AQA GCSE Maths 209732 AQA Higher Teacher Pack title page.indd 1 3/27/15 2:37 PM Higher Student Book Answers Higher Student Book Answers Browse the complete Collins catalogue at www.collins.co.uk William Collins’ dream of knowledge for all began with the publication of his first book in 1819. A self-educated mill worker, he not only enriched millions of lives, but also founded a flourishing publishing house. Today, staying true to this spirit, Collins books are packed with inspiration, innovation and practical expertise. They place you at the centre of a world of possibility and give you exactly what you need to explore it. Collins. Freedom to teach Published by Collins An imprint of HarperCollinsPublishers News Building 1 London Bridge Street London SE1 9GF  HarperCollinsPublishers Limited 2015 10 9 8 7 6 5 4 3 2 1 ISBN 978-0-00-814702-0 A Catalogue record for this publication is available from the British Library Commissioned by Lucy Rowland and Katie Sergeant Project managed by Elektra Media and Hart McLeod Ltd Project edited by Jennifer Yong Answers checked by Amanda Dickson Illustrations by Ann Paganuzzi Cover design by We Are Laura Cover photographs by Procy/Shutterstock (top) and joingate/Shutterstock (bottom) Production by Rachel Weaver All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the Publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd., 90 Tottenham Court Road, London W1T 4LP. Acknowledgements The publishers gratefully acknowledge the permissions granted to reproduce copyright material in this book. Every effort has been made to contact the holders of copyright material, but if any have been inadvertently overlooked, the publisher will be pleased to make the necessary arrangements at the first opportunity. AQA GCSE Maths (4th Edition) 1 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers Chapter 1 – Number: Basic number Exercise 1A 1 a 6000 b 5 cans cost £1.95, so 6 cans cost £1.95. 32 = 5 × 6 + 2. Cost is £10.53. 2 a 288 b 16 3 a 38 b Coach price for adults = £8, coach price for juniors = £4, money for coaches raised by tickets = £12 400, cost of coaches = £12 160, profit = £240 4 (18.81...) Kirsty can buy 18 models. 5 £8.40 per year, 70p per copy 6 £450 7 15 8 3 weeks 9 £250.74 10 Gavin pays 2926.25 – 1840 = £1086.25 11 a Col is correct b Abi has multiplied 30 × 50 as 150 instead of 1500. Baz has lined up the columns wrongly when adding. Instead of lining up the units he has lined up the first digits. Des has forgotten to add a zero on the second line of the multiplication, it should be 1530. Exercise 1B 1 a 4.6 b 0.08 c 45.716 d 94.85 e 602.1 f 671.76 g 7.1 h 6.904 i 13.78 j 0.1 k 4.002 l 60.0 2 a 0.028 b 0.09 c 50.96 d 46.512 3 a 35, 35.04, 0.04 b 16, 18.24, 2.2 c 60, 59.67, 0.33 d 140, 140.58, 0.58 4 a 18 b 140 c 1.4 d 12 e 6.9 5 a 280 b 12 c 240 d 450 e 0.62 6 a 572 b i 5.72 ii 1.43 iii 22.88 7 a Incorrect as should end in the digit 2 b Incorrect since 9 × 5 = 45, so answer must be less than 45 8 300 9 a 27 b i 27 ii 0.027 iii 0.27 10 Mark bought a DVD, some jeans and a pen. 11 Headline A does not give the exact figure so does not convey any useful information. Headline B is accurate and records should be given accurately. Headline C may be correct but without the previous record does not convey any useful information. Exercise 1C 1 a 50 000 b 90 000 c 30 000 d 200 e 0.5 f 0.006 g 0.3 h 10 i 0.05 j 1000 2 a 56 000 b 80 000 c 31 000 d 1.7 e 0.066 f 0.46 g 4.1 h 8.0 i 1.0 j 0.80 3 a 60 000 b 5300 c 89.7 d 110 e 9 f 1.1 g 0.3 h 0.7 4 a 65, 74 b 95, 149 c 950, 1499 5 Elsecar 750, 849; Hoyland 1150, 1249; Barnsley 164 500, 165 499 6 18 to 23 inclusive 7 1, because there could be 450 then 449 8 Donte has rounded to 2 significant figures or nearest 10 000 9 a Advantage – quick. Disadvantage – assumes 3 penguins a square metre which may not be accurate b Advantage. Quite accurate as 5 by 5 is a big enough area to give a reliable estimate. Disadvantage – takes a long time. Exercise 1D 1 a 60 000 b 120 000 c 10 000 d 15 e 140 f 100 g 200 h 0.08 i 0.09 j 45 2 a 5 b 25 c 3000 d 600 e 2000 f 5000 g 400 h 8000 i 4 000 000 3 30 × 90 000 = 2 700 000 600 × 8000 = 4 800 000 5000 × 4000 = 20 000 000 200 000 × 700 = 140 000 000 4 a 54 400 b 16 000 5 1400 million 6 His answer is correct but he had one too many zeros on each value, which cancel each other out. Matt wrote 600,000 rather than 60,000 and 2000 rather than 200. The two mistakes cancelled themselves out due to the zeros involved. 