Answers to Coursebook exercises 1 Integers, powers and roots ✦ Exercise 1.1 Directed numbers 1 a 2 b 4.7 c −5.3 d −0.2 e −2.5 2 a 1.5 b −3.5 c −5.5 d 5.5 e −2.37 3 a 11 b −7.4 c 6.6 d −1.1 e −3.3 4 a 4.1 b 4.7 c −6.1 d −6.9 e −2.13 6 a −2.5 b 0.4 c −6.4 d −9 7 a −8.1 b 2 c 15.4 d −1 5 −1 °C 8 + −3.4 −1.2 5.1 1.7 3.9 −1.3 −4.7 −2.5 9 a 22.08 b −9.6 c −2.3 d 44.16 e 4.8 10 a −8.1 b −0.9 c 1.44 d 13 e −7 11 × 3.2 −0.6 −1.5 −4.8 0.9 −2.5 −8 1.5 12 a 6 b −24 c 1.5 13 a −10 b −7.5 c 9 d −1.3 14 a 5.76 b −2.4 c 4.6 d −35 15 One is 5 and the other is −4. We cannot say which is p and which is q. ✦ Exercise 1.2 1 a 4 < 20 < 5 Square roots and cube roots b 15 < 248 < 16 c 17 < 314 < 18 d 9< c 20 d 5 83 .5 < 10 e 12 < 157 < 13 2 Because 63 = 216 < 305 and 73 = 343 > 305 3 a 13 b 6 3 4 a 4 < 100 < 5 3 b 6 < 222 < 7 5 a 122 = 144 < 160 and 132 = 169 > 160 6 a 35 7 3 3 c 9 < 825 < 10 e 17 3 d 6 < 326 < 7 e 3 < 3 58 . 8 < 4 b 1600 b 11 125 = 5; 102 = 100 so 125 > 10 8 a 25 b 4.5 c 6.8 d 12 e 1.9 9 a 7.42 b 10.39 c 5.85 d 8.57 e 21.54 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 1 Answers to Coursebook exercises F Exercise 1.3 1 a 1 5 Indices 1 625 c 4 a i 1 b a0 = 1 ii 1 iii 1 5 a 13 b 13 1 c 34 6 a 0.2 b 0.04 c 0.1 d 0.01 e 0.001 7 a 2 b 2 c 2 d 2 e 20 8 a 211 b 29 2 a 49 3 a 1 4 4 3 3 F Exercise 1.4 −1 d d 1 d e 343 1 12 e 1 225 1 400 5 −2 −4 c 2−10 Working with indices b 67 c 106 d a7 e 46 2 a 28 b 86 c a5 d 26 e b7 3 a 33 b k c 102 d 5−2 e 72 4 a 2 b 2−1 c 2−2 d 22 e 2−2 5 a 87 b 53 c 46 d 19 e 6 a 2 b −2 c −3 d 3 7 a 117 649 b 49 c 117 649 d 1 8 a 4 ÷4 =4;5–2=3 2 f iv 1 1 a 55 5 2 1 125 c 1 7 c 1 8 b 1 25 b 1 49 b 1 100 3 9 a 3−2 and 3−5 b i 3−7 ii 33 10 a 22 or 4 b a3 7 1 144 e 7 b 2 ÷ 2 = 2 ; 10 – 4 = 6 10 4 6 c d3 d 105 11 a i 24 ii 26 iii 28 m n mn b (2 ) = 2 ; the index is m × n. iv 212 12 a 4K b 32K c K × K or K2 d 1 K 13 a 2 b 2 c 4 v 28 8 Cambridge Checkpoint Mathematics 9 d 6 Copyright Cambridge University Press 2013 Answers to Coursebook exercises Unit 1 End-of-unit review 1 a 3 b −2.5 c −9.2 d −2.1 2 a 16 b −14.7 c 12 d 0.1 3 a 11.25 b 2.5 c −5.625 4 a −3.6 b −6.3 c 2.5 5 a 10 < 111 < 11 b 18 < 333 < 19 c 4 < 3 111 < 5 d 6 < 3 333 < 7 6 a 14 b 6 c 0.25 d 0.04 7 15.8 8 8.2 9 31 < 1000 < 32; 3 1000 = 10 and 3 × 10 = 30 10 a 0.5 11 a b 0.25 1 9 b 1 8 c 1 6 d 1 144 12 a 5 1 b 1001.1 13 a 102 b 103 c 10−2 d 10−3 e 100 14 a 95 b 83 c 73 d a−2 e n−1 15 a 2−1 or 1 b 152 c 202 d 5−2 16 a a6 b a−2 c a2 d a5 17 a 43 b a−2 c n 18 a 0 b −1 c 7 4 2 Copyright Cambridge University Press 2013 e a−2 d 2 Cambridge Checkpoint Mathematics 9 3 Answers to Coursebook exercises 2 Sequences and functions F Exercise 2.1 1 a b c d e f g h i Generating sequences linear, term-to-term rule ‘add 4’ linear, term-to-term rule ‘add 10’ non-linear, term-to-term rule ‘add 1, add 2, add 3, add 4, …’ non-linear, term-to-term rule ‘subtract 2, subtract 3, subtract 4, subtract 5, …’ linear, term-to-term rule ‘subtract 5’ non-linear, term-to-term rule ‘subtract 3, subtract 6, subtract 9, subtract 12, …’ linear, term-to-term rule ‘add 0.5’ non-linear, term-to-term rule ‘add 2, add 4, add 8, add 16, …’ linear, term-to-term rule ‘subtract 8’ b 2 1 , 4, 5 1 , 7 2 a 8, 3, −2, −7 2 2 c 4, 5, 7, 10 d 24, 12, 6, 3 c 5, 8, 13, 20 d 3, 12, 27, 48 3 42. Check students’ methods. 4 243. Check students’ methods. 5 a 6, 7, 8, 9 6 a i 21 b 2, 5, 8, 11 ii 41 iii 81 b i 23 ii 98 iii 398 7 C 8 Question 1: Answer: Question 2: Answer: term = 2 × position number + 7 2nd term = 11 3rd term = 13 term = 5 × position number 2 1st term = 5 3rd term = 45 F Exercise 2.2 1 Finding the nth term 3, 6, 9; 30 2 5, 4, 3; −14 3 11, 15, 19; 207 4 2, 12, 22; 992 5 A vi, B iv, C i, D v, E ii, F iii 6 a 2n + 1 f 15 − 4n b 3n + 2 g 3 − 5n c 5n − 1 h 9n − 12 d 8n − 6 i 12n + 11 e 10 − 2n 7 a 201 f −385 b 302 g −497 c 499 h 888 d 794 i 1211 e −190 (2 ) 8 No. The term-to-term rule is ‘add 1 ’, so although Jake got the first part correct 1 n , the first term 1 + 4 is not 4, 2 2 so he got the second part wrong. The correct answer 1 n + 3 1 . 