Uploaded by nott.srw

pdfcoffee.com answers-to-coursebook-mathematicspdf-pdf-free

advertisement
Answers to Coursebook exercises
1 Integers, powers and roots
F Exercise 1.1
1
a −3
Arithmetic with integers
b −11
c −6
d −17
e 8
2 a 10
b −180
c −15
d −100
e 5
3 a −2
b −10
c 2
d −12
e −12
4 a 4 + 6 = 10
5 a 9
b −4 + 6 = 2
b −2
6 a
c 16
3
–2
–3
–2
5
2
–10
–6
–4
–7
1
5
–1
–4
7
–6
–8
2
7
Second
−
−4
−2
0
2
4
4
8
6
4
2
0
2
6
4
2
0
−2
0
4
2
0
−2
−4
−2
2
0
−2
−4
−6
−4
0
−2
−4
−6
−8
8 a −20
b −48
c 20
d 60
e −40
9 a −2
b −5
c 3
d 10
e −4
10 a −40
b −4
c −100
d 5
e 48
First
11 a −15 ÷ 5 = −3 and −15 ÷ −3 = 5
12
e 12 + 10 = 22
–12
2
–3
e
3
2
c
–5
1
d −4 + 6 = 2
e 8
–3
–4
–5
d
d 0
b
–6
–2
c 8 + 2 = 10
b 32 ÷ −8 = −4 and 32 ÷ −4 = −8
×
−3
−2
−1
0
1
2
3
3
−9
−6
−3
0
3
6
9
2
−6
−4
−2
0
2
4
6
1
−3
−2
−1
0
1
2
3
0
0
0
0
0
0
0
0
−1
3
2
1
0
−1
−2
−3
−2
6
4
2
0
−2
−4
−6
−3
9
6
3
0
−3
−6
−9
Copyright Cambridge University Press 2013
c −42 ÷ −6 = 7 and −42 ÷ 7 = −6
Cambridge Checkpoint Mathematics 8
1
Unit 1
Answers to Coursebook exercises
13 a
b
100
–36
–6
2
–20
6
–3
–2
–4
c
–5
–1
5
d
64
48
–12
–3
–4
–4
4
–1
–2
–16
2
–8
14 a, b There are six different pairs: 1 and −12; −1 and 12; 2 and −6; −2 and 6; 3 and −4; −3 and 4.
15 a −15
b 2
c 1
d 6
e 16
f
−14
16 a −5
b 12
c −7
d −4
e 4
f
1
F Exercise 1.2
1
Multiples, factors and primes
a 1, 2, 4, 5, 10, 20
e 1, 2, 4, 5, 10, 20, 25, 50, 100
2 a 8, 16, 24, 32
e 33, 66, 99, 132
3 a 24
b 36
b 1, 3, 9, 27
f 1, 2, 7, 14, 49, 98
b 15, 30, 45, 60
f 100, 200, 300, 400
c 28
c 1, 3, 5, 15, 25, 75
c 7, 14, 21, 28
d 60
e 32
f
d 1, 23
d 20, 40, 60, 80
77
4 8
5 a 1, 2, 3, 4, 6, 8, 12, 24
b 1, 2, 4, 8, 16, 32
c 1, 2, 4, 8
d 8
6 a 1, 5
b 1, 2, 3, 6
c 1, 7
d 1, 2, 4, 8
e 1
f
1
7 a 2
b 6
c 10
d 20
e 1
f
15
8 24 and 56
9 37
10 61 and 67
11 Alicia is correct. 91 = 7 × 13
12 1
13 Because 7 will be a factor.
14 a 2, 3
b 3, 5
c 3, 7
15 a Any three from 2, 4, 8, 16, 32, …, …
c Any three from 5, 25, 125, 625, …, …
d 7
e 2, 3, 5
f
7, 11
b Any three from 3, 9, 27, 81, …, …
16 The first one is 16. The next is 25. Any square number has an odd number of factors.
17 The smallest is 30 (2 × 3 × 5). You could also have 42 (2 × 3 × 7), 66 (2 × 3 × 11), etc.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
F Exercise 1.3
1
Unit 1
More about prime numbers
Different trees are possible.
2 a Many different trees are possible. They should end with the same primes as the trees in question 1.
ii 22 × 52
iii 22 × 33
b i 24 × 3
3 20
24
42
50
180
•
•
•
•
•
•
•
•
•
•
22 × 5
2×3×7
22 × 32 × 5
2 × 52
23 × 3
4 a 60
b 54
c 363
d 392
e 144
f
325
5 a 23 × 3
b 2 × 52
c 23 × 32
d 23 × 52
e 3 × 5 × 11
f
23 × 17
6 a i 32 × 5
ii 3 × 52
b 225
c 15
7 a i 2 × 32 × 5
ii 22 × 5 × 7
b 1260
c 10
8 a 1
b 1739
F Exercise 1.4
1
a 9
2 a 100
Powers and roots
b 27
c 81
b 1000
c 10 000
d 243
3 1 000 000 and 1 000 000 000
4 a 35
b 26
5 a 3
b 4
c 45
6 Possible values are 2 and 4.
7 a 3 and −3
b 6 and −6
c 9 and −9
d 14 and −14
e 15 and −15
f
20 and −20
8 256, 289 or 324
9 343
10 a 10
b 20
c 3
d 5
e 10
11 The smallest possible value is 64. Other possible values are 729 and 4096.
12 a 2048
b 4096
c 512
13 a i
ii 3
b 6
9
Copyright Cambridge University Press 2013
c 10
d 15 (Compare the sequence of triangular numbers.)
Cambridge Checkpoint Mathematics 8
3
Unit 1
Answers to Coursebook exercises
End-of-unit review
1
a 2
b −8
c −15
d −10
e −14
2 a 7
b 1
c 17
d 7
e 0
3 a 27
b −2
c −80
d 6
e −2
4
×
−2
3
5
−4
8
−12
−20
−3
6
−9
−15
6
−12
18
30
5 −8 and 32
6 a 1, 2, 3, 6, 7, 14, 21, 42
e 1, 2, 4, 8, 16, 32, 64
b 1, 2, 4, 13, 26, 52
f 1, 3, 23, 69
c 1, 5, 11, 55
d 1, 29
7 a, b, c There are three pairs: 3 and 37; 11 and 29; 17 and 23.
8 a 2 × 32
b 25 × 3
c 23 × 52
d 24 × 3 × 5
9 a 40
b 5
c 288
d 1200
10 a 5 and −5
b 9 and −9
c 13 and −13
d 16 and −16
11 a 8
b 4
12 a 1024
b 2048
e 33 × 5
f
52 × 7
c 4096
13 a Shen worked out 3 × 5 and 5 × 3; both equal 15.
b 35 = 243 and 53 = 125
14 18
4
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
2 Sequences, expressions and formulae
F Exercise 2.1
1
a 1, 6, 11
d 6, 1, −4
Generating sequences
b 20, 16, 12
e −5, −3, −1
c 2, 14, 26
f −3, −9, −15
2 43. Check students’ explanations: e.g. start with 15 and add 7 four times (or 7 × 4).
3 a Yes.
b Check students’ explanations: e.g. 9 more terms with differences of 12 so 9 × 12, then add first term of 3.
c i 77
ii 157
iii 397
4 18. Check students’ explanations: e.g. subtract 7 three times.
5 43. Check students’ explanations: e.g. add 3 nine times.
6
Position number
1
2
3
4
5
10
20
Term
8
9
10
11
12
17
27
7 a 6, 12, 18, 24
8 a
b
c
d
i
i
i
i
15
20
48
25
b −3, −2, −1, 0
ii
ii
ii
ii
iii
iii
iii
iii
25
40
88
75
c 3, 5, 7, 9
d 2, 5, 8, 11
105
200
408
475
9 C. Terms increase by 3 each time; C is the only rule that allows this.
10 No. He has used the term, not the position, to find the last two answers.
F Exercise 2.2
1
Finding rules for sequences
a term = position number + 5
b term = 3 × position number − 2
2 a
b
c
d
e
f
i
i
i
i
i
i
‘add 2’
‘add 5’
‘add 3’
‘add 2’
‘add 4’
‘add 5’
iii
iii
iii
iii
iii
iii
2 × position number
5 × position number
3 × position number + 2
2 × position number + 4
4 × position number + 3
5 × position number + 2
3 a
b
c
d
e
f
i
i
i
i
i
i
‘add 1’
‘add 1’
‘add 1’
‘add 2’
‘add 4’
‘add 5’
iii
iii
iii
iii
iii
iii
term = position number + 3
term = position number + 9
term = position number + 23
term = 2 × position number − 1
term = 4 × position number − 2
term = 5 × position number − 3
4 a 4, 7, 10, 13
b ‘add 3’
c 3 extra blue squares are added to make the next pattern.
d term = 3 × position number + 1
5 a The term-to-term rule is ‘add 2’, so the position-to-term rule will start: term = 2 × position number.
b term = 2 × position number + 2
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 2
Answers to Coursebook exercises
F Exercise 2.3
1
a 7, 8, 9; 16
e 7, 9, 11; 25
2 a
c
d
e
Using the nth term
b −2, −1, 0; 7
f 2, 5, 8; 29
c 4, 8, 12; 40
g 8, 13, 18; 53
d 6, 12, 18; 60
h 1, 5, 9; 37
4, 7, 10, 13
b ‘add 3’
Three extra pink squares are added to make the next term.
term = 3 × position number + 1
second term = 3 × 2 + 1 = 7; third term = 3 × 3 + 1 = 10; fourth term = 3 × 4 + 1 = 13
3 Yes. Check students’ reasoning.
F Exercise 2.4
1
a i
b i
Using functions and mappings
x
1
2
3
4
y
4
5
6
7
ii
ii
x 0 1 2 3 4 5 6 7 8 9 10
iii
1
2
4
6
y
5
7
11
15
x
4
8
10
20
y
7
9
10
15
b i y = 2x + 3
3 a i ‘add 8’
b i y=x+8
6
7
y
1
2
3
4
y=x−3
ii
x
5
y 0 1 2 3 4 5 6 7 8 9 10
y=x+3
2 a i
4
x 0 1 2 3 4 5 6 7 8 9 10
y 0 1 2 3 4 5 6 7 8 9 10
c i
x
ii
iv
x
3
5
9
12
y
8
14
26
35
x
2
4
8
14
y
−2
−1
1
4
iii y = x2 + 5
ii y = 3x − 1
ii ‘multiply by 5’
ii y = 5x
y = x2 − 3
iv
4 Razi. Check students’ explanations: e.g. all of Razi’s work, but only one of Mia’s works.
5 y = 3x + 2 Check students’ explanations.
