Answers to Practice Book exercises
1 Integers, powers and roots
F Exercise 1.1
Arithmetic with integers
1 a 3
b −10
c −10
d 5
e −6
2 a −3
b −10
c 6
d 4
e −13
3 a −4
b −10
c −50
d −10
e −13
4 a 10
b 13
c −5
d 6
e 25
5 a −3
b −1
c 1
d −7
e 2
6
×
−3
−1
2
5
−3
9
3
−6
−15
−1
3
1
−2
−5
2
−6
−2
4
10
5
−15
−5
10
25
7 a −10
b −8
c 11
8 −30 ÷ 6 = −5 and −30 ÷ −5 = 6
9 −5 times −5 is 25.
10 They could be: 1 and −16; −1 and 16; 2 and −8; −2 and 8; −4 and 4.
11 a −10
b −3
c −5
d 7
F Exercise 1.2
Multiples, factors and primes
1 a 12, 24, 36
b 15, 30, 45
c 32, 64, 96
d 50, 100, 150
2 a 40
b 8 and 12
c 9
d 23
3 41, 43, 47
4 a 96, 98 and 100
b 95 and 100
c 100
d 100
e 97
f 96
5 No. 67 is prime but 57 is not because 3 and 19 are factors.
6 a True. 84 = 7 × 12.
d False. It is 18.
7 a 12
b 60
8 a 1, 3, 9, 27
b False. 75 and 90 are multiples of 15 but not 75.
e False. It is 25.
c 100
b 1, 2, 4, 7, 14, 28
c True. It is 97.
d 42
c 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
9 a 2
b 2 and 3
c 2, 5 and 7
d 3 and 11
10 a 3
b 6
c 12
d 1
d 1, 2, 41, 82
e 1, 31
e 19
11 Because 17 is a factor of each of them. If 221 was prime, the only factors would be 1 and 221.
12 There are lots of possible answers. One is 4 and 9. Another is 10 and 21.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 1
Answers to Practice Book exercises
F Exercise 1.3
1 a
More about prime numbers
b
88
8
11
9
4
2
c
135
3
260
26
15
3 3
2
5
10
13 2
5
2
2
2 a Several trees are possible. The end numbers are always 2, 2, 2, 2, 5.
3 a 450
b 432
4 a 2×3
b 2² × 3² × 5 × 7
5 a i 2³ × 3 × 5
ii 25 × 5
6 a 12
b 672
7 a 52
b 312
b 24 × 5
c 484
b 480
c 40
8 2 ×5
4
4
9 a The only common factor of two prime numbers is 1.
10 a 3
b 2 × 7 × 11
4
F Exercise 1.4
c They have no prime factors in common.
Powers and roots
1 a 8
b 27
c 64
d 125
2 a 16
b 81
c 256
d 10 000
3 a 8
b 6
4 a 0
b −100
5 a 1 and −1
6 a
b Multiply the two primes together.
b 6 and −6
e 1000
c 13 and −13
d 16 and −16
33 − 1 = 27 − 1 = 26 = 13 and 3² + 3 + 1 = 9 + 3 + 1 = 13
2
2
2
7 a 64
b 16
8 a 11
b 17
c 20
d 1
9 a 2
b 5
c 3
d 10
10 a 14 641
e 19 and −19
b Both equal 21.
3
c 5 − 1 = 52 + 5 + 1
4
b 11
11 One of the square roots of 25 is −5. That is less that both the square roots of 16, which are 4 and −4.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
2 Sequences, expressions and formulae
F Exercise 2.1
1 a 3, 5, 7
Generating sequences
b 2, 0, −2
c 3, 8, 13
d −1, −6, −11
e −10, 10, 30
f −100, −120, −140
2 35. Check students’ explanations, e.g. start with 10 and add 5 five times (or 5 × 5).
3 195. Check students’ explanations, e.g. start with 5 and add 10 nineteen times (or 10 × 19).
4 7. Check students’ explanations, e.g. start with 23 and subtract 4 four times (or −4 × 4).
5 20. Check students’ explanations, e.g. start with 35 and subtract 3 five times (or −3 × 5).
6 40. Check students’ explanations, e.g. start with 20 and add 4 five times (or 4 × 5).
7 10. Check students’ explanations, e.g. the term-to-term rule is ‘add 3’, so start with 16 and subtract 3 twice
(or −3 × 2).
8
Position number
1
2
4
8
50
Term
0
1
3
7
49
9 a 2, 4, 6, 8
10 a i 110
c i 60
b 11, 12, 13, 14
ii 120
ii 110
c 5, 7, 9, 11
b i 100
d i 40
d 1, 4, 7, 11
ii 200
ii 90
11 C. Sequences A, B, C and D all give the correct third term, but only C gives the correct eighth term.
F Exercise 2.2
Finding rules for sequences
1 a i ‘add 3’
b i ‘add 2’
c i ‘add 6’
iii 3 × position number
iii 2 × position number + 1
iii 6 × position number − 3
2 a term = 6 × position number
b term = 3 × position number + 4
c term = 10 × position number − 2
3 a i ‘add 1’
b i ‘add 1’
c i ‘add 1’
iii term = position number + 1
iii term = position number + 11
iii term = position number + 21
4 a term = position number + 4
b term = position number + 24
c term = position number + 124
5 a 7, 10, 13, 16
b ‘add 3’
c Three extra grey squares are added to get the next pattern (or term).
d term = 3 × position number + 4
6 a
Copyright Cambridge University Press 2013
b term = 2 × position number + 1
Cambridge Checkpoint Mathematics 8
1
Unit 2
Answers to Practice Book exercises
F Exercise 2.3
Using the nth term
1 a 5, 6, 7; 14
d −1, 0, 1; 8
b 2, 4, 6; 20
e 3, 6, 9; 30
2 a
b
c
d
e
c 6, 8, 10; 24
f 1, 4, 7; 28
3, 5, 7, 9
‘add 2’
Two extra grey circles are added to get the next pattern (or term).
term = 2 × position number + 1
2nd term = 2 × 2 + 1 = 5; 3rd term = 2 × 3 + 1 = 7; 4th term = 2 × 4 + 1 = 9
3 a Three extra grey squares are added to get the next pattern (or term).
b term = 3 × position number − 2
4 aCheck students’ explanations, e.g. he put n + 3, it should have been 3 × n and he should have
put +1 at the end, not +3.
b 3n + 1
F Exercise 2.4
1 a i
b i x
Using functions and mappings
x
1
2
3
4
y
6
7
8
9
ii
x
5
6
7
8
y
0
1
2
3
0 1 2 3 4 5 6 7 8 9 10
y 0 1 2 3 4 5 6 7 8 9 10
ii 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 y=x+5
2 a i
iii
ii y = x − 5
x
1
2
3
4
y
7
9
11
13
x
2
6
10
30
y
11
13
15
25
b i y = 2x + 5
3 a i ‘subtract 5’
ii
iv
x
2
5
10
20
y
1
7
17
37
x
5
10
40
50
y
−2
−1
5
7
ii y = 2x − 3
iii y = x +10
iv y = x − 3
ii ‘divide by 2’
b i y=x−5
ii y = x2
2
5
4 y = 2x + 5. Check students’ explanations.
