answers to wjec gcse mastering mathematics intermediate

Answers to WJEC GCSE Mastering Mathematics Intermediate

Number Strand 2 Using our number Algebra Strand 2 Sequences system Unit 3 Linear sequences 89 Unit 5 Using the number system Unit 4 Special sequences 95 effec tively 1 Unit 5 Quadratic sequences 101 Unit 6 Understand ing sta ndard form 6 Unit 6 nth term of a quadratic Unit 7 Calculating with standard form 10 sequence 105 Number Strand 3 Accuracy Algebra Strand 3 Functions and graphs Unit 5 Approxima ting 12 Unit 2 Plotting graphs of linea r func tions 107 Unit 6 Signific anc e 16 Unit 3 The eq ua tion of a stra ight line 118 Unit 7 Limits of ac c urac y 21 Unit 4 Plotting quadra tic and c ub ic g raphs 125 Number Strand 4 Fractions Unit 5 Find ing equa tions of stra ight lines 135 Unit 6 Divid ing frac tions 25 Unit 6 Perpend ic ula r lines 142 Number Strand 5 Percentages Algebra Strand 4 Algebraic methods Unit 4 App lying perc enta ge inc reases Unit 1 Trial and improvement 146 and dec reases to amounts 30 Unit 2 Linear inequalities 148 Unit 5 Find ing the perc entage c hange Unit 3 Solving pairs of equations by from one amount to another 34 substitution 152 Unit 6 Reverse perc entages 38 Unit 4 Solving simultaneous equations by Unit 7 Repea ted perc entage inc rease/ elimination 154 dec rease 40 Unit 5 Using graphs to solve simultaneous equations 157 Number Strand 6 Ratio and proportion Unit 2 Sharing in a given ratio 42 Algebra Strand 5 Working with Unit 3 Working with proportional quadratics quantities 46 Unit 1 Fac torising quadra tic s 166 Unit 2 Solving equa tions by fac torising 170 Number Strand 7 Number properties Unit 4 Index notation 50 Geometry and Measures Strand 1 Units Unit 5 Prime factorisation 53 and scales Unit 6 Rules of indices 57 Unit 7 Converting approximately Unit 7 Frac tiona l ind ic es 61 between metric and imperial units 172 Unit 8 Bearings 174 Algebra Strand 1 Starting algebra Unit 9 Scale drawing 178 Unit 4 Working with formulae 65 Unit 10 Compound units 183 Unit 5 Setting up and solving simple Unit 11 Dimensions of formulae 187 equations 69 Unit 12 Working with compound units 189 Unit 6 Using brackets 73 Unit 7 Working with more complex Geometry and Measures Strand 2 equations 79 Properties of shapes Unit 8 Solving equations with brackets 81 Unit 5 Angles in triangles and Unit 9 Simplifying harder expressions 83 quadrilaterals 191 Unit 10 Using complex formulae 86 Unit 6 Types of quadrilateral 193 Unit 7 Angles and parallel lines 198 Unit 8 Angles in a polygon 202 Unit 9 Congruent triangles and proof 206 Unit 10 Proof using similar and congruent triangles 211 Unit 11 Circle theorems 215

Answers to WJEC GCSE Mastering Mathematics Intermediate Geometry and Measures Strand 3 Statistics and Probability Strand 1

Measuring shapes Statistical measures Unit 4 Area of circles 219 Unit 4 Using grouped freq uenc y tab les 295 Unit 5 Pythagoras’ theorem 223 Unit 5 Inter-qua rtile range 301 Geometry and Measures Strand 4 Statistics and Probability Strand 2

Construction Statistical diagrams Unit 3 Constructions with a pair of Unit 3 Pie charts 307 compasses 226 Unit 4 Displaying grouped data 313 Unit 4 Loci 234 Unit 5 Scatter diagrams 321 Unit 6 Using lines of best fit 328 Geometry and Measures Strand 5

Transformations Statistics and Probability Strand 3 Unit 3 Transla tion 242 Collecting data Unit 4 Reflec tion 246 Unit 2 Designing questionnaires 334 Unit 5 Rota tion 251 Unit 6 Enla rgement 256 Statistics and Probability Strand 4 Unit 7 Simila rity 262 Probability Unit 8 Trigonometry 264 Unit 2 Single event probability 338 Unit 3 Combined events 343 Geometry and Measures Strand 6 Unit 4 Estimating probability 350

Three-dimensional shapes Unit 5 The multiplication rule 353 Unit 3 Volume and surfac e a rea of Unit 6 The addition rule and Venn diagram c uboids 270 notation 359 Unit 4 2D rep resenta tions of 3D shapes 274 Unit 5 Prisms 281 Unit 6 Enla rgement in two and three d imensions 285 Unit 7 Construc ting p lans and eleva tions 289

© Hodder & Stoughton Ltd 20151

Unit 5 Answers

Practising skills (pages 6–7)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 6 500 000

c less; more3 a 4

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

2

9781471856211_WJEC_Maths_Intermediate.indb 1 13/04/16 10:49 AM


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

2

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




3

Developing fl uency (page 7)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




4

Problem solving (pages 8–9)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.




5

Reviewing skills (page 9)1 a 0.82

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



1

6

Unit 6 Answers

Practising skills (pages 11–12)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

2



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

7

Developing fl uency (page 12)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 �




8

Problem solving (page 13)1 Venus, Mars, Mercury, Sun, Jupiter, Saturn, Uranus, Neptune2 5 (questions, 3, 4, 6, 8 and 10)




9

Reviewing skills (page 13)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

4 b 3.2 × 106



Unit 7 Answers

Practising skills (pages 15–16)1 a 7.8 × 105

b 1.7 × 10−2

c 8.7 × 103

d 8.4 × 103

e 4 × 10−3

f 3.1 × 10−2

2 a 1.18 × 106

b 7.66 × 105

c 5.32 × 106

d 2.6 × 105

e 6.74 × 106

f 3.88 × 10−1

3 a 6 × 1012

b 8 × 108

c 1 × 108

d 9 × 102

e 1 × 10−1

f 6.3 × 104 a 3 × 102

b 2 × 10 or 2 × 101, although first is better. c 3 × 102

d 1.5 × 104

e 5 × 10f 2.5 × 10−3

5 a 5.99 × 105 to 6.01 × 104

b 6.01 × 10−3 to 5.99 × 10–2

c 7.11 × 102 to 7.09 × 103

d 7.09 × 10−6 to 7.11 × 10–7

6 If the numbers multiplied together give a result greater than 10, then will need to add an additional 1 to the index.

Developing fl uency (pages 16–17)1 a 3.604 × 102

b 2.804 × 102

c 1.28 × 102

d 8.01 × 102

2



Strand 2 Using our number system Unit 7 Calculating with standard form Band h

11

2 9 × 1015

3 a Jupiter, Saturn, Earth, Mercuryb 3.17 × 103

c 5.76 × 103

d 1.82 × 104 a 2.33 × 1018

b 1.07 × 10c 2.33 × 1018 (NB this is the same as a)

5 7 × 1021

Problem solving (pages 17–18)1 Yes, total mass = 5.424 × 105 kg2 a 6.25 × 107

b 13.6 cm3 a 1 × 10−4 m

b 1.783 × 10−2 kg4 a 18 seconds

b 1 km. The time taken for the light to travel this distance is assumed to be zero.5 Australia has the greatest area of land/person. UK has the least.

Country Area (in km2) Population Area (in km2) per person

Australia 7.7 × 106 2.2 × 107 0.350

Brazil 8.5 × 106 2.0 × 108 0.043

China 9.6 × 106 1.4 × 109 0.007

Germany 3.6 × 105 8.3 × 107 0.0043

UK 2.4 × 105 6.4 × 107 0.0038

USA 9.8 × 106 3.2 × 108 0.031

6 327.06 (330 – 2.94) metres

Reviewing skills (page 19)1 a 9.29 × 104

b i 84 800ii 8400iii 300

c 84 800 + 8400 − 300 = 9290 = 9.29 × 104

2 a 9 × 1012

b 3.2 × 10−7

c 2.6 × 106

d 2 × 10−3

e 1.52 × 102

f 3.8 × 103

3 c 3.004 × 10–26 kg



3

12

Unit 5 Answers

Practising skills (pages 22–23)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 5 Approximating Band g


Developing fl uency (pages 23–24)1 a 1000

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




Problem solving (pages 24–26)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




Reviewing skills (page 26)1 a ≈ 1000

b ≈ 20c ≈ 1600

2 a ib iv

3 a ≈ 280b ≈ 3850c ≈ 6750

4 ≈ £46.80



3

16

Unit 6 Answers

Practising skills (page 28)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 6 Signifi cance Band f


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




Developing fl uency (page 29)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 Dewib Ada, Ben and Cainc Ben and Cain




Problem solving (page 30)1 Ami. Ifor has rounded down instead of up, Milly thinks that the zero is not significant and Iwan 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.




Reviewing skills (page 31)1 a 1

b 0.01c 1000d 1 000 000

2 a 6.4b 20c 0.0052d 0.010

3 a 0.3068c 515 300d 2.0

4 a 30 000



3

21

Unit 7 Answers

Practising skills (page 33)1 a 79.5, 80.5

b 75, 85c 299.5, 300.5d 250, 350

2 a 4999.5 m, 5005.5 m.b 4995 m, 5005 mc 4950 m, 5050 md 4500 m, 5500 m

3 a 595 m, 605 mb 597.5 m, 602.5 mc 550, 650d 575, 625

4 5.5 � m � 6.55 8.5 � l � 9.56 a 2.5, 3.5

b 55, 65c 0.35, 0.45d 0.065, 0.075

7 a 23.5, 24.5b 355, 365c 0.825, 0.835d 0.0185, 0.0195

8 Number Lower limit Upper limit

a 4 3.5 4.5

b 70 65 75

c 600 595 605

d 0.3 0.25 0.35

e 0.06 0.055 0.065

f 80 km 75 km 85 km

g 68 mg 67.5 mg 68.5 mg

h 0.032 0.0315 0.0325



Strand 3 Accuracy Unit 7 Limits of accuracy Band g

22

Developing fl uency (page 34)1 1475 ml � V � 1525 ml2 52.5 km3 182.25 cm2 � A � 210.25 cm2

4 No. The minimum in one bag is 245 g. If all 3 bags are at the lower limit she will only have 735 g.5 225 m � p � 255 m, 2812.5 m2 � A � 3612.5 m2

6 iii, all the others are to the nearest 5, iii is to the nearest 10.7 a n = 680+/−5

b 675 � n � 685c 685 and 675d 10

Problem solving (page 35)1 6 (Max = 652 members)2 No, only 8 per page.3 a If each person’s mass is given to the nearest kg, their total mass could be 652 kg.

b 80 kg each4 48

Reviewing skills (page 36)1 a 39.5, 40.5 ml

b 35, 45 mlc 650, 750 kgd 695, 705 kg

2 c 625 and 6753 315 g � m � 325 g4 a 850 cm, 950 cm

b 1500 km, 2500 kmc 0.15 g, 0.25 gd 0.0045 m, 0.0055 m

5 a 7050 m, 7150 mb 48.5 cm, 49.5 cmc 515 mm, 525 mmd 0.00275 km, 0.002 85 km

6 Lowest limit of a side is 19.5 cm. If each side is the lowest limit the perimeter will be 78 cm and the ribbon would be too short.



3

23

Unit 8 Answers

Practising skills (page 39)1 a 3.5 b 3.55 c 3.555 d 7.5415 e 12.000 052 a 6.5 kg and 7.5 kg b 6.5 m and 7.5 m c 6.5 litres and 7.5 litres3 a 52.5 cm and 53.5 cm b 52.95 cm and 53.05 cm c 5.25 cm and 5.35 cm4 a 14.5 kg and 15.5 kg b 15.65 kg and 15.75 kg c 15.355 kg and 15.365 kg5 149.9875 kg and 150.0125 kg

Developing fl uency (page 40)1 0.55 cm2 1384.25 cm2 and 1304.25 cm2

3 a 193.375 and 390.625 b 245.700 (245.699 875) and 258.400 (258.400 125)4 a 5048.125 m2 and 4987.125 m2

b It is correct to 2 significant figures5 61.7π and 61.9π6 a 6378.136 995 km and 6378.137 005 km b 40 075.016 654 km and 40 075.016 717 km7 The upper bound implies that the value is less that that upper bound, not less than or equal to.



Strand 3 Accuracy Unit 8 Upper and lower bounds Band i

24

Problem solving (pages 41–42)1 Each of the boxes has a height of 25 cm to the nearest centimetre. This means that the upper bound for

the height is 25.5 cm. 4 × 25.5 cm = 102 cm = 1.02 m. This is 2 cm more than the height available and so the boxes may not fit.

2 The piece of wood is 80 cm long to the nearest 2 cm and so the longest the shelf can be is 81 cm. Therefore the maximum length the space can be is 1 cm.

3 a 8.5 cm and 9.5 cm, 5.5 cm and 6.5 cm, 18.5 cm and 19.5 cm b She can definitely do this.4 205.56 cm � A � 211.69 cm (to 2 decimal places)5 a 0.5 / (25.5/60/60) = 70.59 mph = 71 mph to 1 d.p.

b Lower bound of Viv’s speed = ×0.49525.75 3600 = 69.2 mph

6 8.5074 m/s � speed � −8.5867 m/s7 a upper bound = 53.1, lower bound = 48.7 b 50 (1 significant figure)8 Malcolm is definitely wrong. Lower bound = 325.1 ml, upper bound = 339 ml

Reviewing skills (page 42)1 No. the upper bound of the combined masses is 452 kg and the lower bound of the lift’s limit is 450 kg.2 166 CDs3 Upper bound 490.25 cm2. Lower bound 446.25 cm2.



4

25

Unit 6 Answers

Practising skills (pages 45–46)1 a 7

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


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




Developing fl uency (pages 46)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




Problem solving (pages 47–48)1 5 children2 No, can make 96 pieces3 a No, only 33 glasses

b 17

c 17

4 ₤ 6.03





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

5 a 542


5


Unit 4 Answers

Practising skills (page 53)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

31

Developing fl uency (pages 54–55)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 Elen, by £1.13




32

Problem solving (pages 55–57)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 £1506 a 13 000 €

b 15% of (12 500 – 3000) = 1425 €c 15% of 10 000 + 50% of (45 850 – 13 000) = 1500 + 16 425 = 17 925 €

7 a 250 550 – (250 550 × 0.86 × 0.86) = 65 243 Rubles depreciation (to the nearest Ruble.)b Dimitri pays 100 000 + (23 × 7000) = 261 000 Rubles. The value of the car in 2 years’ time is

(250 500 × 0.86 × 0.86) = 185 307 Rubles (to the nearest Ruble). Therefore Dimitri makes a loss of 261 000 – 185 307 = 75 693 Rubles.

8 1.19 × 0.83 = 0.9877. So Carys would pay 0.99 € per litre (to the nearest cent).9 a £507.19 (to the nearest penny)

b Mr Evans spent £315 in total on insulting his house but he has only saved £285.30 over the two years. He has not yet recouped the cost of his outlay.




33

Reviewing skills (page 57)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


5


Unit 5 Answers

Practising skills (pages 59–60)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

35

Developing fl uency (pages 60–61)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 Garth’s calf10 Glenda’s house11 50%




36

Problem solving (pages 61–63)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 reducingOther roads 3.5%; 5.2%; 5.3%; 8.9%; 0.3% respectively; increasing up to 2001

3 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 Silverstone8 Raffi borrows £150 000. The payments he makes (excluding the deposit) cost (1000 × 12 × 25) = £300 000.

The difference in these is 300 000 – 150 000 = 150 000. 150 000150 000 = 1 = 100%, so Raffi is paying back 100% of the amount he borrowed in interest.

9 a The overall cost with Reef bank is 36 000 + (855 × 12 × 20) = 241 200. The overall cost with Rhudd building society is 27 000 + (925 × 12 × 20) = 249 000. Therefore Marged is better off with Reef bank.

b Marged borrows £180 000. With Reef bank she will be paying back £205 200. 205 200 – 180 000 = 25 200.

25 200180 000 = 0.14 = 14%. So Marged will pay back 14% of the amount she borrowed in interest.

10 a 8.50 × 10 = £85b 85 – 50 = 35

3550 = 0.7 = 70%. Therefore Rita pays back an extra 70% of the amount that she borrowed in interest.




37

Reviewing skills (page 64)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)


5


Unit 6 Answers

Practising skills (page 66)1 £402 £153 £304 a £115

b £235 a 70 cm

b 128 mc 450 kmd 5 me 850 km

6 a £230b £173.91c £226.09; no because compound interest is not just adding two percentages together.

Developing fl uency (pages 66–67)1 a £70

b £120c £60d £20

2 a £80b £18c £175d £1250

3 900 ml4 40 hours5 £94006 £180 0007 £448 319 pages9 3640 miles

10 £140



Strand 5 Percentages Unit 6 Reverse percentages Band h

39

Problem solving (pages 68–69)1 £12.50 each2 £72003 €504 He bid £800.5 Total cut is 7.1%.6 4607 12 years8 10% lower than the average.

Reviewing skills (page 69)1 a £62 a 140 cm on his 13th birthday

b He grew 7 cm.3 a 240

b 364 £22


5


Unit 7 Answers


b 7260c 7986d 33.1%

2 a £7986b 33.1%c The increase is compound and is therefore being applied to an increasingly large amount.

3 a 4000b 3200c 2560d 51.2% of original value

4 a £2560b 48.8% decreasec 2 more years

5 % change Decimal multiplier

20% increase 1.2

60% decrease 0.4

12% increase 1.12

12% decrease 0.88

150% increase 1.5

6 a 300, 180, 162b 19%c 162d They are the same calculation expressed as a percentage and a decimal multiplier.

Developing fl uency (pages 72–73)1 a £30

b £530c £31.80

2 a i £100ii £1000iii £52.50

b i £102.50ii £1221.02iii £54.36

3 a 40%b 16%c 6.4%d 2.56%e 1.024%



Strand 5 Percentages Unit 7 Repeated percentage increase/decrease Band h

41

4 a i £2400ii £600

b i £2400 × 0.8 = £1920ii 7 more years (9 years from purchase)

5 3586 a £92 220

b £97 753.20c 2005

7 a 69.57%b (1 × 0.93)10 = 0.484 × 100 = 48.4%c 2.66%d 0.007%

8 a £690.79b £2203.52c £68 146.10

9 15 km and 12 km

Problem solving (pages 73–75)1 They are both wrong, the reduction equates to 44%.2 a 37 869 822

b 33 366 9383 Decreased by 3.2%.4 Yes, £4 485 000 in two years.5 The Friendly Bank gives £13 more interest.6 The decrease was applied to a higher price so the reduced price was slightly higher than the original.7 17.6 m3

8 Andy (£1610.95) better than Tina (£1607.73).9 No, it has increased by 23.2%.

10 a Year Principal Interest Amount at end of year

1 £800 £64 £864

2 £864 £69.12 £933.12

3 £933.12 £74.65 £1007.77

b 1007.77 = 800 × (1+ 8

100)3

c It applies the interest as a compound rate. d i £1292.97

ii £757.49iii £119.07

Reviewing skills (page 75)1 a Simple interest = £5280, compound interest = £5948.07

b Simple interest = £28 800, compound interest = £50 540.362 a £6800

b £55083 No, after 3 months his weight will be 17.33 stones.


6


Unit 2 Answers


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

43

Developing fl uency (pages 79–80)1 £562 723 £1804 63 cm and 35 cm5 21




44

Problem solving (pages 80–81)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




45


b 15

c 45

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

3 214

hours


6


Unit 3 Answers

Practising skills (page 84)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

47

Developing fl uency (pages 85–86) 1 £112 2 £6.80 3 £5.66 4 a 6 kg for £14.70

b 150 ml for £24 c 60 g for £12.06

5 No. He will charge £14.60 6 The maximum Roland can buy is 400 × 153.24 = 61 296 Krona. Since the bank only has 10 Krona notes, this means

that they will give him 61 290 Krona. 61 290 ÷ 153.24 = 399.96, therefore it will cost Roland 399.96 €. 7 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.

8 a i She bought 660 Zloty.ii It cost her £149.32

b 660 − 84.40 = 575.60 Zloty left. The shop bought back 575 Zloty, giving her £127.49 in return. 9 £65.6010 2 hours 6 minutes11 19 hours12 22 hours




48

Problem solving (pages 87–89) 1 a £537.50

b 6 hours 2 13 (brown sugar is the limiting ingredient) 3 No; small = 7.4p per chocolate, large = 6.9p per chocolate, medium = 6.8p per chocolate 4 Harvey’s company at 50p/mile (Albert’s is 49.5p/mile) 5 Yes, she can do it for £14.56 6 Using these exchange rates, £1 is worth 1.27 Euros. 7 41 cents 8 Not enough flour, she needs 168 g; she has enough milk as she needs 1.704 litres 9 £8.50 in 2p coins, £42 in 5p coins10 a In 2014 the room cost 135 Euros per night. To calculate how much it cost in 2015 we multiply the price by

1.11 (111%): 135 × 1.11 = 149.85 Euros per night.

Now we must find out how much the room cost per night in pounds in 2014: 135 Euros1.12 = £120.54

We also need to know how much the room cost per night in 2015: 149.85 Euros

1.32 = £113.52Finally we work out the difference in prices for the 10 nights: (10 × 120.54) – (10 × 113.52) = 70.20, i.e. the room cost £70.20 less in 2015 than it did in 2014 for the 10 nights. The reason for this is because although the room cost more in Euros, the exchange rate was better in 2015.

b The exchange rate in 2016 is £1 = 1.22 Euros. The cost of the room in 2016 is 149.85 × 1.02 = 152.847. The total cost of the room for 10 days is therefore 10 × 152.847 = 1528.47 Euros. We need to find this in pounds: 1528.47

1.22 = 1252.84Therefore the Davies family need to budget £1252.84 for their trip in 2016.