y
x
5
1
+2
8
2
×3
11
3
F Exercise 2.5
Constructing linear expressions
b x+8
c
2 a 6n + 1
b n +5
4
c 2n − 3
3 a $(c + 3s)
b $(3c + 4g + 6s)
1
a x−7
x
2
d 2x + 1
d
n +7
10
4 C. Check students’ explanations: e.g. to multiply n − 3 by 2 the n − 3 must be in brackets.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
F Exercise 2.6
1
a 2
g −21
2 a 21
g 1
Unit 2
Deriving and using formulae
b −2
h 4
c −18
i 23
d −5
j −7
e 3
k −3
f
l
−7
2
b −15
h 54
c 45
i 3
d −15
j −44
e 16
k 8
f
l
51
200
b 1
c 29
e 160 cm
f
120 cm
3 a −3 × −3 = +9, not −9
4 a She should have worked out the value of the brackets first.
b −40
c −54
5 a i months = years × 12
ii m = 12y
6 a 125
b 158
c 200
7 a 12
b 54
c −32
8 a 145 cm
b 157.5 cm
c 132.5 cm
b 96
d 175 cm
9 Prism B, by 18 cm3.
10 a i −5.8 °C
ii 9.2 °C
b i 54 = 5F − 160
iii 31.4 °C
ii 162 = 5F − 160
iii
270 = 5F − 160
End-of-unit review
1
a 7, 10, 13
b 11, 6, 1
c 8, 16, 24
d 1, 5, 9
2 B. Rules B, C and D give the correct 3rd term, but only B gives the correct 8th term.
3 a i ‘add 6’
ii
Position number
1
2
3
4
Term
6
12
18
24
iii term = 6 × position number
b i ‘add 5’
ii
iv Look for evidence of students’ checks.
Position number
1
2
3
4
Term
6
11
16
21
iii term = 5 × position number + 1
c i ‘add 1’
ii
iv Look for evidence of students’ checks.
Position number
1
2
3
4
Term
8
9
10
11
iii term = position number + 7
iv Look for evidence of students’ checks.
4 Y
es. Check students’ explanations: e.g. term-to-term rule is ‘add 3’, so rule starts 3n. 3 × 1 + 3 = 6,
3 × 2 + 3 = 9, 3 × 3 + 3 = 12 and 3 × 4 + 3 = 15
5 a i
x
1
2
5
8
y
10
11
14
17
ii
b i y=x+9
ii
x − 10
3
6 a 4x
b 2x + 7
c
7 a −5
b −22
c −17
x
1
2
5
11
y
–1
1
7
19
y = 2x − 3
d 5(x + 4)
d 40
e −1
f
32
8 150
9 No. 42 = 32 + 2 × 7 × s simplifies to 16 = 9 + 14s.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Coursebook exercises
3 Place value, ordering and rounding
F Exercise 3.1
1
a i
c i
Multiplying and dividing by 0.1 and 0.01
ii
ii
1000
10 000 000
b
d
one thousand
ten million
i
i
ii one hundred thousand
ii ten
100 000
10
2 a 102
b 107
c 104
d 1010
3 a 6.2
e 0.37
b 5
f 6
c 12.5
g 7.5
d 0.32
h 0.04
4 a 70
e 200
b 45
f 850
c 522
g 32
d 6.7
h 722.5
5 a 1.8
b 0.236
c 6
d 450
6 a ÷
b ×
c ×
d ×
e ÷
f
÷
7 a 0.01
b 0.1
c 0.01
d 0.1
e 0.1
f
0.1
8 B.
9 125
10 a Multiply by any negative number.
F Exercise 3.2
1
Ordering decimals
a 2.06, 5.49, 5.91, 7.99
d 8.9, 9.09, 9.4, 9.53
g 6.17, 6.178, 6.71, 6.725
2 a
c
e
g
b Use any number less than 1.0.
b 2.55, 2.87, 3.09, 3.11
e 23.592, 23.659, 23.661, 23.665
h 11.02, 11.032, 11.1, 11.302
c 11.82, 11.88, 12.01, 12.1
f 0.009, 0.084, 0.102, 0.107
780 g, 1950 g, 2.18 kg, 2.3 kg
b 0.8 cm, 9 mm, 12 mm, 5.4 cm
0.5 m, 53 cm, 650 cm, 12 m
d 95 ml, 450 ml, 0.55 l, 0.9 l,
780 m, 1450 m, 6.4 km, 6.55 km
f 50 kg, 0.08 t, 0.15 t, 920 kg
0.009 km, 9800 mm, 0.85 km, 920 m, 95 000 cm
3 a <
g >
b >
h <
c >
i <
d >
j <
e >
k >
4 a ≠
f ≠
b ≠
g ≠
c =
h =
d ≠
i =
e =
f
l
<
<
5 a 25 km, much further than other distances
b Yes, 0.2 km × 8 = 1.6 km and her furthest is more (1.64 km)
c Shen: all his lengths are multiples of 25 m; some of Mia’s are not.
6 a A: 2.5, B: 2.4, C: 2.3, D: 2.1, E: 2.25, F: 2.45
b 2.1, 2.25, 2.3, 2.4, 2.45, 2.5
F Exercise 3.3
1
a 40
g 30 000
2 a 75
g 9.45
Rounding
b 160
h 130 000
c 200
i 500 000
d 500
j 1 400 000
e 4000
k 8 000 000
f
l
13 000
25 000 000
b 10
h 12.92
c 20
i 0.08
d 11.5
j 146.80
e 0.9
f
125.9
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 3
Answers to Coursebook exercises
F Exercise 3.4
1
a 14.59
e 28.72
i 8.28
Adding and subtracting decimals
b 36.81
f 26.27
j 72.715
c 13.21
g 23.62
k 10.428
d 29.28
h 133.17
l 20.176
2 a 2.21
e 35.87
i 71.23
b 14.43
f 30.78
j 7.44
c 11.29
g 56.84
k 26.13
d 12.73
h 38.07
l 1.062
3 a 20.35
e 15.24
b 44.24
f 37.34
c 73.55
g 48.94
d 222.51
h 216.82
4 66.84 m
5 Yes, 2.69 m > 2.67 m
F Exercise 3.5
1
a 29.7
e 125.6
2 a 1.88
e 1.27
Dividing decimals
b 13.1
f 197.3
c 9.3
g 16.1
d 8.1
h 91.7
b 1.82
f 1.43
c 0.25
g 0.27
d 0.14
h 0.23
3 6.24 g
F Exercise 3.6
1
a 0.496
f 0.203
Multiplying by decimals
b 0.528
g 1.168
c 2.088
h 1.359
d 4.635
i 3.04
e 0.2508
j 10.74
2 a Multiplying by 0.06 is the same as multiplying by 6 then dividing by 100.
b i 0.854
ii 2.142
iii 0.696
iv 0.536
3 a 86.4
b 8.64
c 0.864
d 0.00864
4 0.6 × 6839.5 kg = 4103.7 kg = 4.1037 t = 4.1 t to one decimal place.
F Exercise 3.7
1
Dividing by decimals
a 160
f 500
k 1350
b 150
g 800
l 435
c 25
h 700
m 870
d 78
i 700
n 42
2 a 108.3
b 8.7
c 207.1
d 92.14
3 a 0.6
b 60
c 6
d 600
e 765
j 850
o 2240
e 13.17
4 39.74
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
F Exercise 3.8
1
Unit 3
Estimating and approximating
$45
2 a $115
b 4 hours 15 minutes
3 $72
4 $325
End-of-unit review
1
a 10 000
b ten thousand
2 10
8
3 a 4.1
b 0.23
c 72
d 24
4 a 10.09, 10.8, 10.9, 10.98
b 0.7 m, 77 cm, 7 m, 750 cm
5 a >
b <
c >
6 a ≠
b =
c ≠
7 a 6700
b 240 000
c 8 000 000
d 64
8 a 57.02 m
b 2.44 m
9 a 13.7
b 92.7
10 a 1.41
b 0.97
11 a 0.624
b 1.41
c 28.8
d 7.12
12 a 420
b 7
c 900
d 70
e 12.6
f 7.57
13 35.52
14 i $796
ii 18 × $15 + 12 × $28 + 5 × $38 = $270 + $336 + $190 = $796
iv 20 × $15 + 10 × £30 + 5 × $40 = $300 + $300 + $200 = $800
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Coursebook exercises
4 Length, mass and capacity
✦ Exercise 4.1
1
Choosing suitable units
a m
b mm
c g
d kg
e l
f
ml
2 a m2
b km2
c cm2
d m3
e km3
f
mm3
3 a T
b T
c F
d T
4 Possible if she has a huge house, but probably not sensible.
5 Yes, any sensible reason, e.g. a standard egg weighs about 60 g, so a large egg may weigh 75 g; two eggs
weigh about the same as an apple which could be 150 g.
6 No, he would not drive at 200 km/h.
7 9 kg
8 16 l
9 1 to 2 kg
10 Yes, 500 kg ÷ 8 = 62.5 kg and most adults would weigh more than 62.5 kg.
11 9 × length of car (3 m to 5 m) = 27 m to 45 m
12 1.7 m × 8 = 13.6 m or 1.8 m × 8 = 14.4 m
✦ Exercise 4.2
1
a T
Kilometres and miles
b F
c F
d T
e F
2 Yes, a kilometre is shorter than a mile.
3 a 40 miles
b 25 miles
c 35 miles
4 a 15 miles
b 30 miles
c 60 miles
5 a 88 km
b 32 km
c 136 km
6 a 16 km
b 160 km
c 200 km
d 110 miles
d 288 km
7 70 miles; 104 km = 65 miles or 70 miles = 112 km
8 152 km; 152 km = 95 miles or 90 miles = 144 km
9 a 75
b 168
10 a 1392 km
b $278
c 184 km = 115 miles
Copyright Cambridge University Press 2013
d 140 miles = 224 km
Cambridge Checkpoint Mathematics 8
1
Unit 4
Answers to Coursebook exercises
End-of-unit review
1
a m
b mm
c kg
d g
2 a m2
b mm2
c cm3
d m3
e ml
f
l
3 Possible if she has a very small house, but probably not sensible as a door is 2 m high.
4 4m
5 8 × (70 to 80 kg) + 6 × (30 to 60 kg) = 740 to 1000 kg
6 6 × (1.7 to 1.8 m) = 10.2 to 10.8 m, rounded to 10 or 11 m
7 a T
b F
8 a 70 miles
b 130 miles
9 a 72 km
b 328 km
c T
10 300 miles; 472 km = 295 miles or 300 miles = 480 km
11 a 235 miles
2
b $94
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
5 Angles
✦ Exercise 5.1
1
Parallel lines
a p and t, q and u, s and w, r and v
2 a b
b d
3 a q, r, u
b p, s, t
4 a corresponding
b alternate
5 a E
b H, N, W
b q and w, r and t
c CQX
d BPY
6 If they were parallel, then the angles XSA and XTC would be equal. This is not the case.
7 a b, f, j
b, c c and e; c and i
8 a i and q; i and k
b o and j; o and t
9 a neither
b corresponding
✦ Exercise 5.2
c corresponding
d alternate
e
neither
Explaining angle properties
Alternative explanations are possible for some questions.