5 Maha. Check students’ explanations, e.g. all of Maha’s work, but only one of Hassan’s works.
6 y = 5x − 3. Check students’ explanations.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
F Exercise 2.5
Unit 2
Constructing linear expressions
1 a c−2
b c + 10
2 a 2n + 7
b n3 + 6
3 a $(4f + r)
b $(12f + 3s + 2r)
c
c
2
d
3c or 3 c
4
4
e 2c + 5
4 D. Check students’ explanations, e.g. to multiply n + 4 by 3 the n + 4 must be in brackets.
5 2(n – 5)
F Exercise 2.6
Deriving and using formulae
1 a 10
e −2
i −22
b 2
f 7
j −2
c −9
g 25
k −5
d −7
h −2
l 12
2 a
e
i
m
b
f
j
n
c 25
g 8
k 32
d −11
h 640
l 100
10
48
6
19
−6
501
−25
−40
3 a i number of seconds = 60 × number of minutes
b 1800
ii s = 60m
4 64
5 8
6 24
7 Neither. Volume of pyramid A = 32 cm3, volume of pyramid B = 32 cm3.
8 477.25 = 12.55h + 38
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Practice Book exercises
3 Place value, ordering and rounding
F Exercise 3.1
Multiplying and dividing by 0.1 and 0.01
1 a i 100
c i 100 000 000
ii one hundred
ii one hundred million
b i 10 000
d i 1 000 000 000
ii ten thousand
ii one billion
2 a 101
b 106
c 103
d 107
3 a 3.3
e 0.77
b 99.9
f 0.7
c 3
g 7
d 0.87
h 0.07
4 a 50
e 500
b 56
f 560
c 556
g 5560
d 5.5
h 55
5 a 2.7
b 0.279
c 2
d 270
6 a ÷
b ×
c ÷
d ×
e ÷
f ×
7 a 0.1
b 0.1
c 0.01
d 0.1
e 0.01
f 0.01
8 D
9 2.34
10 a 0.1, as 0.1 ÷ 0.1 = 1
F Exercise 3.2
b Use any number smaller than 1.00.
Ordering decimals
1 a
c
e
g
3.76, 6.07, 7.36, 7.63
19.42, 19.44, 23.05, 23.4
45.399, 45.454, 45.545, 45.933
31.14, 31.148, 31.41, 31.425
b
d
f
h
2 a
c
e
g
0.2 cm, 7 mm, 27 mm, 4.3 cm
b 19.5 mm, 29 cm, 34.5 cm, 500 mm
2000 g, 3 kg, 5550 g, 75.75 kg
d 0.9 kg, 1.75 kg, 1800 g, 1975 g
100 ml, 0.125 l, 150 ml, 0.2 l
f 0.05 km, 999 m, 2750 m, 25 km
200 g, 50 000 g, 57.725 kg, 359 999 g, 500 kg, 0.75 t, 850 kg, 1.001 t
3.08, 5.99, 8.03, 8.11
1.08, 1.18, 1.3, 2.11
5.009, 5.077, 5.183, 50.44
7.02, 7.052, 7.2, 7.502
3 a <
j <
b >
k >
c >
l <
d >
e >
f <
g <
h >
i >
4 a ≠
b =
c ≠
d ≠
e =
f ≠
g =
h ≠
i ≠
5 a 32 km, since it’s much further than the rest. 1.6 m, since it’s only about two steps.
b No. 0.5 km × 10 = 5 km, but her furthest is only 4 km.
c Honesty. All her distances are multiples of 250 m; most of Frank’s are not.
6 1.23, 1.32, 2.13, 2.31, 3.12, 3.21, 12.3, 13.2, 21.3, 23.1, 31.2, 32.1
F Exercise 3.3
Rounding
1 a 10
f 35 000
j 37 500 000
b 430
g 70 000
k 37 000 000
c 500
h 350 000
l 89 000 000
d 300
i 800 000
e 8000
2 a 83
g 0.05
b 60
h 2.73
c 0
i 60.00
d 523.8
e 37.3
f 1.0
3 a B
b C
c A
d B
e B
f C
4 a No. This is to one decimal place; the correct answer is 17.
b Correct
c Correct
d No. Forgot to change the 5 to a 6, answer is 46.00.
e No. Did not round up, answer is 40.0.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 3
Answers to Practice Book exercises
F Exercise 3.4
Adding and subtracting decimals
1 a 14.72
e 6.15
b 65.65
f 86.267
c 13.52
g 33.197
d 21.1
h 22.179
2 a 3.12
e 5.9
b 19.22
f 40.11
c 41.18
g 11.77
d 14.99
h 3.655
3 a 31.7
b 34.7
c 48.45
d 37.78
4 93.24 m
5 Yes, 0.255 m > 0.23 m.
F Exercise 3.5
Dividing decimals
1 a 16.5
f 130.1
b 14.7
g 113.9
c 13.8
h 101.2
d 7.3
i 13.2
e 151.8
2 a 1.10
f 1.09
b 10.97
g 1.08
c 1.10
h 0.11
d 10.93
i 1.10
e 0.11
3 1.95 m
4 0.43 kg
5 7.43 cm
6 2.1 cm
7 $8.23
8 2.34 kg
F Exercise 3.6
Multiplying by decimals
1 a 4.29 × 3 × 10 should be 4.29 × 3 ÷ 10.
4.29 × 3 equals 12.87, not 12.67
b 1.287
2 a 0.08 = 80 ÷ 100 should be 0.08 = 8 ÷ 100.
31 × 80 ÷ 100 should be 31 × 8 ÷ 100.
b 2.48
3 a 0.46
f 0.3744
b 0.819
g 0.252
c 2.424
h 0.584
d 2.425
i 5.616
e 0.2425
4 a 10.8
f 0.26
b 19.2
g 0.72
c 25.2
h 1.4
d 244.8
i 20.3
e 2.5
5 a 7.38
b 1.036
c 1.316
d 0.046
6 0.2 × 43.6 = 8.72, 96.8 × 0.09 = 8.712, 8.72 > 8.712, so 0.2 × 43.6 is larger.
7 0.4 × 8491.3 m = 3396.52 m = 3.396 52 km = 3.4 km to one decimal place
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
F Exercise 3.7
Unit 3
Dividing by decimals
1 a (24 × 4) ÷ 10 should be 24 × 10 ÷ 4.
b 60
2 a 0.06 = 0.6 ÷ 100 should be 0.06 = 6 ÷ 100.
(35.4 × 100) ÷ 0.6 should be (35.4 × 100) ÷ 6.
b 590
3 a 60
f 54
b 70
g 39
c 60
h 6
d 60
i 765
e 6
4 a 1100
f 60
b 900
g 7100
c 700
h 106
d 300
i 3780
e 20
5 a 58.8
b 31.8
c 29.38
d 20 433.33
6 12.46
F Exercise 3.8
Estimating and approximating
1 a 100
b 100
2 a
b
c
d
i
i
i
i
3 i
$80
600 + 400 = 1000
70 − 50 = 20
900 ÷ 30 = 30
50 × 20 = 1000
c 4
ii
ii
ii
ii
d 15 000
1013 − 424 = 589
28 + 46 = 74
29 × 32 = 928
1128 ÷ 24 = 47
ii total trolleys collected = 401, 401 × $0.20 = $80.20
iv 20¢ = $ 51 , 400 × $ 51 = $80
4 a i $128
ii 3.5 × $28 = $98, add call-out fee $30, total $128
iv estimate $30 call-out fee + 3 hours at $30 per hour = $120
b i 1 hour and 15 minutes
ii $65 − $30 (call-out fee) = $35, 35 ÷ 28 = 1.25 hours = 1 hour 15 minutes
5 i
$2120
iicash price $17 995, payment plan price = $4995 + 36 × $420 = $4995 + $15 120 = $20 115 and
$20 115 − $17 995 = $2120
iv estimate: first payment $5000 + 40 × $400 = $5000 + $16 000 = $21 000; difference $21 000 − $18 000 = $3000
6 i
$19 118.75
ii total muffins = 70 × 5 × 46 = 16 100; payment = 16 100 ÷ 4 × 4.75 = $19 118.75
iv estimate: 400 muffins per week × 40 weeks = 16 000 per year; 16 000 is 4000 batches of 4 muffins;
4000 batches × $5 per batch = $20 000
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Practice Book exercises
4 Length, mass and capacity
✦ Exercise 4.1
Choosing suitable units
1 a m
b cm (or mm)
c t
d g
2 a km2
b cm2
c m3
d cm3
3 a T
b F
c T
d F
e ml
f
l
e F
f
F
4 No. Check students’ explanations, such as: a car is usually less tall than most adults and 2.5 m is much greater than
most adults’ heights.
