answers to ocr gcse mastering mathematics foundation 1

Answers to OCR GCSE Mastering Mathematics Foundation 1

Number Strand 1 Calculating Algebra Strand 2 Sequences Units 1–6 Moving on 1 Units 1–2 Moving on 103 Unit 7 BIDMAS 2 Unit 3 Linea r seq uenc es 104 Unit 8 Multip lying dec ima ls 8 Unit 4 Spec ia l sequenc es 110 Unit 9 Divid ing dec ima ls 13 Algebra Strand 3 Functions and graphs Number Strand 2 Using our number Unit 1 Rea l-life g raphs 116

system Unit 2 Plotting graphs of linea r func tions 122 Units 1–4 Moving on 18 Unit 3 The eq ua tion of a stra ight line 133 Unit 5 Using the number system Unit 4 Plotting quadra tic and c ub ic effec tively 19 graphs 140 Unit 6 Understand ing sta ndard form 24 Geometry and Measures Strand 1 Units Number Strand 3 Accuracy and scales Units 1–3 Moving on 28 Units 1–7 Moving on 150 Unit 4 Round ing to 2 d ec ima l p lac es 29 Unit 8 Bearings 151 Unit 5 Signific anc e 33 Unit 9 Sc a le d rawing 155 Unit 6 Approxima ting 38 Unit 10 Comp ound units 160 Number Strand 4 Fractions Geometry and Measures Strand 2 Units 1–2 Moving on 42 Properties of shapes Unit 3 Multip lying frac tions 43 Units 1–5 Moving on 164 Unit 4 Ad d ing and sub tra c ting frac tions 48 Unit 6 Types of q uadrila tera l 165 Unit 5 Working with mixed numbers 53 Unit 7 Ang les and pa ra llel lines 170 Unit 6 Divid ing frac tions 59 Unit 8 Ang les in a polygon 174 Number Strand 5 Percentages Geometry and Measures Strand 3 Units 1–2 Moving on 64 Measuring shapes Unit 3 Converting frac tions and Units 1–2 Moving on 178 dec ima ls to and from perc entages 65 Unit 3 Circ umferenc e 179 Unit 4 App lying perc enta ge inc reases Unit 4 Area of c irc les 183 and dec reases to amounts 71 Unit 5 Find ing the perc entage c hange Geometry and Measures Strand 4 from one amount to another 75 Construction Units 1–2 Moving on 187 Number Strand 6 Ratio and proportion Unit 2 Sharing in a g iven ra tio 79 Geometry and Measures Strand 5 Unit 3 Working with p rop ortiona l Transformations quantities 83 Units 1–5 Moving on 188 Unit 6 Enla rgement 189 Number Strand 7 Number properties Units 1–3 Moving on 87 Geometry and Measures Strand 6 Three-dimensional shapes

Algebra Strand 1 Starting algebra Unit 1 Moving on 195 Units 1–3 Moving on 88 Unit 2 Understand ing nets 196 Unit 4 Working with formulae 89 Unit 3 Volume and surfac e a rea of Unit 5 Setting up and solving simp le c uboids 211 equa tions 93 Unit 4 2D rep resenta tions of 3D shapes 215 Unit 6 Using b rac kets 97 Unit 5 Prisms 222 Unit 6 Enla rgement in two and three d imensions 226

Answers to OCR GCSE Mastering Mathematics Foundation 1 Statistics and Probability Strand 1 Statistical measures Units 1–2 Moving on 230 Unit 3 Using frequenc y ta b les 231 Unit 4 Using grouped freq uenc y tab les 236 Statistics and Probability Strand 2 Statistical diagrams Units 1–2 Moving on 242 Unit 3 Vertic a l line c ha rts 243 Unit 4 Pie c ha rts 250 Statistics and Probability Strand 4 Probability Unit 2 Sing le event p roba b ility 256 Unit 3 Comb ined events 261

© Hodder & Stoughton Ltd 2015PB 1

Number Strand 1 Moving on Answers (pages 2–4)

1 30p2 £15.863 4.2 cm4 Enough for 24 laptops; 2 students will have to share5 Item Cost price in £ Selling price in £ Profit or loss £

Football boots 35.75 18.50 loss 17.25

Snooker cue 23.50 42.00 profit 18.50

Golf clubs and bag 112.00 108.44 loss 3.56

Set of weights 98.00 72.50 loss 25.50

Cricket bat 15.10 24.30 profit 9.20

Snooker table 58.00 93.60 profit 35.60

Overall profit £16.996 89 seats were vacant7 Outgoings: £14 796; income: £18 612; can save £38168 a yes, 18 boxes cost £270

b 7 tiles left over9 Field = 3600 m2; enough for 28.8 goats, therefore 28 goats can graze, but not 29.

10 £51 37811 a Credits (+) Debits (−) Balance (£)

150

50 200

95 105

85 20

40 −20

100 80

150 −70

25 −95

400 305

b £30512 a Lift A: floor 8; Lift B floor −4

b 12c Lift A (it’s closer to Jenny)

© Hodder & Stoughton Ltd 20152 3

Number Strand 1 Unit 7 Answers

Practising skills (pages 6–7)1 a 21

b 11c 11

2 a 42b 66c 66

3 a 7b 23c 19

4 a 14b 4c 14

5 a 99b 57c 17

6 a 72b 45c 123

7 a 4b 12c 6

8 a 8b 14c 8

9 a 36b 13c 49

10 a 38b 10c 16

11 a 15b 1c 64

12 a 0b 24c 4

13 a 7b 0c 7

© Hodder & Stoughton Ltd 2015

Strand 1 Calculating Unit 7 BIDMAS Band f

2 3

14 a −6b −5c 3

15 a 42b 42c 0



4 5

Developing fluency (pages 7–8)1 a 15

b 10c 0

2 a 4b 0c −5

3 a 8b 10c 0

4 a 15 and 15b 156 and 156c 820 and 820

5 a (6 + 3) × 2 = 18b 5 + (9 − 2) × 2 = 19c (8 + 3) × 2 + 1 = 23d (4 + 3) × (3 + 2) = 35e 6 + 8 − (2 + 1) = 11f 13 − (5 + 4 − 2) = 6

6 a ((100 − 1) × 5) ÷ 3 + 75 or 100 + (75 − 5) × (3 − 1)b (200 × 4) ÷ 5 − 8 − 1 or (200 × (8 − 5)) ÷ 4 + 1 or (200 ÷ 4) × (8 − 5) + 1

7 a 2b 2

c 72

d 2e 6.5f 2

8 a 169b 13c 19d 17e 361f 19

9 a 14

b 1

c 2510

d 1

e 510

f 12



4 5

10 a 4 9 2

3 5 7

8 1 6

b example answers:1 = 4 − 4 + 4 ÷ 42 = 4 − 4 ÷ (4 − 4)3 = (( 4 + 4) × 4) ÷ 44 = 4 + 4 − 4 − 45 = 4 × 4 + 4 ÷ 46 = (4 × ( 4 + 4)) ÷ 47 = 4 ÷ 4 + 4 + 48 = 4 × 4 + (4 − 4)9 = 4 × 4 + 4 ÷ 4



6 7

Problem solving (pages 8–9)1 a 1 0 + 5 × 2

b Joe is correct; the total is £20.c Work out any multiple purchases first.d Yes

2 a (10 − 1) ÷ 3 + 6 = 9b (3 + 7) × (2 − 5) = −30c 2 × (1 + 3)2 = 32

3 Hannah is correct: 4 × 25 − 52 = 48Michelle did 6 × 2 + (3 × 2)2 = 48Finlay did (4 × 3)2 ÷ 3 = 48

4 a 3 + 2 × 4 = 11 b 47 (= 3 + 4 × 11)

5 a 6 × (1 + 7) = 48b 9 ÷ (7 − 4) = 3

6 a 5.114 692 654b 5.36 + 62.87 ÷ 19.86 – 6.52



6 7

Reviewing skills (page 9)1 a 15

b 1c 1

2 a 25b 4c −5

3 a 6.25b −20c 0

4 a 4b 4c 8

5 a 2.5b 0c −2.5

6 a (12 + 4) ÷ (5 − 1) = 4b 7 − 3 × (4 − 2) = 1



Practising skills (page 12)1 a 2.4

b 0.24c 0.024d 2.4e 0.24f 0.24

2 a 4b 0.4c 0.04d 4e 0.4f 0.4

3 a 12b 1.2c 0.12d 0.012e 1.2f 0.12

4 a 100b 10c 1d 0.1e 0.0001f 0.01

5 a 100.7b 10.07c 1007d 0.1007e 1007f 0.000 001 007

6 a 0.06b 0.08c 0.42d 0.72e 0.2f 0.81


Strand 1 Calculating Unit 8 Multiplying decimals Band f

8 9

7 a 1.6b 0.902c 0.010 89d 72e 0.002f 1

8 a 3b 120c 0.006d 0.000064e 0.71f 0.144

9 a 9b 0.09c 0.0009d 0.027e 0.000 027f 0.0081

10 a £3.68b £12.90c £13.08d £14.22



10

Developing fluency (page 13)1 a −1.2

b −0.12c −0.12

2 a −10.8b 10.8c 1.08

3 £78.104 £12.605 £138.066 a 327.5

b 230.57 £240.82 (2 d.p.)8 1.2 m3



11

Problem solving (pages 14–15)1 £3.542 1653.75 cm2

3 £259.924 David (£136.50) earns more than Elaine (£133)5 Annual rail ticket (cheaper by £50)6 £391.207 2.61 m2

8 Portugal since £48 = €57.609 1 minute

12 13© Hodder & Stoughton Ltd 2015


12 13

Reviewing skills (page 16)1 a 1.6

b 4.5c 2.1d 16e 3.9f 0.48g 0.36h 0.0016

2 a 174.84b 17.484c 174.84d 1.7484e 0.0174 84f 1748.4

3 a £7.96b £17.10c £9.20

© Hodder & Stoughton Ltd 201512 1312 13



b 4.85c 3.25d 0.875e 17.6f 4.75g 0.9125h 0.009i 2.004j 7.403 25k 9.871l 0.5672

2 a £9.50b £3.20c £117.50d £48.85

3 a 1.5b 0.15c 0.0015d 150e 1500f 15

4 a 1b 7c 2d 6e 19f 6g 21h 13.5i 0.25j 8.75k 0.35l 35.62


Strand 1 Calculating Unit 9 Dividing decimals Band f

14 15

5 a 370b 1.5c 84.4d 184e 44.5f 14.1 g 700h 0.9i 1035j 0.53k 3560l 72

6 160 ÷ 20



14 15

Developing fluency (pages 19–20)1 72 ÷ 9 = 8

0.72 ÷ 9 = 0.087.2 ÷ 9 = 0.8720 ÷ 0.9 = 8007.2 ÷ 0.09 = 800.72 ÷ 90 = 0.008

2 a −34.5b −1.56c −0.725d −0.0025e 0.625f −0.0005g 0.07h 0.16

3 a 0.2b 0.2c 100d 1e 50 000f 110g 0.000 005h 3000

4 26.25 km5 1.25 kg6 a 0.0625 litres

b 0.286 litres



16 17

Problem solving (pages 20–22)1 £48.202 a 0.4 kg

b 5003 1504 a 90

b £2.655 586 2 pounds7 a Small = 24/£; large = 25/£

b Over 28 bags/£ so best value8 Cuisinaire is £8000; Chucks is £8200, so Cuisinaire is £200 less9 55 mph



16 17


b 24c 1700d 1570

2 a 80b 9c 0.15d 0.0125

3 a 1b 20c 12.5d 0.001

4 a −20b −320c −0.3d 11

5 500 stamps



1 a i 9641ii 1469

b 12 × 30 = 6 × 60 = 4 × 902 a 0.7 < 0.79

b −4 < −3c 0.6 = 0.60

3 1 Greg Rutherford 2 Mitchell Watt 3 Will Claye 4 Michel Torneus 5 Sebastian Bayer 6 Christopher Tomlinson 7 Maura Vinicius Da Silva 8 Godfrey Khotso Mokoena 9 Henry Frayne 10 Marquise Goodwin4 a 0.971

b 0.971 − 0.179 = 0.7925 1 hundreds, 5 tens, 3 units, 4 tenths, 9 hundredths, 8 thousandths6 Yes, it’s 15.7 miles away7 a 3

b car C8 a timekeeper A

b timekeeper C9 £24.28

10 20 by 20 = 400 photos11 a Credits (+) Debits (−) Balance (£)

503.00

60.00 563.00

285.00 278.00

9.00 269.00

250.00 19.00

70.00 −51.00

50.00 −1.00

1500.00 1499.00

12 a No, it reversesb 4 and −2, 2 and −4

13 −32, −2, 28, 58, 88



Practising skills (pages 31–32)1 a i 7 000 000

ii 700 000iii 70 000iv 7000v 700vi 70vii 7

b i 2 468 000ii 246 800iii 24 680iv 2468v 246.8vi 24.68vii 2.468

c more; less2 a i 8

ii 80iii 800iv 8000v 80 000vi 800 000vii 8 000 000

b i 6.5ii 65iii 650iv 6500v 65 000vi 650 000vii 6 500 000c less; more

3 a 4b 60c 0.9d 0.84e 1250f 9930g 62h 51.7


Strand 2 Using our number system Unit 5 Using the number system effectively Band e

20 21

4 a 0.006b 5c 80d 1.45e 24 690f 6130g 3200h 200 000

5 a 0.2259b 0.638c 0.008d 0.0004e 584f 700g 24 900h 81.5

6 a 60b 1.3c 4700d 5290e 0.08f 7650g 500h 0.46

7 a 18 000b 0.023c 691d 70e 0.5f 3200g 0.001 64h 58 990

8 a 600b 0.122c 7d 1.8e 9f 0.0746g 4522.8h 0.003 607 8



20 21

Developing fluency (page 32)1 a £8.60

b 8.60 × 10 = £86c 8.60 × 0.1 = £0.86

2 a 30b 3 ÷ 0.1 = 30

3 a 1.7b 0.265c 7.9d 0.1251e 7f 65g 1124h 7000Order is: 0.1251, 0.265, 1.7, 7, 7.9, 65, 1124, 7000

4 1 correct2 correct answer is 23 correct answer is 254 correct5 correct6 correct answer is 112 1007 correct answer is 0.068 correct answer is 20.4

5 a 10b 0.1c 0.01d 100e 0.1f 0.01

6 a 100b 0.1c 0.001d 0.1e 100f 0.1



22 23

Problem solving (pages 33–34)1 a Divide by 100

Divide by 0.01 Multiply by 100

Multiply by 0.01

b Multiply by 0.001

Multiply by 1000 Divide by 0.001

Divide by 1000

c Divide by 0.0001

Divide by 10 000 Multiply by 0.0001

Multiply by 10 000

2 a 81 × 16 = 1296

162 × 8 = 1296

27 × 48 = 1296

324 × 4 =1296

54 × 24 = 1296

9 × 144 = 1296

648 × 2 = 1296

108 × 12 = 1296

18 × 72 = 1296

3 × 432 = 1296

1296 × 1 = 1296

216 × 6 = 1296

36 × 36 = 1296

6 × 216 = 1296

1 × 1296 = 1296

b You start to get decimals in the calculations, e.g. 2592 × 0.5 is the leftmost box.c They are square numbers, the first number is divisible by 3 and the second is divisible by 2.



22 23


b 13c 0.04d 0.008e 0.063f 0.009g 0.201h 0.0007

2 a 28b 3000c 80d 200e 6000f 0.4g 1 000 000h 10

3 10 000

24 2524 25© Hodder & Stoughton Ltd 2015


Practising skills (pages 36–37)1 a i 5.12 × 103

ii 5.12 × 102

iii 5.12 × 101

iv 5.12 × 10−1

v 5.12 × 10−3

vi 5.12 × 10−4

b 5.12 × 100

2 a 500b 80 000c 2600d 190 000e 8170f 90 500g 74 000 000h 10 040

3 a 6 × 102

b 7 × 104

c 8.9 × 103

d 8.16 × 102

e 1.33 × 105

f 4 × 106

g 9.5 × 107

h 4 × 109

4 a 0.068b 0.005c 0.0299d 0.0007e 0.104f 0.000 086g 0.000 005h 0.032 27

5 a 6.9 × 10−1

b 5.2 × 10−2

c 1.14 × 10−2

d 7 × 10−4

e 3.8 × 10−3

f 6 × 10−6

g 9.55 × 10−1

h 9 × 10−5


Strand 2 Using our number system Unit 6 Understanding standard form Band h

24 2524 25

Developing fluency (page 37)1 1600, 0.8 × 103, 9 × 1003

2 a 6 × 104

b 1.08 × 105

c 1.5 × 108

d 3 × 10−3

e 2.6 × 10−6

3 a 9000, nine thousandb 2100, two thousand one hundredc 680, six hundred and eightyd 922, nine hundred and twenty twoe 10 800, ten thousand eight hundredf 70, seventyg 0.7, seven tenthsh 0.03, three hundredths

4 a 6 × 103

b 7.4 × 10c 8.1 × 102

d 2.015 × 103

e 4 × 10−1

f 3 × 10−2

g 2.24 × 10−6

h 5.108 × 106

i 6.78 × 107

j 2.3 × 107

k 4 × 109

l 7.001 × 10−9

5 7.95 × 102, 7.09 × 103, 7100, 6.8 × 104, 9 × 104

6 0.04, 3.9 × 10−2, 3.82 × 10−2, 2.2 × 10−3, 2 × 10−3

7 a ,

b ,

c =d ,

e .

f .