1 a 125°
b 40°
c 48°
2 a 72° and 73°
b 145° and 107°
3 a Draw a line from R parallel to PQ; x = p, corresponding angles; y = q,
alternate angles; the exterior angle is angle SRQ = x + y = p + q; this is the
required result.
b x + y + r = 180, angles on a straight line; hence p + q + r = 180, which is the
required result.
4 a alternate angles
b alternate angles
c angle XAB + angle BAC + angle YAC = 180°, angles on a straight line;
angle ABC + angle BAC + angle ACB = 180°. This proves the result.
S
R x°
y°
r°
p°
P
q°
Q
5 Draw HF to divide the quadrilateral into two triangles. Show that the six triangle angles are the four
quadrilateral angles.
6 a alternate angles
b corresponding angles
c
x=a+y=a+c
7 a x is the exterior angle of triangle PQR.
b y=d+e
c x + y + c + f = 360, angles at a point; hence a + b + d + e + c + f = 360. These are the angles of the quadrilateral.
8 a alternate angles
b corresponding angles
c angle CBD = angle XDY, corresponding angles; angle BCD = angle CDX, alternate angles. The six angles
round D add up to 360°. The result follows from this.
✦ Exercise 5.3
Solving angle problems
Alternative explanations are possible in some questions.
1 Because 30° and 20° are opposite angles and should be equal. Similarly, 150° and 160° are opposite angles and
should be equal.
2 a = 136°, alternate angles; b = 136°, corresponding angles; c = 180° − 136° = 44°, angles on a straight line;
d = 44°, alternate angles.
3 a d + b = 180°, angles on a straight line and b + a + c = 180°, angle sum of a triangle, so d = a + c
b e = a + b; f = b + c
c d + e + f = 2(a + b + c) = 360
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 5
Answers to Coursebook exercises
4 angle BAC = 180 – (2 × 68) = 44°, isosceles triangle; angle EDC = 44°, corresponding angle
5 S how that the angles of the triangle and the quadrilateral together make the angles of the pentagon. The sum
of the angles is 180° + 360°.
6 T
he angles at A and D are equal (corresponding angles); the angles at B and E are equal (corresponding angles);
the angle at C is common to both triangles.
7 Angle BAC = q, alternate angles; r = angle BAC + p, exterior angles. The result follows.
8 a w = a + c, exterior angle of a triangle; y = b + d, exterior angle of a triangle. The result follows.
b w + y = the sum of two angles of the quadrilateral; x + z = the sum of the other two angles of the quadrilateral;
w + x + y + z = the angle sum of the quadrilateral = 360°.
9 a exterior angle of a triangle
b exterior angle of a triangle
c a + x + y = 180°, angle sum of a triangle; hence a + (b + d) + (c + e) = a + b + c + d + e = 180°.
End-of-unit review
1
a e
b f
c c
d d, f, b or h
2 a = 45°, corresponding angles; b = 45°, vertically opposite angles or alternate angles; c = 45°, vertically opposite
angles; d = 135°, angles on a straight line.
3 a and b, or f and g
4 8 2° + 27° = 109° so the angle between 82° and 27° is 180° – 109° = 71°; hence a = 71°, alternate angles.
b = 27°, corresponding angles.
5 a = 125° − 41° = 84°, external angle. b = 84° − 35° = 49°, external angle.
6 a corresponding angles
b alternate angles
c
corresponding angles
d alternate angles
7 A
ngle ADB = angle ABD, isosceles triangles; angle CDB = angle CBD, isosceles;
Angle B = ABD + CBD = ADB + CDB = angle D.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
6 Planning and collecting data
F Exercise 6.1
1
a experiment
e experiment
Collecting data
b observation
f observation
c survey
g survey
d survey
2 All. There are only 38 members, a sample would be too small.
3 a Cheaper, quicker, easier.
b 86
4 34
5 95
6 a B
b C
c B
7 a i Not enough, should have at least 24.
ii Not good, has not given numbers. People will have different opinions of how often ‘sometimes’ is.
iii It seems to be true, but he would need to ask more people, to be sure.
b i Students’ data collection sheets must include non-overlapping numerical values that allows for zero and
extreme data.
ii, iii Check students’ results and conclusions.
8 a i
ii
About 10%, and can be done fairly easily, so is a good decision.
Confusing and has overlapping numbers of pets – someone with three pets could be put in two different
categories.
iii It depends on what you mean by ‘lots’.
b i Students’ data collection sheets must include non-overlapping numerical values that allows for zero and
extreme data.
ii, iii Check students’ results and conclusions.
F Exercise 6.2
1
a discrete
f continuous
Types of data
b continuous
g discrete
c continuous
h continuous
d discrete
i discrete
e discrete
j discrete
2 No. Shoes are sold in whole and half sizes, no other. This is discrete data.
3 No. Age, like any time, is continuous data.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 6
Answers to Coursebook exercises
F Exercise 6.3
1
a
Using frequency tables
Height, h (cm)
Tally
Frequency
150 < h ≤ 160
////
4
160 < h ≤ 170
////
5
170 < h ≤ 180
//// //
7
180 < h ≤ 190
///
3
190 < h ≤ 200
/
1
Total
20
b 3
c 11. Add up last three frequencies; all are taller than 170 cm.
d 16. Add up the first three frequencies; all are shorter than 180 cm.
2 a
Time, t (seconds)
Tally
Frequency
25 < t ≤ 30
//
2
30 < t ≤ 35
//// /
6
35 < t ≤ 40
//// ////
9
40 < t ≤ 45
//// //
7
45 < t ≤ 50
b 27
3 a
c 7
3
27
d 19
e 8
Height, h (cm)
Tally
Frequency
10 ≤ h < 18
//// ///
8
18 ≤ h < 26
////
5
26 ≤ h < 34
//
2
34 ≤ h < 42
b 18
4 a 4
///
3
Total
18
c 5
d 15
e 5
b 6
c 30
d 14
5 a
Maths
Science
English
Other subject
Total
Girls
8
4
5
1
18
Boys
6
5
1
2
14
Total
14
9
6
3
32
b 5
c 3
6
2
///
Total
Car
Bus
Bicycle
Total
Male
7
8
5
20
Female
10
9
3
22
Total
17
17
8
42
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
Unit 6
End-of-unit review
1
a experiment
b observation
c survey
2 All. A 10% sample would be too small.
3 99 or 100 for a 10% sample.
4 a C
b C
5 a discrete
6 a
b continuous
Weight, w (g)
Tally
Frequency
150 < w ≤ 170
/
1
170 < w ≤ 190
////
5
190 < w ≤ 210
//// //
7
210 < w ≤ 230
b 5
7
c 10
///
3
Total
16
d 13
e 16
A
B
C
Total
Maths
4
9
5
18
Science
5
2
3
10
Total
9
11
8
28
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Coursebook exercises
7 Fractions
✦ Exercise 7.1
1 a 1
Finding equivalent fractions, decimals and percentages
b 0.4
4
2 a i 0.14
ii
c i 0.24
ii
3 a i 34%
ii
c i 68%
ii
c 80%
14 = 7
100 50
24 = 6
100 25
17
50
17
25
4 a i 0.36
c i 0.04
ii 36%
ii 4%
5 a 12.5%
g 25.5%
b 87.5%
h 1.5%
✦ Exercise 7.2
1
a 0.68
e 0.6
74 = 37
100 50
8 = 2
100 25
3
50
81
100
b i 0.74
ii
d i 0.08
ii
b i 6%
ii
d i 81%
ii
b i 0.35
d i 0.95
ii 35%
ii 95%
c 7.5%
i 66.5%
f
d 47.5%
j 94.2%
e 3.2%
k 3.4%
1
5
g
7
10
h 75% = 0.75
f 53.6%
l 1.8%
Converting fractions to decimals
b 0.55
•
1
2
d
•
c 0.125
d 0.3125
• •
• •
e 0.90625
• • •
•
•
2 a 0.6
b 0.1
c 0.63
d 0.39
e 0.123 or 0.123
3 a 0.385
b 0.857
c 0.762
d 0.514
e 0.436
4 Yes. Both 1 and 4 have one number that is recurring and both 1 and 7 have two recurring decimals.
15
15
22
✦ Exercise 7.3
Ordering fractions
1
3 , 5 , 11
4 6 12
5, 3, 5
8 4 6
b
4 , 1, 3
11 3 10
17 , 9 , 19
20 11 25
b
a
e
2 a
e
3
4
f
f
1, 4, 9
2 7 14
1, 4 , 7
6 15 10
c
5 , 11 , 2
9 18 3
d
4 , 11 , 8
7 20 15
17 , 11 , 32
18 12 35
c
18 , 5 , 2
61 18 9
d 11 , 3 , 12
16 5 21
22
3, 4, 9
4 5 10
1 , 11 , 5 , 4
3 27 12 9
1 is smaller than 1 , so 5 is closer to one than 4 , so is bigger. Same reasoning for 4 and 3 , etc.