5 Yes. Check students’ explanations, such as: some of my friends weigh about this much.
6 No. Check students’ explanations, such as: he could not walk as fast as 10 km per hour.
7 67.5 kg
8 10 kg
9 3–6 kg
10 a 6.5–7.5 m
b 11–13 m
✦ Exercise 4.2
1 a T
b F
Kilometres and miles
c F
d T
e F
2 No, a kilometre is shorter than a mile.
3 a 10 miles
b 32 ÷ 8 = 4, 4 × 5 = 20 miles
c 80 ÷ 8 = 10, 10 × 5 = 50 miles
4 a 55 miles
b 45 miles
c 75 miles
5 a 24 km
b 25 ÷ 5 = 5, 5 × 8 = 40 km
c 55 ÷ 5 = 11, 11 × 8 = 88 km
6 a 48 km
b 480 km
c 72 km
d 125 miles
d 7200 km
7 128 km. 128 km = 80 miles and 75 miles = 120 km.
8 296 km. 180 miles = 288 km and 296 km = 185 miles.
9 a 65 miles
b 152 km
c, d 105 miles = 168 km, 304 km = 190 miles
10 $11 077.40, possibly rounded to $11 000 or $11 100
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Answers to Practice Book exercises
5 Angles
F Exercise 5.1
Parallel lines
1 a They are vertically opposite angles.
b
c
x°
y°
2 a = 75°, vertically opposite angles; b = 75°, corresponding to the given angle; c = 105°, angles on a straight line;
d = 105°, alternate angle to c.
3 a g and i
b c and e
4 a i BEF
ii DEB
iii EBC
b No
5 Lines l and n are parallel because corresponding angles (80° and 100°) are equal. The angles are not the same for
line m so that is not parallel to the other two.
6
t°
s°
120°
s = 120°, vertically opposite angles; s = t, corresponding angles; hence t = 120°.
7 Yes. The angle vertically opposite 50° is also 50°. That angle and the 75° add up to 125° and that makes an angle
alternate to the 125° marked.
8
a°
b°
c°
F Exercise 5.2
a = c, corresponding angles; b + c = 180°, angles on a straight line; hence a + c = 180°.
Exploring angle properties
There are alternative explanations for some of the answers in this exercise.
1 120°
2 Each one is 165°.
3 Angle WXV = angle XYZ, corresponding angles; angle VXZ = angle XZY, alternate angles; angles
WXV + VXZ + ZXY = 180°, angles on a straight line; hence angle X + angle Y + angle Z = 180°.
4 Alternate angles; alternate angles; angles on a straight line.
5 Divide the shape into two triangles. Show the angles of the two triangles are the same as the four
angles of the quadrilateral.
6 a = c + d, exterior angle; e = g + h, exterior angle; a + e + f + b = 360°, angles at a point;
hence c + d + g + h + f + b = 360° and these are the angles of PQRS.
7 a The six angles of the two triangles add up to 2 × 180 = 360°. The four angles A, B, C and D must total
less than 360°.
b The 120° angle is the exterior angle of both triangles. Angles at A and D add up to 120° and so do angles
at B and C. Hence A + B + C + D = 240°.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 5
Answers to Practice Book exercises
F Exercise 5.3
Solving angle problems
There are alternative explanations for many of the answers in this exercise.
1
a°
b° c°
a + b = 180°, angles on a straight line, so a = 180 − b; b + c = 180°, angles on a straight line, so c = 180 − b;
hence a and c are equal.
2 a
a° y°
x° b°
c°
z°
x = a + c, exterior angle; y = b + c, exterior angle; z = b + a, exterior angle; x + y + z = a + c + b + c + b + a =
2(a + b + c) = 2 × 180 = 360°.
b
a° w°
x° b°
y° c°
d° z°
a + w = b + x = c + y = d + z = 180°, angles on a straight line; hence (a + b + c + d) + (w + x + y + z) =
4 × 180 = 720°; but w + x + y + z = 360°, angles of a quadrilateral; Hence a + b + c + d = 720° – 360° = 360°.
3
A
w° x°
D
B
y°
z°
C
a x = y, alternate angles; w = z, alternate angles; hence x + w = y + z or angle A = angle C.
b Prove that angle B = angle D in a similar way, by drawing the diagonal BD.
4 a = 110 – 40 = 70°, exterior angle of a triangle; b = 110 – 55 = 55°, exterior angle; c = 110 – 70 = 40°,
exterior angle.
5
x°y°
b°
a°
118°
74°
a = 118 – 74 = 44°, exterior angle; so x = 44°, vertically opposite angle.
b = 74°, alternate angle; y = 180 – 44 – 74 = 62°, angles on a straight line.
6 Divide the hexagon into two quadrilaterals by joining two opposite vertices. Show that the angles of the
quadrilaterals make the angles of the hexagon, so the angle sum is 2 × 360 = 720°.
7
a°
c°
b°
c = a, corresponding angles; c + b = 180°, angles on a straight line; hence a + b = 180°.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
6 Planning and collecting data
✦ Exercise 6.1
Collecting data
1 a experiment
f observation
b survey
g survey
2 a Cheaper, quicker, easier.
c survey
h experiment
d observation
i survey
e experiment
b 41
3 A sample. It would take too long to ask 394 people.
4 a No
b 48
5 a Yes
b 89
6 a population
b population
c sample is much easier, 43
7 a B (unless you are at a very large school)
b C
c A
d population
d C
8 a About 10%, and can be done fairly easily – a good decision.
b Confusing and has nowhere for zero or for more than 10 pairs of shoes. It has overlapping numbers of pairs
of shoes – someone with three or four pairs of shoes could be put in two different categories.
c It depends on what you think of as ‘lots’.
d The data-collection sheet should include non-overlapping numerical values that allow for zero and extreme
data. It is better to use tallies for counting.
9 a
b
c
d
Not really enough, he should have at least 30.
Not good as it has not given numbers. People will have different opinions of what ‘often’ means.
It depends on what you think of as ‘a lot’.
The data-collection sheet should include non-overlapping numerical values that allows for zero and
extreme data.
✦ Exercise 6.2
1 a discrete
f discrete
Types of data
b discrete
g continuous
c continuous
h discrete
d continuous
i discrete
e discrete
j continuous
2 No, height is a measurement, which is continuous data.
3 Yes, but because weight is a measurement which is continuous data, not because some of the values
given are in halves.
✦ Exercise 6.3
1 a T
2 a
Using frequency tables
b F
c F
Length, l (cm)
d T
Tally
Frequency
1<l≤5
///
3
5<l≤9
////
4
9 < l ≤ 13
///
3
13 < l ≤ 17
//// /
6
17 < l ≤ 21
////
4
Total
20
b 6
c 13. Add up the last three frequencies; all in these groups are longer than 9 cm.
d 10. Add up the first three frequencies; all in these groups are shorter than 13 cm.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 6
3 a
Answers to Practice Book exercises
Height, h (cm)
Tally
90 ≤ h < 110
////
4
110 ≤ h < 130
////
4
130 ≤ h < 150
//// //
7
150 ≤ h < 170
//
2
170 ≤ h < 190
/
1
Frequency
18
Total
b 18
4 a 7
c 3
d 8
b 8
c 30
5 a
Arabian
Morgan
Thoroughbred
Other
Total
Female owner
42
18
55
4
119
Male owner
26
44
25
8
103
Total
68
62
80
12
222
b 44
c 142
6
AAA
AA
C cell
Total
Ordinary
3000
6000
15 000
24 000
Rechargeable
1000
4000
1500
6500
Total
4000
10 000
16 500
30 500
Chocolate bar
Ice cream
Popcorn
Total
Boys
3
3
8
14
Girls
10
2
4
16
Total
13
5
12
30
7
2
d 11
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
7 Fractions
F Exercise 7.1
1 a 30%
b
e 20%
f
2 a i
0.99
ii
c i
0.16
ii
3 a i
98%
ii
c i
12%
ii
4 a i
c i
0.45
0.85
5 a i
d i
g i
0.375
0.136
0.625
6
Finding equivalent fractions, decimals and percentages
2
5
3
4
99
100
16 = 4
100 25
c 0.8
d 10%
g 0.5
h 90% = 9
10
88 = 22
100 25
4 = 1
100 25
b i
0.88
ii
d i
0.04
ii
b i
78%
ii
d i
5%
ii
ii 45%
ii 85%
b i
d i
0.06
0.96
ii 6%
ii 96%
ii 37.5%
ii 13.6%
ii 62.5%
b i
e i
h i
0.025
0.525
0.602
ii 2.5%
ii 52.5%
ii 60.2%
49
50
3
25
39
50
1
20
c i
f i
ii 8%
ii 40%
0.08
0.4
75 = 0.075 = 7.5%, not 75%.