26 27

Problem solving (page 38)1 Venus, Mars, Mercury, Sun, Jupiter, Saturn, Uranus, Neptune2 5 (questions, 3, 4, 6, 8 and 10)



26 27

Reviewing skills (page 38)1 a 200 800

b 2 450 000c 7 803 000 000d 645 000 000e 0.9f 0.000 000 207g 0.006 145h 0.1007

2 a 2.025 × 104

b 2.3 × 107

c 6.547 × 102

d 2.562 487 × 104

e 3 × 10−1

f 7 × 10−2

g 2.04 × 10−3

h 9.9 × 10−2

3 a 7 × 109

b 8 × 10−3



1 a £135b £130

2 a ten thousands and also tensb 85 000

3 a 46 millionb ten timesc 6800 million

4 For example:take 2.22 kg from the 25.71 kg bag to make it 23.49 kgtake 0.17 kg from the 23.66 kg bag to make it 23.49 kgthese two bags will now round to 23 kg.put the 2.39 kg into the 19.08 kg bag to make it 21.47 kgleave 21.28 kg bagleave 22.54 kg bag

5 By rounding: 4 cm + 8 cm + 4 cm + 11 cm = 27 cm, a lot less than Salome’s answer




b 2.7c 0.9d 12.8e 60.2f 0.7g 40.7h 0.1i 8.0j 5.6k 115.0l 247.0

2 a 5.13b 8.29c 0.02d 12.99e 17.00f 0.05g 706.10h 2.67i 0.01j 52.00k 90.00l 1.08

3 a £24.31b £61.59c £9.80d £0.70e $210f $13.06g €0.50h €30

4 a 6.25b 7.95c 2.44d 0.7e 12.495f 29.995g 0.215h 1.4995


Strand 3 Accuracy Unit 4 Rounding to 2 decimal places Band e

30 31

Developing fluency (page 44)1 Number Nearest whole

numberTo 1 decimal

placeTo 2 decimal

places

a 8.431 8 8.4 8.43

b 6.918 7 6.9 6.92

c 14.277 14 14.3 14.28

d 0.8063 1 0.8 0.81

e 63.592 64 63.6 63.59

f 109.711 110 109.7 109.71

g 799.498 799 799.5 799.50

h 8069.515 8070 8069.5 8069.52

i 99 999.9069 100 000 99 999.9 99 999.91

j 699 999.999 700 000 700 000.0 700 000.00

2 0.03 a i 1097

ii 1096.8iii 1096.79iv 1096.794

b i 81ii 81.0iii 81.05iv 81.047

c i 1ii 1.0iii 0.96iv 0.965

d i 511ii 510.8iii 510.80iv 510.801



30 31

Problem solving (pages 44–45)1 33.67 cm, 33.7 cm and 33.60 cm2 0.08 has just one significant figure; 0.079 m2 3 a 4.5 kg

b 4.5 kgc yes, 0.046 kg

4 The times are only measured to the nearest tenth of a minute, so 18.5 minutes would have been a sensible answer.



32 33


b 0.1c 30.0d 500.0

2 a 0.20b 1.00c 57.33d 19.98

3 a $160.06b £1008.90c $10.10d £77 000.00

4 a 0.010b 0.0095c 0.009 75d 0.009 875

5 a i 3ii 3.1iii 3.14iv 3.142

b i 2ii 2.2iii 2.24iv 2.236

32 33© Hodder & Stoughton Ltd 201532 33


Practising skills (page 47)1 a 2

b 4c 1d 2e 3f 3g 4h 5

2 a 30b 50c 400d 900e 7000f 20 000g 20h 60i 0.9j 0.7k 0.02l 0.02

3 a 870b 920c 620d 710e 700f 3300g 5100h 19 000i 73 000j 8000k 0.64l 0.60

4 a 400 000b 380 000c 384 000d 384 000e 384 030


Strand 3 Accuracy Unit 5 Significance Band f

34 35

5 a 8b 8.0c 8.00d 8.000e 8.0000

6 a 0.008b 0.0081c 0.008 11d 0.008 106e 0.008 106 0

7 a 20b 0.60c 71000d 4e 6.51f 27.00



34 35

Developing fluency (page 48)1 Number Round to 1

significant figureRound to 2

significant figures

a 742 700 740

b 628 600 630

c 199 200 200

d 4521 5000 4500

e 3419 3000 3400

f 8926 9000 8900

g 8974 9000 9000

h 36 294 40 000 36 000

i 0.2583 0.3 0.26

j 0.079 61 0.08 0.080

k 0.000 3972 0.0004 0.000 40

l 0.001 023 0.001 0.0010

2 a 20b 10c 51d 0.048e 17 600 000f 100g 300h 1.01i 677

3 a 0.2b 0.48

4 a trueb truec falsed false

5 a Ada and Cain, Ben and Daveb Ada, Ben and Cainc Ben and Cain



36 37

Problem solving (page 49)1 Ami. Dan has rounded down instead of up, Milly thinks that the zero is not significant and Bob has changed the size

of the number.2 a C = 24 cm

b C = 24.8 cmc C = 25.12 cmd C = 25.136 cm

3 a £590b £591.80c Real value is £587.475. Therefore, to 1 significant figure, a is more accurate. It is also easier.

4 a 0.004 cmb Measure the height of a number of the same book stacked in a pile or use a more accurate measuring device.



36 37


b 0.01c 1000d 1 000 000

2 a 6.4b 20c 0.0052d 0.010

3 a 0.3068c 515 300d 2.0



Practising skills (pages 52–53)1 a true

b falsec falsed truee truef false

2 a 90b 800c 1500d 15e 900f 11 000

3 a iiib iic iii

4 a, c, f5 No, she does not have enough money.6 a, d, f


Strand 3 Accuracy Unit 6 Approximating Band g

38 3938 39


b 70c 3600d 1600e 1000f 121

2 a iii, 5002 = 250 000b iv, 70 × 7 = 490c i, 42 + 33 = 43

3 £10004 £3605 50 mph6 £20007 4 months8 a wrong

b rightc wrong

9 a 100 timesb 4 timesc 400 times



40 41

Problem solving (pages 54–56)1 a Less; normal pay ≈ £200, overtime ≈ £50, Sunday ≈ £39

b Yes; ≈ £12 000 wages + ≈ £2000 holiday pay2 a 0.25 seconds

b 9000 pots × 30 hours = 270 000 pots in a week. 270 000 ÷ 100 pots/carton = 2700 cartons

3 a Yes, with about £3 to spareb ≈ £2 or £3

4 Catch the 08:20 train from London, the 09:00 does not allow for any delay.Catch the 18:12 train from Stoke, the 17:50 does not allow for any delay. Leave home at about 8 a.m., arrive back at about 8 p.m. (12 hours).

5 £15 to £166 5 tins



40 41

Reviewing skills (page 56)1 a ≈ 1000

b ≈ 20c ≈ 1600

2 a ib iv

3 a ≈ 280b ≈ 3850c ≈ 6750

4 ≈ £46.80


Number Strand 4 Moving on Answers (page 58)

1 a 45

b

c 45

, with explanation

2 25

, 23

, 710

, 34

3 No, he has 2 oz butter but needs 2.5 oz

4 a Wayne = 13

; Andy = 14

; John = 512

b Give £4 to Andy.



Practising skills (page 61)1 a B

b Ac Ed Ce D

2 a 16

b 130

c 316

d 532

e 245

f 920

g 3554

h 2132

3 a 712

b 1865

c 310

d 511

e 519

f 225

g 910

h 45


Strand 4 Fractions Unit 3 Multiplying fractions Band e

44 45

4 a 18

b 118

c 320

d 221

e 720

f 914

g 23

h 89

5 a 16

b 116

c 13

d 23

6 a 110

b 415

c 12

d 116

e 140

f 563

g 58

h 532



44 45

Developing fluency (pages 62–63)1 × 1

223

3

16

112

19

12

310

320

15

910

27

17

421

67

2 110

3 a trueb truec falsed truee false

4 a 14

b 18

5 215

6 a 25

b 320

c 920

7 13

of 45

= 415

= 830

25

of 34

= 310

= 930

25

of 34

is greater

8 Friday 110

= 550

Saturday 15

of 910

= 950

She eats more on Saturday



46 47

Problem solving (pages 63–64)1 a 1

8b no, it is 78 800 km2

c 112 m

2 a 2140

b Yes, only 500 left3 £424 a 40 m

b Theoretically she will always be half the previous distance from home and never actually reach home. However after 10 days, she would be 2.5 metres away.



46 47


35

b 47

c 2572

d 120143

2 a 415

b 1229

c 23

d 34

3 a 925

b 12

c 415

d 5584

4 a 16

bucket

b 34

gallon

5 35




5

b 37

c 89

d 1113

e 47

f 25

g 23

h 1

2 a 23

b 57

c 23

d 511

e 25

f 27

g 819

h 1921

3 a 34

b 45

c 47

d 58

e 712

f 310

g 1120

h 914


Strand 4 Fractions Unit 4 Adding and subtracting fractions Band f

48 4948 49

4 a i 13

= 412

14

= 312

ii 13

+ 14

= 412

+ 312

= 712

iii 13

− 14

= 412

− 312

= 112

b i 23

= 1015

15

= 315

ii 23

+ 15

= 1015

+ 315

= 1315

iii 23

− 15

= 1015

− 315

= 715

c i 58

= 3556

27

= 1656

ii 58

+ 27

= 3556

+ 1656

= 5156

iii 58

− 27

= 3556

− 1656

= 1956

5 a 920

b 1924

c 310

d 512

e 2435

f 1130

g 118

h 910



50 51

Developing fluency (pages 67–68)1 11

12

2 720

3 + 1

416

49

37

1928

2542

5563

25

1320

1730

3845

320

25

1960

107180

4 a 712

b 1115

c − 38

d 815

5 930

can



50 51

Problem solving (pages 68–70)1 Example: fills C from A leaving 1

8 in A, pour this into B, now 29/40 full

2 Mr Bader

3 110

4 a 58

b 58

5 a 1 and 16

pints

b Fill the 12

pint cup with cream and from this, fill the 13

pint cup; 16

pint now remains in the 12

pint cup.6 £577.507 a 3

16b 36

8 a No, 13

full.

b Student’s diagram showing fuel gauge 13

full



52 53


10

b 111

c 1

d 1112

2 a 12

b 99100

c 13

d 2140

3 a 13

b 3756

c 1724

d 940

4 1320




3

b 94

c 72

d 75

e 114

f 316

g 209

h 487

2 a 114

b 212

c 313

d 234

e 325

f 356

g 889

h 1234

3 a 534

b 156

c 6

d 412

e 2 110

f 412


Strand 4 Fractions Unit 5 Working with mixed numbers Band f

54 55

4 a 416

b 414

c 31415

d 7 310

e 5 712

f 62740

5 a 1124

b 145

c 11720

d 21924

e 2 512

f 12990

6 a 823

b 845

c 634

d 637

e 14

f 3035

7 a 778

b 729

c 219

d 1023

e 3 524

f 1238

8 a 6 320

b 21724

c 212

d 418

e 6 512

f 4 940



54 55

Developing fluency (pages 75–76)1 11

5 = 6

5135

= 235

123

= 106

212

= 208

134

= 314

2 a >b =c <d >

3 a 223

b 4

c 513

d 623

e 8

f 913

4 538

km

5 a trueb falsec falsed true

6 a 152324

kg

b 2 712

kg

c 20 73120

kg

7 a 234

b 21316

c 21924

d 71724

e 5f 7



56 57

8 a 214

b 614

c 1 916

d 338

e 10 81125

f 75881

9 a 125

b 8

c 1 112

10 211115

m



56 57

Problem solving (pages 76–79)1 a 8 km

b 434

km

c 4 516

km

2 No, he needs a further 320

litre

3 a 1 512

cups

b Yes, 5990

left

4 Yes, y is correct but x = 1 118

cm.5 Tom

6 a 1180

b £20 0007 Carpet A (£8.99) and Carpet B (£9.99). This assumes the carpet bought is a rectangular piece from which there will

be wastage.



58 59


5

b 203

c 899

d 514

2 a 458

b 567

c 1335

d 61720

3 a 4b 5

c 513

d 814

e 134

f 5 215

4 a 412

b 634

c 9

d 1114

e 1312

f 1534

g 18

h 2014




b 75

c 120

d 23

e 821

f 629

g 831

h 956

2 a 110

b 112

c 320

d 15

e 754

f 940

g 140

h 1100

3 a 6b 8c 15d 3e 81

3f 8g 72h 72


Strand 4 Fractions Unit 6 Dividing fractions Band f

60 61

4 a 49

b 512

c 49

d 821

e 23

f 1516

g 14

h 4

5 a 512

b 423

c 1 111

d 11315

e 57

f 11213

g 32327

h 1556

6 a 521

b 427

c 2110

d 57

e 548

f 940

g 152

h 314

i 2 27

j 1633

k 514

l 218



60 61

Developing fluency (page 82)1 7

24

2 16

3 a falseb truec false

4 × 1

423

15

120

215

56

524

59

5 a 445

m

b 22 110

m

6 a 2930

b 18

c 3

d 125

7 35 miles per hour



62 63

Problem solving (page 83)1 5 children2 No, can make 96 pieces3 a No, only 33 glasses

b 17

c 17



62 63


8

b 43

c 37

d 59

2 a 716

b 112

c 24

d 1823

3 a 910

b 415

c 13

d 21132

4 19 g/cm3



1 562 £93003 £304 £429.605 A and C6 £2245.83




100

b 7100

c 310

d 3150

e 925

f 2125

g 131000

h 67500

2 a 0.34b 0.08c 0.8d 0.002e 0.145f 0.064g 1.64h 0.0075

3 a 910

b 25

c 17100

d 3350

e 1325

f 2131000

g 1250

h 140

4 a 15%b 3%c 60%d 23.7%e 11.7%f 8.6%g 140%h 310.4%


Strand 5 Percentages Unit 3 Converting fractions and decimals to and from percentages Band e

66 67

5 a 0.25b 0.2c 0.1d 0.75e 0.375f 0.875g 0.4h 0.7

6 a 50%b 90%c 27%d 84%e 75%f 30%g 18.75%h 0.7%

7 Fraction Decimal Percentage

a 11100

0.11 11%

b 720

0.35 35%

c 310

0.3 30%

d 325

0.12 12%

e 710

0.7 70%

f 1320

0.65 65%

g 120

0.05 5%

h 2225

0.88 88%

i 1925

0.76 76%

j 27200

0.135 13.5%

k 3125

0.024 2.4%

l 85

1.6 160%



66 67

Developing fluency (pages 91–92)1 Colour Number Fraction Percentage Decimal

Red 8 15

20% 0.2

Orange 10 14

25% 0.25

Blue 7 740

17.5% 0.175

Purple 4 110

10% 0.1

Green 6 320

15% 0.15

Yellow 5 18

12.5% 0.125

Total 40 1.000

2 a 17100

, 15

, 26%, 0.3

b 6%, 12

, 0.55, 0.6

c 0.3, 31100

, 32%, 720

d 710

, 0.715, 72%, 34

e 310

, 0.33, 33.3%, 13

f 0.04, 4.8%, 120

, 4.5

3 a 19100

. 0.18

b 0.6 . 15%

c 725

= 28%

d 0.04 , 4.1%

e 0.114 . 110

f 66.66% , 23

g 1320

. 62%

h 45

. 0.795

i 0.15 . 1.4%

j 2540

= 0.625

4 80%5 14%6 0.4 = 40%, so 61% is 21% greater

7 1st assignment Kate gets 6480

= 80%, 2nd assignment Kate gets 5460

= 90%.

Score improved by 10% points.8 Dave’s Discounts, by 8.33%9 a 26%

b 64%c 90%



68 69

10 Assignment Total mark Pass mark

1 80 60

2 60 45

3 12 9

4 120 90

5 48 36

6 500 375



68 69

Problem solving (page 92)1 a 1

4, 1

40, 4

10, 4

1, 0.14, 0.41, 4.10, 4.01, 1.40, 1.04, 0.14%, 0.41%, 1.04%, 1.4%, 4.01%, 4.1%, 14%, 40%, 41%,

104%, 140%, 401%, 410%

b 410

= 40%, 0.14 = 14%, 0.41 = 41%, 4.01 = 401%, 1.4 = 140%, 1.04 = 104%, 4.10 = 410%

c Yes, there will always be at least one decimal and percentage that equals each other eg digits 111 can be 1.11 = 111%, or 965 can be 9.65 = 965%

2 a 0.666666666b 0.67, 67%c 1

9, 0.1

., 11%; 130

9, 14.4

., 1444%; 2

11, 0.1

.8., 18%; 1.6

11, 0.14

.5., 15%; 100

3, 33.3

., 3333%



70 71


20

b 2425

c 18

d 71000

2 a 0.67b 0.008c 2.3d 0.0428

3 a 750

b 11500

c 147200

d 1125

4 a 58%b 0.9%c 80.9%d 287%

5 a 0.3b 0.55c 0.7d 0.72

6 a 65%b 62.5%c 2.9%d 138%



Practising skills (page 96)1 a £7

b £77c £63

2 a 9 kgb 27 kgc 9 kg

3 a £18b £16.20c £5.85

4 a £52b £41.60c £11.96

5 a reduction = £17, price = £51b reduction = £61, price = £183c reduction = £4.75, price = £14.25

6 a 33b 28c 24d 120e 42f 102g 30h 30i 220


Strand 5 Percentages Unit 4 Applying percentage increases and decreases to amounts Band f

72 73

Developing fluency (pages 97–98)1 A (124 raised by 25%) and F (620 decreased by 75%)

B (220 decreased by 30%) and D (110 increased by 40%)C (750 reduced by 80%) and H (75 increased by 100%)E (130 increased by 20%) and G (390 reduced by 60%)

2 £61.953 a 60 bars

b 75 lollipopsc 200 chews

4 £82805 £190 5506 £322.247 1.38 litres8 150 g reduced by 4%, 224 g reduced by 35.5%, 141 g increased by 2.5%, 245 g reduced by 41%, 115 g increased by

26%, 140 g increased by 6%9 £115.20

10 a £722.40b £1404.32c £191.52d £834.90

11 Sandra, by £1.13



72 73

Problem solving (pages 98–99)1 a £20 000

b £15 000c 25% of £20 000 is more than 25% of £16 000, so the final salary is less than the original.