7 a Value of the money is about 66 000 000 × 0.2 = £13 200 000, so it is enough to buy the yacht. b Weight is 66 000 000 × 5 = 330 000 000 grams = 330 tonnes, so they do not weigh as much as the yacht. 8 1420 000 000 000 ÷ 64 000 000 ≈ 22 000, so the National Debt per person is approximately £22 000. AQA GCSE Maths (4th Edition) 2 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers Exercise 1E 1 a 35 000 b 15 000 c 960 d 5 e 1200 f 500 2 a 39 700 b 17 000 c 933 d 4.44 e 1130 f 550 3 a 1.74 m b 6 minutes c 240 g d 83°C e 35 000 people f 15 miles g 14 m2 4 a 10 b 1 c 3 5 a 8.79 b 1.03 c 3.07 6 82°F, 5 km, 110 min, 43 000 people, 6.2 seconds, 67th, 1788, 15 practice walks, 5 seconds The answers will depend on the approximations made. Your answers should be to the same order as these. 7 a £15 000 b £18 000 c £17 500 8 $1000 9 a 40 miles per hour b 10 gallons c £70 10 a 80 000 b 2000 c 1000 d 30 000 e 5000 f 2500 g 75 h 100 11 a 86 900 b 1760 c 1030 d 29 100 e 3960 f 2440 g 84.8 h 163 12 Approximately 500 13 £1 million pounds is 20 million 5p coins. 20 000 000 × 4.2 = 84 000 000 grams = 84 tonnes, so 5 lorries needed. 14 22.5° C – 18.2° C = 4.3 Celsius degrees 15 a i 27.571 428 57 ii 27.6 b i 16.896 516 39 ii 16.9 c i 18 672.586 16 ii 18 700 16 a 37.5 × 48.6 ≈ 40 × 50 = 2000 21.7 ×103.6 ≈ 20 × 100 = 2000 985 ÷ 0.54 ≈ 1000 ÷ 0.5 = 2000 b as both values are rounded down the actual answer must be bigger than 2000. The other two must be less than 2000. c Pete is correct it is not possible to tell. 37.5 × 48.6 = 1822.5 985 ÷ 0.54 = 1824.074 17 149 000 000 ÷ 300 000 = 496.67 ≈ 500 seconds 18 a 58.9 × 4.8 ≈ 60 × 5 = 300 b Lower as both values are rounded up to get the estimate. 19 Macau’s population density is approximately 710 000 times the population density of Greenland. 20 26.8 ÷ 3.1 ≈ 27 ÷ 3 = 9 36.2 ÷ 3.9 ≈ 36 ÷ 4 = 9. Second calculation must be biggest as first is smaller than 27 ÷ 3 and second is bigger than 36 ÷ 4. Exercise 1F 1 a 12 b 9 c 6 d 13 e 15 f 14 g 16 h 10 i 18 j 17 k 8 /16 l 21 2 4 packs of sausages and 5 packs of buns (or multiples of these) 3 30 seconds 4 12 minutes; Debbie will have run 4 laps; Fred will have run 3 laps. 5 1 + 3 + 5 + 7 + 9 = 25, 1 + 3 + 5 + 7 + 9 + 11 = 36, 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 6 a −2 b −7 c −12 d −1 e −30 7 a 1 b 3 c 4 d 2 e −4 8 a 400 b 900 c 2500 d 0.25 e 16 9 a Student’s own explanation b 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105 c Adding consecutive pairs gives you square numbers. 10 Square number Factor of 56 Cube number 64 8 Multiple of 7 49 28 11 2, 3 and 12 12 a 1, 64, 729, 4096, 15 625 b 1, 8, 27, 64, 125 c 3   a a a d Square numbers 13 a 0.2 b 0.5 c 0.6 d 0.9 e 1.5 f 2.1 g 0.8 h 0.7 14 The answers will depend on the approximations made. Your answers should be to the same order as these. a 60 b 1500 c 180 Exercise 1G 1 a 84 = 2 × 2 × 3 × 7 b 100 = 2 × 2 × 5 × 5 c 180 = 2 × 2 × 3 × 3 × 5 d 220 = 2 × 2 × 5 × 11 e 280 = 2 × 2 × 2 × 5 × 7 f 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 g 50 = 2 × 5 × 5 2 a 84 = 22 × 3 × 7 b 100 = 22 × 52 c 180 = 22 × 32 × 5 d 220 = 22 × 5 × 11 e 280 = 23 × 5 × 7 f 128 = 27 g 50 = 2 × 52 3 1, 2, 3, 22, 5, 2 × 3, 7, 23, 32, 2 × 5, 11, 22 × 3, 13, 2 × 7, 3 × 5, 24, 17, 2 × 32, 19, 22 × 5, 3 × 7, 2 × 11, 23, 23 × 3, 52, 2 × 13, 33, 22 × 7, 29, 2 × 3 × 5, 31, 25, 3 × 11, 2 × 17, 5 × 7, 22 × 32, 37, 2 × 19, 3 × 13, 23 × 5, 41, 2 × 3 × 7, 43, 22 × 11, 32 × 5, 2 × 23, 47, 24 × 3, 72, 2 × 52 4 a 2 is always the only prime factor b 64, 128 c 81, 243, 729 d 256, 1024, 4096 e 3, 32, 33, 34, 35, 36; 4, 42, 43, 44, 45, 46 AQA GCSE Maths (4th Edition) 3 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers 5 a 2 × 2 × 3 × 5 b 22 × 3 × 5 c 120 = 23 × 3 × 5, 240 = 24 × 3 × 5, 480 = 25 × 3 × 5 6 a 72 × 112 × 132 b 73 × 113 × 133 c 710 × 1110 × 1310 7 Because 3 is not a factor of 40 so it does not divide exactly. 8 a = 2, b = 7 9 a 2ab a2 4b b 8a3b 4a3b2 Exercise 1H 1 a 20 b 56 c 6 d 28 e 10 f 15 g 24 h 30 2 They are the two numbers multiplied together. 3 a 8 b 18 c 12 d 30 4 No. The numbers have a common factor. Multiplying them together would mean using this factor twice, thus increasing the size of the common multiple. It would not be the least common multiple. 