 gigagrams
9 μg, 9 mg, 9 g, 9 kg, 9 Gg
aA millimetre is a very small measure
of length. It is represented by the
letters mm.
You can also say that there are one
thousand millimetres in a metre or that 1
millimetre is one thousandth of a metre.
b
A microgram is a very small measure of
mass. It is represented by the letters μg.
From the Sun to:
Distance in …
Venus
47.9 Mm
Earth
108 Mm
Jupiter
0.228 Gm
Uranus
1.4 Gm
Neptune
2.9 Gm
10 A and iii, B and v, C and i, D and ii, E and iv
11 aMarcus is incorrect. 1 MW = 1000 kW not
100 kW.
1 MW = 1 000 000 W and 1 kW = 1000 W,
1 000 000 ÷ 1000 = 1000, so
1 MW = 1000 kW
b
12
630 MW = 630 000 kW = 630 000 000 W
Name of
star
Distance
in ly
Distance
in m
Wolf 359
7.78
7.36 × 1016
You can also say that there are one
million micrograms in a gram or that 1
microgram is one millionth of a gram.
Ross 154
9.68
9.16 × 1016
YZ Ceti
12.13
1.15 × 1017
aA kilometre is a very large measure of
length. It is represented by the letters km.
Gliese 832
16.08
1.52 × 1017
1 microgram = 0.000 001 grams which is
the same as 1 μg = 1 × 10−6 g.
1 kilometre = 1000 metres which is the
same as 1 km = 1 × 103 metres.
You can also say that there are one
thousand metres in a kilometre or that 1
metre is one thousandth of a kilometre.
b
A megatonne is a very large measure of
mass. It is represented by the letters Mt.
1 megatonne = 1 000 000 tonnes which is
the same as 1 Mt = 1 × 106 t.
20
a100 cm = 1 m, so
760 cm = 760 ÷ 100 = 7.6 m
c
1 millimetre = 0.001 metres which is the
same as 1 mm = 1 × 10−3 m.
6
8
1 milligram = 0.001 grams OR
1 gram = 1000 milligrams
b
5
7
b
b
4
You can also say that there are one million
tonnes in a megatonne or that 1 tonne is
one millionth of a megatonne.
13 a
12 KB, 936 KB, 42.5 MB, 1.14 GB, 6.3 TB
b
i1 GB can store 178 photos, 16 GB can
store 16 × 178 = 2848 photos.
ii
28 288 photos
iii238 photos. Working: 1.8 MB
can store 476 photos. 1.8
MB × 2 = 3.6 MB, so double the file
size means half as many photos. So,
1 GB can store 476 ÷ 2 = 238 photos.
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
ivLearner’s own working and answer.
For example:
32 GB = 32 000 MB and
32 000 MB ÷ 13 000
photos = 2.46… MB per photo.
d
4
Suggest Sue uses a 2.4 MB file size for
each photo as this will keep her just
below the 32 GB limit.
1
= 0.125 which is a terminating decimal.
8
2
1
c
Yes. Learner’s own explanations. For
example: When a unit fraction is a
terminating decimal then any multiple
of a terminating decimal is also a
terminating decimal.
8
terminating decimal.
3
8
1
= 3 × = 3 × 0.125 = 0.375 which is a
8
terminating decimal.
5
8
5
aSometimes true. Learner’s own
explanations. For example: Apart from
1
= 5 × = 5 × 0.125 = 0.625 which is a
7
8
terminating decimal.
b
14
b
1
= 0.05 which is a terminating decimal.
20
3
1
20
= 3×
20
20
= 5×
1
20
= 3 × 0.05 = 0.15 which is a
20
2
a
b
c
= 9×
= 9 × 0.05 = 0.45 which is a
recurring decimal
.
2
i
= 0.3 recurring decimal
iii
iv
6
3
6
4
6
5
6
.
= 0.6 recurring decimal
b
terminating decimal
ii
iii
2
25
5
25
11
25
= 0.44 terminating decimal
Sometimes true. Learner’s own
1 1 1
, ,
are
5 10 20
1 1 1
,
,
are recurring.
15 35 45
d
Always true. Learner’s own explanations.
For example: Even if the fraction can
be simplified, the denominator will be a
multiple of 3, so will be recurring.
aRecurring decimals. Learner’s own
explanations. For example: When they
are all written in their simplest form,
the denominators are multiples of 3,
so recurring.
b
B and D can be simplified. Learner’s own
explanations. For example: They can both
1
be simplified to 6 . It doesn’t change the
answer to part a, because the denominators
are still multiples of 3, so recurring.
c
Any fraction which has a denominator
which is a multiple of 9, when it is written
in its simplest form, is a recurring decimal.
a
1
terminating 4
b
4
terminating
5
c
8
recurring 15
d
1
terminating
5
= 0.08 terminating decimal
= 0.2 terminating decimal
= 0.5.
Never true. Learner's explanations.
For example: A denominator which is a
multiple of 15 is also a multiple of 3, which
is a recurring, not terminating, decimal.
.
= 0.83 recurring decimal
0.04
i
6
= 0.5 terminating decimal
a
c
21
20
2
c
= 5 × 0.05 = 0.25 which is a
terminating decimal.
.
0.16
ii
3
1
1
terminating, but
terminating decimal.
9
=
explanations. For example:
terminating decimal.
5
Learner’s own answers.
Yes. Learner’s own explanations. For
example: When a unit fraction is a
terminating decimal then any multiple
of a terminating decimal is also a
terminating decimal.
= 2 × = 2 × 0.125 = 0.25 which is a
8
= 0.8 terminating decimal
b
Exercise 8.1
a
25
aTerminating decimals. Learner’s own
explanations. For example: All the
denominators will divide exactly into 10,
100, 1000 or 10 000.
14 140 kg
1
20
iv
7
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
8
9
7
terminating
10
e
1
recurring 6
a
recurring
b
terminating
c
recurring
d
terminating
f
3
A and ii, B and iii, C and i
4
a–d i Learner’s own estimates.
a, b
Number of days off work due to illness
Abi
8
4
=
30 15
Dave
6
1
=
30 5
Bim
5
1
=
30 6
Enid
2
1
=
30 15
Caz
3
1
=
30 10
Fin
9
3
=
30 10
5
ii
6
c
ii
40
10 a
b
c
9 m. Learner’s own answer and
10
explanation.
iv
2 1 13
+ =
5 4 20
8
i
5 2 17
+ =
18 3 18
ii
3 2 37
+ =
5 9 45
9
iii
1 1 1
+ =
6 3 2
iv
2 1 13
+ =
5 4 20
10 a
5
8
b
40
11 a
50
25
1
 2
4
2
3
3+ ×
1
2
2
4
3
22 ÷ − 1
1
1
2
3
8
4
8
8
8
2
3
5
5
6
8
15
=3
2
a
c
22
3
4
3
2
4
4
1
a
4
b
c
4
2×4
8
=
=
5
3 × 5 15
d
8
Indices:
e
22 = 4
3
5
20
5
3
3
20
−1 =
5
40
6
6
b
2
d
5
3
1
15
Division: 4 ÷ = 4 × =
Subtraction:
4
1
Addition: 3 +
c
1
4
Multiplication: ×
4
5
91
×7 =
910
4
5
1  + 1 × 3 or equivalent.
6
6
3
17 2
m
36
5
8 cm2
9
9
b
12 cm
Exercise 8.3
1
1 1
2 + 1 −  Brackets: 1 − = 1 − = 1
8
2
3
2 7
14
÷ = × =
9
3 7
3 3
1
13
2
2
14 91
28 819
1
Addition: + = +
= 47
2
18
18
18
9
29
16 m2
36
3
7
a 15
b 35
c
4
16
2
5
1
 5
1 1 1
+ =
6 3 2
Addition: 2 + 1 = 2 + 1 = 3
b
Division:
iii
Exercise 8.2
a
7
kg
10
7
3 2 37
+ =
5 9 45
10
1
3
56
ii
4
13
c
1 2 5
+ =
6 3 6
4
15
Learner’s own answer and explanation. For
example: He cannot be correct because if
you round both sides up and add them to
6 you get 6 + 5 + 13 = 24. This is nearly 2 m
less than the perimeter, so the third side
must be at least 2 m more than 6 m.
i
2
ii
7
 2
−  4 + 12  or equivalent.
 5
25 
2
(terminating), 3 + 4 = 23 (terminating).
d
14
Multiplication: 6 × 7 =
No. Learner’s own examples. For
example: 1 + 1 = 3 (terminating), 2 + 3 = 31
11
b
50
6
For example:
ii
25
OR
A, B and E are recurring decimals; C, D and F
are terminating decimals.
49
5
8
7
18
b
a
Learner’s own decisions on how to group the
students.
For example: A and F are not unit fractions;
B, C, D and E are unit fractions.
1
4
a
−
f
11
2
5
6
=
29
6
=4
g
5
2
3
3
5
5
6
4
9
3
4
5
8
4
7
6
2
3
a
6
a
1
8
× 12 =
2
3
3
×3× 4 = 2× 4 = 8
× 20 = × 5 × 4 = 3 × 4 = 12
5
× 18 =
× 27 =
× 32 =
5
6
4
9
3
4
5
× 6 × 3 = 5 × 3 = 15
× 9 × 3 = 4 × 3 = 12
× 4 × 8 = 3 × 8 = 24
× 48 = × 8 × 6 = 5 × 6 = 30
× 35 =
8
4
7
× 7 × 5 = 4 × 5 = 20
b
× 20 =
1
4×2
c
18
×4×5 =
1× 5
2
=
5
2
=2
28
1
2
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
3
b
× 25 =
10
c
3
d
5
4
9
× 14 =
× 24 =
3
2×5
3
2×2
5
3× 3
×5 ×5=
×2×7 =
×3×8=
3×5
2
3× 7
2
5×8
3
=
15
=
21
=
2
2
40
3
4
a
5
a
9
b
12
c
15
d
33
e
30
f
6
g
9
1
3
h
i
13
1
5
j
1
3
1
11
5
a
18
35
b
7
15
6
3
3
4
d
e
21
40
f
9
13
g
1
9
7
11
h
1
5
1
9
b
2
1
3
1
e.g. ×
3
9
b
1
3
2
13   + 2 × = 2
 3
3 5
45
2
A and ii, B and v, C and i, D and iii, E and iv
2
a
2
12 ÷
b
18 ÷
3
3
c
20 ÷
8
Sarah is incorrect. 21 cm is the smallest whole
number value for d so that the circumference is
greater than 64 cm.
d
30 ÷
Learner’s own working. For example:
e
24 ÷
1
69 > 64
7
× 22 =
7
1
= 69 cm,
7
7
3
a
15 ÷
6
22
When d = 21 cm, C = × 21 = 22 × 3 = 66 cm,
7
66 > 64
When d = 20 cm, C =
6
62 < 64
22
7
× 20 =
440
7
6
= 62 cm,
a–f i
a
c
ii
e
ii
10 a
1
c
3
3
4
2
4
2
4
3
7
3
7
4
3
4
3
2
5
2
5
4
4
= 20 × = 5 × 4 × = 5 × 7 = 35
= 30 × = 15 × 2 × = 15 × 3 = 45
5
= 24 × = 6 × 4 × = 6 × 5 = 30
7
7
6
3× 2
8
9
9
9
8
4×2
7
= 15 × = 5 × 3 ×
=
5×7
2
=
35
=
27
2
= 17
1
= 13
1
2
12 ÷ = 12 × = 3 × 4 ×
=
3×9
2
2
2
c
3
20
9
4
14
16
2
25
b
1
5
4
2
11
3
1
22
2
ii
d
ii
f
ii
11 2
m
25
b
4 m2
1
5
d
1
5 m2
7
2
3
= 18 × = 6 × 3 × = 6 × 4 = 24
7
Learner’s own estimates.
ii
4
4
3
= 12 × = 6 × 2 × = 6 × 3 = 18
b
7
9
14
1
1
15
484
4
Exercise 8.4
a
When d = 22 cm, C =
23
1
7
22
2
2 2
3
6
= , × =
5 15 7 11
77
Learner’s own answer and explanation. For
example: When the four numbers in the
fractions are all different and are all prime
numbers or 1, then you will not be able
to cancel. When one of the numerators
and denominators are the same, then you
will be able to cancel. When one of the
numerators and denominators are even,
then you will be able to cancel. When one
of the numerators and denominators are
a multiple of each other, then you will be
able to cancel.
3
c
j
12 aLearner’s own examples of two proper
fractions that when multiplied do not
cancel.
1
3
4
19
24
1
16
2
c
35
48
i
11 24 m3
1
= 10
= 13
1
4
2
b
=7
1
6
43 2
m
56
6
15 ÷
10
4
A and iii, B and v, C and ii, D and iv, E and i
5
a
25 b
d
35 e
5
5
5
6
2 ×3
10 × 5
20 ÷
= 20 × = 10 × 2 ×
=
3
=
50
=
39
= 16
2
= 19
1
3
3
d
13
= 15 ×
1
5
13
10
= 3×5×
13
5× 2
=
3 × 13
2
2
3
4
88
1
2
20
15 c
16 f
2
1
4
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
6
15
16
a
8
9
4 c
4
9
1
5
1
11
1
D 1, A 1 , C 1 , B 1
3
16
14
d
7
1
6
b
2 e
f
4 1
b
19
26
3
c
Estimates are given first, then the accurate
answers:
a
1,
25
26
b
4, 3
c
3, 3
5 5
d
e
5, 4
f
1 5
4
=1 ,1
3
3 21
7
1 1
=3 ,3
2 9
2
1
8
4
5
9
5 3 3 1
5
×
=
=
, check
12 5 12 4
12
10 a
3
4
5
⇒ 25 − 5 = 20
2
a
3
a
4 , check
c
13
13 12 12 4
, check × = =
21 13 21 7
21
6
3
1
5
1
7
2
4
1.75 × 2 × 32 ⇒
4.7 × 35 ⇒ 4
7
10
× 35 =
175 cm3
13 2 m
6
a
14 aLearner’s own explanation. For example:
1
3
He rounded 3 to 3 and he rounded 9 to
0.44 × 52 ⇒ 0.44 =
7
15
1
c
1
12
25
3
4
4
8
Completing the working gives
1
1:
15
8
75 4
75
23
×
= 1: = 1: 2
13
26
26
8
2
23
Yes, 1: 2 ≈1 : 3
26
2
1:
3
2
1:1
5
47
10
b
45
44
100
47
× 35 ⇒
4
2
7
× 35
2 10
329
= 164
1
2
c
234
, 52 = 25 ⇒
44
49
1
2
1
4 100
3
2
3
0.9 × 6 ⇒ 0.9 =
× 25 = 11
2
9
2
20
9
20
, 6 =
⇒
×
10
3
3
31
1 10
=6
c
(
)
2.4 × 33 − 7 ⇒ 2.4 =
24
10
, 33 − 7 = 20 ⇒
24
1 10
× 20
2
= 48
4
3
2
1
2
1
÷ = 2 , 1 +3 =5 ,
7 14
3
2
3
6
2
1
3
2 × 1 = 3 , 8 3 − 2 5 = 5 11
5
2
5
4
6
12
7
a
8
4 2
m
5
9
a
b
1
1
2
38 cm2
Exercise 8.5
10 1 m
1
11 8
c
12
b
57
5
8
19 m2
5
a
2
18
× 32
b
9. So 3 : 9 = 1 : 3.
e
2
35
⇒
=
5
d
35
9
12 50 kg
c
5
× =
c
18
b
3
5
= 35 × 4 = 140
a
7
20
4
3
7 3 21
21
× =
⇒
× 12 = 21 × 3 = 63
2 2 4
4
1
4
b
c
81
1 , check 5 × 4 = 20 = 10
11 a
2
1
2
1