2 1.6% profit3 £10 8004 a Car C

b £14 0855 £150



74 75

Reviewing skills (page 99)1 a £16

b £96c £64

2 a reduction £6, price £24b reduction £19.80, price £79.20c reduction £11.10, price £44.40

3 a 126b 216c 145.50

4 £1152



Practising skills (pages 101–102)1 a 30%

b 55%c 25%d 40%e 32%f 28%g 50%h 88.8%

2 a 7%b 90%c 20%d 4%e 15%f 14%g 30%h 17%i 16%

3 Item Cost price Selling price Profit Percentage profit

a Saw £12 £21 £9 75%

b Hammer £10 £17 £7 70%

c Plane £20 £32 £12 60%

d Spanner set £35 £56 £21 60%

4 Item Cost price Selling price Loss Percentage loss

a Book £10 £2 £8 80%

b Saucepan £25 £22 £3 12%

c Dinner set £200 £184 £16 8%

d Armchair £70 £63 £7 10%

e Bicycle £120 £84 £36 30%

f Cushion £2 96p £1.04 52%

5 a 60 mlb 8%

6 a 6b 20%


Strand 5 Percentages Unit 5 Finding the percentage change from one amount to another Band g

76 77

Developing fluency (pages 102–103)1 20%2 24%3 25%4 4%5 15%6 10%7 60%8 a 23.3% increase

b 38.5% decreasec 35.1% increased 28.9% decreasee 7.9% increasef 71.9% increaseg 58.1% decreaseh 266.7% increasei 2.1% decrease

9 Sam’s calf10 Mary’s house11 50%



76 77

Problem solving (pages 103–105)1 15% saving with wall insulation, 20% saving with loft insulation2 a 283%

b Motorways 550% (1961−1971); 100% (71−81); 19.2% (81−91); 12.9%(1991−2001); 2.9% (2001−11); increase is reducing

Other roads 3.5%; 5.2%; 5.3%; 8.9%; 0.3% respectively; increasing up to 20013 8%4 163% increase, so the headline may be a little exaggerated5 a from 2009 to 2010

b 23.9%6 increase by 11.6%7 a 12.6% increase at Brands Hatch

b 11.2% decrease at Silverstone



78 PB

Reviewing skills (page 105)1 a 10%

b 75%c 60%d 80%

2 a 25%b 24%c 42%d 11.2%

3 Item Cost price Selling price Profit Percentage profit

a Dress £80 £84 £4 5%

b Pencil 80p £1.08 28p 35%

c Dressing gown £120 £84 −£36 or £36 loss 30% loss

d Notebook £2 96p −£1.04 or £1.04 loss 52% loss

4 a 3% loss b car B (A = 2% gain B = 4.9% loss C = 4.8% loss)




b £12c £12d £24e £36

2 a 9b 5c 10d 35e 45

3 a 6 b 80 mlc 80 mld 400 mle 320 ml

4 a £24, £36b £80, £16c 60, 100d 98 ml, 28 ml

5 a 310

b 710

c 213

6 a £54 and £36b £15 and £105c 125 and 100d 280 m, 120 m and 80 m


Strand 6 Ratio and proportion Unit 2 Sharing in a given ratio Band f

80 81

Developing fluency (pages 109–110)1 £562 723 £1804 63 cm and 35 cm5 21



80 81

Problem solving (pages 110–111)1 11 different ratios: 1 : 1, 1 : 2, 1 : 3, 1 : 5, 1 : 11, 5 : 7, 2 : 1, 3 : 1, 5 : 1, 11 : 1, 7 : 52 a 30

b hockey = 40°, tennis = 120°, netball = 200°3 a £60

b 20104 a Lesley, 5 more

b Lionel has a better scoring rate; 3040

compared with 3556

5 No, they both have 12 milk chocolates



82 83


b 15

c 45

2 a 60 g, 45 gb 154, 176c 22 g, 44 g, 222 gd 35, 70, 140

3 214

hours



Practising skills (page 114)1 a 24p

b 72p2 a 15p

b £1.203 a £6

b £1384 a i 35p

ii 30piii Dan’s discounts

b i £1.30ii £0.96ii Dan’s discounts

c i £3.72ii £3.84iii Bev’s Bargains

5 Ingredient Quantity for 5 people Quantity for 1 person Quantity for 8 people

Minced beef 900 g 180 g 1440 g

Stock 480 ml 96 ml 768 ml

Onion 2 25 3

15

Tin of tomatoes 1 15 13

5Potatoes 700 g 140 g 1120 g

Worcestershire sauce 40 ml 8 ml 64 ml


Strand 6 Ratio and proportion Unit 3 Working with proportional quantities Band f

84 85

Developing fluency (pages 114–115)1 £1122 £6.803 £5.664 a 6 kg for £14.70

b 150 ml for £24c 60 g for £12.06

5 No. He will charge £14.606 Offer 1 (buy two 40 ml and get one 40 ml free) is the best value, as it is £32.40 for 120 ml, that is £27 for 100 ml.

Offer 2 gives £27.75 for 100 ml.The large bottle of perfume costs £30 for 100 ml.

7 £65.60



84 85

Problem solving (pages 116–118)1 a £537.50

b 6 hours2 13 (brown sugar is the limiting ingredient)3 No; small = 7.4p per chocolate, large = 6.9p per chocolate, medium = 6.8p per chocolate4 Harvey’s company at 50p/mile (Albert’s is 49.5p/mile)5 Yes, she can do it for £14.566 41 cents7 Not enough flour, she needs 168 g; she has enough milk as she needs 1.704 litres8 £8.50 in 2p coins, £42 in 5p coins



86 PB

Reviewing skills (page 118)1 a £2.34

b £21.062 a 2 m for £8.10 is better value, as 2 m for £8.10 is £4.05 per m; 60 cm for £2.40 is £4.15 per m

b 600 g for £5.40 is £9/kg so is better value as 750 g for £7.20 is £9.60/kgc 2 litres for £22.12 is £11.06/l so is better value as 800 ml for £8.92 is £11.15/l

3 a 300 ml for £2.16 is 72p per 100 ml; 400 ml for £2.72 is 68p per 100 ml; 500 ml for £3.45 is 69p per 100 mlb 400 ml


Number Strand 7 Moving on Answers (page 120)

1 a 1, 36, 2, 18, 3, 12, 4, 9, 36b 1, 72, 2, 36, 3, 24, 4, 18, 6, 12, 8, 9c 1, 144, 2, 72, 3, 48, 4, 36, 6, 24, 8, 18, 9, 16, 12

2 a 2b 2c They are the same

3 a 1628, 6336, 5432b 4221, 3249, 3843, 6336

4 a falseb truec trued truee false

5 a 1, 4, 9, 16, 25, 36b (for below 40) 2, 4, 6, 10, 12, 16, 18, 22, 28, 30, 36c (for below 40) 4, 8, 12,16, 20, 24, 28, 32, 36, 404, 16 and 36 have all 3 properties

6 a i and ii both contain square numbersb i doesn’t contain prime numbers as terms all are even; ii does contain prime numbersc yes – 8 is cube in i and 64 is cube in iid i doesn’t contain any factors of 100 (it is multiples of 8); ii does contain factors of 100 (4, 10, 25)e Neither sequence appears to contain a multiple of 21


Algebra Strand 1 Moving on (page 122)

1 A: 2nB: 2n − 2C: n + 2D: 2nE: n − 2F: 2nG: n − 2H: n + 2

2 a Perimeter rectangles (from smallest to biggest): green – 2(a + b); blue – 4(a + b); purple – 6(a + b); orange – 8(a + b)Perimeter L shapes (from smallest to biggest): blue 4(a + b); purple 6(a + b); orange 8(a + b)

b The perimeters are the same for each colour rectangle and L shape


Algebra Strand 1 Unit 4 Answers


b 5c 28d 7

2 a + 3b − 9c − 1d + 14

3 a Turn rightb Stand upc Turn 68° clockwised Walk 6 steps backwards

4 a − 7b + 4c − 129d ÷ 5e ÷ 3.9f × 8g × 0.5 or ÷ 2

5 a − 3b + 9c + 1d − 14

6 a £51.80b × 1.40c 56 l

d i 42 lii 28.5 l


Strand 1 Starting algebra Unit 4 Working with formulae Band e

90 91

Developing fluency (pages 125–126)1 a i 32

ii 48iii 42iv 8

b A: iiiB: iC: ivD: ii

2 a x → × 4 → − 3 → yb x ← ÷ 4 ← + 3 ← yc 17

3 a 14 → × 2 → + 9 → 37b 23c 23 ← ÷ 2 ← − 9 ← 55

4 a Input Output

3 5

10 33

7 21

1 −3

b Input Output

12 41

6 17

20 73

0 −7

5 a i €23ii €43

b 12 kmc €c = 2d km + 3

6 a i 115ii 28.75iii 46iv 2300

b i 20ii 400iii 1600iv 8000

7 a i 540ii 810iii 2232iv 6408

b i 86ii 23iii 142iv 35

8 a 72b 6



90 91

Problem solving (pages 127–129)1 a 6 + 8 = 12 + 2

b 9 edges; Triangular prism2 a 32 cm

b 6 cmc 9 cm

3 a £150b C = 20 + 10nc 15 × 4 = 20 + (4 ×10)

4 a 7b 9c 2n + 1 d 41

5 13 36 or 1:36 p.m.6 a The equivalent formulae in the table have the same colour

b W = FD t = d

st = s

d FINISH

V = IR A = 12

bh F = WD

v = at + u

b = Ah d = st P = VI A = b × h ÷ 2

y = 2x + 3 P = FA A = b × h s = d

t



92 93

Reviewing skills (page 129)1 a d → × 60 → + 30 → C b i 270

ii 630c d ← ÷ 60 ← − 30 ← C

d i 6ii 9

2 a i 100ii 0iii 35

b i 50ii 86iii 401

3 a i £65ii £155

b 8c 15 is cost per person, 5 is the fixed price addition




b 7c 3d 9

2 a 5b 11c 4d 3

3 a 11b 8c 14d 19

4 a 7b 8c 5d 11

5 a 9b 6c 4d 12

6 a 10b 15c 24d 18

7 a 7b 10c 6d 8e 56f 17

8 Cynthia is correct. Hardip has not divided every term by 4 in the first line.9 a 7

b 4c 6d 2e 3f 9g 2h 4


Strand 1 Starting algebra Unit 5 Setting up and solving simple equations Band f

94 95

Developing fluency (pages 133–134)1 a Total ages is 21

b Andrea is twice as old as Bennyc Benny is 7 years older than Andrea

2 5x + 12 = 47 x = 73 a 2

b −2c 2d −2e −2f 2

4 i x = 12ii x = 2iii x = 11iv x = −3v x = −5vi x = 4

5 a 60b 16c −9d 18e −4f 120g 0h −96

6 8t + 24 = 138 t = £14.257 2a + 28 = 204 a = £888 4a + 64 = 172 a = 27p9 a 7.4

b 1.56c 7.78d 18.8e 5f −2

10 a Fran’s age, f = a + 8b 2a + 18 = 30; a = 6 and f = 14



94 95

Problem solving (pages 134–135)1 30°2 150°3 Ami 10, Ben 20, Ceri 64 l = 125 m. Area = 9375 m2

5 Angles of an equilateral triangle are 60° each.The first two give a value of x to be 10 and the third angle x is 12

6 0.5l7 a C = 40 + 30 × n

b 8 days8 a a + 2(a + 30) = 180

b a = 40c 40, 70, 70

9 a There is an infinite number of possibilities for the right hand side. The final number could be anything that is in the sequence generated by 3x + 2 but x can be positive or negative. Here are some possibilities: −4, −1, 2, 5, 8, 11, 14, 17

b No, 100 − 2 is not a multiple of 3c No all of her guesses aren’t accurate, 4x + 3 = 10 gives x = 74 (or 134 or 1.75) so it isn’t an integer. If x is going

to be an integer then the right-hand side must be odd, but must also be 3 more than a multiple of 4,e.g.: −1, 3, 7, 11, 15, etc.



96 97


b 3c −7d 11e −1f −6g 3h −2

2 a −19.6b −1.2c 92.4d 64

3 3 × 230 + 2t = 1000t = 155 g



Practising skills (pages 138–139)1 a Lexmi:

8 × (30 + 4)= 8 × 30 + 8 × 4= 240 + 32= 272

b Yesc Kabil’s is quicker

2 a 114b 126c 108d 56e 12f 35

3 a 174b 208c 185d 744

4 a 6(3 + 7) = 60b 4(8 − 7) = 4c 3(3 + 8) = 33d 45(3 + 9 + 8) = 900

5 a i 92 cm2

ii 252 cm2

iii 285 cm2

iv 132 cm2

b i 285ii 92iii 132iv 252

6 a i 5(4 + x)ii 55

b i 2(x + 5)ii 24

c i 8(x + 3)ii 80

d i 5(x + 2 + x − 1) = 5(2x + 1)ii 75

Kabil:8 × (30 + 4)= 8 × 34= 272


Strand 1 Starting algebra Unit 6 Using brackets Band f

98 99

7 a 6a + 21b 42 – 24bc 16c – 22d 5 – 40de 8x + 12yf 15e + 6f + 18g 14p + 28qh 40g – 15h + 10



98 99

Developing fluency (pages 139–140)1 5(x – 3) = 5x – 15

6(2x + 3) = 12x + 182(4x – 1) = 8x – 22(4 – x) = 8 – 2x3(5x + 2) = 15x + 65(3x + 1) = 15x + 58(x + 2) = 8x + 164(2x – 1) = 8x – 42(6 – x) = 12 – 2x3(2x + 5) = 6x + 15

2 a 4(x + 2)b 3(y − 4)c 8(2 − f)d 6(2g + 3)e 5(3m − 2)f 7(a + 3b)

3 a 2(x + 6y)b 10a + 5b = 5(2a + b)c 5x + 5y + 5 = 5(x + y + 1)

4 a 90b 0c 5d −40

5 a 5n − 1 is bigger because 5n – 5 is 4 smaller than 5n − 1b 2(3n + 5) is bigger because 6n + 10 is 1 larger than 6n + 9c 5(3n + 7) because 15n + 35 is n bigger than 14n + 35 [unless n is negative]

6 a 8b 5c 4

7 a 6x + 12 + 15x + 5 = 21x + 17b 8x + 24 − 6x − 2 = 2x + 22 = 2(x + 11)c 6x + 18 − 4x + 6 = 2x + 24 = 2(x + 12)

8 a 8x + 20b 4x2 + 10xc 3x2 − 9xd 8x2 + 20xe 3x3 − 6x2

f 18x2 + 12xy



100 101

9 a 4x(x − 1)b 5(2x2 + 1)c 5x(2x + 1)d 6x(x − 2)e 4c(3d – 2)f 2x(3x + 2)g 4x(x − 2y)h 6cd(c + 3d)

10 £31.68



100 101

Problem solving (pages 140–143)1 a 3s + 1.5

b 4s = 3s + 1.5s = 1.5 mSquare 1.5 m, Triangle 2 m.