6
5
6
5
5
4
✦ Exercise 7.4
7
8
g 5
8
2 a 14
9
g 11
4
3 a 7 +
21
b 8 +
30
1 a
4 a
b
h
b
h
15 = 22 ,
21 21
27 = 35 ,
30 30
Adding and subtracting fractions
7
c 1
10
2
1
i 1
3
4
1
c
1
11
3
9
i 21
13
10
2
22 = 1 1 , 7 1
21
21
21
35 = 7 = 1 1 , 14 1
30 6
6
6
85 − 32 = 53 = 2 13
20 20 20
20
d 11
15
7
j
18
d 1 16
45
j 24
15
11
12
k 7
15
e 18
21
k 11
12
e
f
l
f
l
58
99
5
24
1 11
36
15
6
b 55 − 41 = 110 − 41 = 69 = 23 = 5 3
4
6 12
12 12 12
4
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 7
3 78
b
13
8 15
c 10 1
4
g 19
10
h
2 59
i
b
4 18 m
5 a
5m
8
6 a
7
Answers to Coursebook exercises
5
2 14
2 43
j
e 17 21
40
f
10 19
30
k 25
12
l
29
36
3m
4
F Exercise 7.5
1
d 3 17
28
a $9
2 a
3 a
Finding fractions of a quantity
b 4m
c 12 kg
d 25 cm
e 18 ml
9 53 kg
b 15 13 t
c $12 83
d 20 89 mg
e 20 56 mm
5 of 18 m = 10 m, 7 of 24 m = 14 m, 2 of 19 m = 12 2 m, 4 of 30 m = 13 1 m, 5 of 14 m = 11 2 m
9
12
3
3
9
3
6
3
b 12 m
F Exercise 7.6
1
a 15
Multiplying an integer by a fraction
b 24
c 27
d 18
e 63
f 25
b 4 49
c 24 53
d 11 23
e 12 14
f 71
2 a 12 83
2
3 No. Dakarai divided the 78 by 3 and the 15 by 5. The divisors must be the same when cancelling.
F Exercise 7.7
Dividing an integer by a fraction
1
a 28
b 18
2
a 18 23
b 16 12
b 9 12
3 A, 45 ÷ 58
4 a 4 12
F Exercise 7.8
1
a
2 a
3 a
4 a
1
8
3
10
3
8
1 12
b
b
b
b
3
16
1
2
5
6
2 23
c 28
c
49 12
c
3 23
d 20
e 39
f 55
d
42 12
e 57 12
f
22 12
d
9 23
e 8 13
f
10 14
Multiplying and dividing fractions
c
c
c
c
2
15
3
10
21
32
1 14
8
25
4
d 27
d 7 51
d 1 13
d
e
e
e
9
28
1
4
3
3 10
e 2
f
f
f
f
14
27
2
11
7
2 10
1 16
5 MENTAL MATHS IS FUN
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
Unit 7
End-of-unit review
1
Fraction
3
4
4
5
1
5
3
10
2
5
1
2
Decimal
0.75
0.8
0.2
0.3
0.4
0.5
Percentage
75%
80%
20%
30%
40%
50%
2 a 0.32
b
3 a
b
6%
32 = 8
100 25
6 = 3
100 50
4 a 0.16
b 16%
5 a 0.375
b 0.364
6
7
8
1 , 11 , 3 , 5
2 20 5 8
a 78
1 m
a 110
b
1
2
b
81 m
c 0.415
8
c 1 21
5
d 12
1
e 6 12
1
24
19 14
d 12
35
e
d 17 12
e 38
f
117
18
f
1 11
21
10
9 a $18
b 21
c
10 a 9 13 kg
b 10 54
c
5
6
11 A, 32 × 53
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Coursebook exercises
8 Shapes and geometric reasoning
F Exercise 8.1
1
a AC
Recognising congruent shapes
b DF
c HI
d KM
2 D, G
3 a i
b i
3.1 cm
23°
ii
ii
6.5 cm
62°
iii
iii
4 a i
b i
FG
∠FGH
ii
ii
EH
∠EFG
iii AB
iii ∠ABC
7.8 cm
95°
iv CD
iv ∠BCD
5 No. The angles are both 90°, but not corresponding. ∠LKN and ∠PSR (not ∠SRQ) are corresponding.
6 No. Although the angles in two equilateral triangles will all be 60°, the sides of the two equilateral triangles can
be of different lengths.
F Exercise 8.2
1
Identifying symmetry of 2D shapes
a
b
c
d
e
f
g
h
i
j
k
l
2 a 2
g 2
b 2
h 1
c 1
i 1
d 4
j 2
e 2
k 1
f
l
1
2
3 a 6
b 0
c 8
d 0
e 8
f
5
g 4
h 0
4 a 6
b 1
c 8
d 1
e 8
f
5
g 4
h 2
5
Square
Rectangle
Rhombus
Parallelogram
Kite
Trapezium
Isosceles
trapezium
Number
of lines of
symmetry
4
2
2
0
1
0
1
Order of
rotational
symmetry
4
2
2
2
1
1
1
Shape
6 a i
3
ii
3
b i
1
Copyright Cambridge University Press 2013
ii
1
c i
0
ii
1
d i
1
ii
1
Cambridge Checkpoint Mathematics 8
1
Unit 8
Answers to Coursebook exercises
7 a
b
c
8
F Exercise 8.3
1
2
a square
Classifying quadrilaterals
b parallelogram
2 a J
b H
c M
3 a (3, 4)
b (3, 3)
c (4, 2)
c kite
Cambridge Checkpoint Mathematics 8
d L
d rectangle or parallelogram
e P
f
N
e isosceles trapezium
g K
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
F Exercise 8.4
1
Unit 8
Drawing nets of solids
There are many possible nets; these are examples.
a
b
2 A, B, D, G
3
or
4 Students’ nets must be accurate to ± 2 mm.
a cube
b cuboid
3 cm
3 cm
6 cm
3 cm
3 cm
4 cm
4 cm
4 cm
3 cm
3 cm
3 cm
4 cm
Diagrams not full size
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Unit 8
Answers to Coursebook exercises
c triangular prism (isosceles triangle)
5 cm
d triangular prism (right-angled triangle)
45 mm
27 mm
45 mm
5 cm
5 cm
5 cm
6 cm
7 cm
65 mm
5 cm
5 cm
Diagrams not full size
36 mm
27 mm
27 mm
5 a E
b L
c H
d F
e J
6 a Students’ nets must be accurate to ± 2 mm.
b 24.8 cm ± 5 mm
f
45 mm
I
8 cm
4 cm
5 cm
5 cm
5 cm
4 cm
5 cm
Diagram not full size
F Exercise 8.5
1
a 180 m
Making scale drawings
b 8 cm
2 a 6.5 m
b 10 cm
3 a i 3m
b 2 cm
ii 1.5 m
c 7 cm
4
iii 1 m
iv 0.5 m
v 2m
vi 2 m
10 cm
3 cm
1.5 cm
5.5 cm
1.5 cm
8.5 cm
Diagram not full size
4
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
Unit 8
5 a
15.4 cm
9 cm
12.5 cm
Diagram not full size
b 3.08 m (allow 3.04 m to 3.12 m)
6 26.4 m (allow 26.1 m to 26.7 m)
End-of-unit review
1
PQ
2 a i 4.2 cm
b i 80°
ii 7.1 cm
ii 30°
3 a i
2
2
ii
4 a (2, 3)
iii 7.6 cm
iii 70°
b i
1
b (4, 4)
ii
1
c i
3
ii
d i
3
0
ii
4
c (3.5, 4)
5 There are many possible nets, these are examples.
6 a E
b
F
c
7 a 4.5 m
b
7 cm
d
G
L
e
D
f
K
8 a Students’ scale drawings must be accurate to ± 2 mm.
12 cm
2 cm
9 cm
A
B
7 cm
2 cm
10 cm
Diagram not full size
b 15 m
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
5
Answers to Coursebook exercises
9 Simplifying expressions and solving equations
✦ Exercise 9.1
1
Collecting like terms
b m+4
h 6jk
a 8n
g 2v + 4
c 8h
i 6m + 2p
d 5ef
j 5c − 2a
e 5a + 2b
k 3 − 4pq
f 5 − 3d
l −8ab − 3xy
2 a 2 + n2, 2n + 2, n × 2 + 2; −2 + 2n, 2 × n − 2 × 1, n2 − 2; 2m + 2n, 2 × n + m2; 2n2, 2 × n × n, n2 × 2
b i −2 + 2 × n × m
ii 2mn − 2
3 a 13x
g 10cd + 2de
b 11y
h 3v − 7
4
c −3z
i 7u − 3t
d 8a + b
j 7x2 + 6x
e 5c − 3d
k 6y2 − 3y
f 3ab
l 7a + 2
24a + 18b
13a + 10b
5a + 7b
11a + 8b
8a + 3b
3b + 2a
3a + 4b
5
3a + 5b
5a – b
6b – 2a
7cd – 7ef
3cd – 8ef
4cd – 4ef
3cd + 4ef
4cd + ef
–cd – 4ef
cd – 8ef
5cd + 5ef
–2cd + 4ef
7cd + ef
6 a 1. 7ab and 2ac can’t be simplified by adding them together as the algebra terms are different.
2. 4xy − yx can be simplified by subtraction to 3xy as the algebra terms are the same.
b 1. 7ab + 2ac
2. x2 + 3xy
✦ Exercise 9.2
1
a 4x + 24
g 40 − 5b
m 4p + 6q
Expanding brackets
b 3y + 21
h 36 − 6d
n 20c + 16d
c 7z − 14
i 6a + 24
o 54t − 18s
d 2w − 8
j 48b + 36
p 6ab + 9c
e 2a + 10
k 10c − 5
q 42xy − 14z
f 8g + 72
l 18 − 24e
r 10x + 5y + 20
2 a 5x + 18
b 8y + 24
c 23z + 44
d 4w + 3
e 12v + 2
f 9a + 19b
3 a 3xy + 2x
g a − 3ab
m 2x2 + 6xy
b y2 + 8y
h 5c − cd
n 15y2 + 18y
c 2wz − z
i 2e2 + 7ef
o 24b2 − 8ab
d m2 − 4m
j 7g2 + 3gh
p 18h2 + 6h
e 2n2 + 5n
k 2h2 − 5hk
q 30km − 40k2
f 9n − 8n2
l 3cd − 5de
r 4f 2 + 2fg − 6f
4 a 2x2 + 7x
b 6z2 + 6z
c u2 + 2u
d 2w2 + 20wx
5 a 1.
2.
3.
b 1.
He wrote −6x + 21 instead of −6x − 21.
ac + 3bc can’t be simplified by adding them together as the algebra terms are different.
He worked out x(3x + 4y) = 9x2 + 4xy, instead of 3x2 + 4xy.