1
 + 1.5 + 9 ⇒  + 1  = ( 2) = 4 ⇒ 4 + 9 = 13
2
2
2
24
b
80
b
24 1 4
× =
5 6 5
b
d
 1

 1 3
52 −  4 + 0.75 ⇒  4 +  = 5 ⇒ 52 = 25
 4

 4 4
3.5 × 1.5 × 12 ⇒
Sofia is incorrect. Learner’s own examples. e.g.
1 1
2 3 10
÷ = 2, ÷ =
2
3
3
 3

 3 3
 2 − 0.6 − 3 ⇒  2 −  = ( 2) = 8 ⇒ 8 − 3 = 5
5
5 5
10
11
3
4
12 Terms are 12, 15, 18. nth term rule is 3n + 9, so
50th term = 159.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
13 a
b
c
V = 154 cm3
c
−6, −5.6, −5.2, −4.8, −4.4, −4, −3.6
22
22 × 12
3V = 3×154, πh = × 12 =
7
7
1
7
3V
3 × 154
3 × 154 × 7
49
= 22 × 12 =
=
πh
4 and
1 22 × 12 4
7
49 7
r=
= = 3.5
4
2
d
7.5, 6.25, 5, 3.75, 2.5, 1.25, 0
a
B
b
The 6th term, which is 2390. (Sequence is
3, 10, 38, 150, 598, 2390, …)
3 × 27
=
22
22
× 154
17
r=
81
=
222
92
9
=
222
22
Check:
1
1
3
V = π r 2h =
3
71
1
22
9
9
× 1 ×
×
× 154 = 3 × 9 = 27
3
22 1 22 1
7
1
2
3
4
5
25
a
5, 7, 9, 11, 13
b
0, 3, 6, 9, 12
c
11, 9, 7, 5, 3
d
2, 2.5, 3, 3.5, 4
e
210, 190, 170, 150, 130
a
4, 5, 7, 10, 14
b
5, 7, 11, 17, 25
c
20, 17, 13, 8, 2
a
non-linear
b
linear
c
linear
d
non-linear
a
linear
b
non-linear
c
linear
d
linear
e
non-linear
f
linear
a
9, 5, 1, −3, …
b
12, 17, 22, 27, …
c
3, 4, 6, 9, …
d
10, 9, 7, 4, …
e
64, 40, 28, 22, …
f
8, 10, 14, 22, …
c
17, 15, 11, 5, …
d
32, 24, 12, −4, …
11 a
A 343; B 64; C 179
b
64, 179, 343 or B, C, A
If ✱ = 6, then the sequence is −2, −2, −2,
−2, …
5, 6, 17, …
b
−1, 2, 5, …
c
4, 1, 4, …
d
−2, 0, 4, …
a
3, 4 , 5 , 7, 8 , 9 , 11
b
10, 9 , 8 2, 7 , 6 , 6, 5
2
3
1
3
3
5
2
3
4
5
If ✱ = 5, then the sequence is −2, −3, −22,
−10 643, …
So, as long as ✱ is greater than 6, there will be
positive numbers in the sequence.
13 Timo’s method is incorrect. Learner’s
own explanation. For example: He has
reversed the order of the operations, but
he hasn’t used inverse operations to reverse
the actual operations. Correct solution
is: 4th term = 72 + 8 = 80, 80 ÷ 2 = 40,
3rd term = 40 + 8 = 48, 48 ÷ 2 = 24.
14 16
a
5
15, 19, 26, 36, …
If ✱ = 7, then the sequence is −2, −1, 6,
223, …
7
1
5
b
If ✱ = 8, then the sequence is −2, 0, 8,
520, …
A and iv, B and ii, C and i, D and iii
1
3
7, 8, 11, 16, …
If ✱= 9, then the sequence is −2, 1, 10,
1009, …
6
8
10 a
12 Sofia is incorrect. Learner’s own explanation.
For example:
Exercise 9.1
1
9
1
5
15 Two of the terms in the sequence are negative.
Learner’s own working. For example:
Sequence is −4, 2, −10, 86, …
16 First term = −10. Learner’s own working. For
example: 4th term = 512, 3rd term = 0, reverse
the function so rule is cube root and subtract ?.
3
512 − ? = 0
8−? = 0
?=8
The reverse function is cube root and subtract 8.
3
3
2nd term = 0 − 8 = −8, 1st term = −8 − 8 = −10
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 9.2
1
a1st term = 3 × 1 = 3 2nd term = 3 × 2 = 6
3rd term = 3 × 3 = 9 4th term = 3 × 4 = 12
b
2
8
1
1
1st term = × 1 =
2nd term = 22 = 4
4th term = 42 = 16
d
1st term = 13 = 1 3rd term = 33 = 27
2nd term = 23 = 8
4th term = 43 = 64
b
1st term = 12 − 1 = 0 2nd term = 22 − 1 = 3
3rd term = 32 − 1 = 8 10th term = 102 − 1 = 99
c
1st term = × 1 = 2nd term = × 2 =
1
1
1
2
2
2
2
2
d
10
2
2
=1
They are the same.
b
i
=5
1
2
2
1
n
The sequence n is the same as .
3
3
1
n
The sequence n is the same as .
5
5
3
3n
The sequence n is the same as .
4
4
iv
4
A and iv, B and i, C and v, D and vi, E and iii,
F and ii
5
a
15, 23, 31, …, 87
c
−3 , −3, −2 , …, 1
e
1
1
2
2
b
−1, 6, 13, …, 62
d
1 2 3
, , , …, 1
10 10 10
21, 24, 29, …, 120
6
A and iv, B and iii, C and i, D and ii
7
a
Learner’s own answers.
b
B has the smaller value.
A 11th term = 121 − 33 = 88,
B 120th term = 2 × 120 + 7 = 87.
c
quadratic
e
neither
f
linear
a
n2
b
n2 + 20
c
n2 − 2
d
n2 + 7
n
13
12 a
A 3, B 5, C
b
5
8
n
9
n
12
c
4
7
b
4
7
3
5
C ,A ,B
5
8
13 aNo. n2 + 34 = 292, n2 = 258, n = 258 =
16.06…, which is not a whole number.
b
14 a
Yes. 5832 = 18 , which is a whole number.
5832 is the 18th term in the sequence.
3
1
n+8
1
3
1
4
2
b
12.8 − 0.3n
−2 − n
d
3.5 − 3.5n
15 a
2, 7, 14, …
b
119
16 a
9, 17, 27, … 153
b
1, 10, 25, …, 298
c
9, 12, 19, …, 180
b
64 and 125
c
n
The sequence n is the same as .
iii
d
11 a
2
1
2
2
2
3
1
10
3rd term = =1 10th term = = 5
2
2
2
a
neither
1
1st term = 2nd term = = 1
ii
26
=1
1
10th term = × 10 =
3
4
=1
a1st term = 6 × 1 + 1 = 7
2nd term = 6 × 2 + 1 = 13
3rd term = 6 × 3 + 1 = 19
10th term = 6 × 10 + 1 = 61
2
c
10 Learner’s own explanations. For example:
For the sequence n2 − 10, the first term will be
negative, so 7 cannot be the first term.
1st term = 12 = 1 3rd term = 32 = 9 2
quadratic
4
c
3rd term = × 3 =
b
9
4th term = × 4 =
3
linear
1
2 1
×2= =
4
4 2
1
4
2nd term =
4
4
3
1
3rd term = × 3 =
4
4
1
a
3
3
Exercise 9.3
1
a
16 and 25
2
a
i
ii
x
2
y
4 16 25
x
2
4
y
8
64 125
4
b
They are the same.
c
i
y = x2
ii
y = x3
5
5
d i
x 0 1 2 3 4 5 6 7 8 9 10 11 12 13
y 0 2 4 6 8 10 12 14 16 18 20 22 24 26
3
Learner’s own answers.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
ii
x 0
1
2
3
4
5
6
7
8
9 10 11 12 13
y 0 10 20 30 40 50 60 70 80 90 100 110 120 130
3
a
i
ii
x
3
6
8
y
10 37
65
x
1
10
3
7
a
x
1
8
1
4
1
2
2
y
1
16
1
4
1
16
i
x
2
−4
−6 −10
y
2
8
18
50
x
5
−3
−15
−7
y
64
0
144
16
i
y=
x2
2
ii
y = (x + 3)2
ii
y −1 25 998
4
b
i
a
i
ii
x −2
1
3
y 16
4
36
x −1
3
4
y
iii
5
a
x −1
2
y = 4x2
ii
y = (3x)2
iii
y = (x + 1)3
x −2
y
ii
iii
6
27
8
9
12
36 16
64
x −3
2
y 14
9
x
−3
1
2
i
y = (x − 4)2
ii
y=x +5
iii
y = x3 −
x
×2
Sofia is correct. Learner’s own explanation.
For example: When you square the positive
and negative of the same number you get
the same answer, e.g. 22 = (−2)2 = 4 and
52 = (−5)2 = 25.
Zara is incorrect. Learner’s own explanation.
For example: When you add 1 to the positive
and negative of the same number you get
different answers, so when you square these
answers, your final answers will be different,
e.g. (2 + 1)2 = 32 = 9 and (−2 + 1)2 = (−1)2 = 1.
4
0
y −27
b
y=x −2
0 27 125
i
i
b
3
9 81 144
y
b
ii
y=x +1
2
a
b
1
2
5
−
1
2
−
5
8
c
1
4
1
2
−
d
3
8
e
2
1
2
a
2
y
f
i
y = x4
ii
x=± 4 y
iii
Learner’s own check.
i
y = x5
ii
x= 5 y
iii
Learner’s own check.
i
y = (3x)2
ii
x=±
iii
Learner’s own check.
i
y = x3 − 10
ii
x = 3 y +10
iii
Learner’s own check.
i
y =  
ii
x=±4 y
iii
Learner’s own check.
i
y=
ii
x = 3 2y
iii
Learner’s own check.
y
3
x
 4
2
x3
2
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
10 a, b Learner’s own answers for grouping the
functions: For example:
•One step functions: B, F, G, L
Two step functions: A, C, D, E, H, I, J, K
•Contains a power: A, C, D, F, H, J
Contains a root: B, E, I, K
Contains no powers or roots: G, L
•Contains the number 4: A, D, E, L
Contains the number 9: G, H, I, K
Contains the number 2: C, J
Contains no numbers: B, F
•Contains fractions: C, D, E, I
Contains no fractions: A, B, F, G, H, J,
K, L
11 Sofia and Zara are both correct. The table of
values works for both equations.
This is because
x = ± 4y = ± 4 × y = ± 2 × y = ± 2 y
12
3
x
×10
y
Exercise 10.1
1
a
20 + 15 × 4 = $80
c
y = 15w + 20
2
a
3
4
5
6
1
4
−3
1
4
5
32
−270
1
y
Learner’s own explanation. For example:
Start by working out the missing
number in the function machine using
the first pair of values in the table.
20 kg
b
2x + 4y = 22
a
a + b = 36
b
23
c
b = 3a
a
24
b
i
a
84 minutes
c
36 minutes
a
85
c
15 (with two pentagons)
His conjecture is incorrect for the second
function y = 5 − x5. If x5 is greater than 5, the
y-value will be negative. For example: when
x = 2, y = 5 − 25 = 5 − 32 = −27.
28
b
t = 2g − 20
b
5p + 6h = 100
y = 10 + x because all the other functions are
equivalent.
9
a
r + b = 18
c
15 red and 3 blue
b
r + 4b = 27
Exercise 10.2
1
2
1 1
5
5
 1
  × ? = 1 , × ? = , ? = × 8 = 10, so the
2
4 8
4
4
missing number in the function machine is 10.
His conjecture is correct for the first
function y = (x − 5)4. When you work out
x − 5, if the answer is positive or negative,