2 60°3 a x = 9

b 20 cm4 R + R + 6 + 3(R + 6) = 84; 5R + 24 = 84; R = 12. Tom is 545 a (7000 + 5m) pence or equivalent in £

b 1200 miles6 Area of the square = 4x(x + 6) = 4x2 + 24x

area of triangle = ½ × 2x × 4(x + 6) = 4x2 + 24x, so the areas are the same7 a 2H + 2W – 4. This is because otherwise the corners would be counted twice.

b 2(W − 2 + H − 2 + 1) + 2 = 2W + 2H – 6 + 2 = 2W + 2H – 4. This expression works out half of the shape first, then doubles, then adds in the top left and bottom right corners.

c 2(H + W) – 4 = 2H + 2W – 4. This calculates half the shape (but counting the top right corner twice), doubles and then subtracts 4 for the corners counted twice

d H – 1 + H – 1 + W − 1 + W – 1 = 2H + 2W – 4. Counts squares unique to each side and adds theme 2H + 2W – 4 = 2(W − 2 + H − 2 + 1) + 2 = 2(H + W) – 4 = H – 1 + H – 1 + W − 1 + W – 1



102 PB


b 270c 252d 90e 0f 1089

2 a 16 983b 738c 1308d 9801

3 a 5(8 + 12) = 100b 7(3 + 2 − 5) = 0c 11(13 + 11 − 4) = 220

a i 8(x + 11)ii 128

b i x(2x + x + 1) = x(3x + 1)ii 80

5 a 10a + 15b + 8a + 2b = 18a + 17bb 15a + 18b – 6a – 12b = 9a + 6b = 3(3a + 2b)c 8a − 8b + 18a + 12b = 26a + 4b = 2(13a + 2b)d 12a + 8b − 5a + 15b = 7a + 23b

6 a 3f + 9g = 3(f + 3g)b 8k − 8m = 8(k − m)c 8x + 9


Algebra Strand 2 Moving on (pages 145–146)

1 a 72.5, 36.25, 18.125b −3.75, 1.875, −0.9375c −648, −972, −1458

2 10:503 a 1, 4, 5, 8, 9 and 2, 3, 6, 7, 10

b Keep switching numbers from the top sequence to the bottom4 a

b

c i C: 6nii A: 9n + 1 is total number of shapes (tiles) Assumes that the black square is one square tile and not two

triangles, though the large black triangles are triangle tilesB: 3n + 1 is number of black shapes (tiles)

d i 58

ii 58

5 a odd numbersb even numbersc divisible by 3

d i divisible by 6ii 1, 2, 3 and 6 are factors of 6, the number that you are adding. The first term in the sequence dictates what

each term can be divided byiii 1 and 5



Practising skills (pages 148–149)1 a 4 and +7

b 6 and +5c 38 and −3d 12 and +3

2 a 3, 6, 9, 12, 15b 6, 10, 14, 18, 22c 27, 29, 31, 33, 35

3 a A 38 46, B 81 78b A 830, B −216c A 8, B 87

4 a i 23 and +4ii 39 and 43

b i 38 and +6ii 62 and 68

c i 46 and −3ii 34 and 31

d i −4 and +2ii 4 and 6

e i 6 and −5ii −14 and −19

5 a i 4, 7, 10, 13, …, 301ii 4 and +3iii The difference between the terms is the multiple of n, take away the difference between the terms from the

first term to get the position-to-term formula. b i 4, 10, 16, 22, …, 598

ii 4 and +6iii The difference between the terms is the multiple of n, take away the difference between the terms from the

first term to get the position-to-term formula. c i 7, 11, 15, 19, …, 403

ii 7 and +4iii The difference between the terms is the multiple of n, take away the difference between the terms from the

first term to get the position-to-term formula.6 a 3n + 2

b 2n + 2c 4n + 1

7 a 9, 12, 15, 18, 21, …, 66b 7, 9, 11, 13, 15, …, 45c 4, 11, 18, 25, 32, …, 137


Strand 2 Sequences Unit 3 Linear sequences Band f

104 105104 105

8 a i 4n + 7ii 407

b i 10n − 8ii 992

c i 7n + 4ii 704



106 107

Developing fluency (pages 149–151)1 3, 6, 9, 12 = 3n

7, 9, 11, 13 = 2n + 56, 11, 16, 21 = 5n + 1

2 a Input, n 2 3 5 10 12

Output 30 45 75 150 180

15nb Input, n 6 12 14 15 32

Output 27 33 35 36 53

n + 21c Input, n 1 2 5 10 20

Output 11 15 27 47 87

4n + 7d Input, n 1 6 10 20 23

Output 4 49 85 175 202

9n − 53 a

Pattern 5

Pattern 6

b Number of pentagons 1 2 3 4 5 6

Number of matchsticks 5 9 13 17 21 25

c 29, term to term is +4d 4n + 1

e i 41ii 81

f 254 a 5n − 2

b 3n + 8c 6n − 5

d 12

n + 3



106 107

5 a i 22ii +4

b 4n + 2c 162d 87th

6 a i 27ii +6

b 6n − 3c 237d 45th

7 a

Pattern 4

Pattern 5

b Pattern number 1 2 3 4 5 6

Number of matchsticks 4 13 22 31 40 49

c i 58ii Term to term is + 9 or position to term is 9n − 5

d 9n − 5 e i 85

ii 175f Because they all share one matchstickg 28 squares



108 109


b 4n + 2c 4n + 2 = 77; n = 18.75

So 19 tables with 4 × 19 + 2 = 78 chairs; 1 empty chair2 a 31 and 28 b iii 43 – 3n

c 2 = 43 − 3n; 3n = 41, n = 13.667So 2 is not in the sequence as n needs to be a whole number

3 4m − 3: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 43, 49, 55, 6163 − 7n: 56, 49, 42, 35, 28, 21, 14, 7, 02 terms in common

4 a D is position 4 of 4, A will say 97, B will say 98, C will say 99, D will say 100. 100 is divisible by 4 giving 25b 5th position

c i 4th positionii 2nd position



108 109

Reviewing skills (page 152)1 a +8

b 8n + 1c 401d 56

2 a i −4n + 84ii −316

b i −5n + 105ii −395

c i −2n + 62ii −138

d i 6n + 74ii 674

3 a 13, 17b d = 4n − 3

c i 357ii 477

d 162



Practising skills (page 155)1 a i even numbers

ii

iii 2n b i square numbers

ii

iii n2

c i cube numbersii

iii n3

2 a

17 262 5 10

+3 +5 +7 +9

b

142−1 7

+3 +5 +7 +9

23

c

3 2110

+11 +15+7 +19

5536

d

2 8 18 32

+10 +14+6 +18

50

3 a 11, 14, 19, 26, 35b 0, 7, 26, 63, 124c 1, 3, 6, 10, 15 [triangular numbers]


Strand 2 Sequences Unit 4 Special sequences Band f

110 111110 111

4 a n2

b n2 + 1c n2 – 1d 2n2

5 a n2 + 10b n3

c n3 + 5d 2n3



112 113

Developing fluency (pages 156–158)1 a

b Pattern number 1 2 3 4 5

Number of red triangles 1 3 6 10 15

Number of green triangles 0 1 3 6 10

Total number of triangles, T 1 4 9 16 25

c Number of red triangles and number of green triangles are triangular numbersTotal number of triangles are square numbers

d 100i 55ii 45

e T = n2

2 a i Double the previous termii the difference between one term and the next is 1 more than the difference between the previous 2 terms.

b i 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16 384,32 768, 65 536

ii 1, 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79, 92, 106, 121c 2, 4 and 16d 4, 16, 64, 256, 1024, 4096, 16 384, 65 536

3 a


Number of black tiles 4 4 4 4 4

Number of blue tiles 1 4 9 16 25

Total number of tiles, T 5 8 13 20 29

c 104d 15th patterne T = n2 + 4

f i No, because 400 is a square number, and each number in the sequence is 4 more than a square number.ii 19th pattern; 35 tiles left over

4 a i 3ii 5iii 7



112 113

b 9c 100d n2

e Pattern 7f 10 000

5 a 2, 6, 12, 20, 30

b i nth term = 12

n(n + 1)

ii Triangular numbersiii

Pattern 1 Pattern 2 Pattern 3 Pattern 4

Pattern 5

iv 820 c i

Pattern 4 Pattern 5

ii n(n + 1)iii Number of red circles = 1

2n(n + 1)

Number of green circles = 12

n(n + 1)

iv The number of red and green circles are triangular numbers.6 a


Number of matches, M 4 12 24 40 60

c 144 d i 1, 3, 6, 10, 15

ii Number of matches = 4 × triangular numbersiii 840iv M = 2n(n + 1)



114 115

Problem solving (pages 158–159)1 a 21, 34, 55, 89, 144

b 2, 3, 5, 13, 89c 8 = 23, 144 = 24 × 32

2 1443 a i 3

ii 5iii 8

b Fibonaccic For example, there are two ways to make rectangles using 6 dominoes. We can use all of the ones we made using

5 dominoes and can put a single vertical domino on the front:

There will be 8 of these.Alternatively, we can use all of the ones we made using 4 dominoes and can put two horizontal dominoes on the front:

There will be 5 of these.To work out the next one, therefore, we need to add the previous two numbers: 5 + 8 = 13 It is only putting one vertical or two horizontal dominoes in front of the previous numbers as three dominoes horizontally is too high



114 115

Reviewing skills (page 159)1 a 6, 9, 14, 21, 30

b 1, 7, 17, 31, 49c 4, 11, 30, 67, 128d 0, 2, 6, 12, 20

2 a n2 + 2b n2 – 3c n3 + 1d 3n2

3 a Pattern number 1 2 3 4 5 n

Number of blue squares 1 4 9 16 25 n2

Number of red squares 2 6 12 20 30 n2 + n

Total number of squares 3 10 21 36 55 2n2 + n

b n2

c 110d n2 + ne 500 = n2 + n; n2 + n – 500 = 0. Cannot solve for n whole number so no pattern will have 500 squaresf T = 2n2 + n = n(2n + 1). A composite number



Practising skills (pages 163–164)1 1 D Cooling curve of liquid cools quickly at first, then slowly to ambient temperature 2 A Constant speed means constant distance travelled for each time period, therefore straight line 3 C Doubling is exponential increase, represented by an increasing curve 4 B Constant speed should have constant fuel use, therefore fuel level decreases over time as a straight-line graph 5 A The conversion rate is the constant, m, in a y = mx equation, represented by a straight-line graph 6 E The ball will be accelerating towards the ground due to gravity, so the loss of height increases per unit time

represented by a downward curve 7 A Current housing market has a constant increase in house prices over time, shown by a straight-line graph 8 D Diets tend to have people lose weight quickly at first, levelling off to a constant weight after a long time =

downward curve graph with plateau2 a Wind speed is steady

b Wind speed decreases uniformlyc Wind speed increases uniformlyd Wind speed increases uniformly, stays constant and decreases uniformlye Wind speed increases then decreases uniformly and repeats this patternf Wind speed decreases uniformly and then increases uniformly

3 a iib iiic vd iiv

Dis

tanc

e

Time


Strand 3 Functions and graphs Unit 1 Real-life graphs Band f

116 117116 117

Developing fluency (pages 164–167)1 a 0.6 m b i 4.1 m

ii 6.2 m c i 2.4 or 3rd year

ii 6.8 or 7th year d Added line joins (0,0.6) to (10,7.6):

0 2 4 6 8 100

1

2

3

4

5

6

7

8

Hei

ght

of t

ree

(m)

Years after planting

y

x

2 a €89 b £45

c i For example, read off at £50 and ×3 ii €168

d £405 e £80 = 89 euros so 90 euros is higher/more expensive

3 a i 2 breaksii 10:12 for 24 min and 14:00 for 30 min

b i 54 kmii 90 km iii 84 km

c 11:48–11:50 and 15:58–16:00



118 119

4 a R

oubl

es (

RU

B)

Pounds (£)0 10 20 30 40 50 60 70 80 90 100

0

1000

2000

3000

4000

5000

6000y

x

b i £8ii 4800–4900 RUB iii £25–£26iv 24300–24400 RUB

5 a 10:36 and 8 minb 40 km c 11:04

d i Added line joins (10, 90) to (11.26, 0) )

0

10

20

30

40

50

60

70

80

90

100

10:00 10:30 11:00 11:30Time

Dis

tanc

e fro

m A

vonf

ord

(km

)

y

x

ii 10:38 in the station6 a 560 SEK

b £81 c i For example read off at £50 and ×5

ii 2800 SEK d £594 e £2 is an inaccurate reading, £90 = 1000 SEK so dividing by 90 and £1 = 11 SEK roughly

7 a 27–28 km b 2.8 hours c 4 hours after starting out for 1 hour

d i 13:45 ii 13:27 and 15:21



118 119

8 a Added line from (0, 0) to (20, 6) and then from (20, 6) to (35, 10)D

ista

nce

run

(km

)

Minutes after Jim’s start0 10 20

Henry

30 400

1

2

3

4

5

6

7

8

9

10y

x

b Jim leads Henry up to 34 minutes when Henry overtakes and finishes 1 minute earlier



120 121

Problem solving (pages 168–169)1 a 15 minutes

b 145 kmc 60 km/h

2 a i £60ii £42

b There is a fixed cost of £15 and then a charge of £9 per day c 10 days

3 A 4B 3 C 1D 5E 2

4 a €14.52b £50c €417.45d £452

5 a £16.25b £1.25/dayc Added line from (0, 0) to (6, 30)

30

25

Cos

t (£

) 20

15

10

5

0 2 4 6 8 10 12Days

y

x

Equipment hire

Pumps are Us

d For 5 days, cheaper with Pumps are Us6 a i 31.5 litres

ii 5.3 gallonsb 550 gallons



120 121

Reviewing skills (page 170)1 a 51 kg

b 4.6–4.8 stonesc 160 kgd Yes, he weighs 96 kg

2 a

Time

Hei

ght b

Hei

ght

Time

c

Hei

ght

Time3 a 2.8 km

b 1.5 hoursc 12:30d Quicker to friend’s house

To sports centre: speed = 3.7333 km/h, to friend’s house = 4.8 km/he 8 km/h



Practising skills (pages 174–175)1 a x 0 1 2 3 4 5 6

3x 0 3 6 9 12 15 18

+ 1 1 1 1 1 1 1 1

y = 3x + 1 1 4 7 10 13 16 19

b

14

16

18

20

y

x

12

10

8

6

4

2

0 1 2 3 4 5 6

c i 3.5ii 17.5


Strand 3 Functions and graphs Unit 2 Plotting graphs of linear functions Band f

122 123122 123

2 a x 0 1 2 3 4 5 6

2x 0 2 4 6 8 10 12

− 3 −3 −3 −3 −3 −3 −3 −3

y = 2x − 3 −3 −1 1 3 5 7 9

b

10

0

12

y

x

8

6

4

2

0

–21 2 3 4 5 6

c i 4ii 2

3 a x −4 −3 −2 −1 0 1 2 3 4

5x −20 −15 −10 −5 0 5 10 15 20

− 2 −2 −2 −2 −2 −2 −2 −2 −2 −2

y = 5x − 2 −22 −17 −12 −7 −2 3 8 13 18

b

1

2

−2−4−6−8

−10−12−14

468

1012

14

0−1−2−3 2 3x

y

c i −1.6ii 15.5



124 125

4 a x 0 1 2 3 4 5 6

4x 0 4 8 12 16 20 24

− 3 −3 −3 −3 −3 −3 −3 −3

y = 4x − 3 −3 1 5 9 13 17 21

b

321

8

2

−2

46

101214161820

y

x04 5

c i 4.5ii 19

5 a x 0 1 2 3 4 5 6

2x 0 2 4 6 8 10 12

+ 5 5 5 5 5 5 5 5

y = 2x + 5 5 7 9 11 13 15 17

b

3210

8

246

101214

1618

20

04 5

x

y

c i 5.5ii 12



124 125

Developing fluency (pages 175–176)1 a Weight (kg) 1 2 3 4 5 6 7 8

40W 40 80 120 160 200 240 280 320

+ 20 20 20 20 20 20 20 20 20

Time (T = 40W + 20) 60 100 140 180 220 260 300 340

b

3210

80

204060

100120140

T = 40W + 20160180200

04 5 6 7

220240260280300320340

T

W

c i 20ii The extra 20 min

d 280 mine 4.5 kg



126 127

2 a Number of hours (t) 1 2 3 4 5 6 7 8

3t 3 6 9 12 15 18 21 24

+ 4 4 4 4 4 4 4 4 4

C = 3t + 4 7 10 13 16 19 22 25 28

b

3210

8

246

101214

C = 3t + 41618

20

04 5 6 7

2224262830

C

t

c £17.50d 2.5 hours

3 a Weight of apples (kg) 5 10 15 20

Cost (£) 6 12 18 24

b

6420

8

246

101214 C = 1.2W

1618

20

08 10 12 14 16 18

2224C (£)

W (kg)

c £9.50d 17.5 kg



126 127

4 a Weight of potatoes (kg) 1 2 3 4 5 6

Cost (£) 0 0.8 1.6 2.4 3.2 4

b

321

2

1

–1

3

W (kg)

04 5 6

C (£)

c £2.80d 3.5 kg

e i −0.8ii The free first kilo of potatoes

5 a x −2 −1 0 1 2 3 4 5

4x −8 −4 0 4 8 12 16 20

− 2 −2 −2 −2 −2 −2 −2 −2 −2

y = 4x − 2 −10 −6 −2 2 6 10 14 18

b

1

2

−2−4−6−8

−10

468

1012

14

16

0 2 3 4−1x

y

y = 4x – 2

c 0.5



128 129

6 a x −2 −1 0 1 2 3 4 5 6 7 8 9 10

− x 2 1 0 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10

8 8 8 8 8 8 8 8 8 8 8 8 8 8

y = 8 − x 10 9 8 7 6 5 4 3 2 1 0 −1 −2

b

642

4

2

6

8

0 8x

y

y = 8 – x

c 9d 9

7 a x −2 −1 0 1 2 3 4 5 6 7 8

− 2x 4 2 0 −2 −4 −6 −8 −10 −12 −14 −16

12 12 12 12 12 12 12 12 12 12 12 12

y = 12 − 2x 16 14 12 10 8 6 4 2 0 −2 −4

b

321

8

6

4

2

10

12

14

0 4 5 6 7–1

–2

x

y

y = 12 – 2x

c −1.5



128 129

Problem solving (page 177)1 a n 0 200 400 600

20 20 20 20 20

0.05n 0 10 20 30

C = 20 + 0.05n 20 30 40 50

b Straight line drawn from (0, 20) to (600, 50)

100 150 200 250 300 350 400 450 500 550500

20

51015

253035

C = 20 + 0.05n4045

50

0

55C

nc 2000

2 a n 0 200 400 600

5 5 5 5 5

n20

0 10 20 30

T = 5 + n20 5 15 25 35

b Straight line drawn from (0, 5) to (600, 35)

100 150 200 250 300 350 400 450 500 550500

10

5

15

T = 5 + 20

25

30

0

T

n

n20

c 700



130 131

3 a x −3 −2 −1 0 1 2 3

2x −6 −4 −2 0 2 4 6

− 1 −1 −1 −1 −1 −1 −1 −1

y = 2x − 1 −7 −5 −3 −1 1 3 5

b Straight line drawn from (–3, –7) and (3, 5)