11 1 hour 50 minutes




49

Reviewing skills (page 90)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


7


Unit 4 Answers


b 35

c 34

d 37

2 a 53

b 26

c 22 × 52

d 75

e 113 × 172

f 38

g 25 × 19h 53 × 73

3 a 125b 1024c 2401d 1e 729f 7776g 20.25h 2 097 152i 29j 50 625k 65 536l 1296

4 a 24

b 24

c 24

d 26

e 28

f 22

g 22

h 28

5 a 38

b 33

c 35

d 36

e 36

f 36

g 39

h 34

i 35



Strand 7 Number properties Unit 4 Index notation Band f

51

6 a 56

b 58

c 512

d 512

e 536

f 521

g 520

h 518

i 532

Developing fl uency (pages 94–95)1 210 = 45, 83 = 29, 93 = 36

2 a 28b 500c 337d 216

3 35, 44, 43 × 6, 73, 29

4 a 12b 120c 64d 990e 7f 17g 72h –1

5 a 22 × 73

b 5 × 63

c 32 × 54

d 24 × 33

e 2 × 63 × 72

f 32 × 53 × 72

6 a 27

b 24

c 28

d 215

e 210

f 29

g 28

h 28

7 a 27

b 29

c 24

d 28

e 210

f 213

g 25

h 2



Strand 7 Number properties Unit 4 Index notation Band f

52

8 a 6b 3c 6d 7

9 128

Problem solving (pages 95–96)1 a 6.4 mm

b There would be a very large number (216 = 6.5536 metres) of tiny pieces, too small for Bill to cut up.2 2 × 54 = 1250 pigs3 a £29, (= £512)

b 21 questions

4 (25 × 84) × 162

46 = 8192 = 213

5 a 45 = 1024b Round 1 has 820 – 68 = 752 players. So there are 188 games with one winner each. In round 2 there are 188 + 68

= 256 which is the correct number for three further rounds = 5 rounds total.6 LHS = RHS = 2.1 × 109


b 22 × 53

c 34 × 52

d 2 × 33 × 112

2 c 25

3 a 128b 180c 32d 1

4 a 12b 6c 6d 20



7

53

Unit 5 Answers


b 110c 12d 18e 75f 250g 735h 4725

2 a 22 × 3 × 5b 2 × 32 × 7c 22 × 52

d 2 × 33

e 32 × 52

f 2 × 7 × 11g 3 × 5 × 7h 32 × 5 × 11 i 22 × 3 ×52

j 34 × 5k 22 × 53

l 24 × 3 × 133 a 1, 2, 3, 4, 6, 12

b 1, 2, 4, 5, 10, 20c 1, 2, 4d 4

4 a 2 b 5c 6d 9e 8f 26g 5h 18i 12j 1k 22l 17



Strand 7 Number properties Unit 5 Prime factorisation Band g

54

5 a 6, 12, 18, 24, 30, 36, 42, 48, 54, 60b 8, 16, 24, 32, 40, 48, 56, 64, 72, 80c 24, 48 d 24

6 a 18b 40c 70d 60e 42f 30g 80h 65i 120j 300k 54l 180

Developing fl uency (pages 100–101)1 a 2 × 32 × 72

b 24 × 3 × 52

c 3 × 53 × 11d 26 × 7 × 13

2 a 22 × 3 × 5b 2 × 32 × 7c

60 126

25

23

37

d 6e 1260




55

3 a 180 = 22 × 32 × 5b 63 = 32 × 7c

180 63

225

33

7

d 9e 1260

4 a HCF = 4, LCM = 420b HCF = 15, LCM = 180c HCF = 14, LCM = 29 400d HCF = 35, LCM = 36 750

5 8.30 p.m. (12 hours later)6 a i 111

ii 1111iii 1001iv 11111

b 3 × 7 × 11 × 13 × 37

Problem solving (pages 101–102)1 Yes, 144 attend the party, food costs £34.80 + £22.50 + £56.80 = £114.10 So £5.90 change.2 72 000 miles3 3 laps4 9 (at 6 km intervals)5 7 times6 None of Nomsa’s factors are prime numbers.

209 = 11 × 19; 119 = 7 × 17; 187 = 11 × 17; 133 = 7 × 19 So either way 24 871 = 7 × 11 × 17 × 19




56


b 400c 2310d 23 100e 3 210 000

2 a 2 × 32 × 5b 3 × 5 × 11c 2 × 5 × 7 × 11d 32 × 7 × 13e 23 × 52 × 7f 2 × 53 × 7g 24 × 32 × 11h 25 × 7 × 13

3 a 54 = 2 × 33, 60 = 22 × 3 × 5b 6c 540



7

57

Unit 6 Answers

Practising skills (pages 105–106)Index form In full Ordinary number

25 2 × 2 × 2 × 2 × 2 32

24 2 × 2 × 2 × 2 16

23 2 × 2 × 2 8

22 2 × 2 4

21 2 2

20 1 1

2−112 0.5

2−21

(2 × 2) 0.25

2−31

(2 × 2 × 2) 0.125

b i 128ii 1

24 = 1

16 = 0.0625

Index form In full Ordinary number In words

103 10 × 10 × 10 1000 One thousand

102 10 × 10 100 One hundred

101 10 10 Ten

100 1 1 One

10−1 110

0.1 One tenth

10−2 1(10 × 10)

0.01 One hundredth

10−3 1(10 × 10 × 10)

0.001 One thousandth

b 1 000 000, one millionc 0.000 001, one millionth



Strand 7 Number properties Unit 6 Rules of indices Band h

58

3 a i 32 × 33

3 × 3 × 3 × 3 × 3 35

check: 9 × 27 243 35 = 243

ii 54 × 52

5 × 5 × 5 × 5 × 5 × 5 56

check 625 × 25 15 625 56 = 15 625

iii 103 × 104

10 × 10 × 10 × 10 × 10 × 10 × 10 107

check 1000 × 10 000 10 000 000 107 = 10 000 000

b am × an = a(m + n)

4 a 26, 64b 36, 729c 53, 125d 53, 125

5 a i 36 ÷ 34

32

check 729 ÷ 81 = 9 32 = 9

ii 54 ÷ 53

51 = 5 check 625 ÷ 125 = 5

iii 105 × 102

103

check 100 000 ÷ 100 = 1000 103 = 1000

b am

an = a(m – n)

6 a 23, 8b 31, 3c 103, 1000d 5−3, 25




59

7 a i (24)3

2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 212

check 16 × 16 × 16 = 4096 212 = 4096

ii (32)5

3 × 3 × 3 × 3 3 × 3 × 3 × 3 3 × 3 310

check 9 × 9 × 9 × 9 × 9 = 59 049 310 = 59 049

iii (103)2

10 × 10 × 10 × 10 × 10 × 10 106

check 1000 × 1000 = 1 000 000 106 = 100 000

b (am)n = a(mn)8 a 210, 1024

b 210, 1024c 107, 1 000 000 (one million)d 1012, 1 000 000 000 000 (one billion)


b 32c samed 25e 26f 1−2

g 2−3

h same2 a 1 and 9, 2 and 8, 3 and 7, 4 and 6, 5 and 5; 5 pairs3 a 313

b 212

c 106

d 56

4 a 69

b 69

c 622

d 69

5 a 35

b 35

c 35

d 35




60

6 a 24

b 29

c 26

d 11

7 i, a, g, b, c and d, f e h j8 Marged is correct. Each value is larger than the previous one, but her estimated values are too low.

Problem solving (page 107)1 a Various combinations, check students’ answers.

b 11c no


b 51

c 51

2 a 10−2

b 10−1

c 10−2

3 a 22

b 71

c 100

4 232 × 23−2 = 1 = 230



7

61

Unit 7 Answers

Practising skills (page 110) 1 Missing number is 7 1

3

3 12 3

3 17 37

7 12 72

2 13 23

7 13 73

2 17 27

2 a 8 b 11 c 107 d 39

e 11312

f 1818

3 a 12 12

b 18 12

c 1513

d 313

e 318

f 7 13

4 a 8, 3 and 2 b 2, 2 and 196 c 64, 4, 645 a 58 b 2 c 10 d 5 e 11 f 15



Strand 7 Number properties Unit 7 Fractional indices Band i

62

Developing fl uency (page 111)1 Missing number is 7 23 .

372 37

3 27 327

7 32 732

2 73 273

7 23 723

2 37 237

2 a = =4 64 832 2

b = =8 64 423 3

c =625 54

d = =36 46656 21632 2

e = =1 1 179 9

f = =32 1024 425 5

3 a 12 52

b 18 32

c 15 43

d 3 83

e 3 38

f 7143

4 a 8 23

8 423= =

b 196 36 196 1436= =

c 4 32

4 2 832 3= = =

5 a 4 53

b 4 35

6 a 4 12

or 4 36

b 9 12

or 924

or 936

or 9 48

c 263

or 2 84




63

Problem solving (pages 112–113)1 c, a, e, d, b

2 a = =b b bbb

b a = 2, b = 4 or a = 4, b = 2

3 a side = =V V313

cube SA = 6 face SA = × = ×6 side 6223 V

b V A6

3

( )=

4 a 1 year b 11.8 years c 109 million kilometres5 powers of 2

26

23 23

22 21 22

21 21 20 22

powers of 4

624 or 43

41

41

41

40

324

324

124

124

124

powers of 8

82

81 81

80

238

238

238

138

138

138

powers of 16

160

6416

3416

3416

2416

1416

1416

1416

1216

1216

powers of 32

320

6532

3532

2532

2532

2532

1532

1532

1532

3532




64

powers of 64

640

641

3664

2664

3664

1364

1364

1664

1664

1664

Reviewing skills (page 113)1

512 3

52 5

23 3

25 3

12 2

23 2

53 5

32 2

12

5 35 523 325 3 223 253 53 2

2 a 4 b 27 c 4

3 a 8 213 =

b 25 25 12532

3( )= =

c 32 835 =



1

65

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

66

Developing fl uency (pages 118–119)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




67

Problem solving (pages 119–121)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




68

Reviewing skills (page 121)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



1

69

Unit 5 Answers


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

70

Developing fl uency (page 125)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 Ffion’s age, f = a + 8b 2a + 18 = 30; a = 6 and f = 14




71

Problem solving (pages 126–127)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 = 7

4 (or 13

4 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.




72


b 3c −7d 11e −1f −6g 3h −2

2 a −19.6b −1.2c 92.4d 64

3 3 × 230 + 2t = 1000t = 155 g

4 c 16.4



1

73

Unit 6 Answers

Practising skills (pages 130–131)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 92 b i 252 cm2

ii 252 c i 285 cm2

ii 285 d i 132 cm2

ii 1326 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

74

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




75

Developing fl uency (pages 131–132)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




76

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 £31.6811 a 37b − 21c

b 15x – 23yc 8x + 11yd 18a + 37b

12 a 28t – 30hb 9g – 6hc 6d + 11ed 15k – 6m




77

Problem solving (pages 132–135)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




78


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



1

79

Unit 7 Answers

Practising skills (page 138)1 a 3x = 18, so x = 6

b 2x + 4 =16, so x = 6c 5x + 4 = 3x + 14, so x = 5d 8x + 12 = 3x + 42, so x = 6

2 a 4.5b 7c 5d 60e −2f −2.5g 6h 6

3 a −4b 3c 1d 9

4 a −7b −3

c 12

d − 12

Developing fl uency (pages 138–140)1 a 6a + 20 = 4a + 64

b 222 a She has not reversed the operations so it should be 11 − 5 = 6x + 4x.

b 11 − 5 = 6x + 4x leading to 6 = 10x, so 0.6c 11 − 4 × 0.6 = 6 × 0.6 + 5 11 − 2.4 = 8.6 = 3.6 + 5

3 a Lowri: x = 1.50.5, x = 3. Tara: 5x = 15, x = 3.

b Tara, because using whole numbers you are far less likely to make mistakes.4 a 2.1

b 8c 2.275d 1.4e 3f 4.5g 100h 2



Strand 1 Starting algebra Unit 7 Working with more complex equations Band f

80

5 a 3.5m + 4.25 = 2m + 11b m = £4.50c Each girl was given £20.

6 a 11w + 108 = 14w + 72b w = 12c 20 rows

7 a 5n − 8 = 12 + 2n + 10b n = 10

Problem solving (pages 140–141)1 a 5n − 3 = 7 + 3n

b n = 52 150 cm2

3 a 350°b x = 120, 12 × 120 – 1320 = 120c The angles are equal so this part alone is a regular polygon. But we do not know the rest of the shape so we

cannot be certain.4 a i 2h

ii h – 40b Harry: £110, Ellie: £220, Tom: £70

5 6a + 108 = 180; a = 12, so base angles are both 72°. Hence triangle is isosceles.


b −6c −0.5d −1

2 a 3b 2.2c −5.5d 1

3 a 4b 1c 3d 7

4 a 5p + 4.25 = 2p + 16.50 + 0.20b p = £4.15c p = 4.15d Zorro had £25.00.



1

81

Unit 8 Answers


b 9c 1.5d 3

2 a 4b 8c 8d 11

3 Instruction Algebra

I think of a number, n n

I multiply it by 5 5n

I add 6 5n + 6

I multiply it by 3 3(5n + 6)

The answer is 123 3(5n + 6) = 123

b 5n + 6 = 41, 5n = 35, n = 74 a 14

b 4c 6d 5

5 a −7b 4c 3d 2

6 a 3b 4c 44

d – 14


b 24c –6d 3e 4, SOLVE

2 a 8(x + 1) + 2(6x – 5)b 8(x + 1) + 2(6x – 5) = 28c 40 cm2



Strand 1 Starting algebra Unit 8 Solving equations with brackets Band f

82

3 a 3(n – 2)b 3(n – 2) = 30 – nc 9d 3 × (9 – 2) = 30 – 9; 3 × 7 = 21

4 a Age now Age in 4 years

Gethin p p + 4

Dad 42 − p 46 − p

b 4(p + 4) = 46 − p

c p = 6

d Gethin is 6 years old and his Dad is 36 years old.

e 6 + 36 = 42. In 4 years’ time, Gethin will be 10 and his dad will be 40; 4 × 10 = 40.5 a i 2x + 1

ii x + 2b x = 1

6 a i 12x + 6ii 6x + 12

b 3(4x + 2) = 2(3x + 6)c x = 1

7 £158 Stem = 33 cm, Bulb = 6 cm

Problem solving (pages 146–147)1 a 3(n + 12) = 60 – n

b 62 a k = 2y k − 2 = 3(y − 2)

b y = 4 k = 83 a c + 20 = h s = 2h c + s + h = 120

b 15 cowsc 35 hens, 70 sheep

4 Shows value of x from one pair of equations does not match the other expression. Sides not equal so can’t be a square.

5 1 hour6 Not parallel because angles won’t be the same.


b −1c 9d 5

4 or 11

4 or 1.25

2 a 1b 2

c 32 or 1

12 or 1.5

d 33 1035 cm2



1

83

Unit 9 Answers


b 25

c 914

d 710

2 a f 2

b g4

c d6

d a24

3 i x × x × x × x × x = x5

ii x × x × x × x × x = x5

iii x × x × xx × x

= x

iv x × xx × x × x = x–1

v x × x × x × x × x × x = x6

vi x × x × x × x × x × x = x6

4 a 20a5

b 2bc 6c14

d 30d8

e 2ef 8f 5g8

g 24m4p5

h 2s3

5 a 3a2

b 4c6

c 5d5

d 3f 5



Strand 1 Starting algebra Unit 9 Simplifying harder expressions Band g

84

6 a i × x 4

x x2 4x

2 2x 8

ii x2 + 6x + 8 b i = x2 + 8x + 15

× x 5

x x2 5x

3 3x 15

ii = x2 + 2x − 24

× x −4

x x2 −4x

6 6x −24

iii x2 − 12x + 35

× x −7

x x2 −7x

−5 −5x 35

Developing fl uency (pages 151–152)1 a middle row x8 x7; top x15

b bottom row a5 a4; middle row a9

2 n12 ÷ n4 = n8; n6 ÷ n3 = n3; n4 × n2 = n6; n3 ÷ n = n2; n3 × n2 = n5; (n5)2 = n10

3 a x2 + 11x + 18b x2 + 2x − 15c x2 – 3x − 4d x2 − 7x + 10

4 a x2 + 14x + 48b x2 + x − 12

5 a Tim is correct, Harry has not multiplied each term in one bracket by each term in the other bracket. b i a2 + 6a + 9

ii b2 + 12b + 36iii c2 − 8c + 16

6 a x2 + 15x + 50b top row x2; 10x; bottom row 5x; 50c They are equal – the small rectangles represent the stages of expanding the (x + 5)(x + 10) bracket.

7 a i 4x2 + 8xii 2x2 + 8x

b red area is 2x × (x + 4) = 2x2 + 8x as well



Strand 1 Starting algebra Unit 9 Simplifying harder expressions Band g

85

Problem solving (pages 152–153)1 w(w + 20) = w2 + 20w2 a2 + 8a3 3

2 b2 + 12b4 (4s)3 – s3 = 60s3

5 a n = 8b n = 6c n = 3

6 a 2x2 + 8x + 8b red is 0.5(2x + 4)(2x + 4) = 2x2 + 8x + 8

Reviewing skills (page 153)1 a a14

b b6

c c0 = 1

d d18

2 a 12a7

b 4b5

c 1d 5g8

3 a a2 + 11a + 30b b2 + 4b − 21c c2 − 4c − 5d d 2 − 8d + 15

4 a x2 + 8x – 48b x = 7: Areas are red rectangle 57 cm2, blue square 49 cm2, small red rectangle 21 cm2.c 84 cm2


1


Unit 10 Answers


b 45c 40d 3

2 a 20b 67c −17d −1.8

3 a 24b 45c 24d 9

4 a 90b 200c −100d 200

5 a x = y + 8

b x = y3

c x = 5y

d x = ( y – 1)

26 a x = y – 4

b x = ( y + 3)

4

c b = a6

d t = pm

7 a 24b 106c 49d 54

8 a 13

b 9 12

c 4 12

d 9 13

9 a −1 12

b 6

c 4 34

d 3 15



Strand 1 Starting algebra Unit 10 Using complex formulae Band h

87


ii 32iii 26

b points – d

3 = wc 5

2 a i 94ii 122iii 156

b c – 2p

1.5 = mc 3 miles

3 a C = 35h + 18 b i 88

ii 228iii 368

c h = (c – 18)

35 d i 4

ii 12iii 27

4 a i 452.389 cm2

ii 1809.557 m2

iii 2 010 619.3 km2

b 514 718 540.4 km2

c r = sπ4

d 4.00 cm5 a i 0

ii 35iii 100iv −40

b F = 9c5 + 32

c i 32ii 95iii 572iv −40

6 a 512 b i M =

hp4t2

ii t = hpM4

iii h = 4Mt2

P c i 50

ii 20iii 7

7 a x = ( y + 5)6 b x =

y5 + 6 c p =

rT4m d p =

(T – m2)4r



Strand 1 Starting algebra Unit 10 Using complex formulae Band h

88

Problem solving (pages 159–160)1 a £165

b h = c – 15

30c 1 hour 15 minutes

2 a £104

b d = c – 2012

c 15 days3 a £49.50

b c = 100(B – 15) – 5t10

c 80 calls

4 a m = dvb 252 g/cm3

5 a h = vπr2

b 10.4 cm6 a 32 cm2

b h = 2a

(a + b)c 5 cmd a = 2a

h – be 14.27 cm

7 a 10π cm2

b l = 4Aπh

c 10 cm

Reviewing skills (page 160)1 a e = d – f

b t = (s – b)

ac g = f h

d w = (v + 3)

4e r =

C2π

f r = Aπ

g r = 100PMt

h t = 3S

2 a i 10ii 0iii –20iv 30

b a = (v – u)

t3 d t = S2 √3

c a = 153 = 3



2

89

Unit 3 Answers

Practising skills (pages 163–164)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

90

8 a i 4n + 7ii 407

b i 10n − 8ii 992

c i 7n + 4ii 704




91

Developing fl uency (pages 164–166)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




92

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

8 a nob yesc yesc yes

9 a yesb yesc nod no




93

Problem solving (pages 166–167)1 a 18

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




94

Reviewing skills (page 167)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



2

95

Unit 4 Answers

Practising skills (page 170)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

96

4 a n2

b n2 + 1c n2 – 1d 2n2

5 a n2 + 10b n3

c n3 + 5d 2n3




97

Developing fl uency (pages 171–173)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




98

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)




99

Problem solving (pages 173–174)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




100

Reviewing skills (pages 174)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



2

101

Unit 5 Answers

Practising skills (pages 177–178)1 a 5, 20, 45, 80, 125

b 6, 9, 14, 21, 30c 1, 7, 17, 31, 49d 7, 16, 31, 52, 79e 5, 35, 85, 155, 245

2 a = ii; b = v; c = iv; d = i3 a 16, n2

b 17, n2 + 1c 18, 2n2

d 35, 2n2 + 34 a Term n2 + 3 1st difference 2nd difference

1 4

2 7 3

3 12 5 2

4 19 7 2

5 28 9 2

b i Term n2 − 1 1st difference 2nd difference

1 0

2 3 3

3 8 5 2

4 15 7 2

5 24 9 2

ii Term n2 + 2 1st difference 2nd difference

1 3

2 6 3

3 11 5 2

4 18 7 2

5 27 9 2

iii Term 22 − 3 1st difference 2nd difference

1 −1

2 5 6

3 15 10 4

4 29 14 4

5 47 18 4



Strand 2 Sequences Unit 5 Quadratic sequences Band g

102

iv Term 5n2 1st difference 2nd difference

1 5

2 20 15

3 45 25 10

4 80 35 10

5 125 45 10

c no5 a = iii; b = v; c = i; d = ii; e = iv6 a and b

i n2 + 4ii no – lineariii 3n2

iv no – cubicv 2n2 + 3

Developing fl uency (pages 178–179)1 a A: 12, B: 36, C: 35, D: 48, E: 47 b i A: 2, B: 2 c i Each term in C is one less than the corresponding term in B.

ii n2 − 1 d i Sequence D = sequence A + Sequence B

ii n2 + 2n e i A and C

ii n2 + 2n − 12 a Drawing of a 5 × 5 pattern.

b Pattern 1 2 3 4 5

Number of tiles 4 9 16 25 36

c Pattern 9d (n + 1)2

e i 50ii 2551iii (n + 1)2 – n or n2 + n + 1

3 a 64b n2

c 104 a 15

b 231c 15d 5050

e i No because the 17th stack needs 153 tins and the 18th stack needs 171 tins.ii 78 and 91




103

5 a i 6ii 12iii 8

b 26

Cube size 2 3 4 5 10 n

1 sticker 0 6 24 54 384 6(n − 2)2

2 stickers 0 12 24 36 96 12(n − 2)

3 stickers 8 8 8 8 8 8

Total number of stickered cubes

8 26 56 98 488 6(n − 2)2 + 12(n − 2) + 8= 6n2 – 12n + 8

Problem solving (page 180)1 a 42

b n(n + 1)c 2(n + 1)d between 50 and 51

2 a Points Number of lines at each point

Total number of lines

2 1 1

3 2 3

4 3 6

5 4 10

6 5 15

7 6 21

8 7 28

n n − 1 n(n − 1)2

b 630 lines3 33 and 244 a 35

b n – 3c n(n – 3)d 11




104

Reviewing skills (page 181)1 a 70, 10n + 10

b 48, 3n2

c There is an error on the Student Book. This is not a sequence.d 177, 11n2 + 1

2 a Term n2 + 5 1st difference 2nd difference

1 6

2 9 3

3 14 5 2

4 21 7 2

5 30 9 2

3 a


Number of green squares 1 4 9 16 25

Number of blue squares 4 8 12 16 20

Number of red squares 4 4 4 4 4

Total number of squares, S 9 16 25 36 49

c i 40ii 4iii 144

d i 4nii 4iii n2

e S = n2 + 4n + 4 or S = (n + 2)2



2

105

Unit 6 Answers

Practising skills (pages 184–185)1 a tn = n2 − 2n + 4 b tn = n2 − 2n + 1 c tn = n2 − 2n + 7 d tn = 2n2 − 4n + 8 e tn = 2n2 − 4n + 3

2 a Ac b Ba c Cb d Dd3 a Un = n2 + 3n − 2 b Un = n2 + 2n − 5 c Un = n2 + 4n + 3 d Un = n2 − n + 1

4 a Un = 2n2 + 3n − 5

b Un = 12 n2 − 2n + 3

c Un = 3n2 − 5n + 2 d Un = 6 − 3n − n2

Developing fl uency (page 185)1 a 11, 13, 17, 23, 31 b No, the 11th term will not be prime. If n = 11, n2 − n + 11 is divisible by 11.2 a = 1.5, b = 43 a = 3, b = 1, c = 194 a 41, 43, 47, 53, 61 b 41st term is not prime5 tn = 1.5n2 + 1.5n

Problem solving (pages 186–187)1 a (teams, matches), (2, 1), (3, 3), (4, 6) b 28 c 1

2 (n(n – 1))

d 380



Strand 2 Sequences Unit 6 The nth term of a quadratic sequence Band i

106

2 a 24 b (n + 1)2 − 1 c Pattern 123 a 14 for shape 2, 27 for shape 3 b 65 c 2n2 + 3n d 2n2 + 6n4 77

Reviewing skills (pages 187)1 a Un = n2 − 2n − 3 b Un = 14 − n − 2n2

2 a = 3, b = 4, c = 5, d = 9, e = 60



3Unit 2 Answers

Practising skills (pages 192–193)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

108

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

−8

−12

−14

−16

−20

−22

−24

−6

−4

−2

2

4

x

y

−2 0 −2−4 −4−6−8−10−12−14

−10

c i −1.6ii 15.5




109

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




110

Developing fl uency (pages 193–194)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




111

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




112

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

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




113

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




114

Problem solving (page 195)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




115

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




116

Reviewing skills (page 196)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




117

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


3


Unit 3 Answers

Practising skills (pages 199–202)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

119

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




120

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




121

Developing fl uency (pages 202–203)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




122

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




123

Problem solving (pages 204–205)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




124

Reviewing skills (page 205)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



3

125

Unit 4 Answers

Practising skills (pages 210–211)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

126

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




127

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)




128

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)




129


–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




130

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




131

Problem solving (pages 213–214)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




132

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




133

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




134

Reviewing skills (page 215)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



3

135

Unit 5 Answers

Practising skills (page 219)1 a and f, b and d, c and g; e is the odd one out.