once you have raised it to the power of 4,
the answer is always positive. For example,
34 = (−3)4 = 81.
2
8
a
3
5
−1
0
1
2
3
4
y
7
12
17
22
27
32
at (0, 12)
c
5 × 5 + 12 = 37, but 5 × 10 + 12 = 62
a
x
−10
0
10
20
30
40
y
8
10
12
14
16
18
b
at (0, 10)
c
11.4
a
x
0
5
10
15
20
25
y
20
15
10
5
0
−5
b
4
x
b
3
13 Marcus is incorrect. Learner’s own
explanations. For example:
ii
4s + 6l = 40
a, b Learner’s own answers.
y = 10x
1
2
$170
7
3
x
b
a
at (0, 20) and (20, 0)
x
0
1
2
3
4
5
6
y
10
8
6
4
2
0
−2
b
Learner’s own graph; A straight line
through (0, 10) and (5, 0).
c
3.5
a
x
0
3
6
9
12
15
y
4
3
2
1
0
−1
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
6
7
8
b
Learner’s own graph; A straight line
through (0, 4) and (12, 0).
c
at (4.5, 2.5)
a
x
0
6
2
5
y
9
0
6
1.5
b
Learner’s own graph; A straight line
through (0, 9) and (6, 0).
c
at (0, 9) and (6, 0)
a
11 aLearner’s own graph; A straight line
through (0, 7) and (14, 0).
i
ii
c
5 and −5
49
1
2
3
Exercise 10.3
y
6
−2 −3
−2
1
6
1
1
d
Yes; (−9) − 3 = 78
a
gradient 10 and y-intercept 20
b
gradient −20 and y-intercept 10
c
gradient 0.5 and y-intercept −2.5
a
gradient
b
gradient 0 and y-intercept 12
c
gradient −30 and y-intercept −45
3
a
1
2
4
a
y =15 − x
b
gradient −1 and y-intercept 15
c
(15, 0)
a
y = 4− x
b
gradient − and y-intercept 4
2
2
x
−3
−2
−1
0
1
2
3
x2 + 1
10
5
2
1
2
5
10
y
8
y = x2 + 1
5
4
y=x −1
2
c
2
0
−2
2
x
0
12
4
8
y
9
0
6
3
Learner’s own graph; A straight line
through (0, 9) and (12, 0).
c
Learner’s own graph; A straight line
through (0, 6) and (8, 0).
d
Learner’s own graph; A straight line
through (0, 3) and (4, 0).
6 × 5 + 5 × 6 = 60
b
at (0, 12) and (10, 0)
c
Learner’s own graph; A straight line
through (0, 12) and (10, 0).
1
10
and y-intercept
3
3
b
−
1
3
c
−2
1
3
1
3
x
0
12
6
3
y
4
0
2
3
d
Learner’s own graph; A straight line
through (0, 4) and (12, 0).
e
Learner’s own checks.
a
4 × 2.5 = 10 and 20 − 10 = 10
b
y=
c
gradient and y-intercept − 2
7
a
C−
8
A and iv, B and ii, C and iii, D and i
9
a
gradient −0.2 and y-intercept 2
b
gradient −2.5 and y-intercept 5
c
gradient −1 and y-intercept 0.4
It is on y = x2 − 1.
x
45
a
6
29
b
0
52 − 3 = 22
10 a
A is y = x2; B is y = x2 − 4
−1
c
b
(2, 6)
−3 −2
Learner’s own graph; A parabola with the
bottom at (0, −3).
a
c
x
b and c
9
Learner’s own graph; A straight line
through (0, 12) and (4, 0).
12 a
b
d
b
6
10 a
1
4
x−2
1
2
1
4
3
5
b
1
2
C2
4 × 0 − 2 × (−6) + 8 = 0 + 12 + 8 = 20
b
4 × 5 − 2 × 4 + 8 = 20 − 8 + 8 = 20
c
y = 0.5x + 3
d
gradient 0.5 and y-intercept 3
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 10.4
1
2
3
4
5
6
30
9
aLearner’s own graph; A line from the
origin through (50, 875).
a
400 m
b
gradient = 8; the speed is 8 m/s
c
y = 8t
d
560 m
a
15 dollars
b
6 dollars/metre
b
75 m
c
d = 6x
c
2.5 m/s
d
51 dollars
d
e
5 metres
y = 300 − 2.5x (learners could use other
letters).
a
i
b
0.35
c
y = 0.35x
d
32.55 dinars
e
2000 dollars
a
Time (hours)
0
5
10
Temperature (°C)
20
17
14
b
y = 17.5x
c
You can exchange 1 Franc for 17.5 Rand.
d
2275 Rand
10 aLearner’s own graph; A straight line from
(0, 300) to (120, 0).
ii
14 dinars
7 dinars
Exercise 11.1
1
aFlour:2 parts = 250 g,
1 part = 250 ÷ 2 = 125 g
Butter: 1 part = 125 g
b
2
Total = 250 + 125 = 375 g
aPeach juice: 3 parts = 450 mL,
1 part = 450 ÷ 3 = 150 mL
b
−0.6; the temperature decreases at a rate
of 0.6 °C/hour
c
y = 20 − 0.6t
d
12.8 °C
a
i
15 000
ii
23 000
Kim: 2 parts = 2 × 13 = $26
iii
27 000
Total they share = 65 + 26 = $91
Pineapple juice: 4 parts = 4 × 150 = 600 mL
b
3
b
400/year or 0.4 thousand/year
c
p = 0.4t + 15
4
Total = 450 + 600 = 1050 mL
Tina: 5 parts = $65 → 1 part = 65 ÷ 5 = $13
a
Benji: 2 parts = $24, 1 part = 24 ÷ 2 = $12
Abdul: 1 part = $12
aLearner’s own graph; A line from the
origin through (25, 42).
b
about 30 dollars
c
The gradient is 42 ÷ 25 = 1.68, so the
equation is d = 1.68l.
d
30.24 is the exact value
e
100.8 dollars
f
40 litres
7
He is not correct. Priya’s speed is 50 km/h and
Mei’s is 60 km/h.
8
aLearner’s own graph; A straight line from
(0, 12) through (20, 40).
b
33 litres
c
1.4 litres/second
d
y = 1.4x + 12
Caen: 3 parts = 3 × 12 = $36
b
Total = 12 + 24 + 36 = $72
5
a
21
b
35
6
a
180 g
b
480 g
7
a
6, 15, 24
b
45
8
aInstead of using $40 = 5 parts (for travel)
he has used $40 = 4 parts (for food).
He has also added up the total number of
parts incorrectly. The total is 16 not 15.
b
5 parts = $40, so 1 part = 40 ÷ 5 = $8
Total number of parts = 4 + 7 + 5 = 16
Total spent = 16 × 8 = $128
9
650 mL
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
10 a
b
$135
Zosia gets $60, Abie gets $75
11 12 : 16 → divide both numbers by 4 → 3 : 4
9 : 12 → divide both numbers by 3 → 3 : 4
12 550 mL vanilla ice cream, 2200 mL grape juice,
2750 mL ginger ale. Learner’s own method.
For example:
Grape juice: 2250 ÷ 4 = 562.5 mL per part,
Ginger ale: 2750 ÷ 5 = 550 mL per part.
Use 550 mL per part as smallest amount.
Ice cream: 1 × 550 mL = 550 mL,
Grape juice: 4 × 550 mL = 2200 mL,
Ginger ale: 5 × 550 mL = 2750 mL
13 0.03 and 0.025 or 0.036 and 0.03
Activity
Child :
staff
ratio
Number
of
children
Horseriding
4:1
20
20 ÷ 4 = 5
so 5
Sailing
5:1
15
15 ÷ 5 = 3
so 3
Rockclimbing
8:1
32
32 ÷ 8 = 4
so 4
Canoeing
10 : 1
28
28 ÷ 10 = 2.8
so 3
Total number of staff = 5 + 3 + 4 + 3 = 15
Exercise 11.2
1
14 22.5 cm
15 a
b
c
90 °, 35 ° and 55 ° or 90 °, 70 ° and 20 °
Two solutions. Learner’s own explanation.
For example: The 20 ° difference could be
between the right angle and one of the
other angles, or it could be between the
two other angles (not the right angle).
2
3
35 ° : 55 ° : 90 ° → 7 : 11 : 18 or
20 ° : 70 ° : 90 ° → 2 : 7 : 9
a
i
b
direct proportion
a
i
$4.40
iii
$8.80
Number
of
children
Horseriding
4:1
22
Number
of staff
a
less time
c
i
×2
2 people = 6 days
4 people = 3 days
a
less than 60 seconds
22 ÷ 4 = 5.5
so 6
b
more than 60 seconds
c
i
×2
Sailing
5:1
17
Rockclimbing
8:1
30
30 ÷ 8 = 3.75
so 4
÷2
Canoeing
10 : 1
26
26 ÷ 10 = 2.6
so 3
d
Learner’s own answer. For example: Move
two children from horse riding to rock
climbing and move two children from
sailing to canoeing. New table is:
b
c
×2
÷2
normal speed = 60 seconds
÷2
2 × speed = 30 seconds
ii
a
more time
1 person = 12 days
inverse proportion
5
160
$6.60
2 people = 6 days
d
Total number of staff = 6 + 4 + 4 + 3 = 17
31
b
÷2
17 ÷ 5 = 3.4
so 4
b
ii
direct proportion
ii
4
iii
120
b
Child :
staff
ratio
ii
80
16 a
Activity
Number
of staff
normal speed = 60 seconds
1
speed = 120 seconds
2
×2
inverse proportion
÷2
×2
×7
4 people = 7 hours
2 people = 14 hours
4 people = 7 hours
8 people = 3.5 hours
4 people = 7 hours
28 people = 1 hour
×2
÷2
÷7
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
7
a
100 g
b
x × y = 180, yes
b
225 g
c
c
400 mL
a
i
3 hours
i and iii
Number of days it takes sheep to
eat a bag of feed
ii
12 hours
iii
4 hours
b
Number of days
6
60 km/h
8
Number
of
students
5
2
3
4
6
7
40
35
30
25
20
15
10
5
0
8
0
1 hour 20 minutes
11 Zara is incorrect. It will take the same amount
of time as 20 minutes is the time the journey
takes. It doesn’t matter how many people are
on the train.
1
3
b
2
3
c
1
4
d
5
12
e
7
12
2
a
0.35
b
0.2
3
a
85%
b
35%
c
20%
b
0.7
c
0.52
169 cm
b
14 a
2 houses
b
12 people
4
a
0.34
c
d
15 days
5
a
i
Houses
b
0.85
Days
165 cm
6
20
4
6
1
120
4
1
30
1
7
a
0.4
b
0.9
6
60
12
8
a
0.01
b
0.98
4
60
8
c
0.5
d
0.95
e
0.89
f
0.81
15 a
32
ii
0.45
5
12
9
10
15
20
30
40
60
Number
of days
(y)
36
x×y
180 180 180 180 180 180 180
18
70
a
1
13 a
5
60
ivAnswer between 25 and 26 days
(accurate answer is 25.7 to 1 d.p.)
12 A = 14, B = 15, C = 49, D = 7.5
Number
of sheep
(x)
30
40
50
Number of sheep
Exercise 12.1
10 4 days
People
20
iiNo, the points do not form a straight
line.
Cost per
student 240 600 400 300 200 171.43 150
($)
9
10
12
9
6