1

1

−1−2−3−4−5−6

234

0−1−2 2x

y

y = 2 x – 1

c y = 2.4d x = –1.75e 2 ≠ 2 × (–1.5) – 1 as 2 × (–1.5) – 1 = –4



130 131

Reviewing skills (page 178)1 a x 5 4 3 2 1 0 −1 −2

2x 10 8 6 4 2 0 −2 −4

− 5 −5 −5 −5 −5 −5 −5 −5 −5

y = 2x − 5 5 3 1 −1 −3 −5 −7 −9

b

1

1

−1−2−3−4−5−6−7

−8

234

0−1 2 3 4x

y

y = 2 x – 5

c i 0ii −1.5

2 a x −3 −2 −1 0 1 2 3 4 5

3x −9 −6 −3 0 3 6 9 12 15

+5 5 5 5 5 5 5 5 5 5

y = 3x + 5 −4 −1 2 5 8 11 14 17 20

b

1

2

−2

468

1012

14

0−1−2 2 3 4x

y

1618

y = 3x + 5

c −0.3



132 133

3 a £0.50b m 0 5 10 15 20 25 30

0.5m 0 2.5 5 7.5 10 12.5 15

+ 4 4 4 4 4 4 4 4

C = 4 + 0.5m 4 6.5 9 11.5 14 16.5 19

c

151050

4

6

8

2

1012141618

C = 4 + 0.5m

m0

20 25

C

d i £13.00ii 14 min



Practising skills (pages 181–184)1 a i A C

ii A B b i x = 6

ii y = 32 a i (2, 1)

ii (3, 1) iii (3, 5)

b 4c 1d 4

3 a For example, the start and end points of each line are:i (3, 0) and (6, 3)ii (2, 0) and (5, 6)iii (0, 0) and (3, 6)iv (0, 2) and (4, 6)

b i 1ii 2iii 2iv 1

c i and ivii and iii

4 a 3; y = 3x − 1b 1; y = x + 1c 0; y = 0d 4; y = 4x + 2e −1; y = −x + 2 f −2; y = −2x + 5 g −3; y = −3x − 3 h −5; y = −5x + 8

5 a For example the start and end points of the lines are:i (0, 0) and (2, 6)ii (2, 0) and (3.2, 6)iii (0, 6) and (6, 0)iv (0, 3) and (6, 0)

b i 3ii 5iii −1iv −0.5

c Negative gradient


Strand 3 Functions and graphs Unit 3 The equation of a straight line Band g

134 135

6 a For example (0, −3) and (2, 1)b

321–1

4

2

–2

– 4

6

8

04

y

xy = 2 x – 3

c i 2ii −3

d Part i refers to the multiple of x, which is 2. Part ii refers to the constant subtraction, −3.

7 a x −2 −1 0 1 2

4x −8 −4 0 4 8

−3 −3 −3 −3 −3 −3

y = 4x − 3 −11 −7 −3 1 5

x −2 −1 0 1 2

5x −10 −5 0 5 10

+2 +2 +2 +2 +2 +2

y = 5x + 2 −8 −3 +2 7 12

b

0–1–2 1 2x

y = 5x + 2

y = 4x – 3

4

2

–2

–4

–6

–8

–10

6

8

10

y

c Gradient of y = 5x + 2 is 5; gradient of y = 4x − 3 is 4 d Intercept of y = 5x + 2 is 2; intercept of y = 4x − 3 is −3e You can see the answers in the equationsf Gradient = 8, intercept = −5



134 135

8 a A:i (0, 8) (4, 0) ii −2iii 8

B:i (4, 8) (1, 0) ii 2.666iii −8/3

C:i (0, 3) (5, 8) ii 1iii 3

D:i (0, 5) (5, 0) ii −1iii 5

E:i (0, 1) (8, 1) ii 0iii 1

b i No line matchesii Diii Aiv Cv E Line B is y = 8

3 x − 8

3



136 137

Developing fluency (pages 184–185)1 B, C, A2 i C

ii A, Fiii A, B iv E v E

3 a

0–1–2–3–4–5–6 1 2 3 4 5 6x

4

2

–2

–4

–6

–8

–10

A

F C

DB

E

6

8

10

11

12

y

b A and E; B and D; C and Fc A and F; B and E

4 a y = 3xb y = 2x + 3c y = 4x + 1d y = x − 2

5 a

– 2

– 8– 6– 4

0

246

–10

–5 –4 –3 –2 –1 1 2 3 4 5 6–6

y = – 3

y = – x + 5

y=–2 x+5

y – 3 = 0

1012

y + 2 x = 4

y + x = 6

x – 2 = 0y

x

x = 4

8



136 137

b i (x = 4) and viii (x − 2 = 0);ii (y = −3) and vii (y − 3 = 0);iii (y + x = 6) and vi (y = −x + 5);iv (2x + y = 4) and v (y = −2x + 5)

c From their gradients (you have to rearrange some equations to y = mx + c format to find gradient, m)6 a C = 0.05m + 20

b £80c 1600 miles



138 139

Problem solving (pages 186–187)1 a C = 5d + 10

b Cost to hire is 5 pounds per day plus £10 c i It is a constant rate per day (i.e. doesn’t get more expensive per day with every day hired)

ii There is a base cost of £102 a B and E; C and D are parallel

b B and C meet at (0, 3)3 a B: C = 25 + 0.3m; C: C = 65

b

604020

20

10

30

40

50

080 100 120 140 160 180 200

60

70

0

80

C

m

C: C = 65A: C = 50 + 0.1m

B: C = 25 + 0.3m

c Company C4 a i 2

ii −3b y = 2x − 3c No, as 12 ≠ 2(8) − 3d y = 2x + 1



138 139

Reviewing skills (pages 187–188)1 A: y = 4x + 3

B: y = 8x + 2 C: y = 2x − 2 D: y = −5x + 4 E: y = −x − 5 F: y = 6x − 3

2 a x = −4 b y = 3x + 1 c y = −x + 4 d y = −1

2x + 3

e y = 33 a C = 40 + 30h b i £40

ii £30 per hourc 7 hours



Practising skills (pages 192–193)1 a x −3 −2 −1 0 1 2 3

x2 9 4 1 0 1 4 9

+ 1 1 1 1 1 1 1 1

y = x2 + 1 10 5 2 1 2 5 10

b

3x

y

210

2

4

6

8

10

–1–2–3

c x = 0d (0, 1)e −1.2 and +1.2


Strand 3 Functions and graphs Unit 4 Plotting quadratic and cubic graphs Band g

140 141140 141

2 a x −4 −3 −2 −1 0 1 2 3 4

12 12 12 12 12 12 12 12 12 12

− x2 −16 −9 −4 −1 0 −1 −4 −9 −16

y = 12 − x2 −4 3 8 11 12 11 8 3 −4

b

–1–2–3 1 2 3x

–20

6

4

2

8

10

12

y

c x = 0d (0, 12)e −3.5 and 3.5

3 a x −1 0 1 2 3 4 5

x2 1 0 1 4 9 16 25

− 4x 4 0 −4 −8 −12 −16 −20

y = x2 – 4x + 2 7 2 −1 −2 −1 2 7

b

2 3 4 51– 1– 1– 2

– 3

x0

21

34567

c x = 2d (2, −2)e 1 and 3



142 143

4 a x −2 −1 0 1 2 3 4 5 6

x − 2 −4 −3 −2 −1 0 1 2 3 4

y = (x − 2)2 16 9 4 1 0 1 4 9 16

b

10–1– 2 3 4 5 6x

4

2

2

10

8

6

12

14

16y

c x = 2d (2, 0)e x = 4.65, −0.65

5 a x −4 −3 −2 −1 0 1 2 3 4

2x2 32 18 8 2 0 2 8 18 32

+ 3 +3 3 3 3 3 3 3 3 3

y = 2x2 + 3 35 21 11 5 3 5 11 21 35

b

–3 –2 –1– 4 1 2 3 4x

105

0

252015

303540

y

c Symmetry x = 0, intercept y = 3, min point (0, 3)



142 143

6 a x −3 −2 −1 0 1 2 3

x3 −27 −8 −1 0 1 8 27

+ 2 +2 2 2 2 2 2 2

y = x3 + 2 −25 −6 1 2 3 10 29

b

–1–2–3 1 2 3x

– 25–20–15

0–5

–10

5101520

3025

y

c Crosses x at −1.26 and crosses y at 2d Rotational symmetry order 2 about (0, 2)



144 145

Developing fluency (pages 193–194)1 a

–1 1 2 3 4 5x

– 25–20–15

0–5

–10

5101520

3025

y

b x = 1, 3c x = −0.6, 4.6d The minimum point is (2, −1), the curve does not go below y = −1

2 a

–1–2 1 2 3

– 25–20–15

0–5

–10

5101520

3025

y

x

b x = 2c when x > 2d mirror images about the line y = 0

3 a x 0 1 2 3 4

x3 0 1 8 27 64

− 6x2 0 −6 −24 −54 −96

11x 0 11 22 33 44

− 6 −6 −6 −6 −6 −6

y = x3 − 6x2 + 11x − 6 −6 0 0 0 6

b

–1 21– 2 3 4 5 6x

105

0

252015

–20–25

–5–10–15

30y



144 145

c x = 1, 2, 3d One value; x = 3.2e x = 3.8

4 a

3 4 5 620x

1

10

50403020

60708090

100y

b 80 mc 6 seconds

5 a

10

15

5

–50

–10

–1–3 –2 21 3x

y

b x = −3 and x = + 2.3 with a crossover at −0.4 AO3, 1b6 h = 10 + 8t − 5t2

a t 0 0.5 1 1.5 2 2.5

10 10 10 10 10 10 10

+ 8t 0 4 8 12 16 20

− 5t2 0 −1.25 −5 −11.25 −20 −31.25

h = 10 + 8t − 5t2 10 12.75 13 10.75 6 −1.25

b

0.4 1.61.20.8 2 2.4 2.8t

–2.5

–5

5

7.5

10

12.5

15

2.5

0

h

c 13.2 md 1.6 seconds

e i After 2.4 secondsii The pebble is under the sea, so its motion won’t be modelled by the same equation; graph doesn’t have any

meaning after 2.4 s



146 147

Problem solving (pages 195–196)1 a −4 ˂ x ˂ 1

b −1 ˂ x ˂ 4 c −2 ˂ x ˂ 2 d (1.5, −6.25) e (−1.5, 6.25)

2 a x < −2 and 0 < x < 2 b x < −2 and 0 < x < 2 c x < −2 and 0 < x < 2 d They are reflections of each other in the x-axis. e Rotational symmetry order 2 about the origin.

3 a check − put values of h and t from table into equationb h

t

202530354045

15105

–5–10–15–20–25–30–35–40–45–50–55–60–65–70–75–80–85–90–95

02 4 6 8

h = 30t – 5t2

c 4 secondsd 80 m



146 147

4 a Length x, width is 12

(80 − 2x) so A = x(40 − x)

b

c 15 cm (or 25 cm) d when x = 20, curve is maximum. 4 × 20 = 80 = perimeter, which is square

5 a v

x

100

150

200

250

300

50

020 4 6 8 10

V = 2x2(10 – x)

b 4.1 cmc approx. 292 cm3 × 5 = 1460 cm3

0

100

200

300

400

500

5 10 15 20 25 30 35

x

y

40



148 149

6 a v 5 10 15 20 25 30 35 40

20v 100 200 300 400 500 600 700 800

6000v 1200 600 400 300 240 200 171.4 150

C = 20v + 6000v

1300 800 700 700 740 800 871.4 950

b

v

c

200250300350400450500550600650700750800850900950

1000105011001150120012501300

100150

50

0 5 10 15 20 25 30 35 40

C = 20v + 6000v

c 17.3 km/h



148 149

Reviewing skills (page 197)1 a x −1 0 1 2 3 4 5 6

5x −5 0 5 10 15 20 25 30

− x2 −1 0 −1 −4 −9 −16 −25 −36

y = 5x − x2 −6 0 4 6 6 4 0 −6

b

–6

–8

–4

–2

2

4

6

8

–1 0 1 2 3 4 5 6

y = 5x – x2

y

x

c Symmetrical about x = 2.5; maximum point (2.5, 6.25), crosses x at 0 and 5 d x = 0, 5

2 a x 0 10 15 20 25 35

0.7x 0 7 10.5 14 17.5 24.5

− 0.02x2 0 −2 −4.5 −8 −12.5 −24.5

y = 0.7x − 0.02x2 0 5 6 6 5 0

b

0

2

4

6

8

5 10 15 20 25 30 35

x

y

y = 0.7x – 0.02 x2

c 30 m and closer


Geometry and Measures Strand 1 Moving on Answers (pages 199–200)

1 £48.932 70 000 m or 70 km3 listed items total = 2.33 kg

1 × fruit cake +1 × jar of jam + 2 × bars of chocolate + 3 × sweets = 2.73 kg4 132 litres5 latest = 5.35 p.m.6 Yes, total height is 20.8 cm7 Yes, he will swim 9 km8 Penny by 2.98 cm or 1 and 11

64 inches

9 Yes, he has a BMI of 20.910 France11 No, the weight of his suitcase is 16.03 kg


Geometry and Measures Strand 1 Unit 8 Answers

Practising skills (pages 203–204)1 a 040°

b 305°c 110°d 250°

2 Bridgetown 053°; Yapton 090°; Littleton 180°; Hopesville 258°; Kings Chapel 335°3 a 180°

b 090°c 045°d 225°

4 a Northb South-Eastc North-Westd West

5 a i student’s own diagramii x = 75°, y = 285°iii 105°iv 285°

b i student’s own diagramii x =50°, y = 310°iii 130°iv 310°

c i student’s own diagramii x =115 °, y = 245°iii 065°iv 245°

6 Student’s own diagrams7 No, the back bearing is 250°


Strand 1 Units and scales Unit 8 Bearings Band e

152 153

Developing fluency (pages 204–205)1 a 067°

b 247°2 a 135°

b 315°3 a 132°

b 197°4 a The three towns lie on a straight line.

b Aneesa, you need to know which town is in the middle to be able to work out the back bearing.5 a student’s own diagram

b 9.4 kmc 017°d 197°

6 a 255°b 075°c 037°d 217°e 125°f 305°



152 153

Problem solving (page 206)1 a student’s own diagram

b 040°2 a student’s own diagram

b 250°c 010°

3 a student’s own diagram, 225°b That the measurements are from the same point in Dalton.

4 See diagram for proof

B

A

100° – 40° = 60°

360° – 140° – 160° = 60°

40°

20°

C

5 4.5 miles6 200°, 290°, 020°



154 155

Reviewing skills (page 207)1 263°



Practising skills (pages 209–210)1 a 1 : 2

b 1 : 3c 2 : 3

2 a 3.5 kmb 20.8 cm

3 a 1 : 20 000b 1 : 2500c 1 : 250 000d 1 : 10 000 000

4 a 20 mb 8 cm

5 a Places Distance on map Distance in real life

Library to Sports centre 6 cm 1.2 km

School to park 2.5 cm 0.5 km

Cinema to supermarket 7.5 cm 1.5 km

Café to cinema 10 cm 2 km

Bowling alley to river 9 cm 1.8 km


Strand 1 Units and scales Unit 9 Scale drawing Band f

156 157

Developing fluency (pages 210–212)1

2.7 cm

3.1 cm

2.3 cm

2.1 cm

2.0 cm

1.6 cm

2 a student’s own diagramb 12.01 kmc 246°d 066°

3

4 a i cylinderii 40 cm

b i 42.5 cmii grey sail: base = 10 cm by height = 17.5 cm

yellow sail: base = 7.5 cm by height = 10 cmiii student’s own diagram

c i 1.7 mii 1.2 miii Yes

Item Plan measurement

True measurement

Length of patio 4.7 cm 9.4 mWidth of patio 1.9 cm 3.8 mLength of lawn 9.0 cm 18.0 mLength of vegetable patch

5.2 cm 10.4 m

Width of pond 1.4 cm 2.8 mLength of pond 2.4 cm 4.8 mWidth of house 1.9 cm 3.8 mLength of shed 1.9 cm 3.8 mWidth of shed 1.4 cm 2.8 mLength of path 5.2 cm 10.4 m



156 157

5 a student’s own diagram b i 8 km

ii 6.2 kmiii 12.7 km

c Tom 6.1 km; Molly 7.4 km; Evan 6.5 km6 Many possible answers, though 1 : 110 is sensible as the maximum diagram dimensions are approximately 27.5 cm

deep and 19 cm wide (with a 1 cm border).7 a i 2 km × 1.7 km

ii 3.4 km² b i 0.5 km

ii 0.25 km²c 0.25 km² / cm² × 13.6 cm² = 3.4 km²d 2.8 cm²



158 159

Problem solving (pages 212–213)1 a student’s own diagram

b 29 m2

Scale: 1 cm to 1 m

3 a student’s own diagramb 15 cm

4 a student’s own diagramb No, 4.2 m

5 a student’s own diagramb 263°, 33.8 miles

6 28 cm



158 159

Reviewing skills (page 214)1 a student’s own diagram

b 270 cm



Practising skills (pages 216–217)1 a 450 km/h

b 32 km/hc 40 km/h

2

1 m/s 3600 m/h 60 m/min 3.6 km/h

÷ 60÷ 60

÷ 1000× 60

× 1000

× 60

3 a 10 m/sb 36 000 m/hc 36 km/h

4 a 40 km/h b i 2 m/min

ii 3.3 cm/sc 0.22 m/s

5 Jamie: £8.50 per hourSarah: £7.80 per hourJamie is better paid


Strand 1 Units and scales Unit 10 Compound units Band f

160 161160 161

Developing fluency (pages 217–218)1 126 mph2 Jake: £1.38 per litre

Amy: £1.42 per litreJake’s petrol is cheapest

3 920 g4 a 2.36 m/s

b 67.5 mph5 a 83 miles

b 66.4 mph6 £4787 a 271 million km

b 19 350 mph8 168


Strand 1 Units and scales Unit 10 Compound units Band g

162 163

Problem solving (pages 218–219)1 UK, by 1.25 mph OR 2 km/h2 31 200 kg3 a 156 miles

b 52 mph 4 3750 seconds5 a 5.4 hectares

b 1000 kg is enough as 993.6 required6 2046 m


Strand 1 Units and scales Unit 10 Compound units Band f

162 163

Reviewing skills (page 219)1 a 65 km/h

b 18.06 m/s2 Toby earns 30p per hour more



1

2 Children = 90°; men = 216°; women = 54°3 45°4 75°5 202.5°6 AOB + 90° + 90° + 2 AOB = 375° ≠ 360°7 a

b For example:

8 165°9 124°10 140°11 120°12 Angle PSQ = 180° − 2a; angle QSR = 180° − 2b. Angle PSQ + angle QSR = 180°; giving 2a + 2b = 180°

Angles of triangle PQR = 180° = a + a + b + b = 2a + 2ba = 45°; b = 45°; angle PQR = 90°

13 60°14 72°



Practising skills (page 226)1 Shape Name of shape How many pairs of

parallel sidesHow many pairs

of equal sidesHow many lines of

symmetry Order of rotational

symmetry

rectangle 2 2 2 2

square 2 2 4 4

parallelogram 2 2 0 2

rhombus 2 2 2 2

trapezium 1 0 0 1no rotational symmetry

isosceles trapezium 1 1 1 1no rotational symmetry

kite 0 2 1 1no rotational symmetry

arrowhead 0 2 1 1no rotational symmetry

2 Rhombus or a square3 Rectangle, square, parallelogram and rhombus4

5 Suzanne is right – A rhombus is a special parallelogram with all sides the same length.Charlie is right – A square is a special rhombus with 4 right angles.