2 2 9 T he sequence is decreasing, so the nth term expression for this sequence cannot start with 6n as this would give an increasing sequence. 10 Y es. Each pattern increases by 3 squares (the term-to-term rule is ‘add 3’), so the nth term will start with 3n. The number of squares in the patterns is 5, 8, 11, 14, and 3 × 1 + 2 = 5, 3 × 2 + 2 = 8, 3 × 3 + 2 = 11, 3 × 4 + 2 = 14. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 2 Answers to Coursebook exercises F Exercise 2.3 1 a y=x−9 2 a x→x−3 Finding the inverse of a function b y=x+1 b x→x+8 a y= x −5 2 x a x → −1 5 b 5 a i x → 10 − x ii iii x → 4 − x iv 3 4 b c y= x d y = 6x 3 c x→ x 4 d x → 3x y= x +7 c y = 2(x − 1) 4 x→ x +7 c x → 5(x + 10) 3 x → x − 1 or 1 − x or 1 − x −2 2 2 x − 3 − x 3 x→ or −4 4 d y = 3x + 4 d x → 4x − 9 b i and iii 6 a x → 4 (x + 13) b 6 − 13 = −11.5 4 End-of-unit review 1 Non-linear; the term-to-term rule is ‘subtract 1, subtract 2, subtract 3, …’ 2 14. Check students’ methods. 3 12. Check students’ methods. 4 i 0, 6, 16, 30 5 a 2n + 3 ii 198 ii 798 b 12 − 2n c 3n − 11 6 The sequence is increasing so it can’t have a −6n term, as this would make the sequence decrease. 7 A nders. Each pattern increases by 2 dots (the term-to-term rule is ‘add 2’), so the nth term will start with 2n. The number of dots in the patterns is 4, 6, 8, 10, and 2 × 1 + 2 = 4, 2 × 2 + 2 = 6, 2 × 3 + 2 = 8, 2 × 4 + 2 = 10. 8 a y=x+2 b y = x8 9 a x→x−1 b x → 4x 10 a x → x − 11 b 4 × −1.2 + 11 = 6.2 11 a x → 2(x + 22.5) b 50 − 22.5 = 2.5 4 2 c y = 5(x − 2) d y = 2x − 1 c x→ x +7 d x → 10x − 2 3 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 3 Place value, ordering and rounding F Exercise 3.1 Multiplying and dividing decimals mentally 1 a 1.6 f 0.3 b 3.6 g 0.36 c 5.6 h 0.66 d 5.4 i 2.4 e 6 j 1.8 2 a 20 f 40 b 30 g 300 c 50 h 400 d 30 i 200 e 600 j 300 3 C, D, I, K (0.015); A, F, H, J (0.15); B, G, L (1.5); E (15) 4 a D b B c C d D 5 a 0.12 f 30 b 1.35 g 9 c 0.072 h 5 d 0.15 i 7 e 0.055 j 40 6 The bottom is 0.12, not 1.2; he wrote the answer with only one decimal place. Answer = 50 7 a 200 b 120 c 300 d 40 8 a i 0.8 b Larger ii 1.6 iii 2.4 iv 3.2 v 4 vi 4.8 9 a i 120 b Smaller ii 60 iii 40 iv 30 v 24 vi 20 F Exercise 3.2 Multiplying and dividing by powers of 10 1 a 1300 e 65 i 0.085 b 7800 f 8000 j 0.45 c 240 g 17 k 0.032 d 85 500 h 0.8 l 1.25 2 a 2.7 e 0.08 i 1800 b 0.45 f 0.0248 j 47 600 c 0.36 g 9 k 70 d 0.017 h 2.5 l 8.5 3 Do not tell anyone the secret! 4 a b 0.8 0.8××10 1011 80 80××10 10–1–1 88÷÷10 1000 0.08 0.08÷÷10 10–2–2 ==88 0.008 0.008××10 1033 32 32÷÷10 1022 3.2 3.2÷÷10 1011 800 800÷÷10 1022 0.32 0.32××10 1000 320 320÷÷10 1033 ==0.32 0.32 32 32××10 10–2–2 3.2 3.2××10 10–1–1 5 a i 400 b Smaller ii 40 iii 4 iv 0.4 v 0.04 vi 0.004 6 a i 0.12 b Larger ii 1.2 iii 12 iv 120 v 1200 vi 12 000 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 3 Answers to Coursebook exercises F Exercise 3.3 Rounding 1 a 4.8 b 8.79 c 0.477 d 0.97 e 3.5998 f 18.350 2 a 25.497 b 25.5 c 25.496 72 d 25.4967 e 25.50 f 25.496 723 8 3 a 100 b 46 000 c 18.7 d 0.09 e 0.79 f 1.41 4 a D b C c B d D 5 a 4510 b 5000 c 4509.0 d 4509 e 4500 f 4509.030 f > ii 17 4 a i She subtracted first; she should have multiplied first. b i She multiplied by 2 after working out the brackets; she should have squared the 5. c i She did not work out the value of the divisor. ii 8 ii 6 96 000 7 0.4 g 8 298 000 000 m/s 9 a i 16 c i 40 ii 16.1 (3 s.f.) ii 42.6 (3 s.f.) F Exercise 3.4 b i d i 700 80 ii ii 713 (3 s.f.) 67.2 (3 s.f.) Order of operations 1 a 20 f 8 b 19 g 24 c 7 