2 a i is the same as v.

2

4

6

8

10

12

14

–2

–4

0–4 2 4 6 8 10

y = 2x + 10

y = 2x + 5

y = 2x + 3

y = 2x – 1

y = 2x – 3

y

x–2

b i 3ii 10iii 1iv −3v 5

c i y = 2x + 3ii y = 2x + 10iii y = 2x − 1iv y = 2x − 3v y = 2x + 5

3 a 3b Substituting values for x and y into the equation gives 11 = 6 + k.c k = 5; y = 3x + 5d (0, 5)

4 a y = 3x − 2b y = 7x − 9c y = 2x − 19

d y = 12x + 1

e y = −12x + 3

5 a 9 − 14 − 2 = 4

b y = 4x − 7c Check: 9 = 4(4) − 7



Strand 3 Functions and graphs Unit 5 Finding equations of straight lines Band h

136

6 i a 1.5b y = 1.5x − 0.5c check: 1 = 1.5 − 0.5, 7 = 1.5 × 5 − 0.5

ii a 1b y = x + 2c check: 3 = 1 + 2, 7 = 5 + 2

iii a 1.5b y = 1.5x − 0.5c check: 4 = 1.5 × 3 − 0.5, 7 = 1.5 × 5 − 0.5

iv a 47

b y = 47x + 4

17

c check: 3 = −87 + 4

17, 7 =

207 + 4

17

v a − 23

b y = 23x + 10

13

c check: 9 = 43 + 10

13, 7 =

103 + 10

13

Developing fl uency (pages 220–221)1 a y = 2x + 4 and y = 3x + 5

b (0, 4) and (0, 5) c i blue

ii blueiii blueiv neitherv blue

2 Gradient is 3 Gradient is −7

y-Intercept is 6 y = 3x + 6 y = −7x + 6

y-Intercept is −5 y = 3x − 5 y = −7x − 5

3 a l: y = − x + 4, m: y = x − 2b

2

4

–2

–4

0–4 –2 2 4 6

y = –x + 4

y = x – 2

x

y

–6

–6

c 90 degrees; they are perpendicular




137

4 a y = x + 1b y = x + 2c y = x + 4d y = x + 4e y = x + 2f y = x + 1

a is same as f; b as e; c as d5 Always negative. If a > b, then point B always has a greater y value, but lower x value than point A − the line

between them has a negative gradient. If b > a the reverse applies, but the line between them still has a negative gradient.

6 a y = x; y = −x + 12; x = 1b

2

4

6

8

10

12

14

16

18

–2

–4

–6

0–6 –4 –2 2 4 6 8 10 12 14 16 18

y

x

c isoscelesd 25 units2




138

7 a

2

4

6

8

–2

–4

0–6–8–10 –4 –2 2 4 6 8 10

y

x

–6

–8

–10

–12

–14

–16

y = –2x + 13

y = –3x + 7

y = 2x – 3

y = 3x – 17

10

12

14

16

18

–18

b A(4, −5), B(2, 1), C(4, 5), D(6, 1)c A kited (4, 1)e AC: x = 4; BD: y = 1

Problem solving (pages 221–222)1 a l : y = 2x − 1; m: y = −3x + 4

b

2

4

6

8

10

–2

–4

–6

–8

–10

0 2 4 6 8 10

y

x–2–4–8–10 –6

y = 2x + 1y = –3x + 4

c (1, 1)




139

2 a p: y = 3x − 7; q: y = −2x − 2b

−2

−4

−6

2

4

6

10 2 3 4−1x

y

Q P

c (1, −4)d −4 = −4(1) so true

3 a r: y = x + 5, s: x + y = 1b (−2, 3)c x + x + 5 = 1 x = −2. Y = 3. They are the same.

4 a

−2

−4

2

4

6

8

−1−2−3 0 1 2x

y

M

L

b l: y = 4x + 6; m: 4x + 3y = 6c (−0.75, 3)d 4x + 3(4x + 6) = 4x + 12x + 18 = 6 16x = −12 x = −0.75




140

5 a

2

4

6

8

10

–2

–4

–6

–8

–10

0 2 4 6 8 10

y

x–2–4–8–10 –6

b (3, 1)c y = 3x − 8d 27 square units

6 a Both lines have gradient −2b

−10

−5

5

10

5 10x

y

−5−10

c y = 12

x + 4

d y + 2x = 9, x = 2, y = 5 so 5 + 2 × 2 = 9

y = 12

x + 4, x = 2, y = 5 so 12 × 2 + 4 = 5

e 5 square units




141

Reviewing skills (page 222)1 a y = 2x + 5

b y = −x + 7c y = 3x − 1

2 a

−2

2

4

6

8

10 2 3 4 5 6−1x

y

P

b (1, 5)c Derive the equation of the line as y = −2x + 7. Insert coordinates P to prove.

3 a Gradient of both AB and CD is 43, therefore they are parallel. Gradient of AD and BC is 0 therefore they are parallel. Gradient AC is 4

9; gradient BD is − 4

3 so the diagonals are not perpendicular and therefore ABCD is not

a rhombus.b AC: y = 4

9; BD y = −4

3x + 8

c Substitute the coordinates into both equationsd

−2

2

4

6

8

10 2 3 4 5 6 7 8 9−1x

y

B

E

A D

C

4 b y = 2x – 5


3


Unit 6 Answers

Practising skills (pages 225–227)1 a A: – 43; B: 38; C: – 14; D: – 2; E: – 83; F: – 4; G: 34; H: 12 b A and G; B and E; C and F; D and H2 a A: 3; B: – 12; C: 3; D: –5; E: 1; F: – 13; G: 2; H: –1

b i A and C ii B and G; A and F; C and F; E and H

3 a Line A i 3 ii (0, −1) iii y = 3x − 1 Line B i −3 ii (0, 2) iii y = −3x +2

Line C i −13 ii (0, 3) iii y = − 13x +3

Line D i 13 ii (0, −1) iii y = 13x − 1

b A and C; B and D4 a i 2y = x + 2 and y = 12x − 3; y – 2x + 4 = 0 and 2y + x – 7 = 0

ii y = 2(2x + 1) and x + 4y = 1

iii y = 3x – 2 and 2x + 5y + 10 = 0

b 3y = x + 95 a Gradient of red line = 12, gradient of green line = 12; so lines are parallel.

−2 × 12 = −1 so lines are perpendicular to blue line.

b Red line: y = 12x + 2, green line: y = 12x − 16 a y = 5x + 1

b y = –15

x + 1



Strand 3 Functions and graphs Unit 6 Perpendicular lines Band i

143

Developing fl uency (pages 228–229)1 a y x= 2

b y x= − +12 5

c (10, 0) d 20 square units2 a, b, c, e

210−1−2−3−4 3 4 5 6x

y

P

1

32

54

76

89

10

c y x= − +12 6

d (0, 6)e y x= +2 6f y x= − +1

272 or y x= − +1

2172

3 a y x= +14 2

b y x= − +4 19 c (0, 19) d 34 square units4 a i y = 2x − 1 and y = − 12 x + 32

ii y = 3x − 2 and y = − 13 x + 43

iii y = 4x − 3 and y = − 14 x + 54

b y mx m 1( )= − − and y m xmm

1 1= − + +

5 y = 43x + 13 and y = − 34 x + 4 346 y = 2x − 107 a Pupils’ own drawings. b (2, −2) c Q y x: 3 4 = − + ; R y x: 2 1 = − −

d Q : y = −3(2) + 4 = −2; R: y = − 22 − 1 = −2




144

Problem solving (pages 230–231)1 a Equations are: y = 12 x + 2; y = 12 x − 12 ; y = −2x + 2; y = −2x − 8

b Area is 10 square units.2 a

1−1

1

0

32

54

76

89

−1−2−3−4−5 2 3 4 5

−2

x

y

M L

b −( )83 ,113

c L: y = 2(83 ) + 9 = 113 ; M y = −(−83) + 1 = 113

3 a y x= − 12 43

b P (0, 4); PQ gradient = − 43 c Q (3, 0). PQ = 5 units d PS is 5 units, base 4 units, height 3 units. S is at point (0 + 4, 4 + 3) = (4, 7)4 (0.8, −0.6)5 Pupils’ own drawings.




145

6 a Pupils’ own drawings.

76

1

32

54

76

89

101112131415

543210 8 9 10 11 12x

y

D

C

A

B

b (11, 2.5)

c y x= − + 125 28.9

d y x= − 5122512

e 84.5

Reviewing skills (page 231)1 a y

1

32

54

−4−5

−3−2−1 210−1−2−3−4−5 3 4 5

x

b i y x= −12 3

2 y x= − +15 5


4

146© Hodder & Stoughton Ltd 2015

Unit 1 Answers

Practising skills (pages 234–235)1 a 3.6

b Robert knew the solution was between 3.5 and 3.6, and once he worked out that 3.55 was too small, it had to be closer to 3.6.

2 When x = 4.65, x3 + 2x = 109.844 625, which is too small. So the answer is closer to 4.7. Jenny is correct.3 a When x = 2, x3 + 2x = 12, which is less than 20. When x = 3, x3 + 2x = 33 which is more than 20.

b x x3 + 2x Comment

2 12 Too small

3 33 Too large

2.5 20.625 Too large

2.4 18.624 Too small

2.45 19.606 125 Too small

x = 2.5 (to 1 d.p.)

4 x x3 + 6x Comment

2 20 Too small

3 45 Too large

2.3 25.967 Too large

2.2 23.848 Too small

2.25 24.890 625 Too small

x = 2.3 (to 1 d.p.)

5 a 82 = 64 and 92 = 81 and 70 lies between them.b 8.32 = 68.89 and 8.42 = 70.56, so he is correct. c 8.352 = 69.7225 so the root of 70 lies between 8.35 and 8.4, so the square root of 70 = 8.4 (to 1 d.p.)

Developing fl uency (pages 235–236)1 x x3 x2 x3 + x2 Comment

4 64 −16 80 Too small

5 125 25 150 Too large

4.3 79.507 18.49 97.997 Too small

4.4 85.184 19.36 104.544 Too large

4.35 82.312 87 18.9225 101.235 375 Too large

The solution is x = 4.3 (to 1 d.p.)

2 x = 4.2 (to 1 d.p.)3 x = 3.94 3.3



Strand 4 Algebraic methods Unit 1 Trial and improvement Band f

147

5 a x = 1.3b When x = −1, x3 – 3x2 + x = −5. Too small. When x = 0, x3 – 3x2 + x = 0. Too large. So there is a solution between

−1 and 0.When x = 2, x3 – 3x2 + x = −2. Too small. When x = 3, x3 – 3x2 + x = 3. Too large. So there is a solution between 2 and 3.

c x = −0.5

Problem solving (page 236)1 x + 2 = 6.3 cm

2 x = −0.8, x = 1.5 and 4.33 x = 0.2 and x = −4.8

Reviewing skills (page 236)1 a When x = 1, x3 + x = 2 which is less than 9. When x = 2, x3 + x = 10 which is more than 9.

b x x3 + x Comment

1 2 Too small

2 10 Too large

1.9 8.759 Too small

1.95 9.364 875 Too large

x = 1.9 (to 1 d.p.)2 x = 4.8


4


Unit 2 Answers

Practising skills (pages 239–240)1 a <

b >c >d <e >f <g >h <

2 a x > 5b x ≥ 5c x ≤ 5d x ≥ 5e x < −5f x ≥ −5g x ≥ 5h x ≤ 5

3 a x ≥ −2b x ≤ 5c −3 < x ≤d −6 ≤ x < 2e −6 < x < −1f −1 ≤ x ≤ 1

4 a 2 9

b 6 8

c –4 –1

d –2 5

5 a 2 < 14 – TRUE b i −9 < 3 – TRUE

ii −6 < 30 – TRUE iii 6 < −30 – FALSE iv −1 < 5 – TRUE v 1 < −5 – FALSE

6 a x < 3b x > −3c x < 5d x ≥ −6 e x > 4f x < 9



Strand 4 Algebraic methods Unit 2 Linear inequalities Band g

149

7 a 3, 4, 5, 6b −2c 1, 2, 3, 4d −5, −4 , −3, −2, −1, 0, 1

Developing fl uency (pages 240–241)1 a £7 ≤ pay ≤ £11

b age ≥ 18c age > 64d 100 < number < 150 (allow equals as well)e number ≤ 16f 12 ≤ age ≤ 17

2 a 6 < 38 < 7b 9 < 90 < 10c 14 < 199 < 15d 2 < 5 < 3

3 a brother: 2x; sister; x + 10 b 4x < 30c he could be 7d brother 14; sister 17e no

4 a 6wb 3 ≤ w < 7c 18 ≤ 2w2 < 98

5 a x ≥ 4b x < 1c x < 25.5d x ≥ 12

6 a x < 3 b x > −3c x < −3 d x < 3

7 Wendy, because if you divide by a negative number you reverse the sign.




150

8 a i and ii

1–1–1

123

–2–3

–2–3–4–5 2

x<3

x=3

3 4 5

y

x

b i

1–1–1

123

–2–3

–2–3–4–5–6 2

x>–3

3

y

x

ii

1–1–1

123456

–2

–2–3 2

y>4

3

y

x

iii

1–1–1

123

–2–3

–2–3 2

y>x

3

y

x

9 a x < −5b x > 7 c x ≤ −3d x ≥ –1.5, 6 – 4x ≤ 12, 4x – 6 ≥ –12, 4x ≥ –6, x ≥ 1.5 e x > 4f x > 6




151

Problem solving (pages 242–243)1 a 2n – 7 < 5

b 1, 2, 3, 4, 52 a 2(n – 8) > 11

b 14, 15, 16, 17, 18, 193 a Sue is x, Ben is 2x, Ceri is x + 4; so 4x + 4 < 28

b Sue could be 3, 4, 5 years oldc Ben 6, 8, 10 years old; Ceri 7, 8, 9 years old

4 a Narinder is x, Rashmi is x + 2, Bhavinder is 2(x + 2) so 4x +6 < 42b x < 9 c 10 years

5 a 4x + 10 > 21; 4x + 10 < 33b Areas 14 to 50 cm2

6 a 0.5m + 80 < 1mb m > 160

7 a i x = 5ii y = 1iii y = 4

b y ≤ x + 1, x ≤ 5, y ≥ 1, y ≤ 4c 0 ≤ l ≤ 34

8 a They reduce to x < 7, x < 2, x > –1, x > 3, B and D cannot both be satisfied.b A, B and C can be true, if x is 0 or 1.c A, B, C; A, C, D

Reviewing skills (page 243)1 a x < 4

b x > 11c x ≤ 2d x ≥ 2

2 a x > 1b x > −2c x > 2d x < 5

3 a Abbi is a; Bobbi is 2a; Cathy is a – 5; 4a – 5 < 35b a < 10c 18 years


4


Unit 3 Answers

Practising skills (page 246)1 a x = 6, y = 3

b x = 12, y = 4 c x = 3, y = 2d x = −1, y = 2

2 a x = −2, y = −4b x = 7, y = 3 c x = −2, y = –1d x = 7, y = 5

3 a x = 3, y = 1b x = −3, y = −3c x = 4, y = 10d x = 1, y = 9

4 a x = 5, y = 1b x = 5, y = 1c x = 5, y = 1d x = 5, y = 1

5 a x = 8, y = 2 b x = −6, y = −2c x = 4, y = 3d x = 2, y = 4

Developing fl uency (pages 246–247)1 a y = 2x, x + y = 15

b x = 5, y = 102 x + y = 6, x – y = 1, x = 3.5, y = 2.5 3 a h = 2p, 3h + 4p = 35

b h = £7, p = £3.50c £24.50

4 a r = m + 2, 3r + 5m = 62b r = £9, m =£7

5 s – c = 7, 4s + 7c = 523, stalls cost £52, circle cost £45 6 a x + y = 42; 40y + 16x = 1080

b adults (y) = 17children (x) = 25

7 a Rearrange second equation to x = y + 7, then x = 9, y = 2b Rearrange second equation to x = 2y + 7 then x = 1, y = –3



Strand 4 Algebraic methods Unit 3 Solving pairs of equations by substitution Band h

153

8 a x = 17 – 3yb 5(17 – 3y) – 2y = 0

85 – 17y = 0y = 5x = 22 + 3(5) = 17 TRUE5(2) – 2(5) = 0 TRUE

Problem solving (pages 247–248)1 a m + n = 27

m + 15 = n, or n + 15 = mb 21 and 6 in either combination

2 a X + Y = 12b Y = 12 – Xc In XY, X is the ‘tens’. In YX, Y is the ‘tens’. Thus the numbers are (10X + Y ) and (10Y + X ). The difference

between the two numbers is 18 so (10X + Y ) – (10Y + X ) = 18.d 9X – 9Y = 18

Substitute for Y: 18X – 108 = 1818X = 126X = 7; Y = 5

e XY is 753 a c = 5 + a

b 15a + 10c = 550c a = 20, c =25

4 a l – s = 100, 6l = 10b s = 150, l = 250c 1100 g or 1.1 kg

5 Form equations, solves to find masses of cans as 240 g and 100 g. Works out 90p can is the better buy.

Reviewing skills (page 249)1 a y = 3, x = 9

b x = 4, y = 8c x = −5, y = −2d x = −3, y = −16

2 a x = −4, y = −2b x = 4, y = 8c x = 4, y = −1d x = 1, y = −1

3 a s = 3f ; 40f + 80s = 560b f = 2; s = 6c 8


4


Unit 4 Answers

Unit 4 Practising skills (pages 252–253)1 a −4x

b xc −2xd 5xe 0f −6g 45h −8

2 a x = 2, y = −1b x = 5, y = 3c x = −3, y = −1d x = 2, y = −3e x = −3, y = −2f x = 1, y = 4

3 a x = 4, y = 1b x = 5, y = −1c x = 2, y =6d x = 1, y = −4e x = 2, y = −2f x = 5, y = −3

4 a x = 3, y = −2b x = 1, y = −5c x = 4, y = 5d x = −3, y = 2e x = 4, y = 3f x = 6, y = −1

5 a because the coefficients are differentb equation 2 by ×2c x = 2, y = 3

6 a x = 4, y = −3b x = 5, y = −2c x = –3, y = 5d x = 8, y = −1e x = 1, y = −6 f x = -4, y = −5



Strand 4 Algebraic methods Unit 4 Solving simultaneous equations by elimination Band h

155

Developing fl uency (pages 253–254)1 a eq. 1 × 2 and eq. 2 × 3 to eliminate x, or eq. 1 × 3 and eq.2 × 4 to eliminate y.

b x = 5, y = 22 a x = 3, y = 2

b x = 6, y = 1c x = 5, y = −4d x = 5, y = −1e x = 3, y = −4f x = 7, y = 4

3 a 5a + 3b = 118, 8a – 2b = 114 b a = 17, b = 11c Area rectangle = (3(17) – 4(11)) × (17 + 3(11)) = 350

Area triangle = 12

× 2(17) × (17 + 2(11)) = 663

Triangle is bigger in area4 a 5s + 3p = 99, 2s + 4p =62

b s = 15, p = 8c 86 km

5 a 2a + 3c = 24, 3a + 5c = 38 b a = 6, c = 4c £138

6 a (Change the signs), −x + 4y = 6(Add) 4x = 24

b By changing the signs you are doing the equivalent of multiplying by −1. Adding negative values is the same as subtracting.