4.5
3
0.65
aT is 3, 6, 9 or 12 and F is 5 or 10 and these
have no common element.
b
i
19
36
ii
2
3
iii
17
36
10 a
i
13
25
ii
12
25
iii
22
25
b
15 is a multiple of both 3 and 5, so the
events are not mutually exclusive. Adding
the probabilities will not give the correct
answer.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 12.2
1
2
3
7
aLearner’s own explanation. For example:
The score on one dice does not affect the
score on the other dice.
b
Yes. For example: The score on one dice
does not affect the score on the other dice.
c
Learner’s own explanation. For example:
If you get an even number you cannot get
an odd number on the same dice and vice
versa.
aLearner’s own explanation. For example:
The thunderstorm increases the
probability that Zara will be late.
b
Learner’s own explanation. For example:
They are not independent. If Arun is late
on Monday he will probably make more
effort not to be late on Tuesday.
a
i
b
1
4
d
Learner’s own explanation. For example:
1
8
ii
1
2
c
iii
8
1
4
a
0.5
c
Learner’s own explanation. For example:
The probability of the letter being in the
word HEAD is the same, whether it is in
the word FACE or not.
d
Learner’s own explanation. For example:
If the letter is in the word FACE, then
P(in EACH) = 0.75. If the letter is not in
the word FACE it is B, D, G or H and
P(in EACH) = 0.25. The probabilities
are different, so the events are not
independent.
b
1
4
ii
2
3
ii
2
3
1
6
of 12 is impossible and P(Z) = 0. The
probabilities are different, so the events
are not independent.
i
1
2
iii
1
3
b
i
1
3
iii
1
3
c
Learner’s own explanation. For example:
9
6
aLearner’s own explanation. For example:
2
If A happens then P(B) = ; if A does not
5
3
happen then P(B) = . These are not the
5
same so the events are not independent.
b
Yes. Learner’s own explanation. For
1
example: This time P(B) = 2 if A happens
and if A does not happen.
33
ii
4
11
b
For all 3 to be the same colour the third
must be red.
The probability of this is
iii
2
11
i
1
C is correct. Learner’s own explanation. For
example: The results of the first three rolls and
of the fourth roll are independent.
5
11
a
P(multiple of 3) = 3 both if the number is
even and if it is not.
5
Learner’s own explanation. For example:
If blue is 6 then P(Z) = P(total of 12) =
P(yellow is 6) = . If blue is not 6 a total
1
a
0.5
aLearner’s own explanation. For example:
If the blue dice is 6 then P(Y) = P(yellow
1
is 6) = 6 ; If the blue dice is not 6 it will
be another number and then P(Y) is
1
again . The same probability implies
6
independent events.
Whether or not R happens, P(F) = 4 .
4
b
2
4
= .
10
5
Exercise 12.3
1
a
0.32
b
0.6
c
0.2
d
0.12
a
b
c
1
6
5
6
1
36
25
36
3
a
0.42
b
0.3
c
0.18
4
a
1
64
b
1
32
c
1
32
d
5
32
e
25
64
2
d
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
a
First
late
0.25
0.75
b
6
i
not late
Outcome
0.4
late
late, late
0.25 × 0.4 = 0.1
0.6
not late
late, not late
0.25 × 0.6 = 0.15
0.4
late
not late, late
0.75 × 0.4 = 0.3
0.6
not late
not late, not late
0.75 × 0.6 = 0.45
ii
0.1
0.45
iii
0.9
iii
0.14
a
First
Second
red
0.6
blue
red, red
0.6 × 0.35 = 0.21
blue
red, blue
0.6 × 0.65 = 0.39
0.35 red
blue, red
0.4 × 0.35 = 0.14
blue
blue, blue
0.4 × 0.65 = 0.26
0.65
b
i
Outcome
0.35 red
0.65
0.4
7
Second
ii
0.21
0.26
a
First
Second
Outcome
3
4
odd
odd, odd
2
3
× 34
=
6
12
=
1
2
1
4
even
odd, even
2
3
× 14
=
2
12
=
1
6
3
4
odd
even, odd
1
3
× 34
=
3
12
=
1
4
1
4
even
even, even
1
3
× 4
1
=
1
12
odd
2
3
1
3
even
b
34
i
1
2
ii
1
12
iii
11
12
iv
5
12
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
8
a
Learner’s own tree diagram. For example:
Outcome
First
Second
3
5
T
TT
3
5
× 35
9
= 25
or 0.36
2
5
F
TF
3
5
× 25
6
or 0.24
= 25
3
5
T
FT
2
5
× 35
6
= 25
or 0.24
2
5
F
FF
2
5
× 5
2
= 25 or 0.16
T
3
5
2
5
F
4
The branches for F and T could be reversed.
b
9
9
or 0.36
25
i
21
or 0.84
25
ii
0.23. Learner’s own method. For example: This could be done with a tree diagram.
0.8 × 0.05 + 0.2 × 0.95 = 0.23
Exercise 12.4
1
a
0.46
2
a
0.45 b
c
Yes. Learner’s own explanation. For example: 0.325 is quite close to 0.35, so there is no reason to
reject the conjecture.
3
a
0.58
4
a
17
= 0.68
25
b
i
c
8 is the best estimate. For example: The relative frequency is tending to 0.8 (1 d.p.) and 0.8 × 10 = 8
a
Total
10
20
30
40
50
60
70
80
Silver cars
2
7
11
16
19
23
27
31
Relative frequency
0.2
0.35
0.367
0.4
0.38
0.383 0.386 0.388
5
6
0.36
0.325
b
0.7 ii
c
0.18
0.785 iii
0.14
0.782
b
Learner’s own graph. Check that the points from the table in part a have been plotted correctly.
c
Learner’s own estimate. 0.38 or 0.39 or 0.4 would be a sensible estimate from the data.
a
b
c
35
b
Flips
20
40
60
80
100
Frequency of 2 heads
5
9
11
17
19
Relative frequency
0.25
0.225 0.183 0.2125
0.19
Learner’s own graph. Check that the points from the table in part a have been plotted correctly.
Flips
20
40
60
80
100
Frequency of 2 heads
4
11
16
20
24
Relative frequency
0.2
0.275 0.267
0.25
0.24
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
d
7
8
Learner’s own graph. Check that the points
from the table in part c have been plotted
correctly.
5
a
N
a, b and c Learner’s own results.
8 cm (80 m)
d
Learner’s own graphs.
85 °
e
Learner’s own comparison of relative
frequencies with 0.5.
145 °
6 cm
(60 m)
a and b Learner’s own results.
c
Learner’s own comparison of relative
1
6
Keri
frequencies with 0.1666 … (that is ).
b
Exercise 13.1
1
2
a
4.5 × 8 = 36 km
b
18 ÷ 8 = 2.25 cm
a6 cm. Learner’s own diagram. Check that
the length of the line is 6 cm and that the
angle between the N arrow and the line
is 120 °.
b
3
Dom
8.5 cm. Learner’s own diagram. Check
that the length of the line is 6 cm and that
the angle between the N arrow and the
line is 35 °.
6
No. Learner’s own diagram and explanation.
For example: The diagram shows that the
jeeps are moving in different directions and
they will not meet at all.
7
a, b Yes. Learner’s own diagram and
explanation. For example: The diagram shows
that the ship is much closer to C than B.
8
a
N
N
Learner’s own diagram. Check that the length
of the line is 4 cm and that the angle between
the N arrow and the line is 95 °.
4
Learner’s own measurement. In the range
70–75 m.
80 °
N
12.5 cm
(125 km)
145 °
8.5 cm
(85 km)
Harbour
Ship
Fabia
3 cm
(300 m)
50 °
Luca
b
Learner’s own measurement and
conversion. In the range 175–180 km.
c
Learner’s own measurement. In the range
283 °–289 °.
Gate
230 °
2 cm
(200 m)
9
N
a
190 °
6 cm
(120 km)
12 cm
(240 km)
Airport
Aeroplane
b
36
N
300 °
Learner’s own measurement and
conversion. In the range 225–230 km.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
c
Learner’s own measurement. In the range
088 °–093 °.
10 a
Greg
N
95 °
Car
14 In the range 72–78 km and 338 °– 342 °.
15 a, b Learner’s own answers. Allow ± 2 ° on the
bearings and ± 2 mm on the distances on the
map. For example:
N
35 °
7 cm
(14 km)
8 cm
(16 km)
Distance
in real
life (m)
Start
A
080 °
3.5
700
A
B
100 °
4.6
920
B
E
140 °
2.4
480
b
Learner’s own measurement and
conversion. In the range 25.8–26.2 km.
E
F
227 °
4.5
900
c
Learner’s own measurement. In the range
245 °–250 °.
F
C
345 °
3.5
700
C
D
255 °
3.6
720
D
Finish
328 °
3.3
660
11 In the range 7.3–7.5 km and 215 °–220 °.
12 a
In the range 45–48 km.
Exercise 13.2
b
In the range 52–55 km.
1
13 a160 °. Learner’s own explanation. For
example: Triangle is equilateral so angle
ABC = 60 °. Line BD is parallel to the
north arrow so angle ABD = 40 ° and so
angle DBC = 60 − 40 = 20 °. Bearing of C
from B = 180 − 20 = 160 °.
b
c
37
Distance
Bearing on map
(cm)
From To
280 °. Learner’s own explanation. For
example: Triangle is equilateral so angle
ACB = 60 °. Line CE is parallel to line BD
so angle ECB = angle DBC = 20 °. Bearing
of C from B = 360 − 20 − 60 = 280 °.
2
3
a
1
1