5 a 168 b 105 c 84 d 84 e 96 f 54 g 75 h 144 6 3 packs of cheese slices and 4 packs of bread rolls 7 a 8 b 7 c 4 d 16 e 14 f 9 8 a ii and iii b iii 9 18 and 24 10 a 6x2y2 b xy Exercise 1I 1 a 7 b –8 c –5 d – 11 e 11 f 6 g 8 h 8 I –2 j –1 k –9 l –5 m 5 n –9 o 8 p 0 2 a –15 b –14 c –24 d 6 e 14 f 2 g –2 h –8 i –4 j 3 k –24 l –10 m –18 n 16 o 36 3 a −9 b 16 c −3 d −32 e 18 f 18 g 6 h −4 i 20 j 16 k 8 l −48 m 13 n −13 o −8 4 a −2 b 30 c 15 d −27 e −7 5 a –9 b 3 c 1 6 a 16 b −2 c −12 7 −1 × 12, 1 × −12, −2 × 6, 2 × −6, −3 × 4, 3 × −4, 8 Any appropriate divisions 9 a −24 b 24 degrees c 3 × −6 10 13 × −6, −15 × 4, −72 ÷ 4, −56 ÷ −8 11 a 32°F and 212°F b –40°C = –40°F 12 –460°F Exercise 1J 1 a –4 b −6 c 4 d 45 e 6 f 6 2 a 38 b 24 c −3 d –6 e −1 f 2 g −25 h 25 i 0 j −20 k 4 l 0 3 a (3 × −4) + 1 = −11 b −6 ÷ (−2 + 1) = 6 c (−6 ÷ −2) + 1 = 4 d 4 + (−4 ÷ 4) = 3 e (4 + −4) ÷ 4 = 0 f (16 − −4) ÷ 2 = 10 4 a 49 b −1 c −5 d −12 5 a 40 b 1 c 78 d 4 6 Possible answer: 3 × −4 ÷ 2 7 Possible answer: (2 − 4) × (7 − 3) 8 (–4)2 = –4 × –4 = +16, –(4)2 = – (4 × 4) = –16 9 (5 + 6) − (7 ÷ 8) × 9 10 –6 Review questions 1 10 weeks 2 16 3 270 4 a 32 × 5 × 7 b 63 5 a 11.412 712 21 b 11.4 6 a 412.603252 b 400.5 7 a iii Prime numbers less than 20 b i 252 ii 3780 iii 18 8 a 10.663 418 78 b 11 9 1200 10 5 11 a 3.141 592 92 b 0.000 009% 12 a 7:25 pm b 4:00 pm on Tuesday 13 a 15 120 b 12 14 a 90 b 240 c 6 15 27 and 36 16 a 2000 b Higher as top values rounded down and denominator rounded up. 17 a p and q are 2 and 5. r is 3 b 15 18 m = 5, n = 3 Chapter 2 – Number: Fractions, ratio and proportion Exercise 2A 1 a 1 3 b 1 5 c 2 5 d 5 24 AQA GCSE Maths (4th Edition) 4 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers e 2 5 f 1 6 g 2 7 h 1 3 2 12 30 = 2 5 3 1 5 4 1 2 5 Jon saves  30 1 90 3 Matt saves,  35 7 100 20 which is greater than 1 3 , so Matt saves the greater proportion of his earnings. 6  13 65 20 100 and  16 64 25 100 so 13 out of 20 is the better mark. 7 3 8 8 11 24 9 3 7 10 9 22 Exercise 2B 1 a 8 15 b 7 12 c 11 12 d 1 10 e 1 8 f 1 12 2 Three-quarters of 68 3 a 47 60 4 b 41 72 c 109 120 1 d 23 30 1 4 a 1 6 b 30 5 No, one eighth is left, which is 12.5 cl, so enough for one cup but not two cups. 6 He has added the numerators and added the denominators instead of using a common denominator. Correct answer is 7 12 3 . 7 Possible answer: The denominators are 4 and 5. I first find a common denominator. The lowest common denominator is 20 because 4 and 5 are both factors of 20. So I am changing the fractions to twentieths. One-quarter is the same as five- twentieths (multiplying numerator and denominator by 5). Two-fifths is the same as eight-twentieths (multiplying numerator and denominator by 4). Five-twentieths plus eight- twentieths = thirteen-twentieths. 8 11 20 of 900 = 495, 2 11 of 495 = 90 left-handed boys. 900 – 495 = 405 girls. 2 9 of 405 = 90 left-handed girls. 180 left-handed students altogether so 180 out of 900 = 1 5 . 9 1 5 + 3 8 = 23 40 , so 17 40 of the counters are yellow. 17 40 of 600 = 255 10 a because 27 40 + 2 5 = 1 3 40 which is greater than 1. b 2 5 of 200 = 80. 5 8 of 80 = 50 women at least 40. 27 40 of 200 = 135 members at least 40. 135 – 50 = 85 men at least 40. 3 5 of 200 = 120, so 120 – 85 = 35 men under 40. 11 a 1 5 is 8 40 . 3 4 is 30 40 . Half-way between 8 and 30 is 19, so the mid-point fraction is 19 40 . b Yes as the mid-point of any two numbers a and b is (a + b) ÷ 2 and adding the same denominator is the same thing as dividing by 2. Exercise 2C 1 a 1 6 b 3 8 c 7 20 d 3 5 e 5 12 f 11 12 2 g 9 10 3 h 1 3 3 2 a 3 4 b 1 15 1 c 5 d 4 9 e 3 5 1 3 a 1 4 b 5 c 8 3 d 4 5 4 a 1 5  b 2 c 9 7  d 5 3 5 3 8 6 1 8 7 40 8 2 5 of 1 2 6 9 £10.40 10 a 9 32 b 256 625 11 After 1 day 7 8 of the water is left. On day 2, 1 8 × 7 8 = 7 64 is lost so total lost is 1 8 + 7 64 = 8 64 + 7 64 = 15 64 , so 1 – 15 64 is left = 49 64 12 50 × 1 2 1 = 75 kg. 120 – 75 = 45, 45 ÷ 1 2 2 = 18, so 18 of the 1 2 2 kg bags are packed. 13 a 77% is about 3 4 . 