Strand 2 Properties of shapes Unit 6 Types of quadrilateral Band e

166 167

6 a

b

c No, the angles must add up to 360°So the fourth angle is also 90°

d Yes



166 167

Developing fluency (page 227)1 a (4, 5)

b (0, 5)2 (1, 5) (The co-ordinates given for point A should be (1, 2))3 a Parallelogram, rectangle. square, kite, arrowhead

b Rhombus (special case square), kite, parallelogram4 a 5

b 75 a Can: square, rectangle, isosceles trapezium, kite, arrowhead. Cannot: rhombus, parallelogram, trapezium

b Cannot draw parallelogram6 a

a

bS P'

M

P R'

Q'

Q

p

q S'

R

b SRP′ is a straight line since PQ is parallel to SR. So a + b = 180°



168 169

Problem solving (pages 228–229)1

2 34°3 a

b

4 Opposite angles are equal so a + b + a + b =360° so a + b = 180°5 56°6 48°



168 169

Reviewing skills (page 229)1 a Rectangle; two pairs of parallel sides; opposite sides parallel; two pairs of equal sides; opposite sides equal; all

angles equal; two lines of symmetry; order 2 rotational symmetryb Kite; two pairs of equal sides; no parallel sides; two lines of symmetry; order 1 rotational symmetryc Equilateral triangle; three equal sides; three lines of symmetry; order 3 rotational symmetryd Square; two pairs of parallel sides; opposite sides parallel; two pairs of equal sides; opposite sides equal; all

angles equal; four lines of symmetry; order 4 rotational symmetrye Rhombus; two pairs of parallel sides; opposite sides parallel; two pairs of equal sides; opposite sides equal; two

pairs of equal angles; opposite angles equal; two lines of symmetry; order 2 rotational symmetryf Parallelogram; two pairs of parallel sides; opposite sides parallel; two pairs of equal sides; opposite sides equal;

two pairs of equal angles; opposite angles equal; no lines of symmetry; order 2 rotational symmetryg Triangle; no equal sides; no equal angles; no lines of symmetry; order 1 rotational symmetryh Arrowhead; two pairs of equal sides; no parallel sides; one line of symmetry; order 1 rotational symmetryi Isosceles trapezium; one pair of parallel sides; one pair of equal sides; opposite sides equal; two pairs of equal

angles; one line of symmetry; order 1 rotational symmetry



Practising skills (page 233)1 a = 70°, b = 110°, c = 123°, d = 67°, e = 67°2 f = 107°, g = 107°, h = 107°, i = 136°, j = 136°, k = 44°, l = 136°, m = 95°, n = 85°, o = 95°, p = 85°,

q = 55°, r = 125°3 Yes, the other angles are either 50° or 130°.4 f = 115° (corresponding), g = 65° (supplementary), h = 115° (vertically opposite), i = 65° (supplementary)


Strand 2 Properties of shapes Unit 7 Angles and parallel lines Band f

170 171170 171

Developing fluency (pages 234–235)1 a a = 72°

b 18°2 a = 53°, b = 37°, c = 53°, d = 37°3 a = 37° (alternate), b = 64° (corresponding), c = 40° (alternate), d = 100° (corresponding),

e = 40° (angles of a triangle)4 a a = 82° (angles on a line), b = 31° (vertically opposite), c = 31° (corresponding), d = 31° (vertically opposite),

e = 67° (corresponding), f = 82° (vertically opposite, or angles in a triangle)b 180°

5 a = 81° (angles in isosceles triangle), b = 99° (angles on a line), c = 99° (outside angle of similar triangle)6 85°

If the line from E is extended to reach the line AC, then angle ACE is 60° (alternate angles)Angle CBX is 25° (angles on a straight line add up to 180°)Angle CXB = 95° (angles in a triangle add up to 180°), so angle x = 85° (angles on a straight line add up to 180°).



172 173

Problem solving (pages 235–236)1 50°

Angle ADE = 65 degrees (corresponding angle and angles on a straight line add up to 180°)Angle ADE = angle AED (isosceles triangle)So angle DAE = 50° (angles in a triangle add up to 180°)

2 8°The other two angles in the small triangle where x is marked as the top angle are 62° (alternate angles) and 110° (symmetry), respectively.Therefore x = 8° (angles in a triangle add up to 180°)

3 Yes, for example by extending one side of the square and using alternate angles4 107°; the angles of an equilateral triangle are each 60°, so x = (60 + 47)° (alternate angles)5 135°6 67°



172 173

Reviewing skills (page 237)1 a p = 85° (vertically opposite angles), q = 95° (angles on a straight line)

b r = 103° (corresponding angles to the 180 − 77 straight line angle)c s = 100° (vertically opposite), t = 80° (alternate angle), u = 80° (vertically opposite) v = 100° (corresponding)

2 a = 115°, b = c = 65°3 A = 70°, B = 40°, so CBF = 70° = DEF (corresponding angles)

CFB = 180 − 40 − 70 = 70° = EFD (angles of a triangle)So angle DEF = 70° = angle EFD, and EDF = 40° (corresponding angles)Triangle EDF is isosceles



Practising skills (pages 239–240)1 a 60°

b 120°2 a 45°

b 135°3

Regular polygon Number of sides Size of each exterior angle

Size of each interior angle

Sum of interior angles

Equilateral triangle 3 120° 60° 180°Square 4 90° 90° 360°Pentagon 5 72° 108° 540°Hexagon 6 60° 120° 720°Octagon 8 45° 135° 1080°Decagon 10 36° 144° 1440°Dodecagon 12 30° 150° 1800°Pendedecagon 15 24° 156° 2340°Icosagon 20 18° 162° 3240°

4 a 90°b 65°c 110°d 75°e 45°

5 a 36b 170° c 6120°

6 a No, a regular pentagon has 5 equal sides and 5 equal angles.b 540°c EDC = 90°, DCB = 90°, ABC = 150°, AED = 150°, EAB = 60°


Strand 2 Properties of shapes Unit 8 Angles in a polygon Band g

174 175174 175

Developing fluency (pages 241–242)1 a 900°

b 360°c 540°d 108°

2 a 4b 720°c 720°d 120°

3 a i 1080°ii 360°iii 720°

b An n sided polygon can be divided into n triangles. The total angle sum of the triangles is n × 180°. The angles at the centre always sum to 360°, so the angle sum of the interior angles is n × 180° − 360°

c n × 180° − 360° factorises to give (n − 2) × 180°4 a 360°

b 12c 150°d 1800°

5 a 60 sidesb 13c 6120°

6 a 6b Yes, because the interior angle of a regular hexagon is 120°, and so you can fit 3 hexagons around a point as

3 × 120° = 360°c No, because the interior angle of a regular pentagon in 108° which is not a factor of 360°

7 a AB – 4 sides, square; BC – 6 sides, hexagon; AC – 12 sides, dodecagonb BC and AC would be sides of 8-sided shapes, octagons

8 a 104°b r = 122°; s = 117°; t = 121°c 59°



176 177

Problem solving (pages 243–244)1 a 72°

b OA = OC as O is the centreAB = BC as the polygon is regular

2 150°3 72°4 12°5 120°6 45°



176 177

Reviewing skills (page 244)1 a 40°

b 100°c 80°

2 a 72b No, the calculation of number of sides does not give a whole number

3 a Question states that a dodecagon has internal angles 144° – this is not true, the interior angles are 150°

Exterior angle = 360°12

= 30°, so internal angle = 150°b 120°c Interior angle of regular hexagon = 180° − 360°

6 = 120° = angle GBC = angle FCB



1 a 84 m2

b £13442 £643 3 packs4 13 tins5 1 cm × 54 cm; 3 cm × 18 cm; 6 cm × 9 cm6 No, it will take up 1

4 of the hall

7 3 packs8 68 slabs (including the 4 at the corners, for a continuous path)9 Yes, the 19.5 m2 area of the sail experiences a force of 624 N



Practising skills (pages 249–250)1 a 25.1 cm

b 28.3 mc 37.7 cmd 18.8 me 39.0 cmf 3.1 km

2 a 14.2 cmb 12.6 cmc 9.4 cmd 11.0 cm

3 5.7 cm4 Radius Diameter Circumference

a 2 cm 4 cm 12.6 cm

b 5 cm 10 cm 31.4 cm

c 6.5 cm 13 cm 40.8 cm

d 2.9 cm 5.8 cm 18.2 cm

e 9.1 cm 18.1 cm 57 cm

f 19.1 cm 38.2 cm 120 cm


Strand 3 Measuring shapes Unit 3 Circumference Band f

180 181

Developing fluency (pages 250–251)1 a 51.4 cm

b 32.1 cmc 55.7 cmd 42.8 m

2 83.7 cm3 70.0 m4 a 301.6 m

b 96 times5 a Minute hand travels 15.71 cm

Hour hand travels 0.92 cmDifference = 14.79 cm

b Minute hand travels 5.24 cmHour hand travels 0.31 cmDifference = 4.9 cm

6 154 tiles



180 181

Problem solving (pages 251–252)1 2272 326 726 cm3 287 mm4 135 m5 0.9 m6 634 cm7 95 cm



182 183

Reviewing skills (page 253)1 a 34.6 cm

b 44.0 cmc 66.0 m

2 a 8.3 cmb 4.1 cm

3 a 100.5 cmb 54.8 cm

4 Dad: 1989.4Cain: 3536.8Cain’s wheels turn more by 1547.4 turns



Practising skills (pages 255–256)1 a 28.3 cm2

b 201.1 cm2

c 50.3 mm2

d 113.1 m2

e 380.1 cm2

f 346.4 mm2

2 a 15.9 cm2

b 12.6 cm2

c 7.1 cm2

d 9.6 cm2

3 3.3 cm4 a 4.4 cm

b 8.7 cmc 27.5 cm

5 Radius Diameter Area Circumference

a 7 cm 14 cm 153.9 cm2 44.0 cm

b 8.5 cm 17 cm 227.0 cm2 53.4 cm

c 5.0 cm 10.1 cm 80 cm2 31.7 cm

d 8.9 cm 17.8 cm 250 cm2 56.0 cm


Strand 3 Measuring shapes Unit 4 Area of circles Band f

184 185

Developing fluency (pages 257–258)1 a 380.1 cm2

b 132.7 cm2

c 56.5 cm2

d 63.6 cm2

e 14.7 m2

f 504 cm2

2 50.1 cm3 71.6 cm2

4 27.5 cm5 Red: 816.81 cm2

Blue: 439.82 cm2

Difference is 376.99 cm2

6 11.8 cm7 351.9 m2

8 a 14 627 cm²b £0.0050/cm²



184 185

Problem solving (pages 258–260)1 76.4 m²2 £75403 Yes, for example cost of 24 bags of chippings is £724 a 159 m

b 207 litres5 £18436 10.725 m²



186 PB

Reviewing skills (page 260)1 a 78.5 cm²

b 38.5 mm²c 18.1 km²

2 a 4.1 cmb 8.1 cm

3 a 112.8 cm²b 195.4 cm²

4 12.6 m²


Geometry and Measures Strand 4 Moving on Answers (page 262)

1 12.4 m2 5.3 m3 254 cm4 18.1 m



1 a y = 10b x = 0 (The co-ordinates of point S should be (2, 4))

2 a 6 units, this is a complete reflectionb 6 units, as the line on both sides of the reflecting line are reflected

3 a Movement (4, 0)b Movement (8, 0)



Practising skills (pages 267–268)1 a and g; b and i; c and j; d and f; e and h; k and l2 a

C

b C

c

C

d C

3 a Enlargement scale factor 2, centre (0, −2)b Enlargement scale factor 2, centre (5, −3)c Enlargement scale factor 2, centre (3, 4)d Enlargement scale factor 3, centre (4, 1)

e Translation

5–1

f Translation

36

g Enlargement scale factor 12, centre (0, −2)

h Enlargement scale factor 12, centre (5, −3)

i Enlargement scale factor 13, centre (4, 1)


Strand 5 Transformations Unit 6 Enlargement Band f

190 191

4 a Two identical kite shapes K and L, which consist of 3 kites, from smallest to largest, green, blue, pink. A larger kite shape, which consists of two kites, smallest to largest: red, purple

b i 2ii 3

c A translationd Enlargement scale factor 3e Enlargement scale factor 1

2



190 191

Developing fl uency (pages 269–272)1 a and b

10

123456789

10

x

y

2 3 4 5 6 7 8 9 10

P

A

B

c Translation

–2–5

2 a and b

0

123456

2 3 4 5 6– 1– 1– 2– 3– 4– 5– 6

– 2– 3

– 5– 6

y

x1

– 4

W

A

22 33 4

2

22 4

22

22 4

22B

c Translation

–1–5

3 a, b and c

10

123456789

10

x

y

2 3 4 65 7 8 9 10

TA

CB

d C is an enlargement of B, scale factor 2



192 193

4 a Scale factor f = 2, centre (−1, 1)b Scale factor g = 1.5, centre (5, 3) c h = 3d Scale factor 3, centre (0.5, 0.5)

5 a Enlargement scale factor 3, centre (5, −3)

b Enlargement scale factor 13

, centre (5, −3)

c Enlargement scale factor 2, centre (2, −6)

d Enlargement scale factor 12

, centre (2, −6)

6 a and b

A

B

0

123456

2 3 4 5 6– 1– 1– 2– 3– 4– 5– 6

– 2– 3

– 5– 6

y

x1

– 4

P

c Enlargement scale factor 12

, centre (6, 5)

7 a Any enlargement, with at least 1 scale factor not 2 b Scale factor k × m

8 a, b, c

A

Y

CZ

B

X

d Regular tetrahedron, triangle based pyramide The resulting 3D shape would not be a regular tetrahedron



192 193

Problem solving (pages 272–273)1 No, e.g. 15 ÷ 6 = 2.5, 13 ÷ 4 = 3.252 92–94 m2

3 6 m4 a Student’s own drawing

b 1 : 45 a (1, 1)

b (7, 7), (10, 7) and (10, 4)

6 a 25

b The co-ordinates of the other vertices are (1, 3), (5, 3) and (5, 1)



194 PB

Reviewing skills (page 274)1 a

C

b

D

2 a and b (In a, the co-ordinates of the centre of enlargement are (−5, 6))

0

123456

2 3 4 5 6– 1– 1– 2– 3– 4– 5– 6

– 2– 3

– 5– 6

y

x1

– 4

12

22 311

22111

33

122

33

A

B

Q

c scale factor 2, centre (−5, 6)

d scale factor 12

, centre (−5, 6)

PB 195© Hodder & Stoughton Ltd 2015


1 I have 4 vertices – tetrahedron; I have 3 faces – cylinder; I have 8 faces – hexagonal prism; 1 have 16 vertices – octagonal prism; 5 of my faces are rectangles − pentagonal prism