1000
F Exercise 7.2
1 a 0.875
•
Converting fractions to decimals
b 0.4375
••
c 0.35
•• •
2 a 0.2
b 0.02
c
3 a 0.429
b 0.273
c 0.231
d 0.28
•
•
0.002 or 0.002
d
e 0.175
••
e
0.18
d 0.214
f 0.0875
••
f
0.06
e 0.136
• •
0. 03
f 0.115
4 No. Ahmad must not double 0.006. Double 0.00666666... is 0.0133333...
5 No. Some calculators round the display.
F Exercise 7.3
Ordering fractions
1 a 3 × 5 = 15, so 1 = 1 should be 3 × 5 = 15, so 2 + 10 = 12
3 × 5 15
5 + 10 15
2
+
10
12
5 + 10 = 15, so
should be 3 × 5 = 15, so 2 × 3 = 6
=
5 × 3 15
5 + 10 15
b 4 , 1, 2
15 3 5
2 a
3 1 3
4, 2, 8
b
5 1 1
12 , 4 , 6
c
3 11 5
4 , 16 , 8
d
2 3 1
5 , 10 , 4
e
4 18 7
5 , 25 , 10
3 a
3 1 2
7, 2, 3
b
3 6 1
10 , 19 , 3
c
13 , 5 , 19
21 8 30
d
5 , 11 , 7
9 19 12
e
10 , 7 , 5
13 9 6
f
67 , 23 , 31
90 30 40
4 5 , 12 , 17 , 3
8 17 24 4
5 The numerators are all the same number. The bigger the denominator, the smaller the fraction.
She must be correct.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 7
Answers to Practice Book exercises
F Exercise 7.4
1
c 5
d 1
e 1
2
4
3
6
2 a 11
b 7
c 11
d 11
e 17
14
6
24
12
24
8
5
13
13
2
1
3 a 5 + 4 = 20 + 20 = 20 , 7 20
3 10 13 13
1 5
1
1
1
b 4 + 6 = 12 + 12 = 12 , 12 = 112 , 9 + 112 = 10 12
1 a
3
4
Adding and subtracting fractions
23 −
3
−
b 42
5
5 a 31
2
1
f 1
2
4 a
b
21 = 115 − 63 = 52 = 3 7
5
15 15 15
15
59 = 84 − 59 = 25 = 5 = 2 1
10 10 10 10 2
2
5
11
b 4
c 2
d 2 13
12
12
16
5
1
g 15
h 2
i 6 42
10
55
6
e 69
32
23
j 1
24
5 .
6 B, 3 5 + 4 11 = 8 1 ; the other two equal 8 18
9
18
6
19
7
7 A, 4 20 − 2 10 = 2 14 ; the other two equal 2 13 .
8 a 2 1 km
8
b 19 3 km
8
F Exercise 7.5
Finding fractions of a quantity
1 a $7.50
b 9 mm
c 20 km
d 15 kg
e 20 cm
f 10 g
2 a 6 2 cm
3
b 23 1 ml
4
c $ 20 5
d 10 2 kg
9
e 9 9 mm
10
f
6
1 13
20 m
3 A, 7 of 24
8
4 B, 3 of 25
4
5
4 of 24 = 19 1 , 7 of 28 = 19 3 , 11 of 27 = 19 4
5
5 10
5
5 15
F Exercise 7.6
Multiplying an integer by a fraction
1 a 25
b 45
c 25
d 20
e 33
f
10
2 a 14 2
3
b 12 4
5
c 10 2
7
d 77
9
e 84
11
f
7
5 13
3 a 13 1
2
b 17 1
2
c 13 3
4
d 13 1
3
e 10 1
2
f
19 14
4 No. Oditi divided the 68 by 4 and the 12 by 3. The divisors must be the same when cancelling.
Then 85 ÷ 4 = 21 14 , 20 54 gives the same decimal (21.25), but it has not been simplified enough.
The correct answer is
3
2
5 × 68 17, 5 × 17 = 85, 85 ÷ 3 = 28 1 .
3
12
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
F Exercise 7.7
Unit 7
Dividing an integer by a fraction
1 a 28
b 24
c 30
d 28
e 50
f
28
2 a 38 1
2
b 12 1
2
c 46 1
2
d 56 2
3
e 33 1
3
f
24 43
3 C, 26 ÷ 6 ; the other two have whole-number answers.
11
F Exercise 7.8
Multiplying and dividing fractions
1 a
1
12
b
9
16
c
5
42
d
9
20
e
6
35
f
5
12
2 a
2
5
b
2
5
c
3
10
d
1
2
e
7
11
f
2
5
3 a
3
4
b
5
12
c
6
7
d
35
54
e
18
25
f
7
30
e 47
12
f
2 10
21
e 3
f
4 23
4 a 11
2
b 11
20
c 11
6
d 15
9
5 a 11
4
b 11
3
c 11
2
d
6 a
720
5040
b
3
5
1 × 2 × 3 × 4 × 5 × 6 =1
5
4
2
3
6
7
7
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Practice Book exercises
8 Shapes and geometric reasoning
F Exercise 8.1
1 a WX
Recognising congruent shapes
b PQ
c TV
2 The angle ABC is not 90°.
3 C, F
4 a i 10.3 m
b i 96°
ii 5.1 m
ii 25°
iii 12 m
iii 59°
5 a i WX
b i ∠XYZ
ii XY
ii ∠YXW
iii PQ
iii ∠RSP
F Exercise 8.2
iv PS
iv ∠SPQ
Identifying symmetry of 2D shapes
1 a
b
c
d
e
f
g
h
i
j
k
l
2 a 2
h 1
b 1
i 1
c 1
j 2
d 2
k 3
e 6
l 1
f 1
g 4
3 a 2
b 5
c 0
d 2
e 2
f 8
g 8
h 4
4 a 2
b 5
c 2
d 2
e 2
f 8
g 8
h 4
5 a
Copyright Cambridge University Press 2013
b
c
Cambridge Checkpoint Mathematics 8
1
Unit 8
Answers to Practice Book exercises
6
F Exercise 8.3
Classifying quadrilaterals
1 a rhombus
b kite
c rectangle
2 a square: 2
e trapezium: 1
b rhombus: 8
f isosceles trapezium: 1
c rectangle: 7
g kite: 2
3 a (2, 4)
b (3, 3)
c (4, 5)
F Exercise 8.4
d parallelogram: 3
Drawing nets of solids
1 There are many possible nets, e.g.
a
b
2 C, D, F
3 There are several possible nets, e.g.
4 Students’ nets must be accurate to ±2 mm.
5 a D
b F
c B
d H
6 a Students’ nets must be accurate to ± 2 mm.
b 25.5 cm (allow 24.6 cm to 26.3 cm)
F Exercise 8.5
Making scale drawings
1 a 80 m
b 24 cm
2 a 1.2 m
b 11 cm
3 a Check students’ scale drawings are accurate, with scale marked as 1 : 200 or equivalent.
b i 24 m
ii 6 m
iii 8 m
iv 10 m
v 16 m
vi 16 m
c 5.5 cm
d 1.5 cm
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
4 a
Unit 8
b 3.30 m (allow 3.23 m to 3.37 m)
8 cm
21 cm
5 122 m (allow 119 m to 125 m)
6 a Students’ drawings must be accurate to ±2 mm.
b 7.8 m (allow 7.65 m to 7.95 m)
A
12 cm
14 cm
B
4 cm
14 cm
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Answers to Practice Book exercises
9 Simplifying expressions and solving equations
F Exercise 9.1
1
a 8a + 7
g 3g + 10
Collecting like terms
b b+8
h 3hw
2 a 3a2: A, E, G
3a + 3: B, D, L
3a – 3: C, H, I
3a + 3b: F, J
b i 3–a×3
ii 3 – 3a
3 a 11a
g 12gh – 2ab
b 2b
h h+9
4
c 3c
i 8i + 4v
d 5dy
j 5u – 2j
e 11e + 6x
k 3 – 9kt
f 5f − 3
l –3ls – 4qr
c –2c
i 14i – 13j
d 3d + e
j 5j 2 + 12 j
e 2f – 4g
k 5k 2 – 2k
f 7ef
l 2l 2 + 6
25x + 10y
16x + 2y
8x
9x + 8y
8x + 2y
2y + 3x
x + 6y
5x – 2y
5
3x + 4y
2y – 2x
15ab – 6cd
10ab
8ab + 3cd
7ab + cd
5ab – 6cd
2ab – 3cd
ab + 2cd
3ab – 3cd
ab – 5cd
2ab + 2cd
For questions 6 and 7, check students have drawn two pyramids. Each adjacent pair of blocks in the bottom row
must add to give the block directly above them.