243 is about 240, so 3 4 of 240 = 180. b Lower, as both estimates are lower than the original values. AQA GCSE Maths (4th Edition) 5 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers Exercise 2D 1 a 1 11 20 b 1 1 4 c 1 63 80 d 11 30 e 61 80 f 167 240 2 a 12 1 4 miles b 3 1 4 miles 3 a 6 11 20 b 8 8 15 c 11 63 80 d 3 11 30 e 7 61 80 f 4 277 396 4 a – 77 1591 b Answer is negative 5 18 2 5 cm 6 5 12 (anticlockwise) or 7 12 (clockwise) 7 a 3 5 b 27 128 c 5 2 5 d 5 1 7 e 3 9 32 f 11 18 8 a 8 11 20 b 65 91 100 c 52 59 160 d 2 17 185 e 2 22 103 f 7 881 4512 9 18 5 12 m² 10 3 11 a 6 × (1 3 4 )2 = 18 3 8 cm² b 14 144 25 25 34 6   , 144 12 2 25 5 5 2   cm 12 22 ÷ (2 × 22 7 ) = 7 2 , 22 7 × 7 2 × 7 2 = 38 1 2 cm² 13 Volume cuboid = 11 24 22 cm³, 11 24 22 ÷ ( 22 7 × 4 3 ) = 343 64 , 343 3 64 = 1 3 4 cm 14 After 1 day 7 8 is left, after two days 49 64 and after three days 343 512 is left 15 120 × 1 2 4 = 540. 175 × 1 2 1 = 262 1 2 . 540 – 262 1 2 = 277 1 2 . 277 1 2 ÷ 2 1 2 = 111 bags. Exercise 2E 1 a 1.1 b 1.03 c 1.2 d 1.07 e 1.12 2 a 0.92 b 0.85 c 0.75 d 0.91 e 0.88 3 a 391 kg b 824.1 cm c 253.5 g d £143.50 e 736 m f £30.24 4 a 731 m b 83.52 g c 360 cm d 117 min e 81.7 kg f £37.70 5 448 6 No, as the total is £101. She will save £20.20, which is less than the £25 it would cost to join the club. 7 £29 425, −7% pay rise is an increase of £2425 per year which is better than £150 × 12 = £1800 8 a £6.125 (£6.13) b x × 0.025 c y ÷ 1.175 × 1.2 9 Offer A gives 360 grams for £1.40, i.e. 0.388 pence per gram. Offer B gives 300 grams for £1.12, i.e 0.373 pence per gram, so Offer B is the better offer. Or Offer A is 360 for 1.40 = 2.6 g/p, offer B is 300 for 1.12 = 2.7 g/p, so offer B is better. 10 c Both the same as 1.05 × 1.03 = 1.03 × 1.05 11 a Shop A as 1.04 × 1.04 = 1.0816, so an 8.16% increase. 12 £425.25 13 0.9 × 1.1 = 0.99 (99%) 14 Area of original circle = 200.96 Enlarged area = 200.96 x 1.6 = 321.536 Enlarged radius = 321.536 3.14  = 10.1192885125 % increase = 2.11928/8 × 100 = 26.49% 15 a Let r = 10. Approx formula gives V = 4000, actual gives V = 4188.79, 188.79 ÷ 4188.79 = 0.045 which is 4.5% b The value is lower as 4 3 × π is greater than 4 as π is 3.14. Exercise 2F 1 a 25% b 60.6% c 46.3% d 12.5% e 41.7% f 60% g 20.8% h 10% i 1.9% j 8.3% k 45.5% l 10.5% 2 32% 3 6.5% 4 33.7% 5 a 49.2% b 64.5% c 10.6% 6 4.9% 7 90.5% 8 Stacey had the greater percentage increase. Stacey: (20 − 14) × 100 ÷ 14 = 42.9% Calum: (17 − 12) × 100 ÷ 12 = 41.7% 9 Yes, as 38 out of 46 is over 80% (82.6%) 10 Let z = 100. y = 75, x = 0.6 × 75 = 45, so x is 45% of z 11 Let z be 100, x = 60. If x is 75% of y, y = 80, so y is 80% of z. 12 30% of 4800 = 1440. 1.2 × 4800 = 5760. 70% of 5760 = 4032. (4032 – 1440) ÷ 1440 = 1.8, so the AQA GCSE Maths (4th Edition) 6 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers increase in numbers owning a mobile phone is 180%. 13 31 ÷ 26 = 1.19 which is a 19% increase. 31% is 5% more of the total votes cast than 26% Review questions 1 £572 2 a 36 seconds b i 25.2 seconds ii Eve iii Eve 3 £120 4 £576 5 a £9 b £13.20 6 a 0.875 b 11 35 c 1 5 5 7 £322 8 19 40 9 1 12 4 10 5 11 a 221 71 , 22 7 , 312 99 , 54 17 b 22 7 12 28% 13 77% 14 25% 15 For bag A P(red) = 0.1875 and for bag B P(red) = 0.186 so Tomas is wrong. 16 13% 17 a 150 men, 100 women b 12% Chapter 3 – Statistics: Statistical diagrams and averages Exercise 3A 1 a b 16 c 42 2 Pie charts with these angles: a 36°, 90°, 126°, 81°, 27° b 90°, 108°, 60°, 78°, 24° c 168°, 52°, 100°, 40° 3 a Pictogram with suitable key b Bar chart correctly labelled c Vertical line chart correctly labelled d Pie chart with these angles: 60°, 165°, 45°, 15°, 75° and correctly labelled e Vertical line chart. It shows the frequencies, the easiest one to draw and comparisons can be made. 4 a 36 b Pie charts with these angles: 50°, 50°, 80°, 60°, 60°, 40°, 20° c Student’s bar chart. d Bar chart, because easier to make comparisons. 5 a Pie charts with these angles: 124°, 132°, 76°, 28° b Split of total data seen at a glance. 6 a 55° b 22 c 33 1 3 % 7 a Pie charts with these angles: Strings: 36°, 118°, 126°, 72°, 8° Brass: 82°, 118°, 98°, 39°, 23° b Overall, the strings candidates did better, as a smaller proportion obtained lower grades. A higher proportion of Brass candidates scored very good grades. 