h 17 d −12 j 100 e −2 k 50 2 a = b < c > d < e < 3 a , 12 b c , −3 50 o. Oditi forgot that 5b means 5 × 3, not 53. Shen did not use BIDMAS rules for working out the 5 N term in brackets. Shen added a + 5 before the multiplication 5b. Answer = 38 6 a 22 b 64 c 36 d 72 End-of-unit review 1 a 2.1 f 20 b 6 g 70 c 0.63 h 300 d 0.36 i 60 e 1 j 500 2 a 0.16 f 4 b 0.45 g 0.7 c 0.088 h 4 d 0.1 i 5 e 0.0016 j 80 3 a i 0.4 b Larger ii 0.8 iii 1.2 iv 1.6 v 2 4 a i 150 b Smaller ii 75 iii 50 iv 37.5 v 30 5 a 900 f 534 b 3700 g 2 c 240 h 1 d 5.55 i 0.62 e 0.075 j 76 6 No. Check students’ examples. 2 7 a 2.8 g 100 b 11.86 h 230 c 0.555 i 0.65 d 0.30 j 0.02 e 0.1235 k 1.00 8 a 4000 f 3893.0 b 3900 g 3893.01 c 3890 h 3893.010 d 3893 i 3893.0096 e 3893.0 j 3893.009 56 9 a 4 f −1 b 33 g 10 c 37 h 14 d 20 j 25 e 11 k 0 10 a 16 b 38 c 121 d 490 Cambridge Checkpoint Mathematics 9 f l 112.000 1.0 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 4 Length, mass, capacity and time F Exercise 4.1 Solving problems involving measurements 1 b 10.8 g a 45.6 g 2 8 days 3 4 hours and 40 minutes 4 328 km 5 a 43.5 cm b 3 6 a 1.5 tonnes b $433.50 7 a 80 b 3 (He will need another partial fill.) F Exercise 4.2 Solving problems involving average speed 1 c $200 29 km/h 2 280 km 3 13 minutes and 20 seconds 4 12 noon 5 10.3 km/h 6 a 1 hour and 4 minutes b 9.375 km/h c 10 hours and 40 minutes 7 a i 68.75 mph ii 56.25 mph iii 62.5 mph b No. 75 mph is about 120 km/h; the limit is 130 km/h. c Yes. 55 mph is about 88 km/h; the limit is 80 km/h. iv 31.25 mph 8 89 m/s 9 253 m/s F Exercise 4.3 1 Using compound measures First train. Speed of first train = 168 km/h, speed of second train = 163 km/h 2 Second. Speed for first part = 88 km/h, speed for second part = 91.2 km/h 3 a Tuesday = 0.133 km/min, Friday = 0.15 km/min b Friday 4 a 4 pack = 57 cents each, 9 pack = 55 cents each b 9 pack 5 a 750 ml = 0.24 cents/ml, 1.4 litre = 0.25 cents/ml b 750 ml bottle 6 500 g bag. 500 g bag = 0.128 cents/g, 2 kg bag = 0.1325 cents/g 7 1.5 litre bottle. 330 ml bottle = 0.127 cents/ml, 1.5 litre bottle = 0.11 cents/ml 8 a 18 numbers = 15 seconds each, 32 numbers = 12 seconds each b The puzzle with 32 numbers 9 a i 88 km/h ii 94.5 km/h b Second c 83.5 km/h 10 a i 36 km/h ii 28.8 km/h b First c 32 km/h Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 4 Answers to Coursebook exercises End-of-unit review 1 12 2 a 15 b $180 3 13.5 km/h 4 9.3 km/h 5 31 m/s 6 Cyclist B. A = 23.6 km/h, B = 24.1 km/h 7 a i 8.8 km/h 2 ii 9.5 km/h b Second part Cambridge Checkpoint Mathematics 9 c 8.3 km/h Copyright Cambridge University Press 2013 Answers to Coursebook exercises 5 Shapes F Exercise 5.1 Regular polygons 1 a i Square b i 90° and 90° ii Equilateral triangle ii 60° and 120° 2 a 360° ÷ 6 = 60° b 180° − 60° = 120° Check students’ reasons. 3 a 360° ÷ 8 = 45° b 180° − 45° = 135° Check students’ reasons. 4 a 180° − 144° = 36° b 10 5 360 ÷ 30 = 12 6 360 ÷ 18 = 20 7 180 − 168 = 12, 360 ÷ 12 = 30 8 360 − 36 = 324, 324 ÷ 3 = 108, 180 − 108 = 72, 360 ÷ 72 = 5; the shape is a pentagon 9 360 − 120 − 90 = 150, 180 − 150 = 30, 360 ÷ 30 = 12 10 a No. 360 ÷ (180 − 110) is not a whole number. b Yes, 6 c No d Yes, 9 e Yes, 12 11 128.6° to one decimal place 12 a 72 b 180 F Exercise 5.2 More polygons 1 a 900° b 1260° c 1440° 2 a 170° b 90°, 80°, 70°, 60°, 50° and 10°. The sum is 360°. 3 104° 4 It is an octagon so they add up to 1080°. 5 They add up to 530°. For a pentagon they should add up to 540°. 6 (N − 2) × 180 = 1800 → N − 2 = 10 → N = 12. It has 12 sides. 7 a 120° b No. If four angles are 90° they add up to 360°. Then the fifth must be 540° − 360° = 180°. This is impossible. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 5 Answers to Coursebook exercises F Exercise 5.3 1 Solving angle problems a a = 180 − (35 + 40) = 105 b b = 40, corresponding angles 2 a c = 68 + 54 = 122, exterior angle of triangle b d = 122 − 86 = 36, using exterior angle of triangle 3 T hird angle of triangle = 180° − (25° + 40°) = 115°. Angles of square and equilateral triangle are 90° and 60°. d = 360 − (115 + 90 + 60) = 95. 4 O AC is isosceles so a = (180 − 54) ÷ 2 = 63. The 54° angle is the exterior angle of isosceles triangle OBC so b = 54 ÷ 2 = 27. 5 The angles at Y and X are 90° and 105°. The sum of the five angles is 540°. w = 540 − (90 + 90 + 105 + 105) = 150. 6 a x = 73, alternate angles b y = 46, alternate angles c z = 73, corresponding angle to x 7 The angles of the pentagon and the hexagon are 108° and 120°. a = 360 − (108 + 120) = 132 8 Sum of angles = 360° → 5a = 360 → a = 360 ÷ 5 = 72 9 a = 33, isosceles triangle. Angle WYZ = a° + 33° = 66°, exterior angle of triangle XWY. b = 180 − 2 × 66 = 48, angle of isosceles triangle. c = 66 + 48 = 114, exterior angle of triangle WYZ. 10 Extend AB. e = d, alternate angles. a + e = 180, angles on a straight line, so a + d = 180. Similarly, c = f, alternate angles; b + f = 180, so b + c = 180. A e° a° d° D F Exercise 5.4 B b° f ° c° C Isometric drawings 1 Other diagrams are possible. a b c 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises Unit 5 2 b a 3 a 9 cm and 9 cm b 15 cm and 25 cm 4 5 a green Other diagrams are possible. b brown 6 7 a 3 by 4 Other diagrams are possible. b Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 3 Unit 5 Answers to Coursebook exercises F Exercise 5.5 1 a b A B C B C A c d Plans and elevations A A B B C C 2 3 a 4 b Other diagrams are possible. Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises F Exercise 5.6 Unit 5 Symmetry in three-dimensional shapes 1 a b c 2 a Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 5 Unit 5 Answers to Coursebook exercises b 3 a 8 faces b and c Other diagrams are possible. 4 a Other diagrams are possible. b 6 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 5 a and b Unit 5 c End-of-unit review 1 a 24° 2 a b c d b 156° 40° OP and OQ are the same length because O is the centre of the shape. 70° The interior angle is twice the size of b and 2 × 70 = 140. 3 128° 4 a 46°, corresponding angles b 152°. Angle ECD = 28°, corresponding angles, and b = 180 − 28, angles on a straight line 5 c = 105 + 33 = 138, exterior angle of a triangle d = 180 − (87 + 33) = 60 6 Other views are possible. 7 a b B C Copyright Cambridge University Press 2013 D Cambridge Checkpoint Mathematics 9 7 Answers to Coursebook exercises 6 Planning and collecting data F Exercise 6.1 1 a b c d Identifying data More men than women watch sport on the TV. Silver is the most popular colour of car that is sold. Girls are better than boys at estimating the masses of different objects. The more you revise, the better your exam result will be. 2 a 1. For example: Boys are better than girls at maths. 2.For example: ‘Are you a boy or girl?’, ‘What maths group are you in?’, ‘What were your percentage marks in the last two maths tests?’ 3. For example: maths test marks of boys and girls 4. For example: survey students, survey maths teachers for opinions 5. For example: 80, 40 boys and 40 girls 6. For example: test marks in percentages b For example: age, maths set, good at other subjects too, like teacher, like school c For example: He won’t know the age, maths set or ability most of the other students in the sample, so it will be difficult to compare fairly. 