6
7d + 4ef
Students’ own
7
answers
9ef + 5mn
6ef + 2mn
Students’
3ef + 3mn
own
answers
8 a 1. You can’t simplify 10xy and 5xz by adding them together as the algebraic terms are different.
2. You can simplify 4cb – bc by subtraction to 3bc as the algebraic terms are the same.
b 1. 10xy + 5xz 2. 2a 2 + 3bc
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 9
Answers to Practice Book exercises
F Exercise 9.2
1
a
g
m
s
6a + 36
36 – 6g
54a + 48m
3a – 9b – 3
Expanding brackets
b
h
n
t
5b + 35
35 – 5h
35b + 30n
5x – 5y – 5z
c 7c – 56
i 56i + 63
o 49c – 56x
d 6d – 54
j 42j + 48
p 54px + 48y
e 5e + 40
k 30k – 35
q 35qy – 30x
f 7f + 49
l 56 – 63l
r 49r + 7s + 56
d 48d + 7
e −20e − 33
f
d 3dx − 3d
j 3aj − 7j 2
p 18p 2 + 18p
e 3e 2 + 6e
k 3k 2 − 6kx
q 30q 2 + 36qx
f 5f 2 + f
l 3lx − 6lz
r 9r 2 − 3rx − 9r
d 2d 2 − d
e 9e − e 2
f
2 No. 4(a − 7) = 4a − 28 and 4(7 − a) = 28 − 4a
3a 14a + 114
4 a
g
m
s
b 38b + 92
c 70c + 128
b b 2 − 5b
c 3c 2 + 6c
a2 + a
7g − 3gx
h 6h − hx
i 3i 2 + 7ix
2
2
3m + 9mx
n 18n − 18n
o 24x 2 − 12xy
2
2
4a + 2ab + 6a
t –3x – 3xy – 3xz
5a 2a 2 + 7a
b 5b 2 + 8b
c 8c 2 + 10c
108f + 33g
39fg − 27f 2
6 a 1. He wrote –3(3a + 5) instead of –5(3a + 5), and that 3a – 9a = 6a instead of −6a.
2. He wrote 4pq − 4pq = pq instead of zero.
3pr + pq can’t be simplified by adding them together as the algebraic terms are different.
3.He multiplied the 5b in front of the first bracket with the a in front of the second bracket. He forgot to
include the 15ab when adding the ab terms.
b 1. −12a − 10
2. pr + 2qr
3. 4a 2 + 5b 2 + 21ab
7 Yes. x(x − 3) + x(x + 5) = x 2 − 3x + x 2 + 5x = 2x 2 + 2x and 2x(x + 1) = 2x 2 + 2x.
F Exercise 9.3
Constructing and solving equations
1a a = 2, b = 4
b c = 2, d = 3
c e = 7, f = 3
2a a = 8
b b=5
c c=6
3a x = 5
b x=4
c x=3
4 a 2n + 8 = 20, n = 6
d 5n + 10 = 7n, n = 5
2
b
n − 3 = 2, n = 20
4
d g = 20, h = 7
e 3(n + 1) = 4n − 4, n = 7
Cambridge Checkpoint Mathematics 8
e i = 5, j = 4
f
k = 8, m = 1
c 4n − 6 = 2n + 12, n = 9
f
5(n − 6) = 4(n − 5), n = 10
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
10 Processing and presenting data
✦ Exercise 10.1
Calculating statistics from discrete data
1 a 4
b 3
c
2 a i 51
ii 51
iii 50.6
iv 7
b No. All three averages are over 50.
3 a i 6
ii 8
iii 8.0
iv 5
b 13
4 a 17
b 9
c
1.8
5 a 1, 2 and 3
b 2
c
2.2
2.6
c 13.0
d 5
6 a i 25 minutes
ii 45 minutes
iii 45 minutes
iv 46.5 minutes
b The mode. This is the most popular and it would be interesting to find out why.
c i 20 minutes
ii still 45 minutes
iii still 45 minutes
iv 47.5 minutes.
✦ Exercise 10.2
1 a 20–
Calculating statistics from grouped or continuous data
b 25–
c
29.5
d The maximum possible value for the range is 45 − 20 = 25.
2 a There are two modal classes, 8–10 and 17–19.
b i 13
ii 13
iii 15 or 16 is a reasonable estimate.
3 a 14–
b i 16
ii
18.5
iii About 18
4 The median is in the class 6– and the mean is 6.64. Both of these averages are more than five minutes.
5 a 90–95%
b The mean is 90.7%.
6 a About 45 or 46 is a sensible estimate.
✦ Exercise 10.3
b About 72
c 72.5
Using statistics to compare two distributions
1 The average age of the men is about 9 years more than the average age of the women. The range of the men’s ages
is 48 years and of the women’s ages is 34 years. There is more variation in the men’s ages.
2 Mean: girls 9.3, boys 8.5. Median: girls 9, boys 8.5. Mode is 10 for both. Range: girls 2, boys 4. Girls were better
attenders than boys.
3 a The mean is the best average to use because it uses all the scores.
b The mean for men is 2.3 and for women is 2.7. The women’s average is slightly better but there is little
difference between the two.
4 The new mean and range are about 53.1 and 28. The mean has increased by about 12 and the range of marks has
increased by about 10.
5 a With fertiliser the range has increased by about 10 cm.
b With fertiliser the mean has increased from 21.3 to 28, an increase of about 7 cm.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Answers to Practice Book exercises
11 Percentages
F Exercise 11.1
Calculating percentages
3 , 2 , 1 , 14
10 25 8 5
1 a 0.3, 0.08, 0.125, 1.8
b
2 a 21 m
b 120 people
c 64 kg
d 100
3 a 13.11
b 3569
c $18.59
d $364
4 a 16
b 15.21
c $147
d 10
Only b needs a calculator.
5 The missing numbers are 16.065, 32.13, 96.39 and 128.52.
6 85, 417, 150
7 a 48
b 120
c 16
8 527, 86, 1005
9 996 milllion, 132 million, 24 million
10 Sasha had a better score because 48 out of 65 is 74%.
F Exercise 11.2
Percentage increases and decreases
1 a 3.6
b 33.6 kg
c 26.4 cm
2 a 1900
b 3900 people
c 100 hours
3 a $690
b $3910
4 a 1260
b 6216
c 8736
b $277
c $2115
5 625 g
6 a $770
d $708
7 $20.60, $11.20, $17.70 and $56.90
8 a 3800 km
b 1200 km
9 DVD $9.60, computer game $34, monitor $76
F Exercise 11.3
1 a 72.5%
Finding percentages
b 58%
c 76%
d 71.25%
2 a 34% men, 44% women and 22% children
3 a 75% red and 25% blue
b 83% red and 17% blue
4 a 30.2%
b 69.8%
c 37.9%
5 a 40% increase
e 20% decrease
b 10.8% increase
f 90% decrease
c 85.7% increase
d 2.5% increase
c 87.5% increase
d 100% increase
6 No. The decrease is only 9.2%.
7 a 25% decrease
e 150% increase
b 25% increase
8 First 10%, second 20%, third 15.4%.
9 a 7.9% decrease
b 8.3% increase
c 0.2% decrease
10 a 80%
b 125%
c 85.1%
Copyright Cambridge University Press 2013
d 117.5 %
Cambridge Checkpoint Mathematics 8
1
Unit 11
Answers to Practice Book exercises
F Exercise 11.4
1 a
Using percentages
48 (81.4%)
59
b 127 (78.9%)
161
2 a Highballs 62.5%, Spikers 70%
b Spikers
3 a USA
b China 51%, USA 32.7%, Russia 31.5%
4 a Type A 84%, type B 70%
b Type A
c China
5 a Store X 62.5%, store Y 28.9%
b Store X. The extra number sold was less than Y but the percentage increase was much larger.