 × 3, × 6 = (3 ÷ 3, 6 ÷ 3) = (1, 2)
3
3
b
2
2

 × 3, × 6 = (3 ÷ 3 × 2, 6 ÷ 3 × 2) = ( 2, 4)
3
3
a
1
1

 × 4, × 12 = ( 4 ÷ 4, 12 ÷ 4) = (1, 3)
4
4
b
3
3

 × 4, × 12 = ( 4 ÷ 4 × 3, 12 ÷ 4 × 3) = (3, 9)
4
4
H and iii, I and ii, J and vi, K and i, L and v,
M and iv
Learner’s own accurate sketch and checks.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
4
a
(9, 12)
b
Letter
A
B
C
D
E
F
K
R
W
Position in
alphabet
1st
2nd
3rd
4th
5th
6th
11th
18th
23rd
xcoordinate
1 × 3 = 3 2 × 3 = 6 3 × 3 = 9 4 × 3 = 12 5 × 3 = 15 6 × 3 = 18 11 × 3 = 33 18 × 3 = 54 23 × 3 = 69
ycoordinate
1 × 4 = 4 2 × 4 = 8 3 × 4 = 12 4 × 4 = 16 5 × 4 = 20 6 × 4 = 24 11 × 4 = 44 18 × 4 = 72 23 × 4 = 92
Coordinate
pair
(3, 4)
c
(6, 8)
(9, 12)
(12, 16)
(15, 20)
(18, 24)
(33, 44)
(54, 72)
(69, 92)
iThe x-coordinates are the numbers in the 3 times table. To work out the x-coordinate of any
letter, multiply the position number of the letter in the alphabet by 3.
iiThe y-coordinates are the numbers in the 4 times table. To work out the y-coordinate of any
letter, multiply the position number of the letter in the alphabet by 4.
5
6
a
B (4, 5)
b
C (12, 15)
c
A (3, 2)
d
C (15, 10)
a
(10, 16)
b
(15, 24)
c
Learner’s own explanation. For example: E is the 5th letter of the alphabet, so has coordinates
(5 × 5, 5 × 8) = (25, 40).
d
T is the 20th letter of the alphabet, so has coordinates (20 × 5, 20 × 8) = (100, 160).
e
nth letter of the alphabet has coordinates (5n, 8n).
7
H (28, 36)
8
Difference in x-coordinates = 10 − 1 = 9
Difference in y-coordinates = 13 − 1 = 12
1
3
1
3
×9 = 3
× 12 = 4
E = C(1, 1) + (3, 4) = (1 + 3, 1 + 4) = (4, 5)
9
aDifference in x-coordinates = 7 − 2 = 5
Difference in y-coordinates = 18 − 3 = 15
2
×5= 2
5
2
× 15 = 6
5
H = F(2, 3) + (2, 6) = (2 + 2, 3 + 6) = (4, 9)
b
Learner’s own check by drawing a diagram.
10 aYes. Learner’s own justification. For example:
3
2
1
of the way = 4 × 3 = 12.
3
3
3
1
3
2
y-coordinates: of the way = 10, so of the way = 10 ÷ 2 = 5, so of the way = 5 × 3 = 15.
3
3
3
x-coordinates: of the way = 8, so of the way = 8 ÷ 2 = 4, so
So, B is the point (12, 15).
b
38
No. Learner’s own justification. For example:
OA is
2
2 1
of OB so ratio OA : AB = : = 2 : 1.
3
3 3
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 13.3
11 aLearner’s own diagram. Coordinate grid
with points A (2, 7), B (10, 3), C (5, 3)
and D (11, 5). Line segments AB and
CD drawn. Point where AB crosses CD
labelled ‘E’.
b
E (8, 4)
c
Midpoint of CD =
 5 + 11 3 + 5   16 8 
,