2x + 19
2. ac + 3bc
3. 3x2 + 2y2 + 14xy
✦ Exercise 9.3
Constructing and solving equations
a x = 8, y = 7
b x = 9, y = 5
c x = 7, y = 4
d x = 5, y = 3
e x = 6, y = 3
f
2 a x=2
b x=4
c x = 12
3 a y=7
b y=4
c y = 12
1
Copyright Cambridge University Press 2013
x = 11, y = 7
Cambridge Checkpoint Mathematics 8
1
Unit 9
Answers to Coursebook exercises
b n4 − 8 = 5, n = 52
4 a 3n + 8 = 23, n = 5
d 3n + 7 = 4n, n = 7
c 5n − 4 = 2n + 20, n = 8
e 2(n + 5) = 5n − 14, n = 8
f 3(n − 2) = 7(n − 6), n = 9
End-of-unit review
a 6p
b n+7
c 9bc
2 a 8a + 5b
b 4v − 4
c 2x2 + 12y + 9
1
3
e 5x + 9
f 6a + 3b
15ab + 8bc
8ab + 3bc
3ab + 2bc
7ab + 5bc
5ab + bc
2ab + 4bc
4 a 3x + 12
g 2xy + x
b 8y − 8
h 4n2 + 6n
c 12a + 8
i 8e − de
d 20 − 35b
j 2hk + 8k2
e 6c + 18d
k 6y2 + 18y
f 32xy − 24z
l 6m2 + 3mn − 15m
5 a 8x + 42
b 14w − 14
c 2a + 23b
d 2x2 + 12x
e 2u2 + 2u
f w2 + 16wx
6 a x = 8, y = 7
7 a x = 12
b x = 6, y = 16
b x=9
8 a 5n + 9 = 44, n = 7
c 5n − 10 = 2n + 11, n = 7
2
d 1 − 6u
c x = 5, y = 7
c x=8
b n3 − 7 = 4, n = 33
d 3(n + 2) = 2(n + 5), n = 4
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
10 Processing and presenting data
✦ Exercise 10.1
1
a 4.5 °C
Calculating statistics from discrete data
b 4 °C
c 4.7 °C
d 10 °C
2 a 17
b 10.4
3 a 20
b 1
c 2
d 1.93
4 a 9
b 7.5
c 7.1
d 10
5 a 40
b 0
c 70
6 They are correct if the first is the mean, the second is the median and the third is the mode.
7 a i 12.58
ii 12
iii 12
iv 5
b Not correct. It is true for the three averages but the range will not change.
8 a i 40
ii 1 and 3
iii 2
iv 2.25
b The mean or the median would be best. Every score contributes to the mean. In at least half the matches
the team scored at least the median. The mode is not a good choice.
9 There are lots of possible answers. Here is one.
Number
1
2
3
4
5
Frequency
6
0
0
6
8
In this case the mode is 5, the median is 4 and the mean is 3.5.
✦ Exercise 10.2
1
Calculating statistics from grouped or continuous data
a 21–30
b Because 15.5 is halfway between 11 and 20 (or between 10.5 and 20.5).
c 29.25
2 a 30–
c about 32 or 33 minutes
3 a 89.5 cm
b 30 cm
b halfway between 20 and 25
d 32.5 minutes
c 90.5 cm
4 a 31–40
b Only 18 messages have a length of 20 characters or less. The median is between the 25th and 26th so
it is more than 20.
c There are ten messages at most 10 characters long and one at least 51 characters long. The range must
be at least 51 − 10 = 41 characters.
d 24.9 characters
5 a 57
b between 30 minutes and one hour
c about 45 minutes
d 44 minutes
6 a 70–80
b 120
c 70 seconds
d 68 seconds
e The mean is the best choice because the frequencies from every class are used to estimate it.
✦ Exercise 10.3
1
Using statistics to compare two distributions
a Paper 2. For Paper 2, the median was 5 less and the mean was nearly 7 less than Paper 1.
b Paper 1 because Paper 1’s range was greater than Paper 2’s range.
2 A was better than B because the mean grade for A was 3.37 and for B it was only 2.75. The mean is probably
the best average to use because it takes account of all the scores.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 10
Answers to Coursebook exercises
3 B
oth teams have the same median, 2 goals. The mean for Juventus is 2.05 and for AC Milan is 1.82, so Juventus
scored 0.23 more goals per match, on average. The mode is not helpful in this case because there are three modes
for AC Milan: 0, 1 and 2 all have the same frequency.
4 T
he median for the boys is about 132 cm and for the girls is about 135 cm, making the girls about 3 cm taller,
on average. The mean for the boys is 132.3 cm and for the girls is 135.2 cm; again the girls are about 3 cm taller,
on average.
5 T
he mean for May is 8.3 cm and for November is 18.5 cm. The median for May is between 5 and 10 cm and the
median for November is between 15 and 20 cm. Both these show that on average there is about 10 cm more rain
in November than in May. The range for the two months is similar as both spread over five classes.
6 a 45
b You cannot tell. The nine in the classes 80–84 and 85–89 after dieting definitely lost mass, but some of the
others may not have done so.
c The range increased by about 10 kg.
d The mean mass went down from 104.7 kg to 96.2 kg, an average decrease of 8.5 kg.
End-of-unit review
1
a 8 characters
b 9 characters
c 9.3 characters
2 a i 95 cm
ii 100 cm
iii 96 cm
b The mode. Have more of that size in the shop.
3 a 21–25
b 18.9
d 5 characters
iv 30 cm
c It is in the 16–20 class
4 a 20–
b About 31 minutes is a good estimate.
d The estimate should be between 30 and 50 minutes.
c 32.25 minutes
5 a 35 boys and 32 girls
b They have the same median, 10. The mean for the boys is 9.9 and the mean for the girls is 9.5. The boys were
about 0.4 answers better.
6 T
he modal class for the newspaper is 11–15 and for the magazine it is 21–25. The mean for the newspaper is 14.1
and for the magazine it is 18.9. This shows that the sentences in the magazine are longer by about 4.8 words.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
11 Percentages
✦ Exercise 11.1
1
Calculating percentages
1 , 30% = 3 , 37 1 % = 3 , 45% = 9 , 50% = 1 , 60% = 3
5% = 20
2
20
10
5
2
8
2 a 0.15
b 0.05
c 0.9
d 0.065
3 a 15 kg
b 750 litres
c $120
d 84 g
4 a 4.5 cm
b 36 people
c 800
d 5
5 a $11.61
b 159.6
c 3441
d 552
6 a All of them!
e 1.5
b A 60, B 164, C 33, D 63, E 11.5
7 a 2.1
b 5.1
c 32.1
d 35.1
8 a 88.06
b 57.12
c 72.59
d 22.02 (or 22.015)
9 a 336
b 20%
c 84
10 a red 1702, blue 1288, yellow 920
11 a 33 200
b 6800
b 15%
c 17%
12 a copper 28.5 g, tin 1.5 g
13 a 74%
e 15.47
b copper 950 g, tin 50 g
b chromium 25.2 g, nickel 11.2 g
c 36 t of chromium, 16 t of nickel
14 Germany 82 million, France 66 million, Spain 47 million, Sweden 9 million
15 men 2362, women 1350, boys 1181, girls 731 (rounding ‘women’ up has given a total of 5624)
16 17 600
17 All are 15.36 except 18% of 84 and 9% of 168 which are 15.12.
✦ Exercise 11.2
1
Percentage increases and decreases
a $9
b $69
c $51
2 a 2240
b 5440
c 960
3 a 0.38
b 19.38
c 18.62
4 a $264
b $360
c $408
d $480
e $528
5 52 000
6 a 1.809 m
7 a $196, $364, $133 and $301
b $426
8 A $448, B $679, C $421, D $877
9 Ace $15 484, Beta $16 902, Carro $20 961, Delta $23 737
10 a $88
b 10% of $88 is $8.8 so the price will be below $80.
11 a $480
b $96
c $79.20
c $576
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 11
Answers to Coursebook exercises
F Exercise 11.3
1
Finding percentages
a Science 70%, History 85%, Geography 67.5%, English 74%, Maths 84%, Art 57%
2 a 55.6%
b 44.4%
3 a 65%
b 35%
4 a 40%
b 60%
c 48.6% and 51.4%
5 a 61%
b 29%
c 40%
6 a $45
b 18%
b History
7 10%, 7.5% and 6%
8 a China 17.1%, India 40.2%, Indonesia 30.4%, Japan 4.1%, Nigeria 61.2%, United States 22.5%
b 30.2%
c 19.4%
9 a A 6.8% reduction, B 11.6% reduction, C 3.3% increase, D 14.7% decrease
bD had the largest percentage decrease.
10 a 27%
b 34%
F Exercise 11.4
1
c 113%
Using percentages
a 76%, 71%, 74%
b
27 , 69 , 13
38 93 17
2 a New 33%, City 47%, State 41%
3 a 18%, 37%, 52%, 22%
b New College
b the game console
b Friday
4 a Friday 13%, Saturday 18%, Sunday 20%
5 a boys 10%, girls 15%
b boys 13%, girls 24%
cBoys possibly did better. The percentage of distinctions was lower that that of the girls but they had a much
smaller percentage of failures.
b B
c They are all similar. A and C were both 64%, and B was 60%.
6 a A 16%, B 29%, C 27%
d C had the largest percentage of Excellent and the smallest percentage of Poor.
7Men have a larger percentage of overweight (44% against 28%). Women have a larger percentage of underweight
(22% against 13%).
End-of-unit review
b 2
c
2 a 72 m
b 6.45 m
c 18 kg
3 a 83%
b 5976
4 a 106
b 153
5 a 552
b 391
1
a
9
10
5
1
20
d
1
40
d 551
6 No. 20% of 812 is 162 and 812 + 162 = 974.
7 a $8.83
b $22.35
c $53.81
8 a $17.50
b $29.75
c $80.15
9 a 67%
b 84%
10 8.7%
11 a 67% increase
b 8% decrease
c 53% increase
12 X does. 40% are under 25 in town X; in town Y the figure is 30%.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
12 Constructions
F Exercise 12.1
1
Drawing circles and arcs
Check students’ circles, with radii:
a 6 cm
b 3.5 cm
Allow ± 2 mm.
c
d 4 cm
45 mm
e
2.5 cm
f
3 cm.
2 a, b Check students’ accurate drawings, based
on the horizontal line AB 8 cm.
A
C
D
E
B
Diagram not full size
c All of the circles touch the point A.
3 Check students’ drawings of arcs.
a radius 4 cm and angle 50°
F Exercise 12.2
1
b radius 5 cm and angle 85°
c radius 35 mm and angle 120°
Drawing a perpendicular bisector
Check students’ drawings of the perpendicular bisector of AB; all construction lines must be visible.
2 Check students’ constructions of the midpoint of CD; all construction lines must be visible.
3 Question done in pairs; should be checked already.
4 a The two arcs do not have the same radius.
b She used the compasses at one end, but moved the point of the compasses before drawing the second arc.
5 Check students’ drawings of 8 cm by 10 cm rectangle
ABCD. AB and BC must show perpendicular bisector
construction lines and marks at midpoints.
A
100 m
B
80 m
D
Diagram not full size
Copyright Cambridge University Press 2013
C
Cambridge Checkpoint Mathematics 8
1
Unit 12
Answers to Coursebook exercises
F Exercise 12.3
1
Drawing an angle bisector
Check students’ drawings of bisection of a 50° angle ABC. All construction lines must be visible.