c x = 6, y = 3

Problem solving (page 255)1 a 4b + a =150

2b + a =100b Banana is 25p and apple is 50pc 575p

2 a 2w + 2l = 32w + 2l = 26

b l = 10 and w = 63 a 2d + 3c = 315

d + 2c = 175b £2275

4 a 4m + 6s = 5445m + 60s = 570

b 6 mountain bikes and 5 sports bikes5 a 10c + 40t = 7.30

5c + 50t = 7.25b c = 0.25 t = 3

25c £4.25



Strand 4 Algebraic methods Unit 4 Solving simultaneous equations by elimination Band h

156

6 a x + y + (360 − 3x) = 180b x + 2y = 180 (as the triangle is isosceles)c x + y + (360 − 3x) = 180, so 180 + y = 2x Substituting x + 2y = 180, (x = 180 − 2y) we have 180 + y = 2(180 − 2y) 180 + y = 360 − 4y, 5y = 180, y = 36 so x = 108 as 108 = 3 × 36, x = 3yd As x = 3y and x + 2y = 180, then 3y + 2y = 180, 5y = 180, y = 36

Reviewing skills (page 256)1 a x = 2, y = −1

b x = 4, y = 2c x = 3, y = −3

2 a x = 12, y = 5b x = 4, y = −3c x = 5, y = −2

3 a x = 2, y = 0b x = 5, y = 3c x = 8, y = 3

4 a 4p + 3a = 753p + a = 45

b A peach tree is £12 and an apple tree is £9c £99



4

157

Unit 5 Answers

Practising skills (pages 258–259)1 a (3, 1)

b x = 3, y = 1c 3 + 1 = 4, 1 = 3 – 2

2 a (1, 1) b i x = 1, y = 1

ii 16x – 12 = 3x + 1, x = 1, y = 13 a m is y = x, e is 3y = 4x – 1

b (1, 1) c i x = 1, y = 1

ii 3x = 4x – 1 or 3y = 4y – 1, x = 1, y = 14 a f then e

b (1, 2) c 3y = x + 5 and x + 2y = 5d Proof by substitution

5 a x 0 1 2 3

3x 0 3 6 9

−2 −2 −2 −2 −2

y = 3x – 2 −2 1 4 7

b x 0 1 2 3

6x 6 6 6 6

−x 0 −1 −2 −3

y = 6 – x 6 5 4 3

c

20–2 4 6 8

5

10

15

20

25

–5

–10

y

x

d (2, 4)e x = 2, y = 4

Substitution method: 6 – x = 3x − 2



Strand 4 Algebraic methods Unit 5 Using graphs to solve simultaneous equations Band h

158

6 a

5

10

–5 50 10

y

x

b (−1, 2)7 a x 0 1 2 3

y = 8 − x 8 7 6 5

b x 0 1 2 3

y = x + 4 4 5 6 7

c

1–1–1

123456789

2 3 4 5

y

x

d (2, 6) x = 2, y = 6e x = 2, y = 6 via substitution, x + 4 = 8 − x

8 a

1–1–1

123

–2–3

–2–3 2 3

y

x

b (0, −1), x = 0, y = −1c x – 1 = 3x – 1, x = 0, y = −1




159

Developing fl uency (pages 260–261)1 a 4x = 2x2 4 = 2x x = 2 y = 8

b

–5

–10

5

10

–5–10 50 10

y

x

c check graph2 a x + 3 = 4x – 3

x = 2, y = 5b

1–1–1–2–3–4

12345678

2 3

y

x

c check graphd 5 = 8 – 3, 5 = 2 + 3




160

3 a (1, 5)

2–2–2

2468

–4–6–8

–4–6–8–10 4 6 8 10

y

x

b 5 = 3 + 3, 1 + 5 = 64 a There is no solution, 2x + 3 can never be equal to 2x – 1.

b The coefficients of x are the same, the coefficients of y are the same.c

1–1–1–2–3–4

12345678

2 3

y

x

d The lines are parallel so there is no solution, they will never intersect.5 a

20

123456789

10

4 6 8 10 12

2x + 3y = 24

y = x + 3

y

x

b x = 3 and y =6c Students’ own check.d 40.5 square units




161

6 a i C = 30 + 18uii C = 50 + 16u

b

15

100

200

300

x

y

105

c i u = 10; £2.10ii Students’ own check.

d Sparkle7 a

−1−1−2−3−4−5−6−7−8−9

−10

1

123456789

10

2 3 4 5 6 7 8 9 10x

y

−2−3−4−5−6−7−8−9−10

b (−4, 5), (4, 7),( 6, −1) and (−2, −3)c Students’ own check.d Square




162

Problem solving (pages 261–263)1 a C = 50m, C = 20m + 600

b

500

20406080

100120140160180200

100

150

200

250

300

350

400

y

x

c Cars 2 go2 a x 0 1 2 3 4 5

3x 0 3 6 9 12 15

−3 −3 −3 −3 −3 −3 −3

y = 3x – 3 -3 0 3 6 9 12

x 0 1 2 3 4 5

3x 0 3 6 9 12 15

−9 −9 −9 −9 −9 −9 −9

y = 3x – 9 −9 −6 −3 0 3 6

x 0 1 2 3 4 5

−3x 0 −3 −6 −9 −12 −15

+15 +15 +15 +15 +15 +15 +15

y = −3x + 15 15 12 9 6 3 0

x 0 1 2 3 4 5

−3x 0 −3 −6 −9 −12 −15

+9 +9 +9 +9 +9 +9 +9

y = 3x + 9 9 6 3 0 −3 −6

b y

x

5

0 5

c P: (3, 0), Q: (2, 3), R: (3, 6), S: (4, 3)d Students’ check their answer.e rhombus




163

3 a

1

123456789

10

2 3 4 5 6 7 8 9 10x

y

y = x –3

2y = x + 4

x + 2y = 12

b i (4, 4), (6, 3), (10, 7)ii A: x = 4 and y = 4; B: x = 6 and y = 3; C: x = 10 and y = 7

c i

1

123456789

10

2 3 4 5 6 7 8 9 10 11 12x

y

ii x = 4, x = 10, y = 3, y = 7iii 24 square units

d 6 square units




164

4 a i 30m + 8t = 500ii 20m + 12t = 500

b

10

20

30

40

50

60

70

10 20 30x

y

Q mobile

Pear

c 10 minutes of calls, 25 texts d i

10

20

30

40

50

60

70

10 20 30x

y

ii Daisy would have to top up 60p.iii Chloe would have 60p credit left.

5 (0.5, 2) (−1, −1.5) (please note that the answers have been rounded)6 (−1, 2) (2.5, −2) (please note that the answers have been rounded)




165

Reviewing skills (page 263)1 a

1–1–2

2468

10

–4–6–8

–10

–2 2 3 4 5 6

y

x

b (2, −1)c –1 = 4 – 5; –1 = 6 – 7

2 a

1 –1–2

2468

10

y

x

–4–6–8

–10

–223456

b 1 = –3 + 4; 1 = –1 + 2c Students’ check their answer.

3 a i C = 150 + 2mii C = 100 + 4m

b

10

100200300400500

20

Equation 1: y = 150 + 2xEquation 2: y = 100 + 4x

30 40 50 60 70 80 90 100x

y

c i 25 miles; £200ii Students’ own check.

d i Gwilym’s Coaches ii £110


166

5


Unit 1 Answers

Practising skills (pages 267–268)1 a x2 + 3x

b x2 – x c x2 – 25d x2 + x – 6 e x2 – 3x – 40f x2 – 5x + 4g x2 – 4

2 a added 7; multiplied 12b added 8; multiplied 12c added 5; multiplied 4d added 4; multiplied −5e added 6; multiplied 9f added 6; multiplied −16g added −7; multiplied 12h added 0; multiplied −4

3 a 4 and 2b 4 and −2c −4 and 2d −4 and −2e 8 and 1f 8 and −1g −8 and 1h −8 and −1

4 a −3 and −2b 3 and 2c −3 and 2d 3 and −2e 6 and 1f 6 and −1g −6 and −1h −6 and 1

5 a 2(a + 5) + 7(a + 5) = 9 (a + 5)

b 12(b – 6) – 3(b – 6) = 9 (b – 6)

c 8(c + 2) + (c + 2) = 9 (c + 2)

d –2(d – 7) + (d – 7) = –1 (d – 7)

6 a x(x + 5) + 7(x + 5) = x + 7 (x + 5)

b x(x – 6) – 3(x – 6) = x + 3 (x – 6)

c x(x + 2) + (x + 2) = x + 1 (x + 2)

d –x(x – 7) + (x – 7) = – x + 1 (x – 7)



Strand 5 Working with quadratics Unit 1 Factorising quadratics Band h

167

7 a x2 + 2x + 4 x + 8

x(x + 2) + 4 (x + 2)

x + 4 (x + 2)

b x2 – 3x – 5 x + 15

x(x – 3) – 5 x – 3

x – 5 x – 3

c x2 + 4x – 6 x – 24x x + 4 – 6 x + 4

x – 6 x + 4

d x2 + 2 x – 1x – 2x x + 2 – 1 x + 2

x – 1 x + 2

8 a (x + 3)(x + 5)b (x – 4)(x − 5)c (x + 4)(x + 4)d (x + 7)(x − 3)e (x + 2)(x − 3)f (x + 16)(x + 1)g (x + 1)(x − 16)h (x − 10)(x + 2)

9 a x(x + 8)b (x − 10)(x + 10)c (x − 16)d (x − 4)(x + 4)e (x + 1)(x + 6)f (x − 1)(x + 1)g (x − 12)(x + 12)h (x − 5)(x + 5)

Developing fl uency (pages 269–270)1 a PKXM = ac, KSNX = ad, MQLX = bc and XLRN = bd, assuming X is the middle apex b i PMNS

ii MQRNiii PQRS

c (a + b)(c + d)




168

2 a

2 cm

3 cm

A

C

B

D

NottoScale

x2 cm2

A is x × 2 cm2

B is 3 × 2 cm2

C is x × x cm2

D is x × 3 cm2

b (x + 3) and (x + 2)c Proof: Area is x2 + 3x + 2x + 6 which equals x2 + 5x + 6 = (x + 3)(x + 2)d Proof: (x + 3)(x + 2) = x2 + 3x + 2x + 6

3 a 2x 8

4xx2

b Sides are x + 2 and x + 4 perimeter = 2[(x + 2) + (x + 4)] = 2[2x + 6] = 4x + 12area = x2 + 2x + 4x + 8 = x2 + 6x + 8

c Proof: Area is x2 + 2x + 4x + 8 = x2 + 6x + 8 = (x + 2)(x + 4)d Proof: x2 + 6x + 8 = x2 + 2x + 4x + 8 = (x + 2)(x + 4)

4 a Figure 1: a2; Figure 2: a(a − b) + b(a − b) = a2 – ab + ab – b2 = a2 − b2 ; Figure 3: a(a − b) + b(a − b) = a2 – ab + ab – b2 = a2 – b2 OR (a + b)(a − b) = a2 – ab + ab – b2 = a2 – b2.

b (a + b)(a − b)c Proof: Area Figure 2 equals area Figure 3. a2 − b2 = (a + b)(a – b)d Proof: Expand (a + b)(a − b) to get a2 – ab + ab – b2 = a2 − b2

5 a i x2 + 11x +10ii x2 + 7x + 10

b 4xc Proof: (2 + 10)(2 + 1) – (2 + 2)(2 + 5) = 36 – 28 = 8 = 2(4) = 4

6 a i x2 + 10x +21ii x2 + 9x + 14

b x + 7c Factorise with (x + 7): (x + 7)((x + 3) – (x + 2)) = (x + 7)(x + 3 − x − 2) = (x + 7)(1) = (x + 7)

Problem solving (page 270)1 642 – 362 = (64 + 36)(64 – 36) = 2800 cm2

2 Let the integer be n, then n2 – 4 = (n + 2)(n – 2), and if n > 3, n – 2 must be at least 2, which means the product is a composite number (with two different factors), it cannot be prime.




169

Reviewing skills (page 270)1 a (x – 4)(x – 2)

b (x – 3)(x + 4)c (x – 4)(x + 3)d (x – 2)(x + 5)e (x – 5)(x + 2)f (x – 4)(x – 4)g (x − 7)(x + 7)h (x – 3)(x + 3)

2 Proof: x2 + 8x + 7 – (x2 + 8x + 12) + x2 – 4 = x2 – 9 = (x + 3)(x – 3)


170

5


Unit 2 Answers

Practising skills (page 273)1 a x = –4 or x = −1

b x = −7 or x = 3c x = 1 or x = 1d x = −2 or x = −1

2 a (x – 3)(x – 4); x = 3 or x = 4b (x + 1)(x – 2); x = −1 and x = 2c (x – 3)(x + 5); x = 3 and x = −5d (x + 5)(x + 1); x = −5 and x = −1e (x + 2)(x – 2); x = −2 and x = 2f x (x – 7); x = 0 and x = 7g (x + 4)(x – 3); x = 3 and x = −4h (x + 3)(x – 3); x = −3 and x = 3

3 a x 2 + 4x + 3 = 0; (x + 3)(x + 1) = 0; x = −1 and x = −3b x2 + 2x – 8 = 0; (x + 4)(x – 2) = 0; x = 2 and x = −4c x2 + x = 0; x(x + 1) = 0; x = 0 and x = −1d x2 – 3x – 4 = 0; (x – 4)(x + 1) = 0; x = 4 and x = −1e x2 – 49 = 0; (x + 7)(x – 7) = 0; x = −7 and x = 7

4 a (x + 8)(x – 8) = 0; x = –8 and x = 8b x2 – 9x – 36 = 0; (x – 12)(x + 3) = 0; x = −3 and x = 12c x = 2 and x = −8d (x + 2)(x – 1) = 0; x = −2 and x = 1e x(x – 9) = 0; x = 0 and x = 9

5 a x2 – 8x + 15 = 0b x2 + 8x + 15 = 0c x2 + 2x – 48 = 0d x2 – 1 = 0e x2 + x – 90 = 0f x2 + 6x = 0

Developing fl uency (pages 273–274)1 a x(x + 4)

b x(x + 4) = 45; x2 + 4x – 45 = 0 c (x + 9)(x – 5) = 0; x = −9 or 5 d Taking positive value for x, 5 cm and 9 cm

2 a x(x – 2) = 48; x2 – 2x – 48 = 0, x = 8 and −6b Taking positive value for x, 8 cm by 6 cm

3 12

x(x + 6) = 8; x2 + 6x – 16 = 0; x = 2 and −8. Taking positive value for x, 2 cm base and 8 cm height.



Strand 5 Working with quadratics Unit 2 Solving equations by factorising Band h

171

4 12

x(x – 2 + x + 6) = 35; x2 + 2x – 35 = 0; x = 5 and −7. Taking positive value for x, 11 cm base, 3 cm for the parallel side and 5 cm height.

Problem solving (page 274)1 a length = (x2 – 64) ÷ (x – 8) = (x + 8)

b x2 – 64 = 36x2 – 100 = 0(x + 10)(x − 10) = 0 x = 10, x = −10Taking positive value for x, length = 18; width = 2

c x – 8 = x – 82

2(x – 8) = (x + 8)2x – 16 = x + 8x = 24

d x =12; therefore perimeter = 482 a 5t(t – 9) = 0; t = 0 or t = 9 seconds

b 5t(t – 9) = 100; t = 4 or t = 5 seconds3 (x + 6)(x – 5) = 26; x2 + x – 30 = 26; x2 + x –56 = 0; (x + 8)(x – 7). Taking positive value for x, x = 7 so length is

13 cm and width is 2 cm.4 (x – 5)2 = 4, so x – 5 = ±2; x is 3 or 75 a (x + 5)2 – 25 = 24; x2 + 10x – 24 = 0; (x + 12)(x – 2) = 0. Then x + 5 = 7, the length is 7 cm.

b The other value of x is –12 and you can’t have a negative length.

Reviewing skills (page 275)1 a x = –2, −3

b x = 2, 4c x = 2, −1d x = 4, −7e x = −1, −4f x = 2, −2g x = 5, −3h x = 10, – 10

2 x2 – 3x – 54 = 0; x = 9 and −6; Taking positive value for x, the rectangle is 9 cm by 6 cm.


172

1


Unit 7 Answers

Practising skills (page 279)1 a 15 litres

b 25 litresc 35 litresd 75 litres

2 a 12.5 mb 7.5 cmc 250 cmd 62.5 cm

3 a 16 kmb 32 kmc 160 kmd 288 km

4 a 40 milesb 25 milesc 375 milesd 1250 miles

5 No, she is 56 inches tall and 130 cm is about 59 inches.

Developing fl uency (pages 279–280)1 Jade is taller by 4 cm or 1.6 inches2 a i 10 yards

ii 60 miles b i 4 pints

ii 1 gallonc 8 gallonsd 19.2 kme 88 lbf 7 pintsg 2.5 mh 50 litres

3 a No b i Approximately 105 m by 70 m

ii Approximately 1110 square metres.4 140 g plain flour

224 g caster sugar112 g oats84 g desiccated coconut126 g butter



Strand 1 Units and scales Unit 7 Converting approximately between metric and imperial units Band e

173

5 a Yes; 65 miles is approx 104 km.b £78.40

Problem solving (pages 281–282)1 Yes, because his BMI is 21.2 £4783 No, as it will take her 4 hours 41 minutes.4 43.755 a 0.28

b 58 0006 9709

Reviewing skills (page 282)1 a 6.6 lb

b 8.8 lbc 15.4 lbd 22 lb

2 Megan by 5.5 cm or 2.2 inches.3 In the UK as the cost is 1.16p per gallon, whereas in the USA it is 1.21p per gallon.


174

1


Unit 8 Answers

Practising skills (pages 285–286)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

175

Developing fl uency (pages 286–288)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°

7 a 30 milesb 130 miles

c i 293°ii 002°iii 105°iv 115°




176

Problem solving (pages 288–289)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°




177

Reviewing skills (page 289)1 263°


178

1


Unit 9 Answers

Practising skills (pages 291–292)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

179

Developing fl uency (pages 292–294)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




180

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




181

Problem solving (pages 294–295)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




182

Reviewing skills (page 296)1 a student’s own diagram

b 270 cm



1

183

Unit 10 Answers

Practising skills (pages 298–299)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 hourAnna: £7.80 per hourJamie is better paid

6 a 1 hour 30 minutes is 112 hours, so it should be written 1.5 not 1.30

b 21 ÷ 1.5c 1 hour 20 minutes is 120

60 or 113 hours, so it should be written 1.33333… not 1.20

d 16 ÷ 1.33333…7 a 2.25 hours

b 3.75 hoursc 5.666… hours d 2.33333… hours



Strand 1 Units and scales Unit 10 Compound units Band g

184

Developing fl uency (pages 299–300)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 f

185

Problem solving (page 301)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 g

186

Reviewing skills (page 301)1 a 65 km/h

b 18.06 m/s2 Toby earns 30p per hour more3 b 12.1 mph



1

187

Unit 11 Answers

Practising skills (page 303)1 a Length

b Volumec Areae Areaf Volumeg Volume

2 b, c and d3 b and d4 a, b and c5 b and c6 d7 a and d

Developing fl uency (pages 304–305)1 a r2h(4 + π)

b 2r(4h + 4r + πh)

2 p3 is a length divided by 3 so remains a length. qr is the square root of an area and so is a length. The expression in the brackets is the sum of 2 lengths which makes another length. This is multiplied by a number, 2π, and so remains a length.

3 a4 a 2

b 1c 1d 1

5 a He needs to use an expression that represents volume, i.e. dimension 3. 4r2 + h2 is dimension 2. Only part of the

expressions 13

r2h + πr2 and r2(h + π) have the correct dimension, so these don’t represent anything. i r2h Q 110 + πR

has dimension 3 so is the right answer.b i Only the first part of the expression has the correct dimension for volume. ii 10 wdh

Problem solving (pages 305–306)1 Expanding the brackets gives πr2 + 2πr2h + πrh. The middle term is a volume, not an area.2 a iii

b i



Strand 1 Units and scales Unit 11 Dimensions of formulae Band g

188

3 a 4πR − 4πr + 2πRh + 2πrhb 2π (R + r)(R + h − r) = 2π (R2 + Rh − Rr + Rr +rh − r2)

= 2π (R2 + Rh + rh − r2)= 2π R2 + 2πRh + 2πrh − 2πr2

4 c


b 1c 2d Incorrect formulae 1f Incorrect formula

2 a, b and c



1

189

Unit 12 Answers

Practising skills (pages 309–310)1 a 1.5 litres of lemonade for 65p

b 12 for 75pc 3 kg of grass seed for £4.20d 1.5 kg for £4.65

2 a 3.57 m/sb 12.86 km/h

3 a 2.11 g/cm3

b 2111.1 kg/m3

4 0.27 g/cm3

5 a nt ms−1

b 3.6nt km/h

6 a 10.5g cm-3

b 84 kg7 a It’s travelling at a constant speed.

b 6 m/s2

c acceleration

Developing fl uency (pages 310–311)1 No, as it will take her 4 hr 41 min.2 In the UK, as the cost is 1.17p per g whereas in the USA it is 1.21p per g.3 £1.174 a A

b Cc B

5 No, change in speed is acceleration. So second car is accelerating, whilst first is at constant speed.6 7.59 g/cm3

7 a Metal A: 2.7 g/cm3 = 2700 kg/m3. Metal B: 8900 kg/m3 = 8.9 g/cm3.b Metal B is heavier.

8 1000 f N/cm2

9 1000mv

kg/m3

10 V × 3600 metres in 1 hour; V × 3600 ÷ 1000 = 3.6 V/kilometres in 1 hour.



Strand 1 Units and scales Unit 12 Working with compound units Band g

190

Problem solving (pages 311–313)1 a 1.728 seconds

b You may have to take the length of the car into account – that would make it a longer time.2 a 10

b 4.67c 3120

3 The 1.8 l because the unit price is 0.45 litre per pound as against the large bottle at 0.43 litre per pound.4 a 900 kg/m3

b Answer yx

5 £165.106 Yes, because his BMI is 21.7 a 60 g copper and 40 g zinc

b 7.81 g/cm3

8 No, as the mass is 989 g which is less than 1000 g.9 a Difference in daily rate is £5 on first day.

b Equation of blue is 5x + 10Equation of red is 4x + 5

Reviewing skills (page 314)1 13.302 54.5 kg/m3

3 a £75b There is an error on the Student Book. This answer cannot be calculated from the information given.