 =  ,  = (8, 4) = E
2
2   2 2
d
1
a, b Learner’s own diagram. Coordinate grid.
Triangle A with vertices (1, 4), (2, 3) and
(2, 5). Triangle B with vertices (4, 3), (5, 4) and
(4, 5). Triangle C with vertices (6, 2), (7, 3) and
(6, 4).
2
a, b Learner’s own diagram. Coordinate grid.
Rectangle A with vertices (1, 4), (3, 4), (3, 5)
and (1, 5). Rectangle B with vertices (3, 4),
(4, 4), (4, 6) and (3, 6). Rectangle C with
vertices (3, 0), (4, 0), (4, 2) and (3, 2).
3
a and ii, b and iii, c and i
4
aA to B is a reflection in the mirror line
x = 3.
3
of the way along AB:
4
Difference in x-coordinates = 10 − 2 = 8
3
×8 = 6
4
Difference in y-coordinates = 3 − 7 = −4
3
× −4 = −3
4
So, A(2, 7) + (6, −3) = (2 + 6, 7 − 3)
= (8, 4) = E.
12 Difference in x-coordinates =
−4 − −9 = −4 + 9 = 5
3
×5= 3
5
Difference in y-coordinates =
−2 − −12 = −2 + 12 = 10
3
× 10 = 6
5
b
Learner’s own working. For example:
d
B to E is a reflection in the mirror line
y = 5.
e
E to C is a rotation 180 °, centre (6, 6).
f
C to F is a translation  −  .
1
 −3
a–d Learner’s own diagram. Coordinate grid.
Triangle A with vertices (−1, −4), (−3, −2) and
(−4, −4). Triangle a with vertices (3, −3),
(6, −3) and (5, −1). Triangle b with vertices
(0, 0), (−3, 0) and (−1, 2). Triangle c with
vertices (0, 2), (−2, 3) and (0, 5). Triangle d
with vertices (3, 3), (6, 3) and (5, 5).
7
Difference in y-coordinates is 5 − −7 = 12.
Points:
P
S
T
U
V
She has only got part b correct. Learner’s own
explanations. For example: The object and its
image are always congruent, so answers are:
x-coordinates
−4 −3 −2 −1
0
1
2
a
Corresponding lengths are the same.
y-coordinates
−7 −5 −3 −1
1
3
5
b
Corresponding angles are the same.
c
The object and the image are congruent.
a
Translation  
b
Translation  0
c
Translation  − 
Q
R
8
39
A to D is a rotation 90 °, anticlockwise,
centre (1, 4).
6
Difference in x-coordinates is 2 − −4 = 6.
There are 6 points after P, so the
x-coordinates increase by 6 ÷ 6 = 1 for
each point, and the y-coordinates increase
by 12 ÷ 6 = 2 for each point.
c
a–c Learner’s own diagram. Coordinate grid.
Triangle A with vertices (6, 4), (7, 6) and
(5, 6). Triangle B with vertices (1, 1), (3, 1) and
(2, 3). Triangle C with vertices (3, 1), (5, 2) and
(3, 3). Triangle D with vertices (5, 0), (7, 0)
and (6, 2).
S (−1, −1)
T (0, 1)
A to C is a translation  2 .
5
So, K(−9, −12) + (3, 6) = (−9 + 3,
−12 + 6) = (−6, −6).
13 a
 6
b
4
 6
 −6
4
 −6
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
9
d
Reflection in line y = 1.
e
Reflection in line x = 0 (or y-axis).
f
Reflection in line y = −2.
g
Reflection in line x = −2.
h
Rotation 90 ° clockwise, centre (0, −4).
i
Rotation 90 ° anticlockwise, centre (0, 0).
j
Rotation 180 °, centre (0, 2).
a
iRotation 90 ° anticlockwise
about (−1, 3).
b
3
Learner’s own diagram. The enlarged rightangled triangle should have the corresponding
vertex on the cross, base length eight squares
and height four squares.
4
 4
 −4
ii
Translation 
iii
Reflection in the line x = −1.
iv
Reflection in the line x = −3.5.
iCheck learners’ own combinations
of at least two transformations. For
example:
5
Rotation 90 ° anticlockwise about
(−2, 3) followed by translation  0
 −5
iiCheck learners’ combinations of
at least two transformations. For
example:
aLearner’s own explanation. For example:
The top vertex of the kite is only two
squares from the centre of enlargement,
not three. All the other vertices are in the
correct position.
b
Reflection in line x = 3, followed by
translation 1
 3
10 Zara is not correct. Arun is correct. Learner’s
own diagrams.
11 a
A to C
b
A to B
c
A to D
d
B to E
12 aLearner’s own diagram. Check that each of
the transformed shapes are drawn correctly.
The final image should have vertices (4, 0),
(5, 0), (5, 1), (5, −1) and (4, 1).
b
 8
 −3
A translation 
6
Exercise 13.4
1
Learner’s own diagram. Enlarged triangle
with vertices (1, 0), (5, 2) and (1, 2).
aLearner’s own diagram. Enlarged square
with vertices (0, 1), (3, 1), (3, 4) and (0, 4).
b
40
scale factor 2
aLearner’s own diagram. Enlarged triangle
with vertices (3, 1), (5, 1) and (3, 5).
b
2
a
Learner’s own diagram. Enlarged triangle
with vertices (0, 2), (3, 2) and (0, 5).
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
b
scale factor 3
f
A = 22 × 12.566 = 50.265 cm2 (3 d.p.)
g
A = πr2 = π × 42 = 16π = 50.265 cm2 (3 d.p.)
15 aArun is incorrect. He has multiplied the
perimeter of H by 3 instead of dividing
by 3.
Perimeter = 36 ÷ 3 = 12 cm
b
c
Area = 54 ÷ 32 = 6 cm2
Exercise 14.1
scale factor 2
1
a
Volume = area of cross-section × length
= 20 × 8 = 160 cm3
b
Volume = area of cross-section × length
= 15 × 6 = 90 cm3
c
Volume = area of cross-section × length
= 12 × 9 = 108 cm3
d
7
aLearner’s own diagram. Shape B with
vertices (6, 0), (10, 0), (10, 2), (8, 4)
and (6, 4).
b
c
8
9
= 30 × 12 = 360 cm3
2
a
Learner’s own diagram. Shape C with
vertices (0, 1), (6, 1), (6, 4), (3, 7)
and (0, 7).
Volume = area of cross-section × length
= 32 × 10 = 320 cm3
b
Learner’s own diagram. Shape D with
vertices (1, 0), (9, 0), (9, 4), (5, 8)
and (1, 8).
1
= 15 × 7 = 105 cm3
c
Area of cross-section = area of circle
= π × r 2 = π × 42
= π × 16 = 50.265… cm2
Area of N = 9 × 4 = 144 cm
2
Volume = area of cross-section × length
= 50.265… × 11 = 552.92 cm3
10 Perimeter of Z = 36 cm
3
a
150 cm3
11 a
Enlargement scale factor 3, centre (6, 2).
b
129.6 cm3
b
Enlargement scale factor 2, centre (3, 5).
c
427.5 cm3
4
Area of
Length
cross-section of prism
Volume of
prism
a
Enlargement scale factor 2, centre (2, 4).
8.4 cm2
20 cm
168 cm3
b
24 cm2
6.5 cm
156 cm3
b
C = πd = 4π = 12.566 cm (3 d.p.)
c
58 m2
5.7 m
330.6 m3
c
C = 2 × 12.56637… = 25.133 cm (3 d.p.)
d
56.85 mm2
62 mm
3524.7 mm3
d
C = πd = 8π = 25.133 cm (3 d.p.)
e
A = πr2 = π × 22 = 4π = 12.566 cm2 (3 d.p.)
12 Enlargement scale factor 3, centre (6, 1).
13 Enlargement scale factor 2, centre (3, 4).
14 a
41
1
Volume = area of cross-section × length
Perimeter of N = 12 × 4 = 48 cm
Area of Z = 72 cm2
Area of cross-section = area of triangle
= 2 × base × height = 2 × 6 × 5 = 15 cm2
(1, 1), (1, 5), (7, 5), (3, 1)
2
Area of cross-section = area of rectangle
= base × height = 8 × 4 = 32 cm2
aLearner’s own diagram. Enlarged shape
with vertices as given in part b.
b
Volume = area of cross-section × length
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
5
6
7
a
480 cm3
c
675 cm
b
480 cm3
b
3
Learner’s own answer and explanation. For
example: Timo has used the diameter instead
of the radius in the volume formula. Correct
answer is 226 cm3 (3 s.f.)
a
754.0 cm
c
42 411.5 mm
b
3
Base: 24 × 8 = 192
Total = 120 + 208 + 192 = 520 cm2
c
492.6 cm
3
3
1

Triangle ends: 2 ×  × 12 × 9 = 108
2

Sloping face: 15 × 12.5 = 187.5
Back face: 9 × 12.5 = 112.5
8
Radius
of circle
Area of
circle
a
7 cm
153.94 cm2
12 cm
1847.26 cm3
b
1.5 m
7.07 m2
2.4 m
16.96 m3
3
c
9 cm
254.47 cm2
7.51 cm
1910 cm3
= π × 32
d
2.19 m
15 m2
3.8 m
57 m3
e
4.55 mm
65 mm
22 mm
1430 mm
9
a
x = 4.3
b
x = 3.3
c
x = 2.1
10 a
729 cm3
b
13 851 g
c
$96 957
Height of Volume of
cylinder
cylinder
2
a
b
2
a
Total = 108 + 187.5 + 112.5 + 150 = 558 cm2
Area of circle = πr2
= 28.27 cm2 (2 d.p.)
Circumference of circle = πd
=π×6
= 18.85 cm (2 d.p.)
Area of rectangle = circumference
of circle × 8
Exercise 14.2
1
Base: 12 × 12.5 = 150
3
11 The smallest. Learner’s own explanation. For
example: In the smallest tin you get 233 cm3
per dollar, compared with 201 cm3 per dollar
for the medium tin and 225 cm3 per dollar for
the large tin.
= 18.85 × 8
= 150.80 cm2 (2 d.p.)
Total area = 2 × area of circle + area
of rectangle
= 2 × 28.27 + 150.80
= 207 cm2 (3 s.f.)
4
a
477.5 cm2
Square base: 10 × 10 = 100
b
401.1 cm2
1