3 a 1. For example: The letter ‘e’ is the most commonly used letter in the book. 2. For example: Is ‘e’ the most commonly used letter in the book? 3. For example: how many of each letter of the alphabet are in the book 4. For example: use a tally chart 5. For example: 12 pages (10% of 120 pages) 6. For example: tally every letter in the pages chosen b For example: Is 12 pages enough? Has she chosen pages with pictures? Has she chosen pages without pictures? Should she bother to tally less usual letters such as q, y, k, z and x? c The tally chart will be very large with lots of information on it; she may get confused and put a tally in the wrong row, although a few errors may not affect the final outcome. 4 aNeed an equal number of boys and girls in the sample. Need to have a wide range of students, not just good mathematicians. b She lives a long way from her school, students from nearby may not use a bus. She needs to ask students who live a variety of distances from her school, choosing 52 students (10% sample) at random. F Exercise 6.2 1 Types of data a b c d Secondary. Sasha can’t measure children from 50 years ago. Primary. Easy to do a survey. Secondary. Impossible for one person to measure the rainfall in the whole of India. Either: Secondary. Can’t find this information for the whole country/world. Or: Primary. Could survey the people in street/school/church, etc. what make of TV they have. e Secondary. There are millions of government employees. f Either: Secondary. Can’t find this information for the whole country/world. Or: Primary. Could survey all 15-year-old students in my area/school g Either: Secondary. Can’t find this information for the whole country/world. Or: Primary. Could ask a sample of the people in street/school/church, etc. about their shoe size. h Either: Secondary. Can’t find this information for the whole country/world. Or: Primary. Could survey the people in street/school/church, etc. how many visits to the dentist they made last year. 2 a For example: People in the USA and Europe are similar, so they would have similar taste in car colour. b For example: Different cars are sold in different parts of the world, so the most popular colours may be different too. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 6 Answers to Coursebook exercises F Exercise 6.3 1 Colour Designing data-collection sheets Tally Frequency Red Yellow Blue Green 2 Make of car Tally Frequency BMW Ford Nissan Toyota Vauxhall Other 3 Number of holidays Tally Frequency 0 1 2 3 4 5 6 4 Score Tally Frequency 2 3 4 5 6 7 8 5 Flavour Stage 7 Stage 8 Stage 9 Stage 10 Stage 11 Vanilla Strawberry Chocolate Raspberry ripple Mint choc-chip Other 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises Unit 6 6 a No ‘less than 20’ category, overlapping values, no ‘over 50’ category. b Age (years) Tally Frequency 10–19 20–29 30–39 40–49 50–59 60+ Total 7 aNo ‘zero’ option, overlapping values, different sized groups, no ‘7 or more’ option, can’t tell whether this is for men or women. b Men Women Number of times Tally Frequency Tally Frequency 0 1–2 3–4 5–6 7+ F Exercise 6.4 1 a Collecting data Number Tally Frequency 1 //// / 6 2 //// 4 3 /// 3 4 //// // 7 5 // 2 6 //// /// 8 Total 30 b The number 6 is the most common number rolled. The number 5 is the least common number rolled. 2 a Number Tally Frequency 0–9 0 10–19 0 20–29 // 2 30–39 //// 4 40–49 //// / 6 50–59 //// 5 60–69 /// 3 Total 20 b The most common score was 40–49 points. 3 a Number Tally Frequency 50–59 //// 4 60–69 //// /// 8 70–79 //// 5 80–89 //// / 6 90–99 / 1 Total 24 b The most commonly found masses were in the 60–69 kg group. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 3 Unit 6 Answers to Coursebook exercises 4 a b c d A suitable question, not requiring measurement of any sort A suitable data-collection sheet for the survey Completed data-collection sheet A valid conclusion 5 a b c d A suitable question, one which requires measurement of some sort A suitable data-collection sheet for the survey Completed data-collection sheet A valid conclusion End-of-unit review 1 Good basketball players are also good at rugby. 