2

191

Unit 5 Answers

Practising skills (pages 318–319)1 a = 35°, b = 50°, c = 40°, d = 3°, e = 68°2 a = 90°, b = 193°, c = 118°, d = 48°, e = 134°3 a = 60°, b = 45°, c = 70°, d = 40°, e = 35°, f = 54°4 a = 42°, b = 32°, c = 148°, d = 70°, e = 40°, f = 135° g = 117°, h = 67°5 a 93°

b 125°c 55°d 55°e 90°f 105°

Developing fl uency (pages 320–321)1 a = 70°, b = 55°, c = 55°, d = 85°, e = 23°, f = g = h = 60°2 30°3 Angles a + d = 180° because angles on a straight line add up to 180°

Angles a + b + c = 180° because angles in a triangle add up to 180°So angles a + d = angles a + b + c because both have a sum of 180°So angle d = angles b + cSo the exterior angle of a triangle equals the sum of interior opposite angles.

4 a 45°b 75°

5 x = 74.7°, 2x = 149.3°6 95o

Problem solving (pages 321–322)1 a No

b 25°2 a p = 36°

b s = 72°3 a + b + c = 180° and p + q + r = 180°, so a + b + c + p + q + r = 360°. Hence a + b + p + c + r + q = 360°. So the

sum of the interior angles of the quadrilateral is 360°.4 a Never true, because if it has 3 exterior acute angles then it would have 3 obtuse interior angles which is

impossible.b Never true, because if it has 2 exterior acute angles then it would have 2 obtuse interior angles which is

impossiblec Sometimes true, because if it has 1 exterior acute angle then it would have 1 obtuse interior angle which is

possible.d Sometimes true, because if it has 0 exterior acute angles then it would have 3 acute interior angle which is

possible, for example in an equilateral triangle.



Strand 2 Properties of shapes Unit 5 Angles in triangles and quadrilaterals Band e

192

5 30° and 60°6 x = 90°

Reviewing skills (page 323)1 a = 42°, b = 32°, c = 148°, d = 70°, e = 40°, f = 135°, g = 117°, h = 67°2 a 37° b i square, rhombus, kite

ii

c 140°140°

40°40°

3 90o



2

193

Unit 6 Answers

Practising skills (page 326)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

194

6 a

b

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

d Yes




195

Developing fl uency (page 327)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°




196

Problem solving (pages 328–329)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°




197

Reviewing skills (page 329)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


198

2


Unit 7 Answers

Practising skills (page 333)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

199

Developing fl uency (pages 334–335)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°).




200

Problem solving (pages 335–336)1 50°

Angle ADE = 65 degrees (corresponding angle and angles on a straight line add up to 180°)Angle ADE = angle AED (is osceles 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°




201

Reviewing skills (page 337)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


202

2


Unit 8 Answers

Practising skills (pages 339–340)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

203

Developing fl uency (pages 341–342)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°




204

Problem solving (pages 343–344)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°




205

Reviewing skills (page 344)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


206

2


Unit 9 Answers

Practising skills (pages 348–350)1 a vertically opposite angles

b angles in isosceles trianglec corresponding anglesd alternate anglese opposite angles of parallelogramf angles in isosceles triangleg opposite angles of parallelogramh alternate anglesi corresponding anglesj vertically opposite angles

2 a equal sides of isosceles triangleb opposites sides of rectanglec radii of a circled opposite sides of a parallelograme sides of an equilateral triangle

3 a p and r; s and qb s and p; p and q; q and r; r and sc p + q + r + s = 360° (angles round a point)

4 a u and r – alternate angles; s and p – alternate angles; q and t must be equal as the triangles are similar (but not congruent) as two angles of the triangle are the same, therefore similar by AAA.

b uts and pqr are angles of a triangle; puq and srt are two angles of a rectangle.5 a SSS

b Not congruent.c ASAd SASe RHSf SASg Not congruent, but similar triangles AAA.h Not congruent as 5 cm side isn’t the same for both triangles.

6 a a = 50° – vertically opp angles; b = c = g = h = 65° – equal angles of equal isosceles triangles as the lines OW, OX, OZ and OY are equal radii of a circle and therefore isosceles triangles OWX and OZY are congruent by SAS; e = k = 130° – angles on a line; d = f = i = j = 25° – equal angles of equal isosceles triangles as the lines OW, OX, OZ and OY are equal radii of a circle and therefore isosceles triangles OXY and OWZ are congruent by SAS.

b Angle bj = angle gf = 65 + 25 = 90° and angle cd = angle hi = 25 + 65 = 90°. WX does not equal WZ as they are sides of different isosceles triangles. WXYZ has four 90° angles and two different sides and is therefore a rectangle.

c Three pairs. OWX and OZY; OZW and OXY; WYZ, WXY, XYZ and WXZ.7 In triangles ABP, ACP: AB = AC (Isosceles triangle); BAP = CAP (given); AP is common. Triangles ABP, ACP are

congruent (SAS); PB = PC (corresponding sides of congruent triangles).



Strand 2 Properties of shapes Unit 9 Congruent triangles and proof Band h

207

Developing fl uency (pages 351–353)1 a AC = CE = 8 cm

ACB = DCE (vertically opposite angles)BAC = CED (alternate angles on parallel lines)So, ABC and CDE are congruent (ASA).

b AB and DE are the same, as they are parallel and the triangles are congruent. BC = DC as they are matching sides of congruent triangles.

2 a In triangles DAB and DBC AB = CB (A rhombus has equal sides)AD = CD (rhombus has equal sides)BD is common to both triangles.So, triangles DAB and DCB are congruent (SSS).

b Parallelogram – two sets of parallel lines, with AC and DB intersecting at right angles.3 a OA = OB (radii)

ON is common to both triangles∠ONA = ∠ONB = 90°So, triangles OAN and OBN are congruent (RHS).

b Point N is in the middle of the line AB as ON bisects AB at 90 degrees, and AN and AB are equal.4 a The angle must be the one between the equal sides.

b The sides are not in corresponding positions.c The hypotenuse on the left is equal to one of the shorter sides on the right.d AAA does not prove congruence. One triangle could be bigger than the other.

5 AD = BC (given)DAB = ABC (given)AB is common.Triangles DAB, CBA are congruent (SAS).BD = AC (corresponding sides of congruent triangles).

6 Let PQ cut AB at X.P

A

X

QB

In triangles AQX, BPX:AQ = BP (given)QAX = PBX = 90° (given)AXQ = BXP (vertically opposite angles)Triangles AQX, BPX are congruent (ASA)AX = XB (corresponding sides of congruent triangles)




208

7 In triangles AMD, BND:AD = BD (given)MD = ND (given)AMD = BND = 90° (given)Triangles AMD, BND are congruent (RHS)CAB = CBA (corresponding sides of congruent triangles)So, ABC is isosceles.

8 BAC = DAC (BAD bisected)

A

D

C

B

BCA = DCA (BCD bisected)AC is common.Triangles ABC, ADC are congruent (ASA)So, AB = AD, BC = CD (corresponding sides of congruent triangles)So, ABCD is a kite.




209

Problem solving (pages 353–354)1 a They are the same length.

b They are the same size.c A

B

CD

E

Statements Reason

AB ≈ BCAE ≈ CD∠A ≈ ∠C

All sides of a regular pentagon are congruent. All interior angles of a regular pentagon are congruent.

BAE ≈ BCD SAS (side angle side)

BE ≈ BD CPCT (corresponding parts of congruent triangles)

d AC, AD, CE Draw in line AD: AED ≈ BCD (SAS), therefore AD ≈ BD (CPCT)

A

B

CD

E

Draw in line EC: EDC ≈ EAB (SAS), therefore EC ≈ EB (CPCT)A

B

CD

E

Draw in line AC: ABC ≈ AED (SAS), therefore AC ≈ AD (CPCT)A

B

CD

E

2 a AB = AC (given), ∠BAD = ∠CAD (given), AD is common (SAS). b i Congruent triangles mean BD = CD, so D is in the midpoint of BC.

ii Midpoint of the line when ∠BAD = ∠CAD means that the line AD is perpendicular to BC so ADC angle is 90°.3 a Consider triangles PQS and PRS, angle QPS = angle RPS (given), angle PQS=angle RPS (given), PS is common

so triangles PQS and PRS are congruent (SAA). Hence PQ = PR. Triangle PQR is isosceles.b From a SQ = SR. So triangle SQR is isosceles, so angle SQR = angle SRQ, so angle SQR + angle SPQ = angle

SRP + angle SRQ, so angle SQR = angle SRQ




210

4 Triangles ABC and BDE are congruent (RHS). So DB = BC, hence AD (= AB − DB) = CE (=BE − BC).5 a ∠PQM = ∠RSN (alternate angles), RS = QP (opposite sides of a parallelogram are equal), ∠RNS = ∠PMS

(AAS).b Triangles RNQ and PMS are congruent. Length SP = length RQ. Hence SM = QN due to similar triangles.

6 a Draw a rhombus with a line from one vertex to the opposite vertex. This creates two triangles that are congruent by side-side-side congruence. We know that all the acute angles formed from drawing the line are equal in measure. By alternate interior angles, each set of opposite lines is parallel.

b Draw the second diagonal from a different vertex to the opposite vertex. By vertical angles, we know the angle in each pair of opposite triangles at the intersection between all four lines is equal. Then, by angle-angle-side congruence, all four triangles are congruent. Then, since the measures of all four angles at the four-way intersection must sum to 360 degrees and they all have equal measures, each angle measures 360

4 degrees = 90 degrees.

Reviewing skills (page 355)1 a none of the above

b alternate/corresponding anglesc angles in an isosceles triangled angles in an isosceles trianglee none of the abovef angles in an isosceles triangle

2 a DAE = EBC = 90° (rectangle properties)AD = BC (rectangle properties)ED = EC (given, isosceles triangle)ADE and EBC are congruent (RHS)

b Point E is in the centre of line AB as AE = EB due to the triangles DAE and EBC being congruent.3 Draw and label the parallelogram. Draw the diagonals. Prove triangles AOB and DOC are congruent using AAS

(corresponding angles and opposite sides of a parallelogram are equal). Therefore lines AO and OC are equal and DO and OB are equal meaning O is at the centre of the lines and the lines bisect each other.

AAA B

O



2

211

Unit 10 Answers

Practising skills (page 358)1 M is the odd one out.

A and N: congruent; isosceles triangles with same sides and angles.B and K: similar; SAS, sides in same ratio.C and U: similar; all equilateral triangles are similar.D and V: similar SAS: two sides in same ratio and included angle is equal.E and S: congruent; all angles the same and same side length.F and Q: similar; all angles the same.G and X: congruent; by Pythagoras’ theorem, sides of 5 cm, 12 cm and 13 cm.H and P: congruent SASI and O: similar; three equal angles as isosceles triangles with angles 30°, 75° and 75° (AAA), no information about side length so can’t state that they are congruent.J and T: similar both right angled isosceles triangles.L and W: congruent; by Pythagoras’ theorem L is right-angled and both have sides of 7 cm, 24 cm and 25 cm.R and Y: similar; by Pythagoras’ theorem all sides are in the same ratio.

Developing fl uency (page 359)1 In triangles FAB, BCD, DEH:

FA = BC = DE (sides of regular hexagon)AB = CD = EF (sides of regular hexagon)FAB = BCD = DEF (angles of regular hexagon)FAB, BCD, DEH are congruent (SAS)So, FB = BD = DF (corresponding sides of congruent triangles)So, BDF is equilateral.

2 a It is a rectangle.b In triangles ABD, BCA:

AD = BC (opposite sides of rectangle)ABD = BCA = 90° (angles of a rectangle)AB is commonTriangles ADB and BCA are congruent (SAS).

c Triangles ADB, BCA are therefore also congruent by RHS.AC = BD (corresponding sides of congruent triangles).

d A rectangle is a parallelogram with four right angles.



Strand 2 Properties of shapes Unit 10 Proof using similar and congruent triangles Band h

212

3 a In triangles ABC and ADC:AB = AD (equal sides of kite) BC = DC (equal sides of kite)AC is a common side to both trianglesSo, triangles ABC and ADC are congruent (SSS).

B

AD

CX

b In triangles ABX and ADX:AB = AD (equal sides of kite)Angle BAX = Angle DAX (corresponding angles of congruent triangles ABC and ADC)AX is a common side to both triangles.So, triangles ABX and ADX are congruent (SAS).

c Triangles ABX and ADX are congruent (SAS), so BX = DX, meaning X is midpoint of BD.d Isoceles triangles means BX = BD. Hence, midpoint.

4 a ABC and DEC are similar.In triangles ABC and DEC:∠ACB = ∠DCE(vertically opposite)∠ABC = ∠DEC (alternate) Therefore the triangles are similar (AA)

b 2 (given in initial question)c AE = AC + CE

AE = 2CE + CE = 3CECEAE

= 13

Therefore, point C is a third of the way along AE from point E.5 In triangles AXY, ABC:

XAY = BAC (common)AXY = ABC (corresponding angles on parallel lines)Triangles AXY, ABC are similar.

So, AYAC

= AXAB

= XYBC

= 12

, so Y is the midpoint of AC and XY = 12

BC.

6 a ∠NAM and ∠BAC are equal. Sides AN and AB are parallel with AB = 2AN; and AM and AC are parallel with AC = 2AM. Therefore these triangles are similar as the ratio of the two sides is the same and the angle between them is equal.

b Since triangle ANM and ABC are similar, angles MAN = BAC; angles ANM = ABC and angles AMN = ACB. Therefore line MN is parallel to BC.By same similar triangles argument, triangle NBL is similar to ABC and triangle CLM is also similar to ABC, and therefore LM is parallel to AB and LN is parallel to BC.∠AMN is equal to ∠ACB (corresponding angles), ∠LMC is equal to ∠BAC (corresponding angles), ∠LMN is thus equal to 180 – ∠LMC – ∠AMN = 180 – ∠ACB – ∠BAC = ∠CBA. By same logic, ∠LNM is equal to ∠ABC and ∠MLN is equal to ∠BAC, so triangles LMN and ACB are similar (AAA).

c Triangles AMN, ABC are similar.N is the midpoint of AB, the ratio between the sides of the similar triangle is 2, so and MN = 1

2BC.

Similarly, X is the midpoint of NM, the ratio between sides of the two similar triangles LMN and XYZ is 2, NX = 12

MN. MN = 2NX so by substitution, 2NX = 12

BC, 4NX = BC. The ratio between the sides of XYZ : ABC is 4 : 1.




213

Problem solving (pages 359–361)1 a DB = BC (given)

BA = BE (sides of equilateral triangle)∠DAB = ∠BEC = 90 (given)Triangles DAB and CEB congruent by RHS.

b Since, triangles DAB and BEC are congruent (RHS), ∠ABD = ∠CBE, so ∠ABD + ∠ABE = ∠DBE = ∠CBE + ∠ABE = ∠CBA, hence result.

c Yes, providing AB = BE.2 a ∠BAC = ∠CFG (alternate angles), ∠ACB = ∠GCF (opposite angles), so the third angles are the same and

equiangular triangles are similar.b 5 : 3

3 a ∠OBP = ∠DBA; ∠OPB = ∠DAB (corresponding angles); ∠BOP = ∠BDA (corresponding angles). Triangles are similar (AAA).

b Any of: AOB and DOC; AOD and BOC; AOP and OQC; OPB and OQB; ABC and ADC; ABD and BDC; ABD and OPB and OQB; ABC and AOP and OQC.

c Triangles OPB and DBA are similar; ∠OBP = ∠DBA and ∠OPB = ∠DAB (corresponding angles) (as proved above). The sides are in ratio 1 : 2 as line OB = 1

2DB is given. Therefore line PB = 1

2 line AB, thus P is the

midpoint of AB.d Triangles BOQ and BDC are similar, so OQ = 1

2DC = PB. Triangles ABC and AOP are similar, so OP = 1

2BC =

BQ. Hence POQB is a parallelogram (opposite sides equal), and line PQ bisects OB. Triangles BPQ and ABC are similar as angle PBQ = angle ABC and ∠BPQ = ∠BAC (corresponding angles on parallel lines). The side PB is in 1 : 2 ratio with AB, so the side PQ = 1

2AC.

4 a SQ = SU. Angle SUQ is 90 degrees due to alternate angles. Hence QU = QR. Therefore triangles are congruent.b U is the point where ST cuts QR, therefore angle TUR = 90° (corresponding angles) which means ST is bisecting

QR, U is the midpoint of QR. QR : UR = 2 : 1.c Angle TUR = 90° (corresponding angles). Angle TRU = PRQ as it’s the shared angle. TUR and PQR are similar

(AA).d TUR and PRQ are similar. QR = 2UR, so the triangles are in ratio 1 : 2. PR and TU are corresponding sides of

similar triangles and are parallel. Therefore PR = 2TU.5 a In triangles DEA and DFC:

∠DEA = ∠DFC (corresponding angles on parallel lines) ∠EDA = ∠FDC (common) So, triangles DEA and DFC are similar (similar triangles have the same angles).

b In triangles BME and CMF:∠BME = ∠CMF (vertically opp angles)∠BEM = ∠DEA (vertically opp angles) = ∠DFC (corresponding angles)Side BM = MC (given)Therefore triangles are congruent (ASA).

c Triangles DEA and DFC are similar. So ratio of EA to CF is the same as DR to DF. Or, EACF

= DEDF

.Since, triangles BME and CMF are congruent, CF = EB so, EA

EB = DE

DF.




214

6 a ∠APB = ∠QPR (shared). Side QR = 2AP and side PR = 2PB. Triangles are therefore similar by same angle and same ratio of sides.

b Shared angle, ∠PQB = ∠PQX, side QX = 2QB (given), and side PQ = 2QA (given). Triangles QAB and QPX are similar. So AB is parallel to PX. So PX is parallel to QR.Similarly AB = 1

2QR and AB = 1

2PX, so PX = QR.

c Proven that one pair of sides of PXRQ is opposite and parallel. In order to prove it is a parallelogram, need both pairs of opposite sides to be parallel. Need to show that PQ is parallel to XR. We know PQ is parallel to XR, since ∠QXR = ∠PQB (alternate angles) and also ∠QXP = ∠XQR (alternate angles), so ∠PXR = ∠PQR and XR is parallel to PQ.So yes, with this information on the diagram we can prove that it is a parallelogram. But without the extra geometry, we don’t know.

Reviewing skills (page 361)1 a In triangles ABX, CDX:

AX = XC (AC is bisected at X)BX = XD (BD is bisected at X)AXB = CXD (vertically opposite angles)ABX, CDX are congruent (SAS)AB is parallel to CD (alternate angles equal)AB = CD (corresponding sides of congruent triangles)BAX = DCX (corresponding angles of congruent triangles)

b ABCD is a parallelogram: AB is parallel to CD (alternate angles equal), ∠DAX = ∠XCB (corresponding angles of congruent triangles) therefore AD is parallel to BC (alternate angles equal). Two pairs of parallel sides means parallelogram.



2

215

Unit 11 Answers

Practising skills (pages 366–368)1 a 104°, angles at centre of circle b 90°, angles in a semicircle c 84°, angles of same segment d 158°, angles at centre of circle e 97°, opposite angles of a cyclic quadrilateral f 63°, angles in a semicircle, angles in an isosceles g 14°, angles in a semicircle, angles in an isosceles h 85°, opposite angles of a cyclic quadrilateral i 170°, angles at centre of circle2 Angle CDO is 62° because base angles in an isosceles triangle are equal.

Angle COD is 56° because angles in a triangle sum to 180°.Angle AOB is 56° because vertically opposite angles are equal.

3 Angle ADB is 53° because the angle in a semicircle is a right-angle.Angle ACB is 53° because angles in the same segment are equal.

4 a 43°, angles of same segment b 103°, opposite angles of a cyclic quadrilateral5 a 52°, angles at centre of circle b 102°, opposite angles of a cyclic quadrilateral c 38°, angles in isosceles triangle with angle BCD6 60°7 112°8 35°

Developing fl uency (pages 368–370)1 Angle ABC = 54°, alternate segment to angle ACF.

Angle ACB = 180 – angle CAB – angle ABC = 54°. Therefore triangle ABC is isosceles, with AC = AB.

2 Angle DAC = angle DBC = 26° (same segment). Angle ACD = 180° – angle AED = 64° (opposite angles cyclic quadrilateral). Angle ADC = 180° – angle ACD – Angle DAC = 180° – 26° – 64° = 90°. Therefore triangle ADC is a triangle of the semicircle, and AC is a diameter.

3 Angle DBC = 12

angle DOC = 50°, so angle ABC = 95°.

Opposite angles of cyclic quadrilateral: angle ADC = 85°.Angle ODC = 40° (isosceles), so angle ADB = 40°, and angle DAB = 95°.Angle DCB = 85° (either opposite angles of a cyclic quadrilateral, or 4 angles of a quadrilateral). Since angle ADC = angle DCB and angle DAB = angle ABC, quadrilateral ABCD is a trapezium with parallel sides AB and DC.



Strand 2 Properties of shapes Unit 11 Circle theorems Band j

216

4 Angle DAC = Angle DBC = angle DEC = 2x.Angle AD = 180 – 2x – x = 180 – 3x. Angle DC = 180 – (180 − 3x) = 3x.Angle ACP = angle DEC (alternate angles parallel lines) = 2x, so angle CPB = 180 – 3x – 2x = 180 – 5x.Therefore angle DPC = 180 – angle CPB = 180 – (180 – 5x) = 5x.

5 Angle CAB = 60° (alternate segments). Angle ACB = 60° (isosceles triangle, which is actually an equilateral).Angle COB = 120° (angle at centre of circle). Angle OCB = 30° (isosceles triangle).Angle OCB = 30° = 0.5 × 60°, so OC bisects angle ACB

6 True. Opposite angles of cyclic quadrilateral add to 180°. In a parallelogram, opposite angles do not add to 180° (but adjacent angles do add to 180°). The only case of a cyclic quadrilateral where adjacent angles add to 180°, and opposite angles add to 180°, is a rectangle, with all angles 90°.

7 a Triangles OAM and OBM are congruent because OA = OB are both radii and AM = BM as M is the midpoint and OM is common to both triangles.i Angle OMA is equal to angle OMB as the triangles are congruent.ii Angle OMA is 90° because angles on a straight line add up to 180°.