8 Work out the angle for ‘Don’t know’ = 40°, so P(Don’t know) = 40 360 ° = 1 9 Exercise 3B 1 a b About 328 million c Between 1980 and 1985 d Rising steeply at first, but then leveling off. Rise in living standards, cheaper flights, more package holidays AQA GCSE Maths (4th Edition) 7 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers 2 a b Smallest difference Wednesday and Saturday (7°), greatest difference Friday (10°) 3 a b about 120 c The same people keep coming back and tell others, but new customers each week become more difficult to find. 4 No, you cannot extrapolate the data or the data is likely to change after 5 weeks 5 All the temperatures were presumably higher than 20 °C. Exercise 3C 1 a 47 b 53 c 55 d 65 2 Mode 3 Three possible answers: 12, 14, 14, 16, 18, 20, 24; or 12, 14, 14, 16, 18, 22, 24; or 12, 14, 14, 16, 20, 22, 24 4 a Median (mean could be unduly influenced by results of very able and/or very poor candidates) b Median (mean could be unduly influenced by pocket money of students with very rich or generous parents) c Mode (numerical value of shoe sizes irrelevant, just want most common size) d Median (mean could be distorted by one or two extremely short or tall performers) e Mode (the only way to get an ‘average’ of non-numerical values) f Median (mean could be unduly influenced by very low weights of premature babies) 5 a i £20 000 ii £28 000 iii £34 000 b A 6% rise would increase the mean salary to £36 040, a £1500 pay increase would produce a mean of £35 500. 6 a Median b Mode c Mean 7 Tom – mean, David – median, Mohammed – mode 8 11.6 9 42.7 kg 10 24 Exercise 3D 1 a i 7 ii 6 iii 6.4 b i 8 ii 8.5 iii 8.2 2 a 1280 b 1.9 c 0 d 328 3 a 2.2, 1.7, 1.3 b Better dental care 4 a 50 b 2 c 2.8 5 a Roger 5, Brian 4 b Roger 3, Brian 8 c Roger 5, Brian 4 d Roger 5.4, Brian 4.5 e Roger, smaller range f Brian, better mean 6 a 40 b 7 c 3 d 2 e 2.5 f the mode, 3 g 2.4 7 5 8 The total frequency could be an even number where the two middle numbers have an odd difference. 9 a 34 b x + 80 + 3y + 104 = 266, so x + 3y = 82 c x = 10, y = 24 d 2.5 Exercise 3E 1 a i 30 < x ≤ 40 ii 29.5 b i 0 < y ≤ 100 ii 158.3 c i 5 < z ≤ 10 ii 9.43 d i 7–9 ii 8.41 2 a 100 < m ≤ 120 b 10.86 kg c 108.6 g 3 a 175 < h ≤ 200 b 31% c 193.3 hours d No the mean was under 200 and so was the mode. 4 24 5 a Yes, average distance is 11.7 miles per day. b Because shorter runs will be run at a faster speed, which will affect the average. c Yes, because the shortest could be 1 mile, the longest 25 miles. 6 Soundbuy; average increases are Soundbuy 17.7p, Springfields 18.7p, Setco 18.2p 7 a 160 b 52.6 minutes c Modal group d 65% 8 The first 5 and the 10 are the wrong way round. AQA GCSE Maths (4th Edition) 8 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers 9 Find the midpoint of each group, multiply that by the frequency and add those products. Divide that total by the total frequency. 10 a Yes, as total in first two columns is 50, so median is between 39 and 40. b He could be correct, as the biggest possible range is 69 – 20 = 49, and the lowest is 60 – 29 = 31. Exercise 3F 1 a good positive correlation, time taken increases with the number of press-ups b strong negative correlation, you complete a crossword more quickly as you get older c No correlation, speed of cars on M1 is not related to the temperature d weak, positive correlation, older people generally have more money saved in the bank 2 a and b c about 19 cm/s d about 34 cm 3 a and b c Greta d about 70 e about 70 4 a b Yes, as good positive correlation 5 a b Little correlation, so cannot draw a line of best fit or predict the value 6 a and b c about 2.4 km d about 8 minutes e you cannot extrapolate values from a scatter diagram or the data may change for longer journeys 7 about 23 mph 8 Points showing a line of best fit sloping down from top left to bottom right Review questions 1 a Grade 7 b 100 360 or 5 18 AQA GCSE Maths (4th Edition) 9 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers c i 48 iii 216 d e.g. pie charts show proportions or they are percentages, not actual numbers or do not know how many students, etc. 2 43.7 matches 3 a 10 < t ≤ 20 b 10 < t ≤ 20 c 19 minutes 4 a because over half the students have more than £10 pocket money, so the mean must be more than £10 b £11.17 5 6 a 100 < m ≤ 150 b 150 < m ≤ 200 c 159 g d e mass, m (grams) Margot’s tomatoes 50 < m ≤ 100 12 100 < m ≤ 150 23 150 < m ≤ 200 34 200 < m ≤ 250 24 250 < m ≤ 300 5 300 < m ≤ 350 2 f mass, m (grams) Margot’s tomatoes mid point x x × m 50 < m ≤ 100 12 75 900 100 < m ≤ 150 23 125 2875 150 < m ≤ 200 34 175 5950 200 < m ≤ 250 24 225 5400 250 < m ≤ 300 5 275 1375 300 < m ≤ 350 2 325 650 totals 100 17150 estimate for the mean = 171.5 g g on average Tom’s tomatoes were generally smaller, but more