Triangular faces: 4 ×  × 10 × 8 = 160
2

2
Total = 100 + 160 = 260 cm
c
8482.3 mm2
5
The tetrahedron has the greater surface area.
Triangular face: × 20 × 17 = 170
Tetrahedron: SA = 448 cm2,
Cylinder: SA = 427.25… cm2
Total = 4 × 170 = 680 cm2
448 cm2 > 427.25… cm2
1
2
Front and back: 2 × (4 × 12) = 96
6
151 cm2 (3 s.f.)
Sides: 2 × (4 × 10) = 80
7
a
3840 mm2
b
1548 cm2
Top and base: 2 × (12 × 10) = 240
Total = 96 + 80 + 240 = 416 cm2
42
1

Triangle ends: 2 ×  × 24 × 5 = 120
2

Sides: 2 × (13 × 8) = 208
8
344 cm2 (3 s.f.)
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
9
Carlos is incorrect. Learner’s own working
and explanations. For example:
5
b
The surface area of the polytunnel =
1
2
( (
)
)
aThere are two vertical and one horizontal
planes of symmetry.
× 2 × π × 4.52 + π × 9 × 27 = 445.32 ... m 2
Carlos will need more than 445 m2 of plastic
to make the polytunnel as the total surface
area is more than 445 m2 and he will need to
allow extra plastic for overlaps at the edges,
wastage, etc.
6
a
2D
regular
polygon
Number
of lines of
symmetry
3D
pyramid
Number of
planes of
symmetry
Learner’s own answers.
triangle
3
triangular
3
b
Prism B has a greater surface area than
prism A.
square
4
square
4
pentagon
5
pentagonal
5
c
Prism A:
hexagon
6
hexagonal
6
octagon
8
octagonal
8
10 a
1
1
2
2
SA = 2 × × π × 52 + × π × 10 × 15 + 10 × 15
= 464 cm2 (3 s.f.)
b
The number of lines of symmetry of a
regular 2D polygon is the same as the
number of planes of symmetry of its
matching 3D pyramid. Learner’s own
explanation. For example: This is because
all the lines of symmetry become vertical
planes of symmetry, but the triangular
sides of the pyramid meet at a point so
there are no horizontal lines of symmetry.
c
i
Prism B:
1
1
4
4
SA = 2 × × π × 82 + × π × 16 × 13 + 2 × 8 × 13
= 472 cm2 (3 s.f.)
Difference = 472 − 464 = 8 cm2
Exercise 14.3
1
a and d
2
a
A and F
c
C
3
Learner’s own diagram. Two vertical
and one horizontal planes of symmetry,
each plane splitting the shape into two
congruent shapes.
b
B and D
7
aLearner’s own diagram. One vertical plane
of symmetry, splitting the shape into two
congruent shapes.
b
ii
12
aLearner’s own diagram. Check that the
drawn part is a reflection of the shape
drawn in the question.
b
One
c
Learner’s own diagram. One plane of
symmetry splitting the shape into two
copies of the shape in the question. The
other plane is perpendicular to the first,
splitting the shape into two congruent
shapes.
Learner’s own diagram. One vertical plane
of symmetry, splitting the shape into two
congruent shapes.
4
8
10
aLearner’s own diagram. Check that the
drawn part is a reflection of the shape
drawn in the question.
b
Cube
c
9
Two like this
Two like this
43
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 15.1
1
2
a
44
Frequency Midpoint
1
6
8
11
14
5
a
Time, t
(seconds)
Tally
Frequency
Midpoint
5
15
25
0 < t ⩽ 10
ll
2
5
10 < t ⩽ 20
llll
5
15
35
45
55
20 < t ⩽ 30
llll lll
8
25
30 < t ⩽ 40
llll
4
35
40 < t ⩽ 50
l
1
45
b
Learner’s own frequency polygon. Same
axes and labels as in the diagram in the
question. Points (5, 1), (15, 6), (25, 8),
(35, 11), (45, 14) and (55, 5) plotted and
joined with straight lines.
a
Height, h (cm) Frequency Midpoint
b
3
Time, t
(seconds)
0 < t ⩽ 10
10 < t ⩽ 20
20 < t ⩽ 30
30 < t ⩽ 40
40 < t ⩽ 50
50 < t ⩽ 60
4
260 ⩽ h < 280
3
270
280 ⩽ h < 300
7
290
300 ⩽ h < 320
9
310
320 ⩽ h < 340
1
330
Learner’s own frequency polygon. Same
axes and labels as in the diagram in the
question. Points (270, 3), (290, 7), (310, 9)
and (330, 1) plotted and joined with
straight lines.
a
32
b
Height, t (cm) Frequency Midpoint
10 ⩽ t < 12
4
11
12 ⩽ t < 14
16
13
14 ⩽ t < 16
7
15
16 ⩽ t < 18
5
17
c
Learner’s own frequency polygon. Make
sure that they use suitable axes labels and
scales. Points (11, 4), (13, 16), (15, 7) and
(17, 5) plotted and joined with straight
lines.
d
20 5
=
32 8
e
Learner’s own answer and explanation.
For example: Zara is incorrect. You
don’t know from the frequency polygon
what the fastest time is. All you can say
is that the fastest time is between 10 and
12 minutes.
5
b
Learner’s own frequency polygon. Make
sure that they use suitable axes labels
and scales. Points (5, 2), (15, 5), (25, 8),
(35, 4) and (45, 1) plotted and joined with
straight lines.
a
50
b
Wednesday
Height, h (cm)
Frequency
Midpoint
120 ⩽ h < 140
4
130
140 ⩽ h < 160
6
150
160 ⩽ h < 180
22
170
180 ⩽ h < 200
18
190
Saturday
Height, h (cm)
Frequency
Midpoint
120 ⩽ h < 140
25
130
140 ⩽ h < 160
16
150
160 ⩽ h < 180
7
170
180 ⩽ h < 200
2
190
c
Learner’s own diagram showing two
frequency polygons on one set of axes.
Make sure that they use suitable axes labels
and scales. Make sure that each polygon is
labelled clearly. Wednesday points (130, 4),
(150, 6), (170, 22) and (190, 18) plotted and
drawn with straight lines. Saturday points
(130, 25), (150, 16), (170, 7) and (190, 2)
plotted and joined with straight lines.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
d
6
aLearner’s own diagram showing two
frequency polygons on one set of axes.
Make sure that they use suitable axes
labels and scales. Make sure that each
polygon is labelled clearly. Falcons
Club points plotted at (2.5, 4), (7.5, 24),
(12.5, 18), (17.5, 12) and (22.5, 10)
and joined with straight lines. Harriers
Club points plotted at (2.5, 10), (7.5, 8),
(12.5, 10), (17.5, 26) and (22.5, 16) and
joined with straight lines.
b
7
d
i
12
iiNo, Sienna cannot fill in the correct
frequencies in her table. Learner’s own
explanation. For example: From the
first table Sienna knows that there are
two girls with a mass between 7.0 and
7.1 kg. However, this does not tell her
how many girls had masses between
7.0 and 7.05 kg and how many girls
had masses between 7.05 and 7.1 kg,
so it is impossible for her to complete
her table. She would have to find the
original data, before it was grouped, in
order to group it the way she wants to.
8
Learner’s own answers. For example:
Learner’s own comments. For example:
The most popular training time for the
Falcons Club was between 5 and 10 hours,
whereas for the Harriers Club it was
between 15 and 20 hours. In the Falcons
Club only 22 athletes trained for more
than 15 hours a week compared with 42
athletes from the Harriers Club.
Time to
solve maths
problem, t
(seconds)
Tally
20 ⩽ t < 30
llll llll
10
25
c
Falcons Club 68, Harriers Club 70.
30 ⩽ t < 40
llll llll llll llll llll
25
35
d
Learner’s own answer and explanation.
For example: Yes, because the number of
athletes surveyed at each club was nearly
the same.
40 ⩽ t < 50
llll llll llll lll
18
45
50 ⩽ t < 60
llll ll
7
55
a
Frequency Midpoint
aLearner’s own frequency polygon. Make
sure that they use suitable axes labels
and scales. Points (7.05, 2), (7.15, 12),
(7.25, 14), (7.35, 9), (7.45, 7) and (7.55, 6)
plotted and joined with straight lines.
b
Learner’s own frequency polygon. Make
sure that they use suitable axes labels and
scales. Points (25, 10), (35, 25), (45, 18)
and (55, 7) plotted and joined with
straight lines.
b
c
Learner’s own comments. For example:
Most students took less than 40 seconds
to solve the puzzle.
i
Mass, m (kg)
Frequency
7.0 ⩽ m < 7.2
14
7.2 ⩽ m < 7.4
23
7.4 ⩽ m < 7.6
13
iiLearner’s own frequency polygon.
Make sure that they use suitable axes
labels and scales. Points (7.1, 14),
(7.3, 23) and (7.5, 13) plotted and
joined with straight lines.
c
45
are more groups so it shows you more
information on the mass of the girls. The
second frequency polygon only has three
groups so less information can be taken
from the graph.
Learner’s own answer and explanation.
For example: On Saturday there were
fewer taller people and more shorter
people. There were only two people with
a height between 180 cm and 200 cm on
Saturday compared with 18 on Wednesday.
There were 25 people with a height
between 120 cm and 140 cm on Saturday
compared with four on Wednesday.
Learner’s own answers and explanations.
For example: The first frequency polygon
gives you better information because there
9
Mass, m (kg)
Frequency
Midpoint
3.6 ⩽ m < 3.8
8
3.7
3.8 ⩽ m < 4.0
12
3.9
4.0 ⩽ m < 4.2
9
4.1
4.2 ⩽ m < 4.4
11
4.3
4.4 ⩽ m < 4.6
5
4.5
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
Exercise 15.2
1
aLearner’s own scatter graph with axes
labelled as in the graph in the question.
Points (3, 5), (11, 10), (18, 18), (19, 20),
(5, 6), (20, 18), (14, 16), (8, 9), (9, 11),
(7, 6), (5, 5), (16, 15), (10, 11), (9, 7) and
(16, 16) marked with crosses.
b
2
4
46
b
5
A. Learner’s own explanation. For
example: The scatter graph is showing a
positive correlation. This happens when
as one value increases, the other value also
increases. In this case, as the French results
increase, the Spanish results also increase.
aLearner’s own scatter graph with axes
labelled as in the graph in the question.
Points (3, 19), (10, 11), (15, 7), (8, 11),
(10, 10), (13, 9), (4, 17), (16, 5), (12, 10),
(8, 14), (17, 2), (11, 9), (5, 15), (20, 4) and
(7, 13) marked with crosses.
b
3
(8, 28), (25, 27), (16, 25), (14, 18), (9, 17)
and (28, 25) plotted.
B. Learner’s own explanation. For
example: The scatter graph is showing a
negative correlation. This happens when
as one value increases, the other value
decreases. In this case, as the art results
increase, the science results decrease.
aLearner’s own scatter graph. Horizontal