8 a i Angle OPQ is the same as angle OQP because triangle POQ is isosceles as OP and OQ are both radii. ii

OQ

R

P

a

a

X

iii angle POQ = 180° – 2a b i Angle ORP = angle OPR as triangle OPR is isosceles with OP and OR radii.

ii

OQ

R

P

X

γ

γ

iii Angle POR = 180° − 2(angle ORP) = 180° − 2γ. (assuming that in bi, the angle ORP is one of the γ angles) c Angle QOR = angle POR – angle POQ = 180° – 2γ - (180° – 2a) = 2a – 2γ. d Angle QPR = angle OPQ – angle OPR = a – 2γ; therefore angle QOR = 2 × angle QPR.




217

Problem solving (pages 371–372)1 x = 45°2 Angle OCA = 90° tangent and radius

angle OAC = 34° angles of triangle OACangle EBA = 108° angles on straight lineangle AEB = 38°, angles of triangle AEBangle FEG = 38°, opposite anglesangle EFG = 90°, triangle in a semicircleangle FGE = 52°, angles of triangle FGEangle COG = 124°, angles on straight lineangle OCG = angle OGC = 28°, angles of isoscelesangle CGF = angle OGC + angle FGE = 80°

3 a angle ACB = 90° (triangle in a semi circle) = angle AXCangle CAX = angle CAB (same angle)angle ACX = 90° − angle CAX = 90° − angle CAB = angle ACBtriangle AXC is similar to ACB (AAA)

b ACAX = AB

AC, similar triangles ratio of sides

therefore, AC 2 = AB × AX4 Angle QPX = angle SRX (alternate angles)

angle QPX = angle QPRangle QPR = angle QSR (same segment)angle QSR = angle XSRangle QPR = angle XSRangle QPX = angle XSRangle SRX = angle XSRSo, triangle SXR is isosceles and SX = RX.

5 55°6 angle BAD + angle ADC = 180

angle BAD + angle BCD = 180So, angle ADC = angle BCD.

7 a When x = 90°, angle AOC = 0°, which is impossible. b i kite

ii arrow iii triangle

8 a 6.8 cmb 90°c 112°d 62°e 6.8 tan(34°) = 4.6 cm




218

Reviewing skills (page 373)1 74°2 a angle DCB = 90° triangle of a semi circle

angle CBD = 32°OB = OA isoscelesAngle AOB = 32°, so OA must be parallel with BC to have alternate angles rule.

b Angle CBA = 74° + 32° = 106° ≠ 90°. Therefore AB is not parallel to CD.



3

219

Unit 4 Answers

Practising skills (pages 376–377)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

220

Developing fl uency (pages 378–381)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




221

Problem solving (pages 379–381)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




222

Reviewing skills (page 381)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



3

223

Unit 5 Answers

Practising skills (pages 383–385)1 a z

z2 = x2 + y2

b mm2 = n2 + e2

c ff 2 = e2 + g2

d rr2 = p2 + q2

2 a x2 = 52 + 122

x2 = 25 + 144x2 = 169x = 169x = 13 cm

b y2 = 62 + 82

y2 = 36 + 64y2 = 100y = 10 cm

c z2 = 242 = 72

z2 = 576 + 49z2 = 625z = 25 cm

3 a 7.2 cmb 8.6 cmc 1.4 cm

4 a 3 cmb 3 cmc 3 cm

The hypotenuse is a square root of a square number and the sides are expressed as square roots of whole numbers.



Strand 3 Measuring shapes Unit 5 Pythagoras’ theorem Band g

224

5 a x2 + 42 = 52

x2 + 16 = 25x2 = 25 – 16x2 = 9x = 9x = 3 cm

b y2 + 152 = 172

y2 + 225 = 289y2 = 289 – 225y2 = 64y = 8 m

c z2 + 122 = 152

z2 + 144 = 225z2 = 225 – 144z2 = 81z = 9 cm

6 a 9.2 cmb 6.7 cmc 6.2 cm

7 a 5 cmb 6 cmc 1 mile

Developing fl uency (pages 385–387)1 Perimeter = 60 + 62.5 + 17.5 = 140 m

Area = 0.5 × 60 × 17.5 = 525 m2

2 B is a right-angled triangle because it obeys Pythagoras’ theorem: 7.22 + 9.62 = 122

3 4 m4 Diagonal = 3.9 m. Total length of timber = 19.2 m.5 6.4 km6 a CD = 5 cm

b 54 cmc 126 cm2

7 Perimeter = 20 mArea = 18 m2

8 120 cm2

9 a 7.1 cmb Each segment = 7.13 cm2

10 x = 6.5 cm and y = 10.4 cm



Strand 3 Measuring shapes Unit 5 Pythagoras’ theorem Band g

225

Problem solving (pages 387–389)1 £15102 £4683 26 km4 Norman is correct because it is 20 460 newtons5 Beth runs 190 m. Ali goes 57 + 76 + 57 = 190 m6 No, she needs 26.6 m. Difference in height between AD and BC is 7 m

Therefore DC is 7.14 m which means AC is 16.6 m. The total cable she needs is 26.6 m so she does not have enough cable.

7 192 mm2

8 x = 5.77 cm or 5.8 cm

Reviewing skills (page 389)1 a 8 cm

b 25 cmc 12 cm

2 a 10.2 cmb 14.3 mc 1.2 cm

3 perimeter = 669.2 marea = 19 200 m2

4 21 + 15.6 + 31.8 = 68.5


226

4


Unit 3 Answers

Practising skills (pages 396–399)1 a 60°, 60°, 60°

b 70°, 70°, 40°c 60°, 38°, 82°d 90°, 62°, 28°e 90°, 60°, 30°f 45°, 30°, 105°

2 a Bisectors meet on the third side, i.e. AC.b Yes, same, bisectors of two sides of a triangle meet on the third side.

A

A

A

A

B

B

B

B

C

C C

C

8 cm

6 cm

8 cm

5 cm

7 cm

7 cm

7 cm9 cm



Strand 4 Construction Unit 3 Constructions with a pair of compasses Band f

227

3 a The circle touches all three points of triangle.P

10 10

O

RQ 8

b i The circle touches all three points of triangle.

O

89

Q

P

R10

ii The circle touches all three points of triangle.R

QP

O88

8

iii The circle touches all three points of triangle.

Q R

P6 5

10

O

iv The circle touches all three points of triangle.

P

RQ

6

8

O

10




228

4 a The circle touches all three points of triangle.

10 ZY

X

88 C

b i The circle touches all three points of triangle.

X Y

Z

C

7 7

5

ii The circle touches all three points of triangle.

9

6 7C

iii The circle touches all three points of triangle.

Y Z10

75

X

C

iv The circle touches all three points of triangle.

Y Z

X

6 10

8

C

5 a BC = a cm, AD = b cm; So the area of the triangle ABC = 12

× a × b cm2 = x cm2

b i 27.8 cm2

ii 15.0 cm2

iii 16.2 cm2




229

Developing fl uency (pages 399–400)1 Kite of sides 8 cm and 5 cm

B

A C

D

T

2 a Rhombus of side 7 cmQ

RP

S

M

b square

Q

RP

S

4 cm

4 cm

4 cm M4 cm

3 a and b

4 cm 6 cm

N

M L8 cm

c 11.6 cm2




230

4 a and b

P O

NM 2 cm Q

4 cm 4 cm

50º 50º

c 25.8 cm2

5 a, b and c

H OB

N

G

M

A

Lw

y

Fvv KE

J

S

D OR

I

QC P

T

d 15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180°, 195°, 210°, 225°, 240°, 255°, 270°, 285°, 300°, 315°, 330°, 345°

6

30°30°

7

45°




231

Problem solving (pages 401–402)1 a i and ii

A

45º 45º

60º

60º

D

D¢

B

C

b ACBD is an arrowhead with and obtuse angle at ACB, ACBD� is a kite with a 90° angle at ACB.2 111 m2

3 a red triangleb green rectanglec blue rhombus




232

4 a Students’ own drawings.

b 6.93 cmc 41.5 cm2

5 a and bP

Q

10 kmS 08:30

9 km

R 08:00

c 9 kmd 18 km/h

Reviewing skills (page 402)1 a BAC 51°, ABC 34°, ACB 96°

b

7 cm

A B9 cm

5 cm

C

c 17.4 cm2




233

2 a

A C

B

M

64

7

b See diagram above.c See diagram above.d See diagram above.e 3.5 cm


234

4


Unit 4 Answers

Practising skills (pages 404–406)1 a

P

b circle2

3 To calculate the distances, build right triangles by drawing a line through P perpendicular to two sides of the rectangle, as shown.

A B

P

C

X

Z

Y

W

D

Distance WP = Distance AY = (AP)2 − (YP)2

Distance XP = Distance BY = (BP)2 − (YP)2

Distance YP = Distance BX = (BP)2 − (XP)2

Distance ZP = Distance CX = (CP)2 − (XP)2

4 a and b

2 cm 2 cm



Strand 4 Construction Unit 4 Loci Band g

235

5 a and bC

D

locus

2.5 cm

2.5 cm

6 a and b

32º

32º

6 cm

6 cmF G

locus

E

c Bisector of the angle EFG.7 a and b

H K5 cm

5 cm

IJ

locus

45º45º




236

8 a and b

A

C D

B

c No

Developing fl uency (pages 406–409)1

2 a and b

P Q

S

6 cm

6 cm

3 cm

locus

T

R




237

3 a XY is the locus of points 1 cm from AB, between 1.5 cm and 7.5 cm from AD and between 1.5 cm and 7.5 cm from BC and 5 cm from CD.

b s c i More than 4.2 cm from AB, more than 4.2 cm from CD, more than 6.3 cm from AD, and more than 6.3 cm

from BC.ii Between 3.3 cm and 4.2 cm from AB, between 3.3 cm and 4.2 cm from CD, between 5 cm and 6.3 cm from

AD, and between 5 cm and 6.3 cm from BC.4 a and b

X

Z

Y

3.5 cm

8 cm

6 cm

locus

3.5 cm

c

X

Z

Y

8 cm

7 cm

3 cm

3 cm

locus

d Y

Z

X

3 cm

3 cm

3.5 cm

3.5 cm

8 cmQ

4 cm




238

5 a Careful to show that when Billy moves to point A, his rope is restricted at A, making a new circle locus of radius 5 m. Equally, when Misty reaches point C, her rope is restricted, making a new circle locus of radius 1 m. Where they can graze is not a simple circle about points B and D.

b

5 m4 mMisty

3m

8m

Billy

A

D

B

C

Shed

8 m

1m

1 m

i red shadingii blue shadingiii green shading

6 a and b

L Mlocus

locusK N

c one pointd To find the point that is equal distance from three sides, you bisect two angles and the bisectors cross at that

point. You cannot bisect three angles and find a point that three bisectors cross in a rectangle.e square




239

7 a First, points that are more than x km from Hen Rock and more than y km from Rooster Rock. Second, more than z km from The Chicks and more than y km from Rooster Rock. Third, points an equal distance from L and M.

b Hen Rock red region: points 0.75 km from the point Hen Rock.Rooster Rock: points 1.25 km from the point Rooster Rock.The Chicks: points 0.5 km from the point The Chicks.

c Any answers such as: Sail due East until you reach an even distance between Hen Rock and Rooster Rock. Sail due south, maintaining equal distance from Hen Rock and Rooster Rock. At point that is an equal distance from The Chicks and Rooster Rock, sail South East. At a point 1.5 km due East of the Chicks, sail due South, maintaining equal distance from L and M.

Problem solving (pages 409–411)1 58.3 m

P Q

RS

X

2 a 16 milesb A

CB

TT

3 a

A B

D C

b 50.28 m2

c The locus of the house is the inside of the path.It misses the fourth side, AD.




240

4 a and b

X

W

Z

Z

c e.g. move the lights at W and at Z 10 m towards X and Z.5 (This is an approximate location for the tree.)

R

Patio

QP

S

6 a and b

P Q

RS

16 m

16 m 16 m

16 m

s r

qp

c 42 md p = 56 m2, q = 56 m2, r = 105 m2, s = 105 m2

e See diagram above.




241

Reviewing skills (page 411)1 a points equal distance from AB and AC

b points equal distance from P and Qc points 2 squares away from the line XYd points 5 squares away from point Q

2 a points within the square and closer to BC than ADb points closer to the line OL than the line OMc points within 2 squares of point Od points more than 3 squares but less than 5 squares away from line UV

3 a

b 5027 m2


242

5


Unit 3 Answers

Practising skills (page 415)1 a–e

FD

2 3 4 5

43

56789

10

210

y

x1 7 8 9 106

B A

CE

2 a 4 to right, 5 upb 6 to right, 1 downc 2 to left, 7 upd 6 to left, 3 downe 7 to rightf 7 upg 8 to lefth 5 up

3 a 82–

b 61

c −

43

d −−

74

e 09

f 60

g –100

h 012–



Strand 5 Transformations Unit 3 Translation Band d

243

Developing fl uency (pages 416–417)1 a–e

y

x

T

1

1 2 3 4 5 6 7 8 9 10

23456789

10

0

Z

WX

Y V

2 a i A translation of 2 units right and 2 units up.ii A translation of 2 units left and 4 units up.iii A translation of 1 unit left and 4 units down.iv A translation of 4 units right and 3 units down.v A translation of 4 units right and 4 units up.vi A translation of 5 units right and 1 unit up.vii A translation of 5 units left and 1 unit down.viii A translation of 7 units down.ix A translation of 6 units left and 7 units up.x A translation of 5 units left and 8 units down.xi A translation of 4 units left and 2 units up.xii A translation of 6 units right.

b i 22

ii 24

−

iii 14

−−

iv 43−

v 44

vi 51

vii 51

−−

viii 07−




244

ix 67

−

x 58

−−

xi 42

−

xii 60

3 a–e

y

x0–1–1

1234

M

56

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6 R

Q

N

O

P

Problem solving (pages 417–418)1 There are 4 translations

R to P 45

P to R −

45–

T to S 35

S to T −

35–




245

2 There are 4 translations

C to F 74

F to C −

74–

A to D −

44–

D to A 44

Reviewing skills (page 419)1 a Translation of 1 unit left and 3 units up.

b Translation of 5 units right and 3 units up.c Translation of 4 units down.d Translation of 4 units right and 2 units down.e Translation of 7 units right and 5 units down.f Translation of 5 units right and 7 units up.g Translation of 5 units right and 5 units down.h Translation of 2 units left and 8 units up.i Translation of 1 unit right and 5 units up.j Translation of 7 units left and 1 unit up.k Translation of 6 units left.l Translation of 8 units right and 8 units down.


246

5


Unit 4 Answers

Practising skills (pages 424–426)1 a–f

a

d e f

b c

Mirror line

Mirrorline

Mirror line

Mirror line Mirror line

Mirror line

2 a–ey

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

A

B

Q

C

RP

3 a x = 2b x = 3c x = 2d x = 5e x = −5f x = −3



Strand 5 Transformations Unit 4 Reflection Band e

247

4 a–by

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

T

V

W

U

y = 3

y = –1

x = 1

5 a–b

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

G K

IJ

H

x = –1 x = 3

y = 1

y = 4




248

Developing fl uency (pages 427–430)1 a–d

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

A

C

D

E

H

G

B

y = –x y = x

F

2 a reflection in x = 3b reflection in x = 3c reflection in x = −1d reflection in x = 2e reflection in x = –2f reflection in x = −4g reflection in x-axis (y = 0)h reflection in x = 4i reflection in y-axis (x = 0)

3 a–d

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

AD

C

G

H

E

F

y = –x y = x

B

12




249

4 a reflection in y-axisb reflection in y = 1c reflection in y = −1d reflection in y-axise reflection in y = xf reflection in y = 1g reflection in y = −xh reflection in y = 4i reflection in y = −11

2j reflection in y = 1k reflection in −31

2l reflection in y = −x

5 y

x0–1

1

2

3

4

5

6

1 2 3 4 5 6–2–3

A

B

Problem solving (pages 431–432)1 a y = 10

b y = x2 a and b 6 units in each case3 b = 44 Translation by

60

, followed by reflection in y = −1

5 a For example: any translation by +2 units parallel to the x-axis followed by a reflection in the dotted line.b For example: any translation by +2 units parallel to the y-axis followed by a reflection in the dotted line.




250

Reviewing skills (pages 432–433)1 a–d

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

TB

P D

C

A

2 a–e

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

F E

A

B C

D



5

251

Unit 5 Answers

Practising skills (pages 437–440)1 full-turn = 360°

three-quarters turn = 270°half-turn = 180°quarter-turn = 90°

2 a Db Ac B, F, Gd C, E, H

3 a–c

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

R

AC

B

4 a–c

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

TA

CB



Strand 5 Transformations Unit 5 Rotation Band e

252

5 a–fy

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

M

F

C

E

D

A

B

Developing fl uency (pages 441–444)1 a–e

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

T

E

B

AC

D




253

2 a–e

y

x0–1–1

123456

–2–3–4–5–6

1 2 3 4 5 6–2–3–4–5–6

PA

E

C

B

D

3 a rotation 90° clockwise about (1, 3)b rotation 90° anticlockwise about (1, 3)c rotation 180° about (0, 0)d rotation 180° about (0, 0)e rotation 180° about (0, 2)f rotation 90° anticlockwise about (−2, −2)g rotation 90° clockwise about (−4, 2)h rotation 180° about (3, −1)i rotation 90° anticlockwise about (−2, 5)

4 a reflection in y = −2b rotation 90° clockwise about (1, 1)

c translation 62

d rotation 180° about (−1, 1)e reflection in the x-axis

f translation 17−

g reflection in y-axish rotation 90° anticlockwise about (−5, −4)i rotation 90° clockwise about (−5, −4)j reflection in the line x = −1

k translation −

68

l reflection in the line y = x




254

Problem solving (pages 445–447)1 a translation by

40

b Because the shape has been rotated by 360°.

c translation by 80

2 a translation by 22−

b translation3 No. For example with T then R, the point (1, 0) is mapped to (5, 2), then (−2, 5). With R then T, the point (1, 0) is

mapped to (0, 1), then (4, 3).

4 a

10

y

x2 3 4

+ +

5 6 7 8 9 10 11 12

b translation by 400

5 translation by 960

6 30 seconds7 12 seconds




255

Reviewing skills (page 447)1 a–e

y

x0–1–1

123456

–2–3–4–5–6–7

1 2 3 4 5 6–2–3–4–5–6

C

B

E

A

D

M

f translation −

22

g translation 08−


256

5


Unit 6 Answers

Practising skills (pages 450–451)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

257

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




258

Developing fl uency (pages 452–455)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




259

4 a Scale factor f = 2, centre (−2, 1)b Scale factor g = 1.5, centre (4, 7) c 2 × 1.5 = 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




260

Problem solving (pages 455–457)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)




261


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)


262

5


Unit 7 Answers

Practising skills (pages 460–463)1 A and P; B and Q; C neither; D and Q; E and P; F and P; G and P; H neither; I and Q; J and P.2 a y = 7.5

b z = 9.6c x = 3d similar

3 a y = 12b z = 16c x = 5d similare Both are Pythagoras triangles.

4 a 2b x = 10c w = 8d The same.e They are similar triangles.

5 a Similar; size ratio of both sides is the same, factor 3.b Not similar; the size ratio of the sides is different 5 × 2 =10 but 2 × 2 does not = 5.c Similar; regular pentagons are similar, factor 1.5.d Not similar; side 4 × 3 = side 12 but the scale factor is not 3 for other sides.e Not similar; scale factor different for each side.f Not similar; angles different.g Similar as angles are the same.

6 a Not similar as the given angle is not the same between the two shapes.b Similar – the given angle is the same, and the sides scale with factor 2.c Not similar – the given angle is the same, but the sides scale with different factors.

Developing fl uency (pages 463–465)1 B and C2 3.75 m3 a 121.5 cm2

b 337.5 cm2

4 a 12b AEC, FBD; AVZ, FYU, BTX, YUC, XET, DVZc 24 5 cm2 or 53.7 cm2

5 a 24 cmb 48 cmc A = 24 cm2 and B = 48 cm2



Strand 5 Transformations Unit 7 Similarity Band h

263

6 a P and A and B and C are congruent.b P and D are similar.

Problem solving (pages 465–467)1 a Using the sum of the angles in a triangle is 180°, the triangles have equal angles and so are similar.

b 32 48 cm3 a Angle C is common, Angle CDE = angle CAB (= 90°) and angle CED = angle CBA (from angle sum of a

triangle = 180°)b 1.4 m

4 63 cm5 3.2 m6 394 cm


b 12.5c 7.2

2 a Unknown angle in T is 40; unknown angle in V is 30. Three equal angles therefore V and T are similar.b 4.5 cmc 5 cm

3 3.36 m


264

5


Unit 8 Answers

Practising skills (pages 471–473)1 a i hypotenuse = a ii opposite = b iii adjacent = c b i h = f

ii o = eiii a = d

c i h = hii o = iiii a = g

d i h = jii o = liii a = k

e i h = oii o = niii a = m

f i h = pii o = riii a = q

2 a sin θ = 35 is true

b cos θ = 45 is true

c tan θ = 34 is true

3 a tan α = 512 is trueb none are truec none are true



Strand 5 Transformations Unit 8 Trigonometry Band h

265

4 i and ii

A

e

A

A

A

H

7 mA

OH

H

H

H

O

O

O

O

10 m

12 m

18 m

15 m x m

30º

48º

43º

20º

50º

x m

x m

x m

x

a

c d

b

iii and iva sin 30 = 10

x ; 10 sin 30 = x; x = 5 m

b tan 50 = 7x; 7 tan 50 = x; x = 8.3 m

c tan 48 = 12x ; 12 tan 48 = x; x = 13.3 m

d cos 20 = 15x ; 15 cos 20 = x; x = 14.1 m

e cos 43 = 18x ; 18 cos 43 = x; x = 13.2 m

5 a 30°b 45°c 60°d 40°e 72°f 82°g 53°h 48°i 61°




266

6 i and ii

H

a

cd

e

f

b

H

H

H

H

H

A

O

O

O O

O

O

A

A

A

A

4 cm

10 cm

7 cm

7.2 m

11.4 m

6.6 m13.2 m

9 m

6 m

4 cm

5 cm

A

θ

θ

θ

θ

θ

θ

3 cm

iii and iva tan θ = 45; θ = 38.7°

b tan θ = 34; θ = 36.9°

c cos θ = 710; θ = 45.6°

d cos θ = 69; θ = 48.2°

e cos θ = 7.211.4; θ = 50.8°

f sin θ = 6.613.2; θ = 30°




267

7 i and ii:

H

a b

cd

e

f

H

H H

HH

OO

O

O

AA

OO

AA30º

70º 68º

61º

33º

50º

8 cm

12 m

9 cm

7.2 m

14.5 m

A

A

3 mx cm

x m

x m

x cm

x

x

iii and iv

a sin 30 = 8x; 8sin 30 = x; x = 16.0 m

b tan 50 = 3x ; 3tan 50 = x; x = 2.5 m

c sin 70 = 12x ; 12sin 70 = x; x = 12.8 m

d cos 68 = 9x ; 9cos 68 = x; x = 24.0 m

e tan 33 = 14.5x ; 14.5tan 33 = x; x = 22.3 m

f sin 61 = 7.2x ; 7.2sin 61 = x; x = 8.2 m

Developing fl uency (pages 473–475)1 a 11.0 m (1dp)

b 44.0 m2 (1dp)c 31.4 m (1dp)

2 perimeter = 53.5 m (1dp) area = 109.8 m2 (1dp)

3 (assuming the diagonal is included) 27.3 m4 a 15.5 m (1dp)

b 32.3°




268

5 5.8 tan(52) = 7.4 miles

PortAura

PortCambria

Buoy218o

52o5.8 miles

Opposite

Adjacent

Hypotenuse

6 a 158.5 m (1dp)b 133.6 m further

7 a i 71.1°ii 12 cm

b 12 cm by x2 + 352 = 372

c yes

8 Lighthouse

Lightship

Yacht2.4 km

1.5 kmHypotenuse Adjacent

Opposite

x

a 2.4 1.52 2+ = 2.8 km

d tan(x) = 2.41.5 gives x as 58°. Therefore the heading from the lightship to the lighthouse is 360 − 58 = 302°.