2 a Falseb Truec Trued False

3 a and b Student’s drawings, for example:

x

yz

4 a 3b 9c Infinited 4

5 12 m6 7 m × 5 cm × 4 cm; 10 cm × 7 cm × 2 cm; 10 cm × 4 cm × 3.5 cm



Practising skills (pages 280–281)1 a i D

ii F b i 6

ii 8iii 12

2 a i Eii C

b i 3iii No, as two of the net edges join to become one cube edge

c i 3iii No, face A folds over the top to join

3 A, C, D, F, H


Strand 6 Three-dimensional shapes Unit 2 Understanding nets Band f

196 197196 197

4 a i For example:

ii 22 cm2

iii The same – surface area is also 22 cm2

b i For example:

ii 40 cm2

iii The same – surface area is also 40 cm2



198 199

5 a Square-based pyramid

b Triangular prism

c Cuboid

d Pentagon-based pyramid



198 199

Developing fluency (pages 282–284)1 A – T – X

B – S – WC – Q – YD – R – UE – P – V

2 i For example:a

b



200 201

ii a

b



200 201

iii a

b



202 203

3 a 24 cm2

b 48 cm2

c 50%d For example:

i 50%

ii 50%

4 a i 8ii Triangleiii 4

b i 4ii Triangle

c No. One face will fold on top of another, leaving one side open5 a Pentagons b i 12

ii 30iii 20

c 12 + 20 − 30 = 2d 19e QP orange, KLM light blue, central blue, CDE light green and FGH orange f XYg Q and Rh 240 cm2



202 203

Problem solving (pages 285–286)1 For example:



204 205

2 For example:



204 205

3 For example:

C

A

6.9 cm

4

A

C

1 cm = 20 cm; length on diagram = 4.5 cm so actual length = 90 cm



206 207

5 a For example:

b Green and light blue; yellow and red; lilac and dark bluec Regular octahedron

d i 8ii 12iii 6



206 207

6 a Rotational symmetry, order 2 about the centre point of square CDGH b i 5

ii 5iii 8

c Gd No, it will form a square-based pyramide Student’s construction of the net, based on these

A B

C

DG

H

EF

f i 3.5 cmii 44 cm2

iii (1 + 3) × 42 = 43.7 cm2



208 209

Reviewing skills (page 287)1 i a for example:

b 62 cm2, surface area is also 62 cm2

ii a For example:

b 102 cm2, so surface area is also 102 cm2



208 209

2 a

b



210 211

3 a

ABV V

b 7 cm



Practising skills (pages 290–291)1 a 60 cm3

b 20 cm3

2 a 60 cm3

b 32 cm3

c 36 cm3

3 a 82 cm2

b 148 cm2

c 36 cm2

d 64 cm2

e 92.8 cm2

4 5 cuboids that create a volume of 72 cm3 for example:

Length Breadth Height Volume1 1 72 72 cm3

2 1 36 72 cm3

4 1 18 72 cm3

6 1 12 72 cm3

8 1 9 72 cm3

5 Length Breadth Height Volume Surface area

a 5 cm 3 cm 2 cm 30 cm3 62 cm2

b 6 cm 2 cm 2 cm 24 cm3 56 cm2

c 5 cm 4 cm 3 cm 60 cm3 94 cm2

d 7 cm 5 cm 1 cm 35 cm3 94 cm2

e 6 cm 4 cm 1.5 cm 36 cm3 78 cm2


Strand 6 Three-dimensional shapes Unit 3 Volume and surface area of cuboids Band f

212 213

Developing fluency (pages 292–294)1 a A and C

b B and C2 a Q, S, R, P b i P

ii Qc The volume of P is double the volume of Q. The volume of R is 4

5 the volume of P.

3 a Wrong unit: it has a volume of 1 cm3

b Wrong unit: it has a surface area of 24 cm2

c It has a volume of 64 cm3

d It has a surface area of 54 cm2

e The volume of a 1 cm sided cube is 1 cm3, but the volume of a 2 cm sided cube is 8 cm3, which is 8 times the size of the 1 cm sided cube

4 a 49 cm2

b 7 cmc 343 cm3

5 a False, there are two units, m and cm; the actual volume is 400 000 cm3

b Truec False, it is 4 times

6 500 cartons7 2 tins8 a 3 cm

b 114 cm2

9 a

30°

2 cm

2 cm4 cm

2 cm2 cm

6 cm

7 cm

5 cm

30°

b 96 cm2

c 44 cm3

d i 16ii 10iii 8

e 10 + 8 = 16 + 2



212 213

Problem solving (pages 294–296)1 Yes, by 9.5 m2

2 Correct, because it will take 2250 min to fill and there are 1440 min in a day3 a 150 l

b That the thickness of the walls is 0/negligiblec 118.5 l

4 1 cm5 4800 cm³6 a 90 m

b 506.25 m2

c 46.4 md Area face = 12 squares, total area = 48 squares = 6075 m2

7 a 4b 4

c i 124 cm2

ii 4 d i cuboid

ii Square-based pyramide 32 cm3

f i 9ii 16iii 9

g 9 + 9 − 16 = 2



214 215

Reviewing skills (page 297)1 a Volume = 60 cm3; surface area = 122 cm2

b Volume = 240 cm3; surface area = 236 cm2

2 a The volume of the box, 500 cm3, is not a multiple of the volume of the lumps, 3 cm3, so there will be wasted space in each box

b 200 lumps



Practising skills (pages 299–300)1 a i ii iii

b ii iiii

c iiiiii

d iiiiii

e iiiiii

f iiiiii

g iiiiii


Strand 6 Three-dimensional shapes Unit 4 2-D representations of 3-D shapes Band f

216 217

h iiiiii

2 a iiiiii

b iiiiii

c i ii iii

d i ii iii



216 217

Developing fluency (pages 300–302)1 a i ii iii

b 10.2 mc 4 or 5 (if one in the loft)

2 a

b

c

d



218 219

3 a Plan = D; front = A; side = Eb Plan = C; front = A; side = Hc Plan = I; front = A; side = Fd Plan = B; front = H; side = Ee Plan = B; front = G; side = A

4 a i

ii

b i

ii

5 a Cone and cylinderb

BaseTop



218 219

Problem solving (pages 302–304)1 a For example:

b 42 For example:

Shape 1 Shape 3

Shape 5

Shape 2 not possible – plan should be 1 squareShape 4 not possible – the side elevation and plan should be the other way around

3 a

b i 90 cm2

ii 54 cm3



220 221

4 a i Triangular prismii Square-based pyramid

b

c



220 221

Reviewing skills (page 305)1 a i ii iii

b i ii iii

c i ii iii

2 a i ii iii

b i ii iii

c i ii iii



Practising skills (pages 308–309)1 a i 18 cm2

ii 144 cm3

b i 20 cm2

ii 240 cm3

c i 16 cm2

ii 176 cm3

2 a i 7.07 cm2

ii 70.7 cm3

b i 50.3 cm2

ii 351.9 cm3

c i 78.5 cm2

ii 628.3 cm3

d i 153.9 cm2

ii 923.6 cm3

3 54 cm3

4 8 cm5 a 14 cm2

b 20 cm6 a i 50 m2; 40 m2; 30 m2

ii 120 m2

b i 12 mii 120 m2

c They are equal d i 6 m2

ii Find the volume = 60 m3

7 a 210 cm3

b 168 cm3

c 96 cm3

d 62.8 cm3


Strand 6 Three-dimensional shapes Unit 5 Prisms Band g

222 223222 223

Developing fluency (pages 310–311)1 3.5 m3

2 a 62.8 cmb 1885 cm2

c 314.2 cm2

d 2513 cm2

3 370 cm3

4 a Qb 301.6 cm3

5 a 2.4 m3

b 8.48 m2

6 a Anna’sb 0.033 m3

7 a Yes, rectangleb Yes, trianglec Yes, circled Noe Nof Yes, squareg No



224 225

Problem solving (pages 312–313)1 a 47.52 m2

b 12.96 m³2 Yes, as 50 bins have a capacity of 154 m³3 a 1676 trips b i 25 133 m2

ii 2482 m3

4 a 2.8 m²b 0.3 m3 or 300 000 cm3

5 a 201 cm2

b 1407 cm2; £1688.926 a i 900 cm2

ii 180 m3 or 180 000 000 cm3

b 420 m3



224 225

Reviewing skills (page 314)1 a i 21 cm2

ii 210 cm3

b i 28 cm2

ii 252 cm3

2 46 cm2

3 a i 113.1 cm2

ii 1018 cm3

b i 63.6 cm2

ii 763 cm3

c i 28.3 cm2

ii 226 cm3

4 189 cm3



Practising skills (pages 317–318)1 a 1 : 2 b i A = 32 cm; B = 64 cm;

ii 1 : 2 c i A = 48 cm2; B = 192 cm2

ii 1 : 42 a 30 cm

b small = 225 cm2; large = 900 cm2

c 1 : 43 a 1 : 3

b 1 : 94 a 64 cm3

b 8000 cm3

c 1 : 5d 1 : 125e 125

5 a Area enlarges by the square of the lengthb 8


Strand 6 Three-dimensional shapes Unit 6 Enlargement in two and three dimensions Band g

226 227226 227

Developing fluency (pages 318–319)1 a 1.5

b x = 6, y = 4.5, z = 8c 240 cm³, 810 cm³d 8 : 27

2 a 6 cm and 10 cmb 3 : 5

3 a 24 m × 16 mb 15.36 cm²c 384 m² or 3 840 000 cm²d 1 : 250 000

4 a i 3.75 cmii 135 cm3

b i 1458 cm2

ii 3645 cm3

5 a 12 cmb 30 cmc 2d 1 : 8



228 229

Problem solving (pages 320–321)1 18 m²2 a 90 cm × 36 cm

b 18c 3240 cm2

3 a 100 cmb Block A 9600 cm2; Block B 60 000 cm2

c Area block A = 6 × 40² = 9600 cm², Area block B = 6 × 100² = 60 000 cm²Ratio is 9600 : 60 000 = 1 : 6.25

4 a 20 cmb Width = 6 m; height = 2.4 mc 115.2 m3

d Width = 15 cm; height = 6 cm; volume = 1800 cm3

e Model = 0.0018 m3; real = 115.2 m3; ratio is 0.0018 : 115.2 = 1 : 64 000 = 1 : 403

5 a 2 : 3b Areas are 96 and 216 cm2 with a ratio of 96 : 216 = 4 : 9 which is not the same as 2 : 3 (=4 : 6)c Masses are 64d and 216d, where d is the density

Ratio is 64d : 216d = 8 : 276 a 24 cm and 36 cm b i 1 : 3

ii 1 : 9c Resulting picture is 32 cm × 44 cm. This is a different ratio of sides to the original (1 : 1.375 vs 1 : 1.5), so it is not

an enlargement of B



228 229

Reviewing skills (page 322)1 a 5 cm × 5 cm × 15 cm

b 56 cm2

c 350 cm2

d 24 cm3

e 375 cm3

f 1 : 6.25g 1 : 15.625

2 112.5 (i.e. 112 cubes)


Statistics and Probability Strand 1 Moving on Answers (page 324)

1 a Geneva mean temp. −2 °C; London mean temp. 0.4 °Cb Geneva median temp. −1 °C; London median temp. −1 °Cc Mean temperature as it shows that London is slightly warmer, which is a true reflection of the data.d Geneva temperature range 8 °C; London temperature range 6 °C

2 a £1.95b £447.58

3 He should put the median donation (£5) as this might encourage more people to donate more.4 a £229.76

b £216.15c The median would be more useful as it eliminates those cars with extremely expensive bills.d The range as it would tell the customer if there were any very expensive bills.


Statistics and Probability Strand 1 Unit 3 Answers

Practising skills (pages 326–328)1 a £3.50 b i 2, 2.50, 2.50, 2.50, 3, 3, 3.50, 3.50, 3.50, 3.50, 3.50, 3.50, 4, 4, 4, 4.50, 4.50

ii £3.50 c i £57.50

ii £3.382 a 25

b 117c 4.68d 4

3 a 39b 6.5c More people don’t work at the weekend and can go shopping.

4 a You have to attempt to take the test at least once to be able to pass (or equivalent comment)b 22c 56d 2.55e 2

5 a 0. This represents seeing no birds, but 21 students did see birds so the mode is not representative of the class. The mode could be 9 students with no gardens or who have gardens that birds don’t like.

b 2c 4.1d Median. The mean is skewed by the two gardens that have 30 and 40 birds.

6 a 35b 0 – 8, 1 – 6, 2 – 7, 3 – 6, 4 – 2, 5 – 2, 6 – 4c 0d 6e 2.286f depends on the opposition


Strand 1 Statistical measures Unit 3 Using frequency tables Band e

232 233

Developing fluency (pages 328–330)1 a i B: 4.7, C: 5.3

ii B: 3, C: 5iii B: 4, C: 5iv B:10, C: 5

b Caroline has the higher average and has a smaller range c Caroline has the higher average score and she is more consistent

2 a i Number of people Frequency

1 3

2 4

3 5

4 6

5 1

6 0

7 0

8 1

ii

1 2 3 4 5 6 7 80

1

2

3

4

Freq

uenc

y

Number of people

5

6

7

b i 3.15ii 4iii 3iv 7

c The mean number of people per house is 3.15, which is not possible for whole people. The mode and median are more realistic measures and are close in size: 4 and 3 respectively. There is a large range in people per house, with just one house with 8 people in it, but no houses with 6 or 7 people.

3 a 70. Traffic/missed train or other reasonable explanationb Not including the outlier, the ranges are: Alex 13, Eddie 11. Alex had the shortest time (15 minutes)

c i A: 21.75, E: 21.9ii A: 21, E: 21iii A: 13, E: 11

d i Alexii mean

4 a 9A:12.68, 9B:11.375b 9A:17, 9B:11c 9A has the higher average but it is more spread out



232 233

5 a 0 6 7 7 8 9 91 2 1 2 3 4 5 7 7 8 3 0 1 1 3 4 6

b A large range with some very small plants and some large. Suggests the compost has very varied effects.c median = 24.5, mean = 21.9. Mean is more representative as it takes into account the whole range.d No, just half have this height. ‘Most’ usually means at least 2/3 so 10 cm or over is better description.

6 a Batch A Stem Batch B

0

2 2 2 2 2 2 3 3 3 4 5 5 6 7 7 8 9 1 4 4 4 4 4 4 4 5 5 5 6 6 6 8 9 9

0 1 1 2 2 2 2 2 2 3 4 4 4 4

b Batch B is marginally more effective because the mean is slightly higher and the range is the same.



234 235


b mean = 2.24, median = 2, mode = 0c median as it takes into account the majority of employees that have few days off

2 a 51 b i £18 000

ii £30 000iii £33 808

c Median. The mean is skewed by the Chief Executive’s salary and the mode misses the employees on middle pay grades.

3 a 29b mean = 432 kg, so 3% (1 horse)

4 a Maximum daily temp. (°C) Frequency

−4 6

−3 5

−2 6

−1 2

0 5

1 3

2 1

3 2

b 7 c i −2 °C

ii −1.4 °C d i 20%

ii 63.3%5 a

Number of children0 1 2 3 4 5 6 7

02468

Freq

uenc

y

101214161820

b It depends on the average that is used. If mode, the average number is 0, so the council shouldn’t provide a play area. If median, the average is 1.5, so the council should provide. If mean, the average is 1.64 so the council should provide.

6 a i 7ii 7.46iii 7iv 8

b Yes. All measures of average are larger for plants treated with Grow-Well, though the mode is the same. 7 9



234 235


b 23 is the outlier, one team may have had a reduced number of players, or were from a different league, or the score may have been written down incorrectly, or any other acceptable comment, and this score does not represent the scores from other matches

c Number of goals 0 1 2 3 4 5

Frequency 3 4 3 3 2 2

d i 2.18ii 1iii 2

e 5f mean = 3.33, mode = 1, median = 2, range = 23; the mean and range are most affected



Practising skills (pages 336–338)1 Height (cm), h Frequency, f Midpoint, m m × f

150 < h , 156 3 153 459

156 < h , 162 6 159 954

162 < h , 168 8 165 1320

168 < h , 174 3 171 513

174 < h , 180 2 177 354Totals 22 3600

Mean height = 360022

= 163.6 cm

2 a Length of call (minutes), l

Frequency, f Midpoint, m m × f

0 < l , 10 1 5 5

10 < l , 20 5 15 75

20 < l , 30 3 25 75

30 < l , 40 5 35 175

40 < l , 50 5 45 225

50 < l , 60 1 55 55Totals 20 610

b 30 ≤ l < 40

c 61020

= 30.5

3 a i 19 and 185 secondsii 166 seconds

b 83.5 seconds

c Time (seconds), t Frequency, f Midpoint, m m × f 0 < t , 40 3 20 60

40 < t , 80 8 60 480

80 < t , 120 9 100 900

120 < t , 160 1 140 140

160 < t , 200 3 180 540Totals 24 2120

d 80 ≤ t < 120

e 212024

= 88.3

f Their marks are above 63 and below 114


Strand 1 Statistical measures Unit 4 Using grouped frequency tables Band f

236 237236 237

4 a Speed (mph), v Frequency, f Midpoint, m m × f 0 < v , 10 1 5 5

10 < v , 20 12 15 180

20 < v , 30 22 25 550

30 < v , 40 4 35 140

40 < v , 50 1 45 45Totals 40 920

Mean speed = 92040

= 23 mph

b 30 mphc 40%

5 a 43445

= 9.64 hours

b 1.61 hours



238 239


20 = 89.55 and 81

b Number of people, n Frequency, f Midpoint, m m × f 0 < n , 50 3 25 75

50 < n , 100 10 75 750

100 < n , 150 5 125 625

150 < n , 200 2 175 350Totals 20 1800

c 50 ≤ n <100d 90e The means are very close, the median in part a is more accurate and is close to both means.f For small samples you would calculate it exactly and for large samples you would estimate it, or any other

correct explanation.2 a Time (minutes), t Frequency, f Midpoint, m m × f

0 < t , 10 0 5 0

10 < t , 20 1 15 15

20 < t , 30 8 25 200

30 < t , 40 14 35 490

40 < t , 50 7 45 315Totals 30 1020

Mean time = 102030

= 34 minutes

b range = 40 minutesc The second club were quicker on average but more spread out / varied in their times.