2 a 1. For example: Boys eat more chocolate than girls do. 2. For example: ‘How much chocolate do you eat per week?’, ‘How many chocolate bars do you eat, on average, per week?’ 3. For example: amount of chocolate eaten by boys and girls 4. For example: survey 5.For example: whole class (if there is an equal gender ratio in the class), or 10% of Maha’s school, with equal number of boys and girls 6. For example: as accurate as possible b Will people they tell the truth? Will they remember chocolate bars but forget individual chocolates they’ve eaten? She needs to find a way of defining the size of chocolate bars. c Some might not want to tell her the truth. Some might not be able to remember accurately. 3 aAmerican women and Canadian women must be fairly similar, so they must have about the same number of shoes. b Different climates mean different footwear, possibly American women have (and spend) more (or less) money than Canadian women. 4 a For example: This average would have been based on a large sample of shop assistants. b For example: Shop assistants in cities might earn more than those not in cities, so the true average might be lower. 5 aNo ‘zero’ option, overlapping values, different sized groups, no ‘7 or more’ option, can’t tell whether this is for men or women. Note: different number of men and women in the sample doesn’t affect her data collection sheet. b Men Women Number of films Tally Frequency Tally Frequency 0 1–2 3–4 5–6 7+ 6 a Number of goals Tally Frequency 0 //// // 7 1 //// 5 2 //// 4 3 /// 3 4 0 5 0 6 / 1 Total 20 b The most common number of goals scored is 0. 4 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 7 Fractions ✦ Exercise 7.1 Writing a fraction in its simplest form b 4 c 3 d 3 5 5 1 b 2 c 2 d 3 5 3 13 = 13 × 12 = 156 , wrong. Ansswer = 13 19 19 × 12 228 18 34 = 34 × 9 = 306 , wrong. Answerr = 35 37 37 × 9 333 38 9 b 13 c 11 d 21 14 18 1 a 2 2 a 3 b c 4 a ✦ Exercise 7.2 1 a 5 9 g 17 15 4 5 3 4 e 2 13 15 e 2 a 10 + 12 = 15 15 b 69 − 46 = 12 12 1 3 a 3 b 8 g 29 h 10 7 11 f f 6 7 3 5 f 3 5 Adding and subtracting fractions b 7 h 3 3 e 5 c 9 d 4 10 14 9 5 i 17 j 3 118 20 12 22 , 22 = 1 7 8 + 1 7 = 9 7 15 15 15 15 15 23 = 1 11 23 12 12 12 23 7 c 5 d 5 13 5 15 36 28 5 i 19 j 43 1 24 14 4 e 2 f 5 e 18 3 f 10 13 3 k 2 15 40 k 45 12 8 l 11 40 30 l 1 23 36 4 e.g. 1 + 1 = 1 , 1 + 1 = 2 4 4 2 3 3 3 5 a 31 m 20 b Check students’ answers. 6 a 23 m 24 b Check students’ answers. ✦ Exercise 7.3 Multiplying fractions 1 a 9 b 20 c 36 d 27 e 84 f 140 2 a 13 1 2 b 17 1 3 c 62 3 d 31 1 2 e 21 2 f 22 1 2 3 a 15 b 3 5 16 k 10 33 f 28 1 g 3 4 a 52 5 e 17 1 4 h b f 5 For example, 3 2 6 a 3 32 b c 18 d 10 21 10 55 3 2 i j 2 15 3 8 c 39 d 41 8 14 16 2 4 1 g 1 h 39 10 7 9 11 × 32 = 9 , 32 < 9 ; 6 × 6 = 36 = 9 = 2 1 , 4 4 4 4 4 16 4 1 4 Copyright Cambridge University Press 2013 e 8 39 l 6 35 6 <21 4 4 Cambridge Checkpoint Mathematics 9 1 Unit 7 Answers to Coursebook exercises F Exercise 7.4 1 a 28 g 22 1 2 Dividing fractions b 35 h 30 1 3 c 63 d 22 i 13 1 2 j 27 1 2 c 1 19 d 11 5 18 36 g 4 h 11 i 11 j 20 3 4 21 5 7 99 32 3 a b 1 c d 6 20 124 39 3 2 e 2 f 2 g 1 h 2 95 8 7 1 1 5 3 3 14 4 For example, 2 ÷ 3 = ; 1 ÷ 2 = 2 2 7 4 8 19 6 14 1 5 a b 2 c 1 d 11 15 7 7 9 2 a 11 20 b 117 F Exercise 7.5 1 a 1 2 8 15 2 9 1 6 1 15 1 6 1 2 5 6 g 2 a g 3 a g 4 a g b h b h b h f 40 k 69 1 3 l 73 1 2 e 15 9 f 11 6 k 11 9 l e 11 27 f 11 11 9 10 Working with fractions mentally b 7 h e 33 8 11 28 1 8 11 20 2 21 1 15 1 3 13 7 c i c i c i c i 7 10 19 45 2 15 3 14 9 20 8 11 5 7 7 8 d 7 e 11 8 j d j d j d j 15 12 1 2 13 28 16 63 20 27 5 8 11 9 6 33 k 40 e 7 10 f 14 l f 5 k 24 l 8 45 k 3 5 e 14 5 k 11 12 f e 5 f e l f l 15 1 112 3 20 5 36 24 65 6 11 1 12 8 9 5 12 6 13 30 5 9 20 8 a 1 3 7 a 3 20 b 2 9 b End-of-unit review b 4 c 3 3 5 4 7 a . Student’s check 9 5 a b 1 c 17 8 3 30 7 a 1 m b Student’s check 20 a 9 b 180 c 38 9 8 1 a b 16 25 a 15 b 24 1 c 62 1 2 2 3 8 4 For example, ÷ = 3 2 9 1 a 1 2 3 4 5 6 7 8 2 Cambridge Checkpoint Mathematics 9 d 2 5 7 d 51 24 d 8 45 d 17 20 7 9 e 2 11 f 3 19 e 2 f e 15 f 11 5 15 3 28 24 1 4 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 8 Constructions and Pythagoras’ theorem F Exercise 8.1 1 Constructing perpendicular lines Check students’ drawings, all measurements ± 2 mm and ± 2°. 