axis labelled ‘Hours reading’. Vertical axis
labelled ‘Spelling test score’. Both axes
shown from 0 to 25. Points (4, 6), (13, 12),
(20, 20), (9, 8), (18, 17), (1, 2), (11, 13),
(8, 10), (18, 19), (2, 3), (15, 16), (10, 12),
(4, 5), (14, 12) and (7, 7) plotted.
b
Positive correlation. The more hours
reading a student does, the better their
spelling test score.
c
Learner’s own line of best fit. Strong
positive correlation.
d
Learner’s own estimate from their line of
best fit.
e
No. 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.
aLearner’s own scatter graph. Horizontal
axis labelled ‘Number of packets of
cookies sold’ and shown from 0 to 30.
Vertical axis labelled ‘Number of packets
of oranges sold’ and shown from 0 to 30.
Points (15, 12), (12, 22), (26, 14), (22, 7),
No correlation. The number of packets
of cookies sold has no relationship to the
number of packets of oranges sold.
aLearner’s own answers. For example:
Negative correlation because learners
are often good at maths and science or
languages and drama, but are not good at
all these subjects.
b
Learner’s own scatter graph. Horizontal
axis labelled ‘Maths result’ and shown
from 0 to 100. Vertical axis labelled
‘Drama result’ and shown from 0 to 100.
Points (72, 27), (34, 62), (81, 19), (57, 41),
(32, 66), (78, 25), (65, 37), (67, 32),
(53, 59), (61, 48), (35, 63), (42, 59),
(55, 40), (79, 35) and (31, 77) plotted.
c
Strong negative correlation. The better the
students’ result in maths, the worse their
drama result.
d
Learner’s own answer.
e
Learner’s own line of best fit.
f
Learner’s own estimate from their line of
best fit.
g
No, because 10% for drama lies outside
the range of the data we are given, so we
cannot predict what will happen.
6
Learner’s own explanation. For example: It is
a coincidence that the graph shows a negative
correlation. While it might be true that if you
have no hair or short hair you need a hat to
keep your head warm or protect it from the
sun, it does not mean that you are going to
buy lots of hats. In this study, the people with
longer hair might have big families, and so they
bought lots of hats for their family members.
The number of hats you need does not depend
on the length of your hair. It depends on
whether you like to wear hats or not.
7
aPositive correlation. The better the score in
algebra, the better the score in geometry.
b
4
c
e
Learner’s own diagram. Scatter graph from
the question with the point (10, 13) plotted.
f
Learner’s own line of best fit. Make sure
the line of best fit passes through the
point (10, 13).
10
d
13
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
g
iLearner’s own estimate using their
line of best fit. For example: 8.
4
a
June
iiLearner’s own estimate using their
line of best fit. For example: 11.
8
aNegative correlation. The further the
house is from the railway station, the
lower its value is.
b
c
a
9 7
3
6 7 8
6 6 6 5 3 3 2 1 0
4
0 3 7
5
0 2 4 5 6 8 8
6
2
8 0
Key: For June, 0 2 means 20 customers
For August, 3 6 means 36 customers
b
i Mode ii Median iii Range iv Mean
June
46
43
48
44
August
58
51
26
49
Key: 0 5 means 5
c
Learner’s own answers. For example: In
August the mode, median and mean are
all greater than in June, showing that
on average there are more customers.
The range, however, is smaller in August
than in June, showing that there is more
variation in the numbers of customers
riding in June.
d
Learner’s own answers. For example: Yes,
because the mode, median and mean are
all greater in August than in June.
a
i–iv
5 7 9 9
5 4 6 0 8 7 9
6 2 5 1 2 7 3 0 4
Key: 0 5 means 5
0
1
2
2
2
y = 6.4 km
0
1
2
b
0
The house that doesn’t fit the trend
is worth $146 000 and is 6 km from
the railway station. Learner’s own
explanation. For example: The house
might not be in a very good state of
repair, which is why it isn’t worth as much
as it should be.
Exercise 15.3
1
5 7 9 9
0 4 5 6 7 8 9
0 1 2 2 3 4 5 6 7
Unordered:
Key: 0 6 means 6
0 6 8 4 6
1 8 2 1 4 7 3 9 0 2
2 1 8 1 0 8 1 2
Ordered:
Key: 0 4 means 4
0 4 6 6 8
1 0 1 2 2 3 4 7 8 9
2 0 1 1 1 2 8 8
5
i Mode ii Median iii Range iv Mean
Girls’
times
27.3
26.05
2.6
26.1
Boys’
times
26.5
27.4
3.6
27.3
b
Learner’s own answer. For example: The
range is larger for the boys, showing that
their times are more varied. The girls have
a lower median and mean which shows
that using these averages they were faster
at solving the puzzle.
c
iThe mode, as the boys’ mode is lower
than the girls’, which makes them
appear faster.
3
Spanish test results for class 9R
9 9 7 5
9 8 7 6 5 4 0
7 6 5 4 3 2 2 1 0
Key for class 9R: 5 0 means 5
47
August
Spanish test results for class 9T
0
1
2
4 6 6 8
0 1 2 2 3 4 7 8 9
0 1 1 1 2 8 8
Key for class 9T: 0 4 means 4
iiThe median or the mean, as the girls’
median and mean are lower than the
boys’, so the girls were faster.
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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
d
6
Location A
e
7
On average, using the median and mean,
class 9R were taller than class 9T. Class 9T
had more variation in heights, and their
modal height was taller than class 9R.
Learner’s own answer. For example: The
girls, as their median and mean are lower,
therefore they were faster than the boys.
Location B
a
1
4
2
3
b
33 %
1
3
0%
c
Range = 305 g
(most variation)
d
Mean = 792.5 g,
Median = 790 g
Exercise 15.4
1
Range = 295 g
i
b
c
8 minutes
1
3
Median = 652.5 g
2
Key: For the top shelf, 4 10 means 104 boxes of cereal
112
123
26
120.5
Middle shelf
139
137
32
134.5
Learner’s own answers. For example: The
sales of cereal were better on the middle
shelf as on average more boxes were sold
(the mean, median and mode were all
greater on the middle shelf than the top
shelf). The sales on the middle shelf were
more varied, but included the largest
number of boxes sold on one day. The
smallest number of boxes sold on one day
were on the top shelf.
8
aMissing numbers from the diagram: top
row 0, second top row 5, third top row 2
and 8, bottom row 5. Missing numbers
from the table: Class 9T row 144 and 145,
Class 9R row 149.
b
48
13
12
13 × 12 = 156
15
10
15 × 10 = 150
17
2
17 × 2 = 34
Totals:
31
417
417
3
= 13 minutes
Learner’s own explanation. For example:
The answers for the range and mean are
only estimates because the data is grouped
and you do not have the individual values
of the data.
a
i
290 ⩽ h < 310
ii
290 ⩽ h < 310
b
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 do not know the
individual values.
c
i
a
150 men, 140 women
b
Top shelf
11 × 7 = 77
d
For the middle shelf, 11 5 means 115 boxes of cereal
Mode Median Range Mean
Midpoint ×
frequency
7
31
Middle shelf
5
0 7 9
0 2 6 8 9 9
2 4 5 7
12 ⩽ t < 14
11
Estimate of mean =
a
10
11
12
13
14
ii
Midpoint Frequency
Mean = 658 g,
Learner’s own answers. For example:
Location A because the mean and
median mass of potatoes was greater than
location B. The range was very similar
showing that the variation in the mass of
potatoes was similar at both locations.
Top shelf
9 4
9 2 2
9 8 7 6 5 4 2 0
0
12 ⩽ t < 14
a
ii
80 cm
290 cm
b
Modal
class
interval
Class interval
Estimate
where the
of mean
median lies
20 ⩽ a < 30
40 ⩽ a < 50
39.6
Women 50 ⩽ a < 60
40 ⩽ a < 50
42.5
Men
c
Learner’s own answer. For example: The
mean age of the women is approximately
three years more than the men. The
median age lies in the same interval for
both men and women. The modal class
interval for the men is a lot younger than
for the women.
Learner’s own comments. For example:
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021
CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE
d
4
a
b
Learner’s own answer. For example:
On average the men are the younger
competitors because their mean and
modal group are both younger than the
women.
c
mean = 15.6, median = 15.5 and mode = 8
Table A
Number
of texts
Tally
Frequency
5–11
llll l
6
12–18
llll ll
7
19–25
llll
5
26–32
ll
2
Table B
Number
of texts
Tally
Frequency
5–13
llll lll
8
14–22
llll llll
9
23–31
lll
3
Class interval
Estimate
where the
of mean
median lies
Table A
12–18
12–18
16.05
Table B
14–22
14–22
15.75
d
iThe estimate of the mean from table
B is closer to the accurate mean.
iiThe accurate median is within both
the class intervals for the medians.
However, the midpoint of 12–18 is 15
and 14–22 is 18, and so the accurate
median of 15.5 is closer to the
midpoint of the median class for table
A than table B.
iiiThe accurate mode is 8. This isn’t in
either of the modal class intervals.
5
a
b
49
i
40 < a ⩽ 50
ii
30 < a ⩽ 40
i
35.25 years
ii
80 years
Frequency
0 < a ⩽ 20
13
20 < a ⩽ 40
8
40 < a ⩽ 60
12
60 < a ⩽ 80
7
d
i
0 < a ⩽ 20
ii
20 < a ⩽ 40
e
i
36.5 years
ii
80 years
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.
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.
c
Modal
class
interval
Age, a (years)
6
Karim is correct.
Estimate of the mean =
5 × 4 + 15 × 14 + 25 × 12 + 35 × 20 + 45 × 10
= 28 emails.
60
7
Mean mass of chicks after hatching =
30 × 4 + 34 × 8 + 38 × 6 + 42 × 2
= 35.2 g.
20
Mean mass of chicks at eight weeks old =
0.8 × 3 + 1 × 6 + 1.2 × 7 + 1.4 × 4
= 1.12 kg.
20
1120 ÷ 35.2 = 31.8 > 30
So, the mean mass of the chicks at eight weeks
old is more than 30 times the mean mass of
the chicks after hatching.
Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021