6 They were similar. Class A was slightly better with a decrease of 16.1%. Class B’s decrease was 15.1%.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
12 Constructions
✦ Exercise 12.1
Drawing circles and arcs
1 Check students have drawn circles, radii:
a 5 cm
b 3.7 cm
c 6 cm
d 4 cm
Allow ±2 mm.
2 a, b Check students’ drawings.
3 Check students’ drawings:
c cone
a radius 6 cm and angle 60°
b radius 6 cm and angle 135°
Allow ±2 mm and ±2°.
4 100°. Allow ±2°.
✦ Exercise 12.2
Drawing a perpendicular bisector
1 Check students’ drawings of perpendicular bisector of AB; all construction lines must be visible.
2 Check students’ drawings of midpoint of CD; all construction lines must be visible.
3 a He has not kept the radius of the compasses the same. He has not put the compass point at the ends of the line
segment AB.
b He measured the line and drew the dot at the midpoint, then used the compasses to make it look like he did it
properly – he cheated!
f It is a rectangle, with all sides half as long as the original.
4 a–e
✦ Exercise 12.3
Drawing an angle bisector
1 Check students’ drawings of bisecting a 60° angle ABC. All construction lines must be visible.
2 Check students’ drawings of bisecting angle ABC – angle to be anything between 20° and 90° [student’s choice],
except 60°. All construction lines must be visible.
3 Check students’ drawings of bisecting angle DEF – angle to be anything between 100° and 170° [student’s choice].
4 Check students’ drawings of landing sector for the javelin, with dotted line bisecting the angle.
✦ Exercise 12.4
Constructing triangles
1 Check students’ accurate drawings of triangles. All construction lines must be visible.
2 Check students’ accurate drawings of triangles. All construction lines must be visible.
3 a, b
Check students’ accurate drawings of two triangles, both using the diameter as a base. All construction lines
must be visible.
c Both are right-angled triangles.
4 a Check students’ accurate drawings of two triangles, both using the diameter as a base. All construction lines
must be visible.
b Both are right-angled triangles.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Answers to Practice Book 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.
b, c
y
2
1
–2 –1
–1
0
1
2
3
4
5
1
2
3
4
2
3
4
5
x
–2
–3
–4
–5
–6
2 a The values of y are −3, −1, 1, 3, 5, 7, 9, 11.
b, c
y
11
10
9
8
7
6
5
4
3
2
1
–3 –2 –1
–1
0
x
–2
–3
3 a The values of y are 7, 6, 5, 4, 3, 2, 1, 0, −1.
b, c
y
8
7
6
5
4
3
2
1
–2 –1
–1
0
1
6
x
–2
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 13
Answers to Practice Book exercises
4 a The values of y are −10, −7, −4, −1, 2, 5, 8.
b, c
y
8
7
6
5
4
3
2
1
–3 –2 –1
–1
0
x
1
2
3
1
2
3
4
1
2
3
4
–2
–3
–4
–5
–6
–7
–8
–9
–10
5 a The values of y are 2, 1.5, 1, 0.5, 0, −0.5, −1, −1.5, −2.
b
y
2
1
–4 –3 –2 –1
–1
0
x
–2
6 a The values of y are 3, 2, 1, 0, −1, −2, −3, −4, −5.
b
y
3
2
1
–4 –3 –2 –1
–1
0
x
–2
–3
–4
–5
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
7 a This is a possible table.
b
Unit 13
y
9
x
−3
−2
−1
0
1
2
3
8
y
−9
−6
−3
0
3
6
9
7
6
5
4
3
2
1
–3 –2 –1
–1
0
1
2
3
x
–2
–3
–4
–5
–6
–7
–8
–9
8 a This is a possible table.
b
x
−4
−3
−2
−1
0
1
2
3
4
y
−5
−3.5
−2
−0.5
1
2.5
4
5.5
7
y
8
7
6
5
4
3
2
1
–5 –4 –3 –2 –1
–1
0
1
2
3
4
5
x
–2
–3
–4
–5
–6
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3
Unit 13
Answers to Practice Book exercises
F Exercise 13.2
1
Equations of the form y = mx + c
a The values of y are −80, −60, −40, −20, 0, 20, 40, 60, 80.
b, c
y
80
70
60
50
40
30
20
10
–4 –3 –2 –1
–10
0
1
2
3
4
1
2
3
4
x
–20
–30
–40
–50
–60
–70
–80
2 a The values of y are −8, −4, 0, 4, 8, 12, 16, 20, 24.
b, c
y
24
20
16
12
8
4
–4 –3 –2 –1
–4
0
x
–8
4
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
3 a
x
−4
−3
−2
−1
0
1
2
3
4
2x + 3
−5
−3
−1
1
3
5
7
9
11
2x – 2
−10
−8
−6
−4
−2
0
2
4
6
b, c, d
e
y
10
Unit 13
a = 63 and b = 58
9
8
7
6
5
4
3
2
1
–4 –3 –2 –1
–1
0
1
2
3
4
x
–2
–3
–4
–5
–6
–7
–8
–9
–10
4 a The values of y are 4, 3, 2, 1, 0, −1, −2, −3, −4.
b
c −3.6
y
d 12.5
4
3
2
1
–40 –30 –20 –10
–1
0
10 20 30 40
x
–2
–3
–4
5 a The values of y are −48, −40, −32, −24, −16, −8, 0, 8, 16.
b
y
20
10
–4 –3 –2 –1
–10
0
1
2
3
4
x
–20
–30
–40
–50
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
5
Unit 13
Answers to Practice Book exercises
F Exercise 13.3
The midpoint of a line segment
1 a (2, 0)
b (4, −3)
c (−2, −3)
2 a (7, 7)
b (1, −5)
c (1, 1)
3 a (0.5, −0.5)
b The midpoint is 4 + –3 , –1 + 0 = (0.5, −0.5) which is the midpoint of AC.
2
2
(
)
4 PQ (12.5, 5), QR (−5, −12.5), PR (−2.5, 7.5)
5 a (−15, 20)
b (−5, 40)
6 a (3.0, 4.3)
b (−0.7, 2.4)
7 (−2.5, −3.5)
8 (10, 5)
9 (6, 5)
F Exercise 13.4
Graphs in real-life contexts
1 a 60 km/h
b 50 km/h
c 4 seconds
2 a twice
b 4 hours
c 10 30
3 a, b
d about 190 km
c 4 km
Distance (km)
6
Asif
5
4
3
2
Nawaz
1
0
0
10
20
30
40
Time (minutes)
4 a, b
c 6 seconds, about 37 or 38 km/h
Speed (km/h)
60
Second car
50
40
30
First car
20
10
0
0
1
2
3
4
5
6
7
8
Time (seconds)
6
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
14 Ratio and proportion
F Exercise 14.1
Simplifying ratios
1 a 1:3
i 10:1
b 1:9
j 4:1
c 1:8
k 5:1
d 1:4
l 11:1
e 1:3
m 70:1
f 1:3
n 1:70
g 5:1
o 1:1
h 9:1
2 a 3:4
i 3:2
b 2:3
j 5:3
c 4:5
k 5:2
d 3:5
l 11:4
e 5:7
m 15:2
f 5:8
n 2:15
g 4:3
o 3:40
h 9:8
3 a 1:5:6
d 5:6:2
b 2:3:4
e 4:1:8
c 4:2:5
f 12 : 3 : 5
4 a 1:4
b 40 : 3
c 21 : 10
5 a 20 : 40 : 3
d 9 : 3 : 40
b 20 : 11 : 4
e 2 : 30 : 1
c 90 : 3 : 1
f 60 : 11 : 50
6 a 1:5
g 5 : 12
b 1:2
h 3:2
c 1:4
i 1 : 6 : 20
d 4:1
e 10 : 7
f
7:2
d 1:5
e 2:1
f
7:5
7 No. 450 g : 550 g : 1100 g simplifies to 9 : 11 : 22.
8 a The ratio shows that route 3 takes longer than route 2, but the notes say it’s the other way around.
b She swapped the times for route 2 and 3 around.