consistent 8 a i Diagram C ii Diagram A iii Diagram B b Diagram A: strong negative correlation, diagram B: no correlation, diagram C: strong positive correlation 8 a/b Student’s graph as follows: Time on horizontal axis from 0 to 20 and Distance (km) on vertical axis from 0 to 10 with the following points plotted: (3, 1.7) (17, 8.3) (11, 5.1) (13, 6.7) (9, 4.7) (15, 7.3) (8, 3.8) (11, 5.7) (16, 8.7) (10, 5.3) and with line of best fit drawn. c/d answers depend on student’s line of best fit Chapter 4 – Algebra: Number and sequences Exercise 4A 1 a 11111 × 11111 = 123 454 321, 111111 × 111111 = 12 345 654 321 b 99999 × 99999 = 9 999 800 001, 999999 × 999999 = 999 998 000 001 2 a 7 × 8 = 72 + 7, 8 × 9 = 82 + 8 b 50 × 51 = 2550, 60 × 61 = 3660 3 a 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25 = 52, 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36 = 62 b 21 + 23 + 25 + 27 + 29 = 125 = 53, 31 + 33 + 35 + 37 + 39 + 41 = 216 = 63 4 a 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64, 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128 b 12 345 679 × 45 = 555 555 555, 12 345 679 × 54 = 666 666 666 5 a 13 + 23 + 33 1+ 43 = (1 + 2 + 3 + 4)2 = 100, 13 + 23 + 33 + 43 + 53 = (1 + 2 + 3 + 4 + 5)2 = 225 b 362 + 372 + 382 + 392 + 402 = 412 + 422 + 432 + 442, 552 + 562 + 572 + 582 + 592 + 602 = 612 + 622 + 632 + 642 + 652 6 a 12 345 678 987 654 321 b 999 999 998 000 000 001 c 122 + 12 d 8190 e 81 = 92 f 512 = 83 AQA GCSE Maths (4th Edition) 10 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers g 512 h 999 999 999 i (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)2 = 2025 7 1 + 500 = 501, 2 + 499 = 501, …. 250 + 251 = 501, 250 × 501 = 125250 Exercise 4B 1 a 21, 34: add previous 2 terms b 49, 64: next square number c 47, 76: add previous 2 terms 2 15, 21, 28, 36 3 61, 91, 127 4 1 3 2 5 3 , , , , 2 5 3 7 4 5 a 6, 10, 15, 21, 28 b It is the sums of the natural numbers, or the numbers in Pascal’s triangle or the triangular numbers. 6 a 2, 6, 24, 720 b 69! 7 364: Daily totals are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78 (these are the triangular numbers). Cumulative totals are: 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364. 8 X. There are 351 (1 + 2 + ... + 25 + 26) letters from A to Z. 3 × 351 = 1053. 1053 − 26 = 1027, 1027 − 25 = 1002, so, as Z and Y are eliminated, the 1000th letter must be X. 9 29 and 41 10 No, because in the first sequence, the terms are always one less than in the 2nd sequence 11 4n − 2 = 3n + 7 rearranges as 4n – 3n = 7 + 2, so n = 9 Exercise 4C 1 a 13, 15, 2n + 1 b 25, 29, 4n + 1 c 33, 38, 5n + 3 d 32, 38, 6n – 4 e 20, 23, 3n + 2 f 37, 44, 7n − 5 g 21, 25, 4n − 3 h 23, 27, 4n – 1 i 17, 20, 3n – 1 j –8, –18, 42 – 10n k 4, 0, 24 – 4n l –1, –6, 29 – 5n 2 a 3n + 1, 151 b 2n + 5, 105 c 5n − 2, 248 d 4n − 3, 197 e 8n − 6, 394 f n + 4, 54 g 5n + 1, 251 h 8n − 5, 395 i 3n − 2, 148 j 3n + 18, 168 k 47 – 7n, –303 l 41 – 8n, – 359 3 a 33rd b 30th c 100th = 499 4 a i 4n + 1 ii 401 iii 101, 25th b i 2n + 1 ii 201 iii 99 or 101, 49th and 50th c i 3n + 1 ii 301 iii 100, 33rd d i 2n + 6 ii 206 iii 100, 47th e i 4n + 5 ii 405 iii 101, 24th f i 5n + 1 ii 501 iii 101, 20th g i 3n − 3 ii 297 iii 99, 34th h i 6n − 4 ii 596 iii 98, 17th i i 205 – 8n ii –595 iii 101, 13th j i 227 – 2n ii 27 iii 99 or 101, 64th and 63rd 5 a   2 1 3 1 n n b Getting closer to 2 3 (0. 6 ) c i 0.667 774 (6dp) ii 0.666 778 (6dp) d 0.666 678 (6dp), 0.666 667 (6dp) 6 a   4 1 5 1 n n b Getting closer to 4 5 (0.8) c i 0.796 407 (6dp) ii 0.799 640 (6dp) d 0.799 964 (6dp), 0.799 9996 (7dp) 7 a £305 b £600 c 3 d 5 8 a 3 4 , 5 7 , 7 10 b i 0.666 666 777 8 ii 2 3 c For n,     2 1 2 2 3 1 3 3 n n n n 9 a 8n + 2 b 8n + 1 c 8n d £8 10 a Sequence goes up in 2s; first term is 2 + 29 b n + 108 c Because it ends up as 2n ÷ n d 79th 11 If there was a common term then for some value of n the expressions would be equal i.e. 2n = 2n – 1, Subtracting 2n from both sides gives 0 = – 1, which is impossible. 12 Difference is 19 – 10 = 9. 9 ÷ 3 = 3 so A = 3. 