2 Check students’ drawings, all measurements ± 2 mm and ± 2°. 3 a Check students’ drawings, all measurements ± 2 mm and ± 2°. b i 60° ± 2° ii 180° – 90° – 30° = 60° 4 a Check students’ drawings, all angles ± 2°. b Check students’ drawings, all angles ± 2°. 5 Check students’ drawings, all angles ± 2°. 6 a Check students’ drawings, all angles ± 2°. b i ∠ABC + ∠BAD = 180° ± 2° F Exercise 8.2 1 a b c d ii 360° – 90° – 90° = 180° Inscribing shapes in circles Check students’ constructions of inscribed square, including construction lines. Check students’ constructions of inscribed regular octagon, including construction lines. Check students’ constructions of inscribed equilateral triangle, including construction lines. Check students’ constructions of inscribed regular hexagon, including construction lines. 2 a Check students’ constructions of inscribed square, including construction lines. All measurements ± 2 mm and ± 2° b 8.5 cm ± 2 mm c Area of circle = 113.04 cm² Area of square: x × x = 8.5* × 8.5* = 72.25** cm2 Shaded area: Area of circle – Area of square = 113.04 – 72.25** = 40.79*** cm2 *Students’ measurements **68.89 to 75.69 inclusive ***37.35 to 44.15 inclusive . 3 a 153.86 − 1168394 = 18.5103 cm2(14.6167 to 22.2783 cm inclusive) 2 113 . 04 − 132665 . = 9.0275 cm2 (5.5107 to 12.4187 cm2 inclusive) b 2 c 153.86 − 132.665 = 10.5975 cm2 (6.4527 to 14.6167 cm2 inclusive) 2 4 a Check students’ constructions of inscribed regular octagon, including construction lines. All measurements ± 2 mm and ± 2°. Check students’ calculations. b A statement to the effect that doubling dimensions quadruples area. c Check students’ constructions of inscribed regular octagon, including construction lines. All measurements ± 2 mm and ± 2°. Check students’ calculations based on diagrams. F Exercise 8.3 1 a 10 cm 2 a 3 cm Using Pythagoras’ theorem b 15 mm c 12.21 cm d 10.30 m b 2.4 m c 12 m d 21.35 cm 3 13 cm 4 14.42 km 5 314 cm2 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 8 Answers to Coursebook exercises End-of-unit review 1 Check students’ drawings, including construction lines, all measurements ± 2 mm and ± 2°. 2 a Check students’ drawings, including construction lines, all measurements ± 2 mm and ± 2°. b i 50° ± 2° ii Angle sum of triangle = 180°, 180° – 90° – 40° = 50°. 3 Check students’ drawings, all angles ± 2°. 4 C heck students’ constructions of inscribed square, including construction lines. All measurements ± 2 mm and ± 2°. 5 C heck students’ constructions of an inscribed equilateral triangle and an inscribed regular hexagon, including construction lines. All measurements ± 2 mm and ± 2°. 6 8.94 mm 7 0.33 m 8 12.53 cm 9 60 cm2 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 9 Expressions and formulae ✦ Exercise 9.1 Simplifying algebraic expressions 1 6 a x g q5 b y h r3 c z10 i t5 d m14 j u2 e n12 k v f l p7 w7 2 a 6x5 g 3q4 b 12y9 h 3r4 c 30z7 i 3t4 d 4m7 j 2u5 e 4n13 k 2v4 f l 8p3 5w 3 a D b B c C d D 9 4 a Group 1: all have an x9 when simplified; 8x6 × x3, 4x5 × 2x4, 12x10 ÷ 2x Group 2: all have an x6 when simplified; 6x3 × 2x3, 12x8 ÷ x2, 2x3 × 3x3 b 3x2 × 4x3. This card doesn’t fit as it has an x5 when simplified. ✦ Exercise 9.2 1 a 7n e 2n + 9 i 2 a b c d Constructing algebraic