55 mins is 0.91666… (or 11 hour), not 0.55
12
1 hour 10 mins is 1.1666… (or 1 1 hour), not 1.1
6
1.1 × 100 = 110, not 11
1.5 × 100 = 150, not 15
11 ÷ 5 = 2.2, not 2
c 11 : 18 : 14
F Exercise 14.2
1
Sharing in a ratio
a $10, $30, $50
2 a $20, $40, $50
b $60, $80, $100
c $300, $500, $200
b $30, $60, $75
c $64, $128, $160
3 a i 560
b 112
c 672
ii 448
iii 1232
4 a i 35
b i 24
ii 7
ii 12
iii 14
d $125, $50, $175
5 Alton $125, Dianne $300, Fredda $100 and Nia $225
6 $750, $1000, $1375, $1625
7 20
8 $30 000
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 14
Answers to Practice Book exercises
F Exercise 14.3
Solving problems
1a $2
b $10
c $16
2a $2.50
b $7.50
c $22.50
3 $36
4 a 288 g plain flour, 24 ml baking powder, 288 g castor sugar, 12 eggs
b 108 g plain flour, 9 ml baking powder, 108 g castor sugar, 4.5 eggs (allow 4 or 5)
c 180 g plain flour, 15 ml baking powder, 180 g castor sugar, 7.5 eggs (allow 7 or 8)
5 a 21
b 35
6 a 75 g
b 375 g
7 a 24, 42
b 120
8 550 ml of vanilla ice cream, 2200 ml of grape juice and 2750 ml of ginger ale
9 650 ml
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
15 Probability
F Exercise 15.1
1 a 0.8
The probability that an outcome does not happen
b 0.3
c 0.9
d 0.95
2 5%
3 a 0.9
b 0.05
4 a 0.4
b 0.7
c 0.8
5 a 5
6
35
36
b
6 a 0.9
b 0.7
c 0.65
7 a 0.9
b 0.4
c 0.3
8 a 40%
b 15%
c 5%
F Exercise 15.2
Equally likely outcomes
1 a
1
20
b 19
c
3
5
d
3
10
2 a
2
3
b 1
c 1
d
1
2
20
3
5
31
ii
4
31
iii
9
31
1
5
ii
4
5
iii
3
5
3
10
ii
4
25
iii
1
50
3 a i
b i
4 a i
5 a
61
100
b 10
c
49
e
2
5
27
31
iv
b i
3
5
ii
b i
5
8
5
12
6 a Names on paper would be one method.
1
3
iii
ii
3
8
1
15
c i
5
7
2
7
ii
7 There are four equally likely outcomes, HH, HT, TH and TT. Two heads is one of these options so
the probability is 1 .
4
8 a Some students are more likely to be first than others.
b Names on paper is one method, or using a random-number generator.
F Exercise 15.3
Listing all possible outcomes
1 a $5 and $10, $5 and $20, $5 and $50, $10 and $20, $10 and $50, $20 and $50
2 a
+
1
2
3
1
2
3
4
2
3
4
5
3
4
5
6
2
9
b i
c
d i
1
3
ii
×
1
2
3
1
1
2
3
2
2
4
6
3
3
6
9
2
9
ii
1
9
iii
5
9
iv 0
iii
5
9
iv
Copyright Cambridge University Press 2013
b i
1
2
ii
1
2
1
9
Cambridge Checkpoint Mathematics 8
1
Unit 15
3 a
Answers to Practice Book exercises
1
2
3
4
5
6
1
0
1
2
3
4
5
2
1
0
1
2
3
4
3
2
1
0
1
2
3
4
3
2
1
0
1
2
5
4
3
2
1
0
1
6
5
4
3
2
1
0
iii
2
9
1
6
b i
5
18
ii
4 a AX, AY, BX, BY, CX, CY
b i
5 a
+
3
4
5
6
7
10
11
12
13
8
11
12
13
14
9
12
13
14
15
b i 1
ii
4
c
d i
e
1
2
iii
1
12
−
3
4
5
6
7
4
3
2
1
8
5
4
3
2
9
6
5
4
3
1
4
ii
1
2
iii
1
18
iv
1
6
ii
iv
1
3
iii
1
2
0
1
12
2
3
6 a
b i
Red
Green
Black
Red
RR
RG
RB
Yellow
YR
YG
YB
Green
GR
GG
GB
1
9
ii
4
9
F Exercise 15.4
iii
7
9
Experimental and theoretical probabilities
1 a green 0.15, blue 0.4, red 0.25, yellow 0.2
b 20 is not enough spins to decide.
c green 0.24, blue 0.35, red 0.17, yellow 0.24
dIt could be biased. The theoretical probabilities are 0.25 if the spinner is fair and the values for blue
and red are not very close to that.
2 a 0.58, 0.54, 0.527, 0.535
b The experimental probabilities become close to the theoretical probability of 0.5. It looks fair.
3 a 6 and 36
b 0.52, 0.57, 0.595, 0.62, 0.6125, 0.622
c 0.62 is a sensible estimate.
4 a There is a lot of variation between the experimental probabilities.
b 0.75
c 0.725
d It is based on the most experiments.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
16 Position and movement
F Exercise 16.1
1
a
Transforming shapes
b
y
7
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
0
1
2 3 4 5 6
mirror line y = 4
7
0
x
0
1
2
3
4
5
6
0
x
7
0
b
b
c
y
6
5
4
4
4
3
3
3
2
2
2
1
1
1
2
3
4
5
6
7
x
0
0
1
2
3
4
4
5
6
7
x
y
6
5
0
3
d
5
1
2
mirror line y = 3.5
c
y
6
0
1
mirror line x = 4
2 a
3 a
y
7
6
0
4
c
y
7
5
6
7
x
0
0
1
2
3
4
5
6
7
x
y
6
A
5
4
3
2
C
B
1
D
0
0
1
2
3
4
5
6
7
x
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 16
Answers to Practice Book exercises
5 Ahmad is correct, Zalika is wrong.
y
6
5
A
4
3
2
B
1
0
0
6 a A to C
b A to B
F Exercise 16.2
c A to D
2
2
3
4
5
6
7
8
9
10
x
d B to E
Enlarging shapes
1 Check students’ answers.
a
scale factor 2
c
1
scale factor 4
Cambridge Checkpoint Mathematics 8
b
scale factor 3
d
scale factor 2
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
e
scale factor 3
f
scale factor 2
g
scale factor 2
h
scale factor 3
i
scale factor 2
Copyright Cambridge University Press 2013
Unit 16
Cambridge Checkpoint Mathematics 8
3
Unit 16
2 a
Answers to Practice Book exercises
b (3, 0), (3, 6), (7, 2)
y
6
5
4
3
2
1
0
0
3 a
1
2
3
4
5
6
7
x
b (0, 5), (4, 1), (8, 1), (4, 5)
y
6
5
4
3
2
1
0
0
4 a
1
2
3
4
5
6
7
x
b (1, 1), (1, 5), (7, 5), (3, 1)
y
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
5 a centre (1, 12), scale factor 2
c centre (24, 3), scale factor 3
4
x
b centre (23, 14), scale factor 3
d centre (7.5, 10), scale factor 3
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
17 Area, perimeter and volume
F Exercise 17.1
The area of a triangle
1 10.8 m2
2 a 44.8 m2
b 9500 cm2
3 a By estimation: half of 4 cm is 2 cm, and 2 cm multiplied by 15 cm is not 65 cm2.
b 8.5 cm
c Dakarai forgot to double the 4.25 cm – the formula is 1 bh, not just bh.