3  5 + b = 10, b = –5 Exercise 4D 1 a Even, + Odd Even Odd Even Odd Even Odd Even b Odd, × Odd Even Odd Odd Even Even Even Even 2 a 1 + 3 + 5 + 7 = 16 = 42, 1 + 3 + 5 + 7 + 9 = 25 = 52 b i 100 ii 56 3 a 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 b because odd + odd = even, odd plus even = odd and even + odd = odd. c i a + 2b, 2a + 3b, 3a + 5b, 5a + 8b, 8a + 13b ii coefficient of a odd and b even, a even and b odd, both odd 4 a Even b Odd c Odd d Odd e Odd f Odd g Even h Odd i Odd 5 a Odd or even b Odd or even c Odd or even d Odd AQA GCSE Maths (4th Edition) 11 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers e Odd or even f Even 6 a i Odd ii Even iii Even b Any valid answer, e.g. x(y + z) 7 a 64, 128, 256, 512, 1024 b i 2n − 1 ii 2n + 1 iii 3 × 2n 8 a The number of zeros equals the power. b 6 c i 10n – 1 ii 2 × 10n 9 a 125, 216 b 1 + 8 = 9, 1 + 8 + 27 = 36, 1 + 8 + 27 + 64 = 100… the answers are square numbers 10 a 28, 36, 45, 55, 66 b i 210 ii 5050 c You get the square numbers. 11 a i If n is odd, n + 1 is even. If n is even, n + 1 is odd. Even times odd is always even. ii 2n must be even, so 2n + 1 must be odd. b Odd Odd Even Even Odd c (2n + 1)2 = 4n2 + 4n + 1 or (2n)2 = 4n2 4n2 + 4n is even so adding 1 makes it odd 4n2 is 2 × 2n2 which is even 12 11th triangular number is 66, 18th triangular number is 171 13 a 36, 49, 64, 81, 100 b i n2 + 1 ii 2n2 iii n2 – 1 14 a 6, 24, 96, 384, 1536 b 21, 147, 1029, 7203, 50 421 c 2, 10, 50, 250, 1250 d 6, 60, 600, 6000, 60 000 e 54, 162, 486, 1458, 4374 15 a 3 × 2n – 1 b 5 × 4n – 1 c 20 × 5n –1 or 4 × 5n d 21 × 3n – 1 or 7 × 3n e 24 × 8n – 1 or 3 × 8n 16 2 as all other primes are odd, so the sum of two of them will be even, so could not be a prime. 17 a There are many answers, 5 + 31 = 36, 7 + 29 = 36, 2 + 47 = 49 etc. b There are many answers, 49 – 36 = 13, 81 – 64 = 17 Exercise 4E 1 a b 4n − 3 c 97 d 50th diagram 2 a b 2n + 1 c 121 d 49th set 3 a 18 b 4n + 2 c 12 4 a i 24 ii 5n − 1 iii 224 b 25 5 a 5, 8, 11, 14 b i 20 cm ii (3n + 2) cm iii 152 cm c 332 6 a i 20 ii 162 b 79.8 km 7 a i 14 ii 3n + 2 iii 41 b 66 8 a i 5 ii n iii 18 b 20 tins 9 a 2n b i 100 × 2n–1 ml ii 1600 ml c Next sizes after super giant are 3.2l, 6.4l and 12.8l with weights of 3.2 kg, 6.4 kg and 12.8 kg, so the largest size is 6.4 litres. 10 The nth term is       3 4 n , so as n gets very large, the unshaded area gets smaller and smaller and eventually it will be zero; so the shaded area will eventually cover the triangle. 11 Yes, as the number of matches is 12, 21, 30, 39, … which is 9n + 3; so he will need 9 × 20 + 3 = 183 matches for the 20th step and he has 5 × 42 = 210 matches. 12 a 20 b 120 13 Alex’s answer gives 4(n + 2) = 4n + 8 Colin’s method gives 4n + 4 Ed’s method gives 4(n + 1) = 4n + 4 Gail’s method gives 2 × n + 2(n + 2) = 2n + 2n + 4 = 4n + 4 Linear sequence is 8 12 16 20 …. Which has an nth term of 4n + 4 so they are all valid methods except for Alex who forgot that the corners overlap and should have taken the 4 overlapping corners away to get 4n + 8 – 4 = 4n + 4 Exercise 4F 1 a i 34, 43 ii goes up 3, 4, 5, 6, etc. b i 24, 31 ii goes up 1, 2, 3, 4, etc. c i 54, 65 ii goes up 5, 6, 7, 8, etc. d i 57, 53 ii goes down 10, 9, 8, 7, etc. 2 a 4, 7, 12, 19, 28 b 2, 8, 18, 32, 50 c 2, 6, 12, 20, 30 d 4, 9, 16, 25, 36 e 2, 8, 16, 26, 38 f 4, 7, 14, 25, 40 3 a 2n + 1 b n c n(2n + 1) = 2n2 + n AQA GCSE Maths (4th Edition) 12 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers d 2n2 + n + 1 4 a n b n + 1 c n(n + 1) d 9900 square units 5 a Yes, constant difference is 1 b No c Yes, constant difference is 2 d No e Yes, constant difference is 1 f No 6 a 4n + 4 b n2 c n2 + 4n + 4 d n2 + 4n + 4 e The sides of the large squares are of length n + 2 so the total number of squares is (n + 2)2 which is the same answer as c. 7 a Table 10, 15, 21; 6, 10, 15; 16, 25, 36 b i 45 ii 100 8 n2 + 2n – 3 = n2 + n + 3, gives n = 6. Substituting gives 45 for both expressions. 9 a Sequences are 4, 7, 14, 25, 40, 59, 82, … and 4, 11, 20, 31, 44, 59, 76, … so 59 is the next common term. b 59 is the 6th term in each sequence so substitute 6 into each expression. This will give 59 in both cases. 