2 Check students’ drawings of bisection of a 120° angle DEF. All construction lines must be visible.
3 Question done in pairs; should be checked already.
4 C
heck students’ accurate scale drawings of shot put circle,
landing area and angle bisector. All construction lines must be
visible. Appropriate scale must be given.
15 m
top half
35°
1.5 m
bottom half
Diagram not full size
5 C
heck students’ accurate scale drawings of roped section
of sea and angle bisector. All construction lines must be
visible.
sea
70°
beach
50 m
Diagram not full size
F Exercise 12.4
1
Constructing triangles
aCheck students’ accurate drawings of triangle ABC.
All construction lines must be visible.
b i 52°
ii 85°
iii 43°. Allow ± 2°.
c 180°
d The three angles in any triangle add to 180°.
2 aCheck students’ accurate drawings of triangle DEF. All
construction lines must be visible.
b 61 mm. Allow ± 2 mm.
c i 45°
ii 45°. Allow ± 2°.
d Isosceles. Angles DEF and EDF are the same.
A
65 mm
75 mm
C
B
95 mm
Diagram not full size
D
86 mm
E 61 mm F
Diagram not full size
3 C
heck students’ accurate drawing of both triangles. All construction lines must be visible.
Sasha’s angle XZY = 46°, Dakarai’s angle XZY = 50°. Allow ± 2°, but not both = 48°.
Sasha is correct.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
Unit 12
End-of-unit review
1
a Check students’ circles, radius 4 cm.
b Check students’ drawings, arc with radius 6 cm and angle 30°.
2 a, bCheck students’ drawings of the perpendicular bisector of AB (7 cm long); all construction lines must be
visible.
3 a, b Check students’ drawings of the bisection of a 65° angle XYZ; all construction lines must be visible.
4 a Check students’ drawings of rectangle ABCD, 7 cm by 3.5 cm.
b i Check students’ drawings of the midpoint of AB. ii Check students’ drawing of the midpoint of CD.
c
A
B
7m
n
pet o
3.5 m Car half
this
s on
Tile half
this
D
C
Diagram not full size
5 a Check students’ accurate drawings of the pendant.
bCheck students’ drawings of bisection of the 30°
angle; all construction lines must be visible.
1.5 cm
30°
5 cm
left right
side side
Diagram not full size
6 C
heck students’ accurate drawings of triangles. All construction lines must be visible.
b
a
4 cm
6.5 cm
78 mm
54 mm
8 cm
Diagram not full size
7 Check students’ accurate drawings of both triangles. All construction lines must be visible.
Hassan’s triangle
Harsha’s triangle
8 cm
6.4 cm
Diagram not full size
8 cm
4.8 cm
Hassan is correct: they are congruent.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Coursebook exercises
13 Graphs
F Exercise 13.1
1
Drawing graphs of equations
a The values of y are −6, −5, −4, −3,−2, −1, 0, 1, 2.
b
y
6
5
4
3
2
1
–4 –3 –2 –1
–1
0
1
2
3
4
1
2
3
4
1
2
3
4
x
–2
–3
–4
–5
–6
2 a The values of y are −6, −4, −2, 0, 2, 4, 6.
b
y
6
5
4
3
2
1
–4 –3 –2 –1
–1
0
x
–2
–3
–4
–5
–6
3 a The values of y are 0, 1, 2, 2.5, 3, 3.5, 4.
b
y
6
5
4
3
2
1
–4 –3 –2 –1
–1
0
x
–2
–3
–4
–5
–6
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 13
Answers to Coursebook exercises
4 a The values of y are 7, 5, 3, 1, −1, −3.
b
y
7
6
5
4
3
2
1
–4 –3 –2 –1
–1
0
1
2
3
5
4
x
–2
–3
–4
–5
–6
–7
5 a The values of y are −7, −5, −3, −1, 1, 3, 5, 7.
b
y
7
6
5
4
3
2
1
–2 –1
–1
0
1
2
3
4
1
2
3
4
1
2
3
4
x
–2
–3
–4
–5
–6
–7
6 a The values of y are 3, 2.5, 2, 1.5, 1, 0.5, 0, −0.5.
b
y
3
2
1
–2 –1
–1
7 a The values of y are 5, 4, 3, 2, 1, 0, −1, −2, −3.
b
0
x
y
5
4
3
2
1
–2 –1
–1
0
5
6
x
–2
–3
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
8 a
x
−3
−2
−1
0
1
2
3
y
−7
−4
−1
2
5
8
11
b
Unit 13
y
12
10
8
6
4
2
–3 –2 –1
–2
0
1
2
x
3
–4
–6
–8
F Exercise 13.2
1
Equations of the form y = mx + c
a The values of y are −40, −30, −20, −10, 0, 10, 20, 30, 40.
b
y
40
30
20
10
–4 –3 –2 –1
–10
0
1
2
3
4
1
2
3
4
1
2
3
4
x
–20
–30
–40
2 a The values of y are −40, −30, −20, −10, 0.
b
y
40
30
20
10
–4 –3 –2 –1
–10
0
x
–20
–30
–40
c If x = 20, y = 5 × 20 – 20 = 80 so (20, 80) is on the line.
3 a The values of y are −35, −20, −5, 10, 25, 40.
b
y
40
30
20
10
–4 –3 –2 –1
–10
0
x
–20
–30
–40
c If x = 5, y = 15 × 5 – 5 = 70 so (5, 80) is not on the line.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Unit 13
Answers to Coursebook exercises
4 a The values of y are 40, 30, 20, 10, 0, −10, −20.
b
y
40
30
20
10
–4 –3 –2 –1
–10
0
1
2
3
4
5
10 15 20
2
3 x
x
–20
–30
c −50
–40
d 60
5 a The values of y are −1, 0, 1, 2, 3, 4, 5, 6, 7.
b
y
7
c 3.6
6
5
4
3
2
1
–20 –15 –10 –5
–1
0
x
–2
6 a The values of y are −100, −60, −20, 20, 60, 100, 140.
b
y
140
120
100
80
60
40
20
–3 –2 –1 0
–20
1
–40
–60
c If x = 10, y = 40 × 10 + 20 = 420 so (10, 420) is on
the line. If x = −10, y = 40 × −10 + 20 = −380
so (−10, −420) is not on the line.
–80
–100
7 a = 3 and b = −2
8 a (0, −10)
b (2, 0)
F Exercise 13.3
The midpoint of a line segment
1
a
b (4, 2)
y
4
A
3
2
1
B
–1
4
0
1
2
3
4
5
6
x
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
Unit 13
2 AB (3, 3); BC (3, 0); CD (−1, −1); DE (−4, 1); EA (−1, 3)
3 a (4, 3)
b (9, 6)
c (5, 9)
4 a (4, 2)
b (−1, 1)
c (4, 4)
5 a (3.5, −4)
b (−0.5, 2.5)
c (−8.5, 7.5)
6 a (35, 20)
b (−10, 10)
c (1, −3)
7 DE (−1, 15); EF (0.5, −10); FD (2.5, −5)
8 a
y
4
A
3
D
2
1
–2 –1
–1
0
1
2
3
4
B
x
C –2
(
) ( )
(
) ( )
b The midpoint of AC is 2 + –1 , 3 + –2 = 1 , 1 . The midpoint of BD is 3 + –2 , –1 + 2 = 1 , 1 .
2
2
2 2
2
(2 2 )
( 2 2)
1 4 + −1
10 No. The midpoints are ( –2 + 5 , 1 + 2 ) = (1.5, 1.5) and ( 0 +
2 , 2 ) = (0.5, 1.5).
2
2
2
2 2
9 The midpoint of PR is 2 + 2 , 5 + –1 = (2, 2). The midpoint of QS is –2 + 6 , 3 + 1 = (2, 2).
11 (6, −3)
F Exercise 13.4
1
Graphs in real-life contexts
a 09 30
b 20 km
c 1 hour
2 a 1 1 hours
b 1 1 hours
c 3 hours
3 a The line is steeper.
b 2 minutes
c They were together at the lap start.
4 a, b
Speed (m/s)
2
2
c 45 seconds
Car
30
20
d about 130 km
Van
10
0
0
10 20 30 40 50 60
Time (seconds)
c about 30 km
Distance from home (km)
5 a, b
40
30
20
Shen
10
Sister
13 00
14 00
15 00 16 00 17 00
Time (24-hour clock)
Copyright Cambridge University Press 2013
18 00
Cambridge Checkpoint Mathematics 8
5
Unit 13
Distance (km)
6 a, b
Answers to Coursebook exercises
4
3
Xavier
2
Alicia
1
0
0
5
10 15 20 25 30 35
Time (minutes)
c The lines are together between 20 and 25 minutes.
d Alicia
Distance (m)
7
400
300
200
100
0
0
10
20
30
40
Time (seconds)
50
150 m from one end and 250 m from the other end.
End-of-unit review
1
A: y = 2, B: x = −4, C: x = 3.5, D: y = x, E: y = −x
2 a The values of y are −4, −2, 0, 2, 4, 6, 8.
b
y
8
7
6
5
4
3
2
1
–3 –2 –1
–1
0
1
2
3
x
–2
–3
–4
6
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
3 a The values of y are 6, 5, 4, 3, 2, 1, 0, −1, −2.
b
Unit 13
y
6
5
4
3
2
1
–2 –1
–1
0
1
2
3
4
1
2
3
5
6
x
–2
–3
c If x = −24, y = 4 − −24 = 28.
4 a The values of y are −10, 0, 10, 20, 30, 40, 50.
b
y
60
50
40
30
20
10
–3 –2 –1
–10
0
x
–20
c If x = 15, y = 10 × 15 + 20 = 170, so (15, 180) is not on the line.