2
F Exercise 17.2
The areas of a parallelogram and trapezium
1 a 160 mm2
b 28.56 cm2
2 a 10.66 m2
b 60 m2
3 a, b A = 24.78 cm2 (6 × 4), B = 16.48 cm2 ( 1 × 10 × 3), C = 20.67 cm2 (5 × 4), D = 14.21 cm2 ( 1 × (4 + 6) × 3).
2
2
c 18.41 cm2
4 155 mm or 15.5 cm
5 6.3 cm
F Exercise 17.3
The area and circumference of a circle
1 a 62.8 cm
b 31.4 m
c 15.7 cm
2 a 78.5 cm2
b 19.6 m2
c 0.8 cm2
3 a 56.52 cm2
b 14.13 m2
c 3.5325 cm2
4 a 30.84 cm
b 15.42 m
c 7.71 cm
5 a 17.9 m
b 10.7 cm
c 12.5 mm
6 She is correct.
Area of the semicircle = 6.28 m2, area of the quadrant = 12.56 m2.
7 He is wrong.
Perimeter of the semicircle = 20.56 m, perimeter of the three-quarter circle shape = 20.13 m.
F Exercise 17.4
The areas of compound shapes
1 a Area A = l × w = 8 × 10 = 80
b Area A = l × w = 6 × 6 = 36
Area B = l × w = 12 × 1 = 12
Area B = 12 × b × h = 12 × 6 × 4 = 12
Total area = 80 + 12 = 92 cm2
Total area = 36 + 12 = 48 mm2
2 a i
7 cm
3 a 104 cm2
ii
b i 3 cm, 6 cm
135 cm2
ii
90 cm2
b 152.54... cm2
5 Xavier is wrong: rectangle = 60 m2, trapezium = 59.625 m2, circle = 59.69... m2.
F Exercise 17.5
The volumes and surface areas of a cuboid
1 a 160 mm3
b 48 cm3
c 7500 cm3
2 a 208 cm2
b 27 cm2
c 30 000 cm2 or 3 m2
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 17
3
Answers to Practice Book exercises
Length
Width
Height
Volume
a
5 cm
6 cm
2 cm
60 cm3
b
20 mm
10 mm
6 mm
1200 mm3
c
4m
3m
6m
72 m3
d
8 cm
4 cm
8 cm
256 cm3
e
5.2 m
7.3 m
10 m
379.6 m3
f
0.3 mm
12 mm
8 cm
288 mm3
4 a 652 mm2
b height of cuboid: 900 ÷ 8 = 112.5, 100 ÷ 20 = 5 mm.
surface area of the cuboid: 2 × 20 × 5 + 2 × 8 × 5 + 2 × 20 × 8 = 600 mm2
5 a 20 m
b 20.125 m2
F Exercise 17.6
Using nets of solids to work out surface areas
1 a i
ii 360 cm2
b i
ii 112 cm2
c i
ii 408 m2
d i
ii 870 mm2
2 Razi is correct; the triangular prism has the smaller surface area.
Surface area of the triangular prism ≈ 2 × ( 1 × 90 × 80) + (90 × 8) + (80 × 8) + (100 × 8) = 9360 cm2.
2
Surface area of the cube ≈ 6 × 40 × 40 = 9600 cm2.
3 Yes.
Surface area of the triangular prism = 2 × 1 × 8 × 3 + 8 × 2.75 + 2 × 5 × 2.75 = 73.5 cm2.
2
Surface area of the cube = 6 × 3.5 × 3.5 = 73.5 cm2.
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
18 Interpreting and discussing results
F Exercise 18.1
Interpreting and drawing frequency diagrams
1 a 12
b 3
c 30
2 a 4
b 40 cm
c 3
3 a
d 15
Number of breakfasts sold
12
Frequency
10
8
6
4
2
0
0–9
10–19 20–29 30–39 40–49
Number of breakfasts
b No month has only 27 days.
c Not really. It could be 49, but you can’t tell from grouped data information; the greatest number of breakfasts
sold could be anywhere from 40 to 49.
d 23. Add the frequencies for the three bars that show number of breakfasts sold were at least 20.
4 a
Height of sunflowers
12
Frequency
10
8
6
4
2
0
1.0
1.2
1.4
1.6
1.8
2.0
Height of sunflowers (m)
b 17
c No. It could not be exactly 1 m as ‘1.0 <’ means that the height could be very close to but not equal to 1.0 m.
d Not really. It could be 2 m, but you can’t tell from grouped data information; the tallest sunflower could be
anywhere from just above 1.8 m to 2 m.
e 28
F Exercise 18.2
Interpreting and drawing pie charts
1 a
None
Pa
vlo
va
Fruit
salad
Ice
cream
Cheese
cake
Ice cream = 81°, Cheesecake = 108°, Pavlova = 45°, Fruit salad = 72°, None = 54°
b 15%
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
1
Unit 18
2 a 60
b
Answers to Practice Book exercises
Monday
Tuesday
Friday
Wednesday
Thursday
Monday = 30°, Tuesday = 48°, Wednesday = 84°, Thursday = 36°, Friday = 162°
c 5 = 1
60 12
3 a Xtreme
b
4 a 180 = 1
360 2
b i
54
3
360 = 20
c 30%
d 75
360
ii 160
iii 720
5 a 60
b 60
c 64. Women = 126, men = 62.
dThere are more women than men in the survey, so when they have the same angles in the pie charts
the women’s sector must represent more than the men’s sector.
F Exercise 18.3
Interpreting and drawing line graphs
1 a i 5 million
ii 6.5 million
b 2007
c 2006 and 2007
d 2010 and 2011 e 2008 to 2009
fThere was a large increase in production from 2006 to 2007, then a small but fairly steady increase
in production from 2007 to 2010; from 2010 to 2011 there was no increase in production.
ii 0
b February
c November and December
2 a i 65
dSki rental starts off high for the first three months of the year (65, 70 and 65), then goes down to zero over the
next three months. Kelly does not rent out any skis from June to September, but rentals improve from October
increasing rapidly to reach 55 in December.
Average price of book ($)
3 a
Book prices
14
12
10
8
6
4
2
0
1990
1994
1998
2002
Year
2006
2010
b The average price steadily falls until 2002 where it levels off. From 2006 it goes up steeply until 2010.
c 2006 to 2010
d $9.50
2
Cambridge Checkpoint Mathematics 8
Copyright Cambridge University Press 2013
Answers to Practice Book exercises
F Exercise 18.4
1 a 19
Unit 18
Interpreting and drawing stem-and-leaf diagrams
b 23 minutes
c 14
2 a 22
b 105 minutes
c 10
e Because one film lasts exactly 2 hours
d i
35 minutes
d 11
f i 113 minutes
ii
37 minutes
iii
26 minutes
ii
121 minutes
iii
32 minutes
3 a 5 | 8 means 5.8 g
5
6
7
8
9
b 17
8
0
1
0
0
9
1
2
2
2
c
9
2
3
5
5
4
3
6
4
5
9
2
25
4
8
9
9
d 44%
c i
6.4 g
ii
7.2 g
iii
3.7 g
4 a 16 | 2 means 162 kg
16
17
18
19
20
b 24
f
i
2
0
0
0
0
175 kg
5 6
2 3
0 8
4 8
7 8
c 4
5
9
9
8
5
5
9
d
6
11
24
ii 180 kg
F Exercise 18.5
9
e 37.5%
iii 47 kg
g 185 kg
Drawing conclusions
1 Yes. $751.78 ÷ 25 = $30.07
2 a Alyson has a higher median (3 compared to 2) and a better mode (3 compared to 1).
b Haito has a better mode (5 compared to 3). Their means are the same (2.5).
3 a
b
c
d
e
26
B. 19 finished in under 15 minutes, but only 15 finished in under 15 minutes on course A.
No. 15 ÷ 26 × 100 = 57.7%.
Yes. 19 ÷ 26 × 100 = 73.1%.
Yes. 34 ÷ 52 × 100 = 65.4%.
Copyright Cambridge University Press 2013
Cambridge Checkpoint Mathematics 8
3