10 a There are many answers, for example a = –3 and b = 1. b The only solution is c = 2 and d = – 3 11 All values of n from 1 to 39 give a prime number. n = 40 gives 1681 which equals 41 × 41 12 a (n + 1)(n – 1) = n2 + n – n – 1 = n2 – 1 b n2 – 1 as 50 × 50 – 1 is easy to work out but 51 × 49 isn’t c (n + 1)(n – 1) as 100 × 98 is easy to work out but 992 – 1 isn’t. Exercise 4G 1 a i 36, 49 ii n2 b i 35, 48 ii n2 – 1 c i 38, 51 ii n2 + 2 d i 39, 52 ii n2 + 3 e i 34, 47 ii n2 – 2 f i 35, 46 ii n2 + 10 2 a i 37, 50 ii (n + 1)2 + 1 b i 35, 48 ii (n + 1)2 – 1 c i 41, 54 ii (n + 1)2 + 5 d i 50, 65 ii (n + 2)2 + 1 e i 48, 63 ii (n + 2)2 – 1 3 a i n2 + 4 ii 2504 b i 3n + 2 ii 152 c i (n + 1)2 – 1 ii 2600 d i n(n + 4) ii 2700 e i n2 + 2 ii 2502 f i 5n – 4 ii 246 4 a 2n2 – 3n + 2 b 3n2 + 2n – 3 c 1 2 n2 + 5 2 n + 1 d 1 2 n2 + 4 1 2 n – 2 e 1 2 n2 + 1 1 2 n + 6 f 1 2 n2 + 1 1 2 n + 2 5 6n2 6 a 26 b 1 1 2 n2 + 1 2 n c 8475 7 a 45 b nth term is 1 2 n2 + 1 2 n so 1 2 × 15 × 15 + 1 2 × 15 = 120, so no. 8 Front face is n2, sides faces are n × (n + 1) = n2 + n so total surface area is 2 × n2 + 4 × (n2 + n) = 6n2 + 4n. 9 Sequence is 1, 7, 19, 37. nth term is 3n2 – 3n + 1 so the 100th hexagonal number is 29 701. 10 a Taking the height first. There are n + 1 strips m feet long. That is m(n + 1) in total. Taking the width. There are m + 1 strips n feet long. That is n(m + 1) in total m(n + 1) + n(m + 1) = mn + m + mn + n = 2mn + m + n b Taking the nails across a width strip. There are n + 1 lots of 2 nails which is 2(n + 1). There are m + 1 width strips, so the total is 2(n + 1)(m + 1). Review questions 1 No. Sequence is 7, 10, 13, 16, 19, 22, 25, 28, … so the first 3 odd terms are prime but 25 is not prime. 2 a 4n + 1 b Not odd c 28th term is 113 3 nth term is 5n + 1. 5  150 + 1 = 751 4 a 6n + 3 b No, 3n + 2 generates the sequence 5, 8, 11, 14, 17, 20, 23, … so the even terms of this sequence are always 1 less than the terms of the original sequence 5 a 2  3n– 1 b Not an even number 6 a 5  6n– 1 b 8th term is 1 399 680 7 a This misprint will be corrected at reprint. The first five terms in the sequence are –27, –21, – 11, 3, 21. Of these terms, 3 is a prime number. b When n2 = 29, the expression can be factorised as 29(2 × 29 – 1) so is not a prime number 8 a 4, 9, 18, 31, 48 b 2, 2, 3, 5, 8 9 n = 1 (n – 1) = 0, n = 2 (n – 2) = 0, n = 3 2(3 – 1)(3 – 2)÷ 5 = 2 × 2 × 1 ÷ 5 = 0.8 10 2n2 – 2n + 3 11 a nth term is n2 + 2n, 12 × 12 + 2 × 12 = 168, so yes enough squares b 40 × 40 + 2 × 40 = 1680 12 2n2 – n, 2 × 202 – 20 = 780 13 The sequence of dots is 5, 15, 30, 50,… n 0 1 2 3 4 c 0 5 15 30 50 a + b 5 10 15 20 2a 5 5 5 AQA GCSE Maths (4th Edition) 13 © HarperCollinsPublishers Ltd 2015 Higher Student Book – Answers a = 2 1 2 , b = 2 1 2 and c = 0, so the nth term is 2 1 2 n2 + 2 1 2 n 2 1 2 × 502 + 2 1 2 × 50 = 6375 Chapter 5 – Ratio and proportion and rates of change: Ratio and proportion Exercise 5A 1 a 1 : 3 b 3: 4 c 2 : 3 d 2 : 3 e 2 : 5 f 2 : 5 g 5 : 8 h 25 : 6 2 a 1 : 3 b 3 : 2 c 5 : 12 d 8 : 1 e 17 : 15 f 25 : 7 g 4 : 1 h 5 : 6 i 1 : 24 3 7 10 4 2 5 5 a 2 5 b 3 5 6 7 : 3 7 2 : 1 8 a Fruit crush 1 6 , lemonade 5 6 Fruit Crush 1.25 1 0.2 0.4 0.5 Lemonade 6.25 5 1 2 2.5 b 0.4 litres c 2.5 litres 9 a 1 2 b 7 20 c 3 20 10 Sugar 5 22 , flour 3 11 , margarine 2 11 , fruit 7 22 11 4 12 1 : 4 13 a 5 : 3 : 2 b 20 14 1 2 13 litres 15 1 : 1 : 1 Exercise 5B 1 a 160 g, 240 g b 80 kg, 200 kg c 150, 350 d 950 m, 50 m e 175 min, 125 min f £20, £30, £50 g £36, £60, £144 h 50 g, 250 g, 300 g i £1.40, £2, £1.60 j 120 kg, 72 kg, 8 kg 2 a 175 b 30% 3 a 40% b 300 kg 4 21 5 a Mott: no, Wright: yes, Brennan: no, Smith: no, Kaye: yes b For example: W26, H30; W31, H38; W33, H37 6 a 1 : 400 000 b 1: 125 000 c 1 : 250 000 d 1 : 25 000 e 1 : 20 000 f 1 : 40 000 g 1 : 62 500 h 1 : 10 000 i 1 : 60 000 7 a 1 2 km or 500m b 78 cm ≈ 39 km. 39 ÷ 15 ≈ 2.6. 2.6 hours = 2 h 36 m. Plus 30 mins is 3 h 06 m so he should be back at about 12.06 pm 8 a Map A 1 : 250 000, Map B 1 : 1 000 000 b 2 km c 1.2 cm d 4.8 cm 9 a 1 : 1.6 b 1 : 3.25 c 1 : 1.125 d 1 : 1.44 e 1 : 5.4 f 1 : 1.5 g 1 : 4.8 h 1 : 42 i 1 : 1.25 10 Diesel : Petrol = 60 : 90. 1 5 of 60 = 12. 4 9 of 90 = 40. Total red cars = 52 which is more than 150 ÷ 3 = 50 so Yes. 11 a 4 : 3 b 90 miles c Both arrive at the same time. 12 0.4 metres 13 13 – 9 = 4. 4 ÷ 5 = 0.8. 2 × 0.8 = 1.6, 9 + 1.6 = 10.6 14 Athos has 3 more parts than Zena. 24 ÷ 3 = 8, so 1 part is 8. Zena has 8 marbles. Exercise 5C 1 a 3 : 2 b 32 c 80 2 a 100 b 160 3 0.4 litres 4 Jamie has 1.75 pints, so he has enough. 5 8100 6 296 7 Kevin £2040, John £2720 8 b C c F d T e T 9 51 10 100 11 40 ml 12 a 160 cans b 48 cans 13 a Lemonade 20 litres, ginger 0.5 litres b This one, in part a there are 50 parts in the ratio 40 : 9 : 1, so ginger is 1 50 of total amount; in part b there are 13 parts in the ratio 10 : 2 : 1, so ginger is 1 13 of total amount. 1 13 ˃ 1 50