5 a (6, −2)
b (−1, 2)
d −40
c (20, −1)
6 a Nisota
b 200 km
7 a 1 hour
b, c
c 400 km
d 200 km
Distance from
Newton (km)
Danville 150
100
50
Newton
13 00
14 00
15 00 16 00 17 00
24-hour clock time
18 00
d Between 90 and 95 km is a reasonable answer.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
7
Answers to Coursebook exercises
14 Ratio and proportion
✦ Exercise 14.1
1
a 1:5
g 2:3
b 1:6
h 3:5
2 a 1:2:3
d 6:5:1
3 a 1:2
g 5:2
Simplifying ratios
c 1:5
i 2:7
b 4:5:6
e 3:1:5
b 3:5
h 2:3
4 a 30 : 50 : 1
d 4:2:1
d 6:1
j 15 : 2
e 3:1
k 18 : 5
f 9:1
l 5:4
c 4:3:5
f 9:2:4
c 1:3
i 4:7
b 3:4:6
e 6 : 5 : 50
d 2:1
e 5:1
f 8:3
c 1:7:3
f 5 : 1 : 25
5 No. The amounts are 750 g : 1500 g which simplifies to 1 : 2, not 2 : 1.
6 a 1:4
g 3:5
b 1:2
h 2:7
c 1:2
i 3:1:2
d 1:3
j 1:5
e 6:1
f 5:1
7 No. 250 : 750 : 1200 simplifies to 5 : 15 : 24.
8 a Her ratio shows that the time on Wednesday is twice that of Monday, but it was less, not more.
b 1 hour 40 minutes = 1.666... hours (or 1 2 hours), not 1.4 hours.
3
50 minutes = 0.8333... hours (or 5 hour), not 0.5.
6
She didn’t divide the 14 by 5 in the last line.
c 2:1:3
✦ Exercise 14.2
1
Sharing in a ratio
a $15, $30, $45
2 a $42, $56, $70
3 a i 95
b 38
4 a i 32
b i 27
b $50, $75, $100
c $144, $240, $48
b $48, $64, $80
c $58.50, $78, $97.50
ii 133
c 38
ii 16
ii 9
d $144, $72, $180
iii 57
iii 24
iii 36
5 Aden = $150, Eli = $100, Lily = $75 and Ziva = $125
6 $300, $600, $750, $900
7 $7.50
8 $8000
9 Share $150 in the ratio 2 : 3 : 1 = $50, $75, $25
Share $126 in the ratio 2 : 6 : 1 = $28, $84, $14
Share $120 in the ratio 3 : 1 : 4 = $45, $15, $60
Share $132 in the ratio 1 : 5 : 6 = $11, $55, $66
✦ Exercise 14.3
1
Solving problems
a $0.50 or 50 cents
2 a $1.50
b $7.50
b $1.50
c $5
c $10.50
3 $27
4 a 200 g butter, 300 g plain flour, 300 g icing sugar, 400 ml honey
b 80 g butter, 120 g plain flour, 120 g icing sugar, 160 ml honey
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 14
Answers to Coursebook exercises
6 a $125
b $200
7 a $24
b $42
8 a 24 and 42
b 120
9 110 g of syrup, 220 g of butter and 440 g of oats
10 1500 ml or 1.5 l
End-of-unit review
1
a 1:4
g 2 : 25
2 a 1:8
b 4:5
h 2:3
c 5:1
i 7:2
b 3:8
c 2 : 11
d 9:8
e 1:5:8
f 2:6:3
3 $72, $108, $180
4 a 21
b 7
c 28
b $12
c $60
5 $495
6 $3500
7 a $3
8 $45
9 a 750 g
10 a 24, 30 and 42
b 1050 g or 1.05 kg
b 114
11 sugar = 50 g, butter =100 g and flour = 400 g
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
15 Probability
✦ Exercise 15.1
1
a 0.3
The probability that an outcome does not happen
b 0.9
2 a 88%
b 86%
c 58%
3 a 0.4
b 0.9
c 0.7
4 a The list does not include all possible makes.
b i 92%
ii 93%
iii 85%
5 a 99.5%
b 42.1%
✦ Exercise 15.2
1
a
2 a
1
6
2
11
4 a
5 a
5
6
9
11
b
b
3 a 0.09
1
40
2
15
c
c
5
6
3
11
2
3
d
d 0
e
6
11
b 0.04
c 0.92
d 0.42
e 0.58
1
5
13
15
364
365
c 18
1
c 30
31
iii 365
d 43
29
d 30
iv 334
365
e
e 0.7
b
b
1
6 a i 365
Equally likely outcomes
ii
e
9
16
3
10
f
3
8
g
15
16
h 0
b ii and iv
7 a 0.1
b 0.9
c 0.3
d 0
8 a 0.01
b 0.99
c 0.81
d 0.19
9 a i
1
ii
2
iii
1
b i
1
4
ii
1
2
iii
1
4
10 a Each pair of coins in Q9 could be combined with a H or a T. Here is a list: HHH, HHT; HTH, HTT;
THH, THT; TTH, TTT.
1
8
b i
1
8
ii
✦ Exercise 15.3
1
a
1
36
b
iii
3
8
3
8
iv
Listing all possible outcomes
25
36
c
5
18
2 a 2
b 12
c The totals are not all equally likely.
3 a 7
b 2 and 12
c
4 a
1
6
b
2
9
1
36
d
1
6
e
5
12
f
1
2
g
5
12
5 a They could be shown in a table like this.
H1
T1
b i
1
12
H2
T2
H3
T3
ii
H4
T4
1
4
Copyright Cambridge University Press 2013
H5
T5
H6
T6
iii
1
6
Cambridge Checkpoint Mathematics 8
1
Unit 15
6 a
Answers to Coursebook exercises
+
1
1
3
2
3
3
5
3
4
4
6
2
9
b i
7 a
×
1
2
3
4
5
6
5
6
6
8
1
3
ii
1
1
2
3
4
5
6
iii
2
2
4
6
8
10
12
b 18
3
3
6
9
12
15
18
ii
5
5
10
15
20
25
30
6
6
12
18
24
30
36
8
9
iv
5
18
v
3
4
iii
2
5
Second pen
First pen
B1
B2
B3
B4
R
B1
X
B1, B2
B1, B3
B1, B4
B1, R
B2
B2, B1
X
B2, B3
B2, B4
B2, R
B3
B3, B1
B3, B2
X
B3, B4
B3, R
B4
B4, B1
B4, B2
B4, B3
X
B4, R
R
R, B1
R, B2
R, B3
R, B4
X
b You cannot take the same pen twice.
9 a
Shen
2
3
R
RR
SR
PR
R
S
P
c
F Exercise 15.4
1
19
36
iii
8 a
b
2
3
iv
4
4
8
12
16
20
24
1
9
c i
5
9
c
i
3
5
ii
1
5
Tanesha
S
P
RS
RP
SS
SP
PS
PP
1
3
Experimental and theoretical probabilities
a 0.6
b Not enough throws
c 0.45
f It is based on the largest number of throws.
d 0.36
e 0.37
2 a 4
b 24
c 0.28, 0.32, 0.42, 0.453, 0.44
d 0.44, 0.48, 0.45, 0.507, 0.47
e They are quite different at first but they get closer together with more throws.
f 0.455
3 a i 0.8
c i 0.8
ii 0.2
ii 0.2
b No, not enough throws to say that.
d Yes, with 100 throws the probabilities should be closer to 0.5.
4 a No, not enough throws to decide.
b No, you should not expect them all to be exactly 50. The experimental probabilities for each score are 0.15,
0.193, 0.16, 0.16, 0.153, 0.183. If the dice is fair the theoretical probabilities are all 0.167. The values seem close,
so there is no evidence that the dice is biased.
5 a 0.6
b i 0.575
ii 0.633
iii 0.6125
iv 0.64
c 0.64 because it is based on all the throws.
d 0.7
e The estimate based on 200 throws is the closest. The estimate based on 100 throws is the next closest.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
Unit 15
End-of-unit review
1
a 0.17
b 0.95
2 a 0.9
b 0.7
3 a Shuffle the cards and place them face down before choosing; take a card without looking.
b 0.7
c 0.7
4 a 0.1
5 a
d
e i
b 0.8
c 0.81
+
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
×
1
2
3
4
1
1
2
3
4
2
2
4
6
8
3
3
6
9
12
4
4
8
12
16
3
16
ii
0
b 5
iii
3
16
c i
iv
13
16
1
8
ii
v
13
16
iii
5
8
1
4
6 a 0
b 0.06, 0.04, 0.045
c Three identical numbers has a small probability. We need a lot of throws to estimate it.
d A 0.025, B 0.015, C 0.005, D 0.035
e 0.025
f It is based on a lot more throws.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Coursebook exercises
16 Position and movement
F Exercise 16.1
1
a
Transforming shapes
b
y
6
5
5
4
4
4
3
3
3
2
2
2
1
1
1
0
0
1
2
3 4 5 6
mirror line x = 4
2 a
7 x
0
1
2
3 4 5 6
mirror line y = 3
b
3 a
y
6
5
0
0
7 x
0
c
b
y
6
c
y
6
5
4
4
4
3
3
3
2
2
2
1
1
1
0
0
2
3
4
5
6
7 x
0
1
2
3
4
2 3 4 5 6
mirror line x = 3.5
7 x
2
7 x
y
6
5
1
1
d
5
0
4
c
y
6
5
6
7 x
0
0
1
3
4
5
6
y
6
5
A
4
c
3
a
2
b
1
0
0
1
5 a A to B
2
3
4
5
6
7
x
b A to C
Copyright Cambridge University Press 2013
c B to D
d C to E
Cambridge Checkpoint Mathematics 8
1
Unit 16
Answers to Coursebook exercises
F Exercise 16.2
1
Enlarging shapes
b
a
Scale factor 2
Scale factor 3
c
d
Scale factor 2
Scale factor 4
e
f
Scale factor 3
Scale factor 4
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
g
Unit 16
h
Scale factor 2
Scale factor 3
i
Scale factor 4
2 a
y
8
b
(4, 4), (10, 4), (4, 7)
b
(1, 2), (9, 2), (5, 6), (3, 6)
7
6
5
4
3
2
1
0
0
3 a
1
2
3
4
5
6
7
8
9
10 11
y
6
x
5
4
3
2
1
0
0
1
2
3
4
5
6
7
Copyright Cambridge University Press 2013
8
9
10
x
Cambridge Checkpoint Mathematics 8
3
Unit 16
Answers to Coursebook exercises
4 a
b
Scale factor 2
Scale factor 3
c
Scale factor 4
5 a Scale factor 2, centre of enlargement at (2, 2)
b Scale factor 3, centre of enlargement at (2, 9)
End-of-unit review
1
a
b
y
6
5
5
4
4
3
3
2
2
1
1
0
0
2 a
1
2
3 4 5 6
mirror line x = 4
0
7 x
b
y
6
5
4
4
3
3
2
2
1
1
0
1
2
3
4
5
6
7 x
Cambridge Checkpoint Mathematics 8
0
1
2 3 4 5 6
mirror line y = 3.5
7 x
0
1
2
7 x
y
6
5
0
4
y
6
0
3
4
5
6
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
3 a
Unit 16
b
Scale factor 2
4
Scale factor 3
y
6
5
A
D
4
B
C
3
2
1
0
0
1
2
3
4
5
6
7
x
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
5
Answers to Coursebook exercises
17 Area, perimeter and volume
F Exercise 17.1
1
a i
b i
The area of a triangle
5.33 cm2
1 × 4 × 3 = 6 cm2
2
ii 48.19 cm2
ii 1 × 10 × 8 = 40 cm2
iii 11.02 m2
iii 1 × 6 × 4 = 12 m2
2
2
2 a By estimating: 1 × 8 × 8 = 32 cm2, quite a way from 40 cm2.