9 2349.7 m10 a Pythagoras – x2 + 22 = 72

b

2

M

OV

√45

θ

x

c 72.7°d 6.4 cm




269

Problem solving (pages 476–477)1 0.350 m2 3.14 m3 a 68.8 cm

b 79.7 cm4 a 19.8 m

b 20.2 m5 a 33.7°

b e.g. the number of steps cancels and you are left with 2030

6 16.16 m

Reviewing skills (pages 477–478)1 a x = 4 cm

i the 8 cm side is the hypotenuseii the side x is the oppositeiii the third side is the adjacent

b x = 19.3 cmi the unlabelled side is the hypotenuseii the side x is the oppositeiii the 9 cm side is the adjacent

c x = 12.3 mi the 20 m side is the hypotenuseii the unlabelled side is the oppositeiii the side x is the adjacent



6Unit 3 Answers

Practising skills (pages 482–483)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

271

Developing fl uency (pages 484–486)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




272

Problem solving (pages 486–488)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




273

Reviewing skills (page 489)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



6Unit 4 Answers

Practising skills (pages 491–492)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 2D representations of 3D shapes Band f

275

h iiiiii

2 a iiiiii

b iiiiii

c i ii iii

d i ii iii




276

Developing fl uency (pages 492–494)1 a i ii iii

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

2 a

b

c

d




277

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




278

Problem solving (pages 494–496)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




279

4 a i Triangular prismii Square-based pyramid

b

c




280

Reviewing skills (page 497)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



6

281

Unit 5 Answers

Practising skills (pages 500–501)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

282

Developing fl uency (pages 502–503)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




283

Problem solving (pages 504–505)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




284

Reviewing skills (page 506)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



6

285

Unit 6 Answers

Practising skills (pages 509–510)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

286

Developing fl uency (pages 510–511)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




287

Problem solving (pages 512–513)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




288

Reviewing skills (page 514)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)



6

289

Unit 7 Answers

Practising skills (pages 517–518)1 a = iv b = vi c = v d = i e = iii f = ii2 a b c d

3 a b

c d

4 Shape Front elevation Side elevation Plan

A 1 2 4

B 9 8 5

C 3 6 7

5 a

front frontside

plan plan

side

b



Strand 6 Three-dimensional shapes Unit 7 Constructing plans and elevations Band g

290

Developing fl uency (pages 519–520)1 a b

2 a b

front frontside

plan plan

side

3 a b

front frontside side

plan

plan

4

front

plan

side




291

5

6 a

b i volume = 12 cm3

ii surface area = 36 cm2

7 a

b




292

c Several possible answers:

Problem solving (page 521)1 a triangular prism

b 24 cm3

c

d 84 cm2




293

2 a

b

3 a Plan

Side elevation Front elevation

b




294

Reviewing skills (page 522)1

2

3

front

plan

side



1

295

Unit 4 Answers

Practising skills (pages 525–527)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

296

4 a Speed (mph), v Frequency, f Midpoint, m m × f0 � v � 20 13 10 130

20 � v � 30 22 25 550

30 � v � 40 4 35 140

40 � v � 60 1 50 50Totals 40 120 870

Mean speed = 87040

= 21.75 mph

b 30 mphc 40%

5 a 41045

= 9 hours

b 1.5 hours




297


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 × f0 � t � 20 1 10 10

20 � t � 30 8 25 200

30 � t � 40 14 35 490

40 � t � 50 7 45 315Totals 30 115 1015

Mean time = 101530

= 33.8 minutes

b range = less than 50 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 × f

0 � d � 2 6 1 6

2 � d � 3 12 2.5 30

3 � d � 4 9 3.5 31.5

4 � d � 5 8 4.5 36Total 35 11.5 103.5

Mean distance jumped = 103.535

= 2.96 metres

b (14 × 3) – 35 = 7 foul jumpsc At most, 8 jumpers qualified.

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




298

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.




299

Problem solving (pages 529–532)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.




300

Reviewing skills (page 532)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



1

301

Unit 5 Answers

Practising skills (pages 536–537)1 a i 27 ii 34, 20 iii 14

b 45

34

27

20

14

Maximum value

Thirdquartile

Median

Firstquartile

Minimumvalue

c Any sensible suggestion, e.g. exam scores.2 a i 14.1

ii 14.8, 13.5iii 1.3iv 3

b Any sensible suggestion, e.g. ages of players in an under 15 football team.3 a 42

40

38

36

34

32

30

28

26

24

22

20

18

16

14

12

10

8

6

b i 42 ii 21.5 iii 31.5, 14.5 iv 17 v 35



Strand 1 Statistical measures Unit 5 Inter-quartile range Band g

302

4 a 47, 56, 60 b

60

50403020Fr

eque

ncy

100

0 10 20 30Height, h

40 50 60

c i 27 ii 23 d 21 m5 a

303540

25201510

Freq

uenc

y

50

0 10 20 30Time, min

40 50 60

b i 27 ii 19–20 c It has increased on average by about 5 minutes and the inter-quartile range has increased so the times are more

spread out. d Some people are living further from work/the traffic has got worse.




303

Developing fl uency (pages 538–540)1 a 100 b median = 72 mph; lower quartile = 65 mph; upper quartile = 79 mph c 79 and 65 mph d 58% e Travelling at 55 or less – 8; travelling at 90 or more – 10.2 a 26 for both, A: 11, B: 7

b Between 1 and 12.c They have the same median, so similar average/total rainfall but Town A has more very wet weeks.

3 a 1 4 12 26 43 50 54b 60

50403020Fr

eque

ncy

10

010.5 10.6 10.7 10.8

Times, s10.9 11 11.1 11.2

c 10.95, 0.15d 17–18

4 a Boys 6 21 40 67 102 123 130; Girls 9 21 49 86 110 118 120

120140

1008060402002.6 2.8 3 3.2 3.4 3.6 3.8 4

Boys Girls

y

x

b Boys 3.38–3.40 and 0.45; Girls 3.25–3.27 and 0.37c Boys are heavier on average by about 0.14 kg and their weights are slightly more spread out.d No, but you can estimate that it is more likely to be a boy.




304

5 a 6050403020

Cum

ulat

ive

freq

uenc

y

100

8 10 12 14Time, t min

16 18 20

b median = 13.8-14 ; IQR = 3.7c 12.2 minutesd The journey back is about 2 minutes quicker on average and has similar spread. They were rowing upstream in

the first part.6 a True. The definition of the quartiles is that half the data lie between them.

b False. The range can be very different to the IQR and is not related to the IQR.c False. Box and whisker divides data into one box of two parts and two whiskers which represent the range.d True. Definition of cumulative is that there is always addition of the frequency.e True, if by midway you mean midway between the frequency and not midway between the variable.

7 a 150 ≤ t < 180 b time, t frequency dogs cumulative frequency dogs frequency owners cumulative frequency owners

90 ≤ t < 120 1 1 0 0

120 ≤ t < 150 5 6 0 0

150 ≤ t < 180 27 33 2 2

180 ≤ t < 210 12 45 5 7

210 ≤ t < 360 3 48 9 16

360 ≤ t < 480 0 48 14 30

c

30354045

25201510C

umul

ativ

e fr

eque

ncy

5090 140 190 240

Time, t s290 340 390

50

440 490

d median = 170 s; IQR = 25 (22 − 27) s e dogs – 147 s; owners – 197 s f Dogs – most of the dogs fall around the median of 170, with 45 of the 48 dogs below 210 s. Owners are more

concentrated to the high times, 23 of the 30 owners are above 210 s. The IQR is very large for the owners.




305

Problem solving (pages 541–544)1 a 100 b 20 c The upper quartile or the 75 percentile is a mark of 15 so James is correct.2 a 3.35 cm b i wing span 2.0 < w ≤ 2.5 2.5 < w ≤ 3.0 3.0 < w ≤ 3.5 3.5 < w ≤ 4.0 4.0 < w ≤ 4.5 4.5 < w ≤ 5.0

frequency 8 18 20 18 14 2

ii 3.36 cm3 a 80 b i median = 13.5 s; upper Q = 14 s; lower Q = 12.9 s ii median = 15.45 s; upper Q = 16.1 s; lower Q = 14.7 s

c man – 11.5 s; woman – 13 sd 29e 12

4 Male range 6500–4500 = 2000, median = 5300, inter-quartile range = 950 Female range = 5000−3200 = 1800, median = 4000, inter-quartile range = 600 The female distribution is negatively skewed. The male distribution is positively skewed. The females’ weight is more consistent as both the range and the inter-quartile range are smaller. The males on average weigh more than the females as shown by the median.5 Both distributions are positively skewed but more prominent to the right in the group that had drugs.

Recovery was slightly less variable in the group that had drugs than in the other group. In the drug group, recovery took 2 to 14 days (range = 12) versus 5 to 18 days (range = 13) for the other group. The inter-quartile range for the drug group was 7 days and for the other group 9 days. So in general the drug group had a more consistent recovery time. The median recovery time was 6 days for the drug group and 13 days for the other group. It appears that the drug had a positive effect on patient recovery.

6 a

20

0 100 200

30

10

40

60

8090

70

50

400300

100

Cum

ulat

ive

freq

uenc

y

Amount spent, A (£)

b i £220 ii £115 c The median amount spent was lower at Christmas as families tended to go out less, however the range is greater

suggesting a couple of families didn’t go out at all or else had one or two expensive outings. The inter-quartile range was lower at Christmas than at Easter.




306

7

100 20 30 40 50 60 70 80 90 100Time (min)

Slowest time 15 minutes

Fastest time 78 minutes

Range 63 minutes

Median 42 minutes

Upper quartile 54 minutes

Lower quartile 32 minutes

Inter-quartile range. 22 minutes


25

0 10 20

50

75

100

4030

125

Num

ber

of e

mpl

oyee

s

Time after 8.30 a.m., t (min)

b 8.47 c i 13 min ii 10 min

d 8.38 a.m. e 5 employees



2

307

Unit 3 Answers

Practising skills (pages 550–551)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 3 Pie charts Band f

308

5 a

Coffee

Tea

Milk

Cola

Orange

b Student’s own opinion and reasoning




309

Developing fl uency (pages 551–552)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.




310

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)




311

Problem solving (pages 553–554)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%; Jones – 7%; Williams – 47%; Roberts – 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




312


8b 225°c

Art club

Football club

Neither

d 140



2

313

Unit 4 Answers

Practising skills (pages 558–559)1 a i discrete

ii continuousiii discreteiv discrete

b i discreteii continuousiii continuousiv continuous

2 A: 150 in two classes. B: 150 not in any class. C: unclear – 150 likely to be in two classes. D: 150 not in any class. 140 � height � 150, 150 � height � 160, etc.

3 a Frequencies are 3, 6, 9, 7, 2.b 60 � w � 70c

0

40 �

w � 50

50 �

w � 60

60 �

w � 70

70 �

w � 80

Weight (kg)

Frequency diagram showing the weight ofTrevor’s sample of people

80 �

w � 90

2

4

Freq

uenc

y

6

8

10

d Chart as it is summarised in order.e

Mass (kg)50400

01234

Fre

quen

cy

56789

10

60 70 80 90

4 a Frequencies are 1, 6, 5, 4, 4, 1.b 30 � n � 40c 7d There are generally between 30 and 70 people in the gym on any day.



Strand 2 Draw and interpret statistical diagrams Unit 4 Displaying grouped data Band f

314

Developing fl uency (pages 560–561)1 a Reaction time, t Frequency

0.2 � t � 0.3 6

0.3 � t � 0.4 5

0.4 � t � 0.5 5

0.5 � t � 0.6 7

0.6 � t � 0.7 4

0.7 � t � 0.8 2

0.8 � t � 0.9 1

b

0

.2 �

t � .3

.3 �

t � .4

.4 �

t � .5

.5 �

t � .6

Reaction time (s)

Frequency polygon showing reaction times of a group of people

.6 �

t � .7

1

Freq

uenc

y

8

234567

.7 �

t � .8

.8 �

t � .9

c 20%d Fairly level until 0.6, with a modal class at 0.5 � t � 0.6, then it falls away.




315

2 a Score, n Frequency

20 � n � 30 1

30 � n � 40 5

40 � n � 50 7

50 � n � 60 5

60 � n � 70 7

70 � n � 80 4

80 � n � 90 6

90 � n � 100 1

b

0

20 �

n � 30

30 �

n � 40

40 �

n � 50

50 �

n � 60

Score

Frequency diagram showing Henry’s computer games scores

60 �

n � 70

1

Freq

uenc

y

8

234567

70 �

n � 80

80 �

n � 90

90 �

n � 10

0

c Table helps to narrow the median down to between 50 and 70 but the original data gives the actual scores so you can work out the median as (58 60)

2+ = 59.

d Fairly even between 30 and 90 with only one score above and one below that. Grouped data.3 a height, h Frequency

140 � h � 150 3

150 � h � 160 10

160 � h � 170 6

170 � h � 180 5

180 � h � 190 5

190 � h � 200 1

b

0

140 �

h � 15

0

150 �

h � 16

0

160 �

h � 17

0

170 �

h � 18

0

Length (mm)

Frequency polygon to show the length of the tails of rats

180 �

h � 19

0

2

Freq

uenc

y

12

4

6

8

10

190 �

h � 20

0

c 11d 150 � h � 160




316

4 a 21b Age, a Frequency

0 � a � 10 4

10 � a � 20 10

20 � a � 30 15

30 � a � 40 19

40 � a � 50 16

50 � a � 60 5

60 � a � 70 2

c Lots of people of working age.d More retired people and less of working age.

5 a Time, t (seconds) Boys Girls

0 � t � 10 3 0

10 � t � 20 2 2

20 � t � 30 5 3

30 � t � 40 4 4

40 � t � 50 2 4

50 � t � 60 1 2

b Boys 20 � t � 30. Using the original data, the median is 25. Girls 30 � t � 40. The median is 35.c

0

Time (s)

Frequency diagram showing how long boys held their breath for

1

Freq

uenc

y

6

2

3

4

5

50 �

t � 60

40 �

t � 50

30 �

t � 40

20 �

t � 30

10 �

t � 20

0 � t �

10

0

Time (s)

Frequency diagram showing how long girls held their breath for

1

Freq

uenc

y

2

3

4

5

50 �

t � 60

40 �

t � 50

30 �

t � 40

20 �

t � 30

10 �

t � 20

0 � t �

10

d Girls – they have a higher average.




317

6 a Discrete – a hair is a hair.b Time is continuous, but discrete values of age are usually easier to work with.c Discrete – you cannot have half a person.

Problem solving (pages 562–563)1 a 150 (Accept: there might be one area of the field which was particularly densely populated. Reject: 150 is so

much higher than all the other values that it is likely to be an error.)b Number of poppies, p Frequency

0 � p � 10 0

10 � p � 20 3

20 � p � 30 1

30 � p � 40 2

40 � p � 50 4

50 � p � 60 5

60 � p � 70 6

70 � p � 80 5

80 � p � 90 3

90 � p � 100 1

c

0

1

Freq

uenc

y

2

3

4

5

6

7

8

Number of poppies0 �

p � 10

10 �

p � 20

20 �

p � 30

30 �

p � 40

40 �

p � 50

50 �

p � 60

60 �

p � 70

70 �

p � 80

80 �

p � 90

90 �

p � 10

0

d The mode is 15 and median is 61. The mode is more representative. Modal class is 60–70 which is also representative of the data.

e 50%




318

2 a

0

2

Freq

uenc

y

4

6

8

10

12

Distance (metres)

0 � m �

10

10 �

m � 20

20 �

m � 30

30 �

m � 40

40 �

m � 50

b Most of them can only swim less than 20 m and few of them can swim further. It is skewed towards the lower end (positively skewed).

c 55%d The better swimmers have slightly increased the distance they can swim underwater; the weaker swimmers have

performed much the same as before.3 a Female chicks:

0123456789

101112131415161718192021222324

Mass (g)

Freq

uenc

y

25 �

m � 30

30 �

m � 35

35 �

m � 40

40 �

m � 45

45 �

m � 50

50 �

m � 55

55 �

m � 60

60 �

m � 65




319

Male chicks:

02468

1012141618202224262830

Mass (g)

Freq

uenc

y

25 �

m � 30

30 �

m � 35

35 �

m � 40

40 �

m � 45

45 �

m � 50

50 �

m � 55

55 �

m � 60

60 �

m � 65

b Diagrams show that male chicks tend to be heavier than female chicks. Modal class for female chicks is 40 � p � 45 and for male chicks is 45 � p � 50.

c i 20%ii 82.8%

4 a Time (t minutes) Frequency

0 � t � 50 5

50 � t � 100 11

100 � t � 150 5

150 � t � 200 5

200 � t � 250 1

250 � t � 300 0

300 � t � 350 0

350 � t � 400 2

400 � t � 450 0

450 � t � 500 5

500 � t � 550 6




320

b

0

1

2

3

4

5

6

7

8

9

10

11

12

Time (t)

Freq

uenc

y0

� t

� 5

050

� t

� 1

0010

0 �

t �

150

150

� t

� 2

0020

0 �

t �

250

250

� t

� 3

0030

0 �

t �

350

350

� t

� 4

0040

0 �

t �

450

450

� t

� 5

0050

0 �

t <

550

c The distribution is bimodal with two distinct groups. Most people don’t spend a lot of time on computer games but there is a group who spend a lot of time on them.

d In August the students are on holiday, so the people in the group that spend less time playing games might spend more time outdoors, but the people who are very keen on playing games will probably play for even longer. The graph will probably be a similar shape, but with lower bars on the left and higher bars on the right.

5 a Prys has used the computer every day, Anna hasn’t. However, when Anna does use the computer she generally spends a long time on it.

b They don’t really need a second computer as the mean time they each spend on the computer is 50 minutes, so there should be plenty of time to use the computer in the evening.

Reviewing skills (page 564)1 a 0.5 kg � n � 1.0 kg

b 1.5 kg � n � 2.0 kg c Weight, n (kg) Frequency

0 � n � 0.5 3

0.5 � n � 1.0 5

1.0 � n � 1.5 7

1.5 � n � 2.0 10

2.0 � n � 2.5 9

2.5 � n � 3.0 6

3.0 � n � 3.5 4

3.5 � n � 4.0 4

4.0 � n � 4.5 1

4.5 � n 5.0 1

d 50



2

321

Unit 5 Answers

Practising skills (pages 566–568)1 a no correlation

b negative correlationc positive correlationd no correlation

2 a negative correlationb

00

1020304050607080

10090

1 2 3 4 5 6

Load carried (kg)

Dis

tanc

e w

alke

d (k

m)

7 8 9 10

c 45 kmd 2 kg

3 a 11b several possible answers, for example:

Mas

s of

car

(kg

)

Length of car (m)

Valu

e of

car

(£)

Age of car (years) Mileage(1000s of miles)

Petr

ol t

ank

capa

city

(litr

es)

c i positive correlation ii negative correlationiii no linear correlation

d i A long low car like a sports car that is light.ii A vintage car that is valuable.



Strand 2 Draw and interpret statistical diagrams Unit 5 Scatter diagrams Band f

322

4 a

0012345678

109

1 2 3 4 5 6

Judge 1 scores

Judges’ marks at a flower showJu

dge

2 sc

ores

7 8 9 10

b positive correlation; both judges give similar scores for both candidates.c See diagram above.d 5

5 a

05

10152025303540455055606570758085

20 25 30 35 40 45 50 55 60 65 70 75 80 85 90Assessment 1

Ass

essm

ent

2

b 45




323


00

102030405060708090

10 20 30 40 50 60

Oral

Comparison of marks in awritten and oral test

Wri

tten

70 80

b positive correlationc 4d See diagram above.

e i 57ii 45

2 a

00

1020304050607080

10 20 30 40 50 60Goals

Goals football teams scored vspoints won

Poin

ts

70 80 90

b positive correlationc See diagram above.

d i 43ii 64




324

3 a

00

20406080

100120140160

10 20 30 40 50 60Police officers

Number of police officers vscrime figures in an area

Cri

mes

70 80 90

180200

b 72 police and 177 crimes.c Negative correlation – the more police officers the less crime.d See diagram above.e 105

4 a

160

150150 160

Mother’s height (cm)

Dau

ghte

r’s h

eigh

t (c

m)

170 180

170

180

b Positive correlation; taller mothers have taller daughters. c i The median is the 8th data point you come to, when you look from left to right. This data point has 7 data

points to the left and 7 to the right. The median height of the group of mothers is 163 cm.ii No, her daughter is 161 cm, and the median height of the daughters is 162 cm.

5 a

1200

510152025

125130135140145150Height

Goa

ls

155160165170

Football players’ heights vsnumber of goals scored

b Lorraine, David and Youssu, because they haven’t scored any goals.c No; no correlation




325

Problem solving (pages 570–572)1 a

251.51.61.71.81.92.02.12.22.3

2.52.4

26 27 28 29 30 31

Time to run 100 m (s)

Hei

ght

(m)

32 33 34 35

b Slight negative correlationYes, it appears the faster they are at running, the higher they can jump.

c 18.15 sd 2 m. No, it is none of these as there are 16 data items so the middle one is between the 8th and the 9th.e 6

2 Neither statement is valid because if you treat the top value (320, 60) as an outlier then there is negative correlation in the scatter diagram, but Llinos’ statement implies positive correlation and Elfed’s statement implies none.