3 a Mean waiting time = 502.535

= 14.4 minutes

b 13 timesc for example, hold-up/accident/power failure and it does not represent a normal journey

4 a Distance (metres), d Frequency, f Midpoint, m m × f0 < d , 1 1 0.5 0.5

1 < d , 2 5 1.5 7.5

2 < d , 3 12 2.5 30

3 < d , 4 9 3.5 31.5

4 < d , 5 8 4.5 36Total 35 105.5

Mean distance jumped = 105.535

= 3.01 metres

b (14 × 3) – 35 = 7 foul jumpsc estimate 4 jumps which could be 2, 3 or 4 jumpers

5 B has a larger mean yield of 854.2 g vs 812.5 g for A



238 239

6 a Current salary Scheme 1 increase Scheme 2 increase Scheme 3 increase

Sandra £8000 £8400 £8870 £8750

Shameet £2200 £2310 £3070 £2950

Comfort £3600 £3780 £4470 £4350

b Scheme 2 benefits Sandra, Shameet and Comfort by the greatest amount. c The MD should choose Scheme 1 because 5% increase in employees’ current salary total is less than 5% of the

mean salary or the median salary so the increases will be smaller and therefore cost the company less.



240 241

Problem solving (pages 341–343)1 Score, s Frequency, f Midpoint, m m × f

0 < s , 10 11 5 55

10 < s , 20 6 15 90

20 < s , 30 12 25 300

30 < s , 40 10 35 350

40 < s , 50 9 45 405Totals 48 1200

a 25b 19c The mean is an estimate, therefore the cut-off for interview is also an estimate, which means the number of

applicants interviewed is also an estimate.2 a Distance travelled, d km Frequency f Midpoint m m × f

20 , d < 30 6 25 150

30 , d < 40 12 35 420

40 , d < 50 20 45 900

50 , d < 60 26 55 1430

60 , d < 70 11 65 715Total 75 Total 3615

b 40 , d < 50c 48.2 kmd 72.3 minutes

3 a 20 , t < 25b 25 minutesc 20.1 minutesd 10% of people are geniuses

4 a 100 peopleb 170 , h < 180c Incorrect, the range is 50 cm, as you take the midpoints of the classes to calculate the ranged 175.5 cme 7%

5 a 30 , c < 40b 40 , c < 50c 53.7d 61.1e 26%

6 Class Y1. Y1 has mean of 56, whereas X1 has a mean of 54. Y1’s median is 70, whereas X1’s median is 50.



240 241

Reviewing skills (page 343)1 a Mean = 51.7

Score, s Midpoint m Frequency f m × f 0 < s , 20 10 6 60

20 < s , 40 30 8 240

40 < s , 60 50 15 750

60 < s , 80 70 14 980

80 < s , 100 90 5 450Totals 48 2480

b 5c Their marks are lower than 50


Statistics and Probability Strand 2 Moving on Answers (pages 345–346)

1 a Tuesdayb 68

c 25

d Wednesday and Fridaye Pizzaf 34g It displays daily totals

2 a Octoberb 9c i 0 hours

ii 24 hoursd 480 hourse To be able to compare the two cities

3 a 19 daysb 13c 14d 27

e i Peter Isobel

0 2 4 8 9 10 9

0 3 4 7 11 2 3 9

7 8 9 12 2 2 6 7 8 9

1 2 13 1 2 5 8

Key: 10 | 0 = 100 minutesii Isobel – there are more entries in the 13- box.

4 There is no y axis label, 3-D makes it difficult to read the bars, bars are different widths.



Practising skills (page 349)1 a

0Jan Fe

bMar Apr May Jun Jul

yAug Se

pO

ctNov Dec

10

20

Month

Num

ber

of g

uest

s

30

40

50

60

b February c August d In winter and spring it is low, rises towards summer, falls in autumn and rises towards Christmas/December.

2 a

0

Max

day

tem

p

Mon Tue Wed ThuDay

Fri Sat Sun

2

4

6

8

10

12

b It shows the differences in the day-to-day temperature more clearly c Wednesday to Thursdayd The midday temperatures started at their maximum for the week, then fell suddenly on Thursday then started to

rise slowly back to the maximum temperature of the week.


Strand 2 Statistical diagrams Unit 3 Vertical line charts Band e

244 245

3 a Bi

llion

s of

peo

ple

01750 1800 1850

Date1900 1950 2000 2050

1

2

3

4

5

6

7

b The first billion increased slowly as it took many years to achieve. The next billions were achieved at a steady rate, as shown by the straightness of the line.

c 2.5 billiond 1980

4 The vertical scale does not go up in even steps. The horizontal axis does not have even steps. The axes are not properly labelled. There is no title for the line chart.



244 245


b Possibly lunch break was between 1 and 2 so no-one was in the shop at 2 p.m., or 2 p.m. is a unusual moment when no one was in the shop – even if there might have been at 1.59 p.m. or 2.01 p.m.

c iii This is the true statement – there is no information for in between the sample points so you can’t say.d The lines represent single sample moments, the times in between are not necessarily connected to the samples

before or after.2 a March and September, as new car registrations

b May and November c

0Jan Feb Mar Apr May Jun

Month

Num

ber

of c

ars

Jul Aug Sep OctNovDec

2

4

6

8

10

12

14

16

18

d The chart as it emphasizes the differences more clearly.3 a

80

Mon1

Mon2

Tue1

Tue2

Wed

1W

ed2

Thu1

Thu2

Fri1

Fri2

82

84

86

88

90

92

% p

rese

nt

94

96

98

100

b Attendance is lower at the start and end of the week – possibly due to the weekend being near, or possibly because of illness at the weekend. Attendance rises during the week from Monday to Thursday as possibly people recover from the illness or the weekend. The second week has marginally higher attendance on each day compared with the first week.



246 247

4 a

Jan0

Month

1 2Number of days Year1 Number of days Year 2

Num

ber

of d

ays

1

2

3

4

5

6

7

8

Feb

Mar Apr May Jun jul Aug Sep

Oct

Nov Dec

b Year 1 – April, May, July and November; Year 2 – May, Junec Year 1 – January and December; Year 2 – January and Decemberd see part a

e i Both years have increased repairs during the winter months.ii In both years the trend of the year is decreasing from a peak in January to a minimum in the summer, and

then an increased number of repairs as the season moves into autumn and then winter.5 a True. The runs increase with age, with a slight dip at age 16.

b False. The number of runs scored is accumulated over the year and is not continuous data. Therefore you cannot make statements of how many runs scored between ages.

c True. That season his runs were above the trend line.d False. Though the trend suggests he will score more than 720, it is not certain.

6 a i 38.4 °Cii 37.6 °C

b 1 p.m., 39.8 °Cc every hour, with every half hour between 12 and 2d The temperature above 37 °Ce Francesca’s temperature gradually rose from 10 a.m. until it peaked at 1 p.m. After 1 p.m. it decreased rapidly,

reaching a stable temperature at 3 p.m.7 Right: the scales are evenly spaced and axes labelled. The line/bars are drawn on the day. Wrong: the line connecting

the tops of the bars is incorrect, as the data is a sample at 3 p.m. and we cannot say what the temperature did between 3 p.m. on successive days.



246 247

Problem solving (pages 353–355)1 a 19 °C

b 12 degreesc 3 °Cd 6 a.m.e 8 a.m.; 20 degrees

2 a i 35ii 0

b 20 c i April, May, June, July, August

ii January, February, October, November, Decemberiii March, September

d Winter/Christmas time/December3 a 350 b i November

ii Januaryc Aprild Between December and January (or, if the year isn’t continuous, between February and March)e Between May and June. No, it just means that the number of books sold and the number of books Alfie has got

into his shop is the same.4 a

0

5

10

15

20

25

30

Mon Tue Wed ThuDays

Tem

pera

ture

(°C

)

Fri Sat Sun

b Whilst the highest midday temperature is on Wednesday, the temperature could be higher at other times of the day that were not measured.

c It was the mean temperature.



248 249

5 a T

ime

(s)

Week

14.51 2 3 4 5 6 7 8

15.015.516.016.517.017.518.018.519.019.520.0

b Week 7c injuryd A gradual decrease in times. Ignore the week 5 outlier, suggests he will run faster in week 9

6 a

Hei

ght

(cm

)

Year

01 2 3 4 5 6 7 8

50

100

150

200

250

Tree A Tree B

b Tree A – its 3rd year (between years 2 and 3); Tree B – its 3rd year (between years 2 and 3)c Tree A: 224–225 cm; Tree B: 108–112 cmd No, they have difference maximum heights and growth rates.



248 249

Reviewing skills (page 355)1 a

Time

37.0

8 a.m

.

9 a.m

.

10 a.

m.

11 a.

m.

12 no

on1 p

.m.

2 p.m

.

3 p.m

.

4 p.m

.

37.5

38.0

38.5

39.0

39.5

Tem

pera

ture

(°C

)

b 8 a.m. to 11 a.m.c around 11 a.m.d 10 a.m. to 11 a.m.e There are no data in between hours, so no.



Practising skills (pages 359–360)1 a University

b 90°

c 14

d Gap year – 42; University – 70; Apprenticeship – 42; Job – 14 2 a Eat anything

b 60°

c 16

d Eat anything – 40; No red meat – 27; Vegetarian – 20; Vegan – 18; Other – 153 a 9

b

Walk

Bus

Taxi

Own car

Lift

c 4204 a 15°

b

Sleep

Prowling

Eating

Grooming

c Add another segment of 7.5°, increase prowling to 97.5° and decrease sleeping to 210°


Strand 2 Statistical diagrams Unit 4 Pie charts Band f

250 251250 251

5 a

Coffee

Tea

Milk

Cola

Orange

b Student’s own opinion and reasoning



252 253

Developing fluency (pages 360–361)1 a Swimming

b 72°

c 15

; 20%

d No sport – 27°; Other sports – 63°; Football – 54°; Swimming – 108°; Netball – 81°; Hockey – 27°2 a 90

b 4°c

Air

Drive

Coach

Live here

Other

d

Air05

101520253035

Drive Coach

Method

Livehere

Other

Freq

uenc

y

e Student’s own opinion3 a 90

b 4°c

Apple

Orange

Peach

Grapefruit

Banana

Others

Pineapple

d Other is the largest category. The fruit seller should have listed more fruit so that he could get more information from the data. The pie chart is hard to see the little difference between fruits, a bar chart might have been clearer.



252 253

4 a Dogs – 135°; Cats – 80°; Rabbits – 90°; Other – 55°b

Dogs

Cats

Rabbits

Others

c Dogs – 108°; Cats – 112°; Rabbits – 88°; Other – 52°5 The pie chart has parts pulled out; the pieces of pie seem to have different radii, there is no scale and there seem to

be pieces of pie missing (there isn’t 360° of pie present)



254 255

Problem solving (pages 362–363)1 Senior is not a half – it is 210°. Adults should be 50°. 2 a

Taylor

Hussain

White

Clift

Treble

b Taylor – 18%; Hussain – 13%; White – 7%; Clift – 47%; Treble – 16% c i The segment is less than half

ii The percentage is less than 50%

3 a 16

b

University

Apprentice

Gap year

c 2404 a

Wrong size

Faulty

Wrong colour

Unwanted gift

Changed mind

b 10005 a £22.50

b £90c £12

6 a 512

b £1600



254 255


8b 225°c

Art club

Football club

Neither

d 140




10

b 310

c 110

d 710

e 0

2 a 16

b 16

c 12

d 12

e 56

3 a i 113

ii 126

iii 152

iv 5152

b i 113

ii 126

iii 152

iv 1213


Strand 4 Probability Unit 2 Single event probability Band e

256 257256 257

4 a 112

b 112

c 16

d 16

e 0

f 112

5 a 19

b 19

c 19

d 0



258 259


36

b 110

c 0

2 a i 110

ii 14

b Not fair, Mark should not accept. Jasmine has less probability that her numbers will win, therefore the share of probability in the syndicate is uneven.

3 a 300

b i 34

ii 320

iii 120

iv 3400

v 1400

c No, she could spend up to £400 to win £20.4 a 250

b Number 1 2 3 4 5 6

Probability 725

19125

950

325

26125

350

c multiply the probability by the total number of throws

d i 350

ii 4750

iii 0

5 a 14

b It would mean that there would be 7.5 red counters and 4.5 blue, which is impossible.c 12 (multiples of 12)

6 No. Each time you spin the spinner there is an equal chance of getting a 1, 2, 3 or 4, ie 14

, but in fact you might

actually get, for example, four ones or two twos and two threes. Events do not always work out like theory.



258 259

Problem solving (pages 371–372)1 62 a 1

12

b 34

c 16

d 56

3 a 0.3b 9c 0.65

4 a 0.1b 0.4c 0.6

5 a 24 or 12b 14 or 7 toffees, respectively c James ate 4 toffees and 0 fudge or James ate 3 toffees and 1 fudge, respectively

6 Spinner with 3 red, 2 yellow, 2 blue and 1 green sectors or 4 red, 2 green, 1 blue and 1 yellow sectors



260 261

Reviewing skills (page 372)1 a i 1

8

ii 34

iii 78

iv 0b Equal probability for all outcomes

2 a i 35

ii 35

iii 15

b 13



Practising skills (pages 376–377)1 a i 3

10

ii 110

iii 25

iv 15

b Total = 1. These are the only ways that Anwar got the batsmen outc 18

2 a PB, PC, PL, MB, MC, ML, RB, RC, RL, SB, SC, SL b 12. Multiply starter by main course: 4 × 3 = 12

c 112

3 a Red die

1 2 3 4 5 6

Blue die

1 1 2 3 4 5 6

2 2 4 6 8 10 12

3 3 6 9 12 15 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

6 6 12 18 24 30 36

b i 118

ii 136

iii 0iv 1

c Nod square numbers


Strand 4 Probability Unit 3 Combined events Band f

262 263

4 a Green spinner

1 2 3 4 5 6

Blue spinner

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

b i 16

ii 0

iii 16

c i 6, 7ii 11, 13, 14, 15, 16, 17, 18, 19iii 5, 6

5 a 2164

b 3164

c 764

d 4364

e 3364

f 1964



262 263

Developing fluency (pages 377–379)1 a i 1

2

ii 310

iii 320

b In this sample this is true, though it is a small sample and cannot be extrapolated to the whole population of cats.2 a G+B; G+G; G+C; R+B; R+G; R+C; B+B; B+G; B+C; C+B; C+G; C+C

b i 112

ii 16

3 a Bag 1

Bag

2

R R B B B

R RR RR RB RB RB

R RR RR RB RB RB

R RR RR RB RB RB

R RR RR RB RB RB

B RB RB BB BB BB

B RB RB BB BB BB

b i 1630

ii 830

iii 630

4 a

12 15

3 FB

B Backs F Forwards

b i 712

ii 512

c

12

3

1 15

B Backs Forwards



264 265

5 a Red spinner

1 0 1 0 1Ye

llow

spin

ner 0 1 0 1 0 1

1 2 1 2 1 2

0 1 0 1 0 1

1 2 1 2 1 2

0 1 0 1 0 1

b i 625

ii 1325

iii 625

c Red spinner

1 0 1 0 1

Yello

w sp

inne

r 0 0 0 0 0 0

1 1 0 1 0 1

0 0 0 0 0 0

1 1 0 1 0 1

0 0 0 0 0 0

d i 1925

ii 625

iii 0



264 265

Problem solving (pages 379–382)1 a 1 2 3 4 5 6

1 1 2 3 4 5 6

2 2 4 6 8 10 12

3 3 6 9 12 15 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

6 6 12 18 24 30 36

b No, P(odd) = 936

= 0.25, so the game is not fair.

c i 19

ii 512

iii 1318

2 a i 0.885ii 0.0125iii 0.609

b Most employees who were late came by bus, so it is likely that the bus was late that day, which is not the employee’s fault.

3 a Cube Cuboid Cylinder Total

Red 14 40 17 71

Green 21 27 12 60

Blue 23 33 13 69

Total 58 100 42 200

b i 0.5ii 0.355iii 0.115

c i 0.27 ii 0.45

4 a Isobel’s spinner Peter’s spinner

1 1

1 2

1 3

2 1

2 2

2 3

3 1

3 2

3 3

b 49

c They win if their spinner shows the higher number. They draw if both spinners show the same number.

P(win) = 13

, P(draw) = 13

, and P(lose) = 13



266 267

5 a i 0.54ii 0.4iii 0.22

b The probability that a women is left-handed is 1127

= 0.41 and the probability that a man is left-handed is

923

= 0.39. As the sample is quite small it is likely that men and women are equally likely to be left-handed.

c i 600ii The estimate is likely to be wrong. The people of one village are more likely to be related and so may share a

genetic tendency to be left-handed. The people in the other villages might not share the same genetic link.6 a

GermanFrench

8

96 40 56

b i 0.68 ii 0.2iii 0.28

7 a i 0.328ii 0.172iii 0.688

b i 1651ii 55 kg



266 267

Reviewing skills (page 382)1 a Green die

1 3 3 4 6 6

Blu

e di

e

2 3 5 5 6 8 8

3 4 6 6 7 9 9

4 5 7 7 8 10 10

4 5 7 7 8 10 10

5 6 8 8 9 11 11

5 6 8 8 9 11 11

b i 236

ii 536

iii 436

iv 3036

v 736

answers to ocr gcse mastering mathematics foundation 1

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