2
c he swapped the 7 and the 9 around.
b 9.7 cm2
F Exercise 17.2
1
The areas of a parallelogram and a trapezium
a 18 cm2
b 390 mm2
c 30.66 cm2
2 a 25 cm2
b 38.5 cm2
c 26.28 cm2
3 a She did not notice that the parallelogram is measured in mm.
b 70.2 cm2
4 a A = 18.81 cm2 ( 1 × (5 + 5) × 4), B = 15.54 cm2 (4 × 4), C = 9.86 cm2 (3 × 3), D = 11.07 cm2 ( 1 × 8 × 3)
2
2
c Any shape that has an area equal to 24.48 cm2.
5 32 mm or 3.2 cm
6 30 mm or 3 cm
F Exercise 17.3
1
a 37.7 cm
d 44.0 cm
The area and circumference of a circle
b 31.4 m
e 28.3 m
c 75.4 cm
f 11.0 m
2 a 28.26 cm2
d 254.34 cm2
b 153.86 m2
e 94.985 m2
c 19.625 cm2
f 32.1536 m2
3 a i 51.4 cm
c i 41.1 cm
e i 22.1 cm
ii 157 cm2
ii 100.5 cm2
ii 29.0 cm2
b i 38.6 m
d i 33.4 m
f i 16.4 mm
ii 88.3 m2
ii 66.3 m2
ii 16.1 mm2
4 Xavier is correct.
area of semicircle = 10.132 cm2, area of quarter-circle = 9.0746 cm2
5 Tanesha is correct.
perimeter of semicircle = 38.55 m, perimeter of quarter-circle = 35.7 m
F Exercise 17.4
1
The areas of compound shapes
a Area A = l × w = 5 × 4 = 20
Area B = l × w = 11 × 2 = 22
Total area = 20 + 22 = 42 cm2
b Area A = 1 × b × h = 1 × 12 × 6 = 36
2
2
Area B = l × w = 12 × 3 = 36
Total area = 36 + 36 = 72 cm2
2 a i 3 cm
c i 7 cm
ii 68 cm2
ii 138 cm2
b i 7 cm, 8 cm
d i 6 cm
ii 98 cm2
ii 180 cm2
3 a 26 cm2
b 55 cm2
c 78 cm2
d 89.25 cm2
4 a 34 cm2
b 34.375 cm2
c 39 cm2
5 No. Area of trapezium shape = 88 cm2, area of circle shape = 87.92 cm2.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 17
Answers to Coursebook exercises
F Exercise 17.5
1
The volumes and surface areas of cuboids
a 189 cm3
b 576 mm3
c 60 m3
2 a 222 cm2
b 432 mm2
c 122 m2
3 a 54 cm3
b 6.3 m3
c 5880 mm3 or 5.88 cm3
4 a 123.6 cm2
b 32.4 m2
c 2716 mm2 or 27.16 cm2
5
Length
Width
Height
Volume
a
4 cm
8 cm
7 cm
224 cm3
b
10 cm
5 cm
6 cm
300 cm3
c
12 mm
9 mm
6 mm
648 mm3
d
8m
2m
6m
96 m3
e
4.2 cm
1 cm
3.5 cm
14.7 cm3
f
3.6 cm
5 mm
12 mm
2160 mm3
6 a A = 34.72 cm3 (4 × 3 × 3), B = 29.92 cm3 (7 × 4 × 1), C = 48.96 cm3 (8 × 3 × 2)
c Sketch of any cuboid that has an volume equal to 24.24 cm3.
7 488 cm2
8 184.5 cm2
F Exercise 17.6
1
2
Using nets of solids to work out surface areas
a i
ii 1620 cm2
b i
ii 264 cm2
c i
ii 756 cm2
d i
ii 390 cm2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
Unit 17
2 a he mixed up the measurements 6 cm and 6.8 cm.
He has not changed the 15 mm to 1.5 cm.
He forgot to add area F.
b 111 cm2
3 No. Surface area of cube = 138.24 cm2, surface area of the triangular prism = 138.54 cm2
End-of-unit review
1
a 66.88 cm2
b 28 cm2
c 160 m2
2 a i
b i
25.1 cm
37.7 cm
ii 50.3 cm2
ii 113.0 cm2
3 a i
b i
2 × 3 × 4 = 24 cm
3 × 12 = 36 cm
ii 3 × 42 = 48 cm2
ii 3 × 62 = 108 cm2
4 15.4 cm
b 57.12 cm2
5 a 29 cm2
6 120 cm2
7 a 200 cm3
b 220 cm2
8 2208 mm2
9 a 5 cm
5 cm
13 cm
b 360 cm2
13 cm
10 cm
12 cm
5 cm
13 cm
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Coursebook exercises
18 Interpreting and discussing results
✦ Exercise 18.1
1
a 8
Interpreting and drawing frequency diagrams
b 7
2 a 13
c 25
b 200–400 g
3 a
c 5
d 50
Number of cups of coffee sold per day
14
Frequency
12
10
8
6
4
2
0
0–19
20–39
40–59
60–79
80–99
Number of cups of coffee sold
b February. The only month which has only 28 days.
c Not really. It could be 99, but you can’t tell from grouped data information; the greatest number of cups
of coffee sold could be anywhere from 80 to 99.
4 a
Speed of cars
14
Frequency
12
10
8
6
4
2
0
50
60
70
80
90
100
Speed of car (km/h)
b 17
c No. It could not be 50 km/h as ‘50 <’ means that the speed could be very close but not equal to 50.
5 a
Heights of plants
Frequency
12
10
8
6
4
2
0
20
25
30
35
40
Height (cm)
b 17. Add the frequencies of the three bars that show heights that are at least 25 cm.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 18
Answers to Coursebook exercises
F Exercise 18.2
1
a
Interpreting and drawing pie charts
Favourite flavours of ice cream Vanilla = 72°, Strawberry = 108°, Raspberry = 60°, Chocolate = 96°,
Caramel
Caramel = 24°
Vanilla
Chocolate
Strawberry
Raspberry
b 20%
2 a Vauxhall
3 a 120
60 = 1
360 6
b
b 135
c 35%
d 40
c No. Men = 180, women = 200.
d More women than men took part in the survey, so when the angles in the pie charts are the same, the women’s
sector must represent a greater number than that in the men’s sector.
4 a
90 = 1
360 4
b i 21
ii 35
iii 180
5 Pembroke School. Pembroke School = 160, Milford School = 154.
F Exercise 18.3
1
a i $1 million
d 2010 and 2011
Interpreting and drawing line graphs
ii $1.5 million
b 2008
c 2007 and 2008
e From 2006 to 2008 profits rise; from 2008 to 2011 profits fall.
2 a i 12
ii 15
b August
c February and March
d From January to August there is a rise in the number of skateboards sold each month. From August to
December there is a fall in the number of skateboards sold each month.
b 2008
c 2004 and 2006
3 a i $120 000
ii $170 000
d From 2000 to 2004 the value went up slowly. From 2004 to 2006 the value went up faster. From 2006 to 2008
the rate of increase in value was slower, then from 2008 to 2010 the value fell fast.
e i $140 000
ii $180 000
4
Number of people
Number of people staying in a hotel each month for a year
30
25
20
15
10
5
0
J
F
M
A
M
J
J
A
S
O
N
D
Month
August and September
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Coursebook exercises
Price of silver (US$)
5
Unit 18
Average price of silver over a 25-year period
25
20
15
10
5
0
1985
1990
1995
2000
2005
2010
Year
2005 to 2010
F Exercise 18.4
1
a 15
2 a 22
Interpreting and drawing stem-and-leaf diagrams
b 45 minutes
c 5
d i 45 minutes
ii 56 minutes
iii 22 minutes
b 6.3 cm
c 8
d i 4.5 cm
ii 4.6 cm
iii 4.0 cm
3 a February. The only month which has 28 days.
b 112
c 15
4 a Key: 5 | 8 means 58 kg
5
6
7
8
9
b 17
8
0
1
0
0
9
1
2
2
2
c
9
2 4 4
3 3 5
5 6 9
5
i 64 kg
4
8
9
9
ii 72 kg
iii 37 kg
5 a Key: 10 | 1 means 101 kb
10
11
12
13
14
15
16
b 10
6 a 44%
7
a 1
2
1
0
5
0
0
1
0
3
5
5
0
5
2
2
c
8 9
7 7
8
1 5 9 9
8
4 5 8
5 6 8
i no mode
b
5 =1
25 5
c 20
iii 67 kb
d 27.88 out of 40 (or 69.7%)
b 25%
F Exercise 18.5
1
ii 137 kb
Drawing conclusions
Yes. $435.20 ÷ 17 = $25.60
2 a Allerton have a higher mode (4 compared to 3).
b Batesfield have a higher mean (2.88 compared to 2.52) and a higher median (3 compared to 2).
3 Yes. 19 ÷ 30 × 100 = 63.333...%
4 a i The level of stock is falling at a steady rate so sales are steady.
iiThe level of stock is falling at a reducing rate, and much more slowly than that of the Scarlets. Sales are
slow and declining.
b No. If the trend continues, they will sell out half way through the week.
c Yes. If the trend continues, they will only sell 1 or 2 shirts and they have 4 in stock.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Unit 18
Answers to Coursebook exercises
End-of-unit review
1
a The data is continuous.
2 a
14
b 13
c 400−600 g
d 4
e 50
Number of MP3 players sold daily over a month
Frequency
12
10
8
6
4
2
0
0–9
10–19
20–29
30–39
40–49
Number of MP3 players sold daily
b No. Several months have 30 days.
c Yes. The maximum is 49.
Number of MP3 players sold 0–9 = 36°, 10–19 = 60°, 20–29 = 144°, 30–39 = 96°, 40–49 = 24°.
daily over a month
40–49
0–9
d
10–19
30–39
20–29
e 40%
3 a Key: 1 | 8 means 18 kg
0
1
2
3
4
5
b 48%
8
2
4
4
3
0
9
8
7
6
5
0
8
8
6
6
c
ei Mode is 28.
4
8
7
9
6
25
8
8
8
9
9
9
d 20
ii Median is 34, mean is 32.
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Download