3 a There is low negative correction correlation, so there is some evidence to support the statement that there is a relationship between the number of hours spent watching television and the exam mark.

b i

Num

ber

of h

ours

spen

t re

visi

ng

Exam

mar

k

Number of hoursspent revising

Number of hoursspent watching TV

ii Yes. As number of hours spent watching TV increases, so exam mark decreases and low exam mark corresponds with fewer hours spent revising. So students who watch more TV spend less time revising.




326

4 a

056789

10111213

1514

1 2

Engine size (litres)

Dis

tanc

e (k

m)

3

b The graph shows negative correlation, meaning that the larger the engine size, the less distance it can cover on 1 l of diesel.

c See diagram above.d 11.7 km

5 a

0012345678

109

1 2 3 4 5 6

Judge 1

Judg

e 2

7 8 9 10

b Yes – when Judge 1 awards a high mark, then so does Judge 2.c See diagram above.d 4e This could be the case as the scatter diagram suggests the mark should be more and be near the line of best fit, so

perhaps 5 or 6. However, Judge 1 could have over marked the dancer, so either Judge could be inconsistent with their marks for the other contestants.




327


00

1020304050607080

10 20 30 40 50 60Age

Age of swimmers vs howmany lengths they can do

Leng

ths

70

b Negative correlation – the older they are, the fewer lengths they can do.c See diagram above.

d i 35ii 40


328

2


Unit 6 Answers

Practising skills (page 575)1 a Least/most fuel consumption: 13.4

31.5. Least/most power: 145

470.

b and c See diagram.

00

–5

5101520253035

100 200 300 400 500 600Power (horsepower)

Power vs Fuel consumption

MPG

700 800

d i The more power a car has, the worse the fuel economy is.ii About 420 horsepower.iii About 13 miles per gallon.

e 3 mpg. The best fit correlates increasing horsepower with lower fuel consumption.

Developing fl uency (pages 576–577)1 D is best – it seems to have half of the points on either side.

B is next best – it seems to split the data fairly evenly, but they’ve forced it through the origin. C is second worst – it doesn’t split the data evenly. A is the worst – they’ve just joined two points and it does not have half the data on either side.



Strand 2 Draw and interpret statistical diagrams Unit 6 Using lines of best fi t Band g

329

2 a 1500 vs 100: negative correlation – the faster you are at 100 m, the slower you are at 1500; 1500 vs jump: positive correlation – the further you can jump, the longer it takes to do 1500 m; 100 vs jump: negative correlation – the longer it takes to run 100 m, the shorter distance jumped.

10.6250

270

290

310

330

10.8 11 11.2 11.4 11.6 11.8

100m time (s)

100m time vs 1500m time

1500

m t

ime

(s)

12

350

2504.0

4.5

5.0

5.5

6.0

6.5

270 290 310 330 350

1500m time (s)

1500m time vs jump distance

Jum

p di

stan

ce (

m)

10.54.0

4.5

5.0

5.5

6.0

6.5

10.7 10.9 11.1 11.3 11.5 11.7 11.9 12.1

100m time (s)

Jump distance vs 100m time

Jum

p di

stan

ce (

m)

b People who tend to run the 100 m faster are better at long jump, and 1500 m runners aren’t very good at 100 m or long jump.

c Increase the sample size – test more athletes. d i The line of best fit suggests this but it is extrapolating from the data and so the trend may not continue; there

could be a physical limit.ii This goes against the correlation of the data and so is untrue.

3 a It shows a positive correlation.b Correlation doesn’t prove causation so both may be wrong. If either is right, it is more likely to be that doing

well at maths makes you good at science, since science is dependent on maths in a way that maths is not dependent on science.



Strand 2 Draw and interpret statistical diagrams Unit 6 Using lines of best fit Band g

330


156080

100120140160180200

20 25 30 35Temperature (ºC)

Num

ber

of s

wim

mer

s

b positive correlationc (34, 144) It might have been too hot for some people to go to the beach. Or, it might be a week day when many

people are at work/school.d See diagram above.e 93. Not very accurate as it is beyond the range of the data. It is likely that very few people will swim once the

temperature drops below a certain level.f The graph shows a correlation which is not the same as causality. Both the number of swimmers and the ice

creams sold depend on a 3rd variable – the temperature.2 a

00

25

20

15

10

5

3020 5040 70 9060 80Age

Hou

rs o

f exe

rcis

e

b There is low negative correlation. Older people tend to exercise less. Middle-aged people may be too busy.c The 45 year old who exercises for 15 hours a week is an outlier.d See diagram above.e About 35 years old, but the estimate should be treated with caution as the correlation is fairly weak.f 0 hours. This is unreliable as it is beyond the range of the data, a healthy 72 year old is probably still exercising.




331

3 a

300

100200300400500600700

40 50 60 70 80 90

Wingspan (m)

Max

land

ing

mas

s (t

onne

s)

b There is evidence to suggest there is positive correlation because when the values in one set increases the values in the other set increases.

c See diagram above.d Approximately 250 tonnes.e No, as it is well outside the range of values plotted. You should not extrapolate this far outside the range of the

values.



Strand 2 Draw and interpret statistical diagrams Unit 6 Using lines of best fit Band g

332

4 a

00

25

20

15

10

5

200100 400300 600 800 900 1000500 700

Mass (g)

Leng

th (

cm)

b and c

00

25

20

15

10

5

200100 400300 600 800 900 1000500 700

Mass (g)

Leng

th (c

m)

d 11.5 cme The estimate would be unreliable as 1 kg is beyond the range of the data. The spring may well break when a mass

of 1 kg is hung from it.f 4.2 cm




333

Reviewing skills (page 579)1 a The graph shows negative correlation.

00

30

25

20

15

10

5

800 1000 1200200 400 600 1400Height (m)

A graph to show the temperature at locationswith different heights above sea level

Tem

pera

ture

(ºC

)

b The 27 °C at 580 m could be an outlier. The weather could have been particularly warm there, or it could have been recorded in a location which is generally hotter than the other areas.

c 19 °Cd −150 me The answer to part c is probably quite reliable since it is interpolated within the data range. The answer to part

d is less reliable since it is extrapolating significantly from our data range. Furthermore the answer seems quite improbable which gives us reason to doubt.



3

334

Unit 2 Answers

Practising skills (pages 583–584)1 a ‘Exercise’ needs to be defined. Some people will interpret it as playing sport or visiting the gym, others might

include walking the dog as exercise.b It needs to be more specific, maybe asking for the number of hours per week spent exercising.

2 Gareth should ask a number of children what their mathematics test scores were and how often they eat fish in a week. By drawing a graph of amount of fish eaten per week against maths test scores he could see if children who eat fish at least twice a week tend to perform better on tests than those who don’t.

3 MALE FEMALE

Time spent, x hours Mon Tue Wed Thu Fri Weekend

Mon Tue Wed Thu Fri Weekend

0 � x � 0.5

0.5 � x � 1

1 � x � 2

2 � x � 3

x � 3

4 She could measure how much water comes out of the tap over a certain amount of time, say 5 seconds. By doing this a number of times she would be able to see whether the same amount of water came out in that time period or not.

5 All people who work in the factory.6 Tick the box that best describes your opinion of burgers.

I think that burgers are…Good for you Neither good nor bad for you Bad for you It depends on the quality of the burger Don’t know

7 a He has not included the under 16s or over 70s.b He has not specified whether he means per week, per month or per year. The categories all overlap at the ends

(£50, £100, £300, £500 and £1000 are all in two categories).

Developing fl uency (pages 584–585)1 A possible answer:

What is the maximum amount you would pay for a round of sandwiches in the high street to take away?Less than £1£1–£1.49£1.50–£1.99£2–£2.49£2.50–£2.99£3.00–£3.49£3.50 or more



Strand 3 Probability Unit 2 Single event probability Band e

335

2 Gill could look at a random sample of music tracks and work out the mean price. By repeating this experiment a number of times she could see whether the mean price is always 69p or not.

3 A possible answer:How many music DVDs did you buy last month?012345 or more

4 Possible criticisms include:The survey is biased as the timing excludes people at work.Question 1 has an error that someone who uses the library twice a month could select the second box or the third box.Question 2 is biased as it asks if people agree, which encourages a ‘Yes’ answer.Question 2 could include a ‘Don’t know’ option or a better range of answers such as very well stocked / quite well stocked / adequately stocked / not very well stocked / very poorly stocked / don’t know.

5 He could test his hypothesis by taking a sample of prisoners, half of which have been taught computer skills and half of which have not and then tracking whether they reoffend after being released. If the numbers of reoffenders are similar then the computer skills make little or no difference. If the number of reoffenders who were taught computer skills is less than those who were not then the prison manager’s hypothesis may be correct.

6 She could ask people who do not go to a gym.She could take the sample only early in the morning.She could ask people who only live in a nearby street, and so on.

7 Height, h cm

x � 140 140 � x � 150 150 � x � 160 160 � x � 170 180 � x � 190 x � 190

Wai

st, w

cm

w � 80

80 � w � 90

90 � w � 100

100 � w � 110

w � 110 170 � x ��180

8 Lily should time how long it takes for sugar to dissolve in different temperatures of water and see if there is a trend of the time taken for the sugar to dissolve decreasing as the temperature increases.

Problem solving (pages 585–586)1 a Yes it is, because sunlight is essential for plants to grow and for the fruit to ripen.

b This could be tested by picking tomatoes at various intervals from a start point and then timing how long it takes them to ripen.

2 a Yes it is, because stress and anxiety around exams can have a negative effect on the body so students may be more likely to get ill.

b Look at the student attendance rates throughout the year and see if they decrease around exam periods.3 Ask a selection of people, half of whom use vitamin C tablets and half of whom do not, how often they get colds. If

those who take the tablets and get colds are fewer than those who don’t take the tablets and get colds then this may indicate that the tablets make a difference.

4 a Has no space for the under 10s or over 30s.b Has overlapping categories; someone who plays 5 or 10 hours a week has a choice of two answers.c Is biased as it assumes that the games should be banned rather than asking if there should be any restriction.




336

5 Possible questions include:Do tall people weigh more than shorter people?Do women exercise more than men?Are men taller than women?

6 a Martin carried out the survey outside the football ground after a game, so nearly everyone will be a football fan. So they will either support Northfield City or the opposition.

b Question 1 has two categories for 18 and 25 year olds.Question 2 does not include possibilities such as ‘Once a fortnight’.Question 3 uses the word ‘support’ which is very vague.

c

1 How old are you?Under 10 years old 10–18 19–25 26–40 Over 40

2 How often do you attend football matches?Never 1–2 games per season 3–6 games per season 7–12 games per season More than 12 games per season

3 Tick all that apply. Do you:Check Northfield’s progress in the league table? Read reports of Northfield’s matches? Buy a Northfield City shirt? Buy other Northfield merchandise? (Please give details) …………………….

7 He should sample the whole population as he needs to provide sandwiches for all the possible needs if he takes on the entire club. If he takes a sample he may miss out some needs, e.g. vegetarians, gluten free, and so on.

8 They should take a small sample of engines and run them until they break down. They cannot test them all as they will not then have any new engines.

9 Company

A B C D E

Distance Fare Distance Fare Distance Fare Distance Fare Distance Fare

Jour

ney

1

2

3

4

5

6

7

8

9

10




337

Reviewing skills (page 587)1 Number of each format bought in the last month

Age Download (single tracks)

Download (albums)

CD albums Vinyl albums

Under 10

10–16

17–25

26–35

36–50

Over 50

2 a It is carried out outside a supermarket so everyone surveyed will use a supermarket.b It excludes people under 21.

30 year olds fit in two age categories.c Some people will use both. For example they might use a supermarket for a major shop at weekends, but top up

from a local shop during the week.



4Unit 2 Answers


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




339

4 a 112

b 112

c 16

d 16

e 0

f 112

5 a 19

b 19

c 19

d 0




340


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.




341

Problem solving (pages 594–595)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 Luc ate 4 toffees and 0 fudge or Luc 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




342

Reviewing skills (page 595)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



4

343

Unit 3 Answers

Practising skills (pages 599–600)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

344

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




345


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




346

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




347

Problem solving (pages 602–605)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




348

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




349

Reviewing skills (page 605)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



4Unit 4 Answers

Practising skills (pages 608–610)1 a 216591 = 72197

b 315 Daily Posts, 142 Gazettes, 365 of The News, 178 Tribunes.2 a i 67

385

ii 215385

= 4377

iii 103385

b Yes, the probability the fault being to do with the power supply is over half.c 870 motherboards, 2792 power supplies and 1338 hard drives.

3 a 0.54, 0.46, 0.5275, 0.486, 0.491, 0.5032.b Yes, because the larger the sample, the closer the relative frequency gets to 0.5, which is what you expect from

an unbiased coin.

4 a i 84189 = 49

ii 72189 = 821

iii 33189 = 1163 b i 3492

ii It probably isn’t that reliable as the sample size was a very small fraction of the population.

5 a Lorry: 1530543572

≈ 0.35; Van: 241143572

≈ 0.055; Bus: 106443572

≈ 0.025; Car: 24 79243572

≈ 0.57.b 5c It may be accurate due to the large sample size used to estimate the probabilities from. However, the sample data

came from just one junction rather than from across the whole city so it might be misrepresentative of the rest of the city.

6 a Total number of throws

10 20 30 40 50 60 70 80 90 100

Total number of sixes thrown

4 9 10 12 15 19 22 26 29 32

Relative frequency of throwing a six

410

920

1030

1240

1550

1960

2270

2680

2990

32100

0.4 0.45 0.33 0.3 0.3 0.32 0.31 0.33 0.32 0.32

b

Rel

ativ

e fr

eque

ncy

of t

hrow

ing

a si

x

Total number of throws

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 10 20 30 40 50 60 70 80 90 100



Strand 4 Probability Unit 4 Estimating probability Band f

351

c The probability of throwing a six seems to settle around 0.32 so the probability of not throwing a six is 1 – 0.32 = 0.68.

d With a fair six-sided die each number has a one in six chance of coming up. The probability of Wyn throwing a six is nearly double that, so it seems like his die is not fair.


40 = 310

ii 1840

= 920

iii 1040

= 14

b 4 red, 5 blue and 3 green. She can make it more reliable by picking out more balls (increasing her sample size).

2 a i 4287

ii 2987

iii 1687

b Yes, we can estimate that there would be 413 students who want to change it.

3 a i 2240 = 1120

ii 1840

= 920

b 29 males, 23 females

4 a i 12

ii 1420

iii 620

b Different probabilities.c Take more samples.

5 a None of them should be right. Sara might be the closest but if it’s a fair die you would expect to roll a six one time in six, rather than one in three.

b i We can’t tell for definite from only 20 throws since the sample size is too small.ii It doesn’t seem to be; 15 times in 100 is nearly the same as 1 in 6.

c Increase the sample size.6 No. Each time he tosses the coin it has a 50 : 50 chance of being a head.7 a i 3

35ii 430

= 215

iii 757

b i Len’s.ii It’s a sentence which uses every letter in the alphabet.

c Take the average over more sentences.d 12. There are in fact 12 Es in a standard Scrabble set.

8 Ask 20% of 3000 = 600 supporters.

20% = 20100 = 15, meaning to ask 1 in every 5 supporters, so asking every 5th supporter.

(Check 600 × 5 = 3000 supporters)9 For example, record the weather data on the 5th of each month.



Strand 4 Probability Unit 4 Estimating probability Band f

352

Problem solving (pages 613–616)1 a i 240

275 = 4855

ii 9275

iii 259275

b 5452 a 0.3, 0.35, 0.15

b The one using the sample size of 100 because of the larger sample size.c 300

3 a about 0.35b 658

4 a i 60ii 38

b 90c 0.6

5 For every 8 tickets they sell they make £1.60 but are likely to lose £1.00 from that, meaning every 8 tickets makes on average £0.60. Hence to make £120 they would need to sell 1600 tickets.

6 a 0.4 as the three results with the largest sample size are clustered around the 0.4 mark.b She would make 48 × £0.08 = £3.84.c She would make 9 × £3.50 = £31.50 before refunds. Assuming from part a that 40%, i.e. 360, of the parsnips are

oversized that means that she will refund 360 × £0.03 = £10.80. Therefore she will make 31.50 − 10.80 = £20.70.7 a i 0.025

ii 0.35iii 0.125

b i 38ii 25

c Too high – Mr Thomas’ prediction is based on his sample of year 9s who have developed more physically than year 7s, so the year 7s probably won’t do quite as well.

Reviewing skills (page 616)1 a i 0.1

ii 0.075iii 0.025

b 3c It probably won’t be that accurate because people might know that it is busier and so turn up earlier than at the

smaller, quieter surgery.



4

353

Unit 5 Answers


b 16

c 16

d 16

e 162 a

16

56

56

16

16

6

Not6

6

Not6

6

Not6

b 136

c 2536

d 1136

e 1036

3 a 452

b 451

c 450

d 349

4 a 112

b 512

c 112

d 512



Strand 4 Probability Unit 5 The multiplication rule Band h

354

5 a

712

512

411

711

511

611

Red Red

Red Blue

Blue Red

Blue Blue

712

611

42132× =

712

511

35132× =

512

411

20132× =

512

711

35132× =

b

712

512

512

712

512

712

Red Red

Red Blue

RedBlue

BlueBlue

712

712 × =

49144

712

512

35144× =

512

512

25144× =

512

712

35144× =

6 a 49b There are 3 red and 5 yellow counters left in the bag.

c i 27

ii 57




355


1329

1629

1528

1328

1628

1228

Boy

Girl

Boy

Girl

Boy

Girl

i 240812

ii 156812

iii 572812

iv 416812

2 a independentb independentc dependentd independent

3 a i 18

ii 16

b i 18

ii 16

c 18

d independent4 a

0.6

0.4

0.5

0.5

0.25

0.75Catch

Drop

Catch

Drop

Catch

Drop

b i 0.2ii 0.35iii 0.45




356

5 a 1 – 0.05 = 0.95b 1 – 0.4 = 0.6c 0.4 × 0.05 = 0.02d 1 – (0.95 × 0.6) = 0.43

6 It is not impossible – each time James tossed the coin it had a 12

chance of being a head, it just never landed on heads.


720

1320

1320

720

1320

720

Red

Blue

Red

Blue

Red

Blue

i 169400

ii 49400

iii 182400

iv 351400

2 a 415

b 215

c 13

3 a 114

× 113

= 1182

b 148

4 a Yes, because 0.3 × 0.8 = 0.24 and 625

= 24100

= 0.24.b 0.62

5 a 31220

or 0.14

b Yes. It is expected that Naela’s train will be late 30% of the time when it’s raining, but it might not have been late at all!




357

6 a

710

310

29

79

39

69

Works

Faulty

Works

Faulty

Works

Faulty

38

Works

Faulty

28

68

Works

Faulty

28

68

Works

Faulty

18

78

Works

Faulty

b i 1120

ii 740

iii 2140

iv 724

c There is a 119120

chance that at least one bulb will work and 70% chance that at least two will.

7 a 715

b 8 – assuming he doesn’t put the socks back into the drawer, if he pulled out 6 blue socks in a row then all that are left are red socks.

8 a First attempt Second attempt Third attempt Fourth attempt Fifth attempt Probability

Red

BlueRed

BlueRed

BlueRed

BlueRed

Blue

1515151515

1

15

14

34

13

122

312

45

b Each of the outcomes is just as likely as any other. The total of the probabilities tells us that there are no other options in outcomes, since they add to 1.

c Instead of the probabilities changing as counters are removed they will stay the same at 15 for a red and 4

5 for a

blue.

d 21013125

≈ 0.67




358


21 ≈ 0.095

b 921

≈ 0.43

c 1021

≈ 0.475

d 1921

≈ 0.905

2 a

0.2

0.8

0.9

0.1

0.85

0.15Rain

Fine

Rain

Fine

Rain

Fine

b 1 − (0.2 × 0.15) = 0.97



4

359

Unit 6 Answers


15b 215

c 13

2 a 160

b 7300

c 125

3 a 57600

= 19200

b 325

c 129600

= 43200

d 471600

= 157200

4 a 13

b 13

c 23

5 a

Football:21

None:8

10 Rugby:11

b 1150



Strand 4 Probability Unit 6 The addition rule Band f

360

6 a

11 7

Own a cat Own a dog None

84

b i 1230 = 25

ii 430

= 215

iii 2330

iv 730

7 a

Multiple of 3Multiple of 2

1

5

7

11

13

15

3

9

612

10

814

2

4

b i 715

ii 13

iii 215

iv 23

v 13


4b 34

c 34

d 852

= 213

e 4452

= 1113

2 The events sun and rain are not mutually exclusive.




361

3 a Wears glasses total: 7. Does not wear glasses total: 22. b i 12

29ii 1529

iii 1329

iv 1029

v 0

4 a 23

b 13

5 a

Odd number4 10 14 8 12

Factor of 18

15

75

1917

1311

13

9

16

20

26

18

b i 12

ii 520

iii 320

iv 920

v 820

6 a

20

44

12 4

W B

All dogs (80)

W = dogs with weak hipsB = dogs with bad eyesight

b Estimated probability = 2080 = 14.It is an estimate because this is based on a sample not on all the dogs of the breed.

c i In the general formula the sets are A and B. Here they are W and B.

p(W) = 3280, p(B) = 2480, p(W and B) = 2080 p(W or B) = p(W) + p(B) – p(W and B)

= 32802480

2080

3680

+ + + = 0.45

ii Using the Venn diagram: the number of dogs with one or more defect is 12 + 20 + 4 = 36. So the probability is 3680

= 0.45. You get the same answer both ways of course.




362


Medicine X:36 Placebo: 24

Medicine Y:25

15

b P(neither medicine) = 24100

2 a

Swimmingpool: 70 Neither: 1

Gym: 1910

b 19100

c 89100

3 a

Bags:65%

Neither:10%

Loyaltycard: 10%

15%

b i 0.1ii 0.65iii 0.9iv 0.1

4 a 512

b 34




363

5 a and b Correct diagram:

Chemistry: 8 Biology: 31

5314

Physics: 6

c 2380

d Neither; only half of the students do one or more science so the probability of choosing one who takes science is 12

.

Reviewing skills (page 634)1 a i 1

2ii 12

iii 56

b The events ‘even’ and ‘3 or less’ are not mutually exclusive as ‘2’ is both even and less than 3.2 a Correct diagram:

History:8

Neither:2

Geography:15

5

b 2330


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