expressions c n–2 g n6 − 4 d 20 – n h n2 100 n b n + 12 f n2 j 2n – 1 k 5(n + 2) l i i i i ii ii ii ii xy 12xy x2 4y2 2x + 2y 8x + 6y 4x 8y 3 a P = 2x + 10, A = 3x + 6 c P = 4n + 8, A = n2 + 4n 8(n – 7) b P = 2y – 4, A = 4y – 24 d P = 10p + 6, A = 4p2 + 12p 4 a i 2 red + 2 yellow = 4 green; both = 8x + 4 ii 3 red + 3 yellow = 6 green; both = 12x + 6 iii 4 red + 4 yellow = 8 green; both = 16x + 8 b n red + n yellow = 2n green (or similar explanation given in words) c i 3 red + 1 yellow = 6 blue; both = 6x + 6 ii 6 red + 2 yellow = 12 blue; both = 12x + 12 iii 9 red + 3 yellow = 18 blue; both = 18x + 18 d 3n red + n yellow = 6n blue (or similar explanation given in words) ✦ Exercise 9.3 1 a 9 e 8 i 5 2 a 21 e 68 i 18 Substituting into expressions b 4 f 0 j 47 c 9 g 8 k –30 d 2 h 30 l –4 b 36 f 64 j –25 c 10 g 3 k –7 d 16 h –18 l 5 3 a For example: Let x = 2; 3x2 = 3 × 22 = 12 and (3x)2 = (3 × 2)2 = 36, so 3x2 ≠ (3x)2 b For example: Let y = 4; (–y)2 = (–4)2 = 16 and –y2 = – (42) = –16, so (–y)2 ≠ –y2 c For example: Let a = 2 and b = 3; 2(a + b) = 3(2 + 3) = 15 and 2a + b = 2 × 2 + 3 = 7, so 2(a + b) ≠ 2a + b Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 9 Answers to Coursebook exercises F Exercise 9.4 1 a S = 60M Deriving and using formulae b S = 900 S c M = 60 c m = 12 d M = 22.5 2 a F = 60 b F = –78 d a = –1.75 3 a v = 87 e t = 10 b v = 125 f a=2 c u = 27 d u = 46 4 a x+2 b T = 2x + 2 c T = 40 d x = T 2−2 5 a 20% b 60% c 125% 6 a 65 kg b 49.1 kg (1 d.p.) c 95.9 kg (1 d.p.) d 57.3 kg (1 d.p.) e x = 23 7 Sasha is correct as 30 °C = 86 °F and 86 °F > 82 °F (or 82 °F = 27.8 °C and 27.8 °C < 30 °C). 8 a She is not underweight as her BMI is 20.05, which is greater than 18.5. b 3.7 kg F Exercise 9.5 1 Factorising a 3(x + 2) b 5(2y – 3) c 6y(x + 2) d x(4x + 1) e 3(3 – 4y) f y(2y – 7) 2 a 2(x + 2) g 5(2 – x) b 3(y – 6) h 7(2 + 3x) c 5(2z + 1) i 2(4 – 5y) d 4(2a – 1) j 6(3 + 4z) e 2(2b + 3) k 3(3 + 5m) f l 4(4n – 5) 10(3 – 2k) 3 a x(3x + 1) g 9(2y – x) b 6y(y – 2) h 3(4y + 3x) c z(z + 4) i 4y(2x – 1) d 2a(2 – a) j 5z(3 + 2y) e 3b(1 + 3b) k 2m(7 + 3n) f l 3n(4 – 5n) 13k(2 – p) 4 a 2(x + 3y + 4) d x(5x + 2 + y) b 4(y – 2 + x) e y(9 – y – x) c 3(3xy + 4y – 5) f 3y(y – 3 + 2x) 5 5(2x + 6) + 2(3x – 5) = 10x + 30 + 6x – 10 = 16x + 20 = 4(4x + 5) 6 6(3y + 2) – 4(y – 2) = 18y + 12 – 4y + 8 = 14y + 20 = 2(7y + 10) ≠ 2(7y + 2) The mistake he has made is when he has expanded. He has done –4 × –2 = –8 and so his expansion is 18y + 12 – 4y – 8 = 14y + 4 = 2(7y + 2). F Exercise 9.6 1 a 2x b 4x 5 3y g 4 2 a Adding and subtracting algebraic fractions 7 7y h 9 a + a = 5a + 2a 2 5 10 10 = 5a + 2a 10 7 a = 10 5 3 a x+y 5 3 a + 4b g 12 4 a A, D, F d x e 2x f i j k l 4 7y 10 3 3y 8 b b + b = 3b + 4b d 5d + 3d = 25d + 18d 6 c x 30 30 = 25d + 18d 30 43 d = 30 b 3x + y 6 12 a + 5b h 30 7 3 12 12 = 3b + 4b 12 7 b = 12 e 5e + 2e = 15e + 16e 8 3 24 24 = 15e + 16e 24 31 e = 24 9 10 a + 9b i 24 b B, C, E 2x 3 3y 14 c 5c − 2c = 25c − 14c 4 c 6x + y 5 9y 25 f d 4x − y j 5 10 8a − 5b 40 35 35 = 25c − 14c 35 11 c = 35 9f 3f 18 f 15 f + = + 10 4 20 20 18 f + 15 f = 20 33 f = 20 e 11x − 4 y 14 9 a − 2b k 30 f l 9x − 8 y 20 20a − 27b 45 c G; the answer is x 3 d You can ignore the letter, work out the fractions, then put the letter back in at the end. 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises F Exercise 9.7 1 Unit 9 Expanding the product of two linear expressions a (x + 4)(x + 1) = x2 + 1x + 4x + 4 = x2 + 5x + 4 b (x – 3)(x + 6) = x2 + 6x – 3x – 18 = x2 + 3x – 18 c (x + 2)(x – 8) = x2 – 8x + 2x – 16 d (x – 4)(x – 1) = x2 – x – 4x + 4 2 = x – 6x – 16 = x2 – 5x + 4 2 a x2 + 10x + 21 d x2 + 4x – 32 b x2 + 11x + 10 e x2 – 9x + 14 c x2 + 2x – 15 f x2 – 14x + 24 3 a y2 + 6y + 8 d a2 – 7a – 18 b z2 + 14z + 48 e p2 – 11p + 30 c m2 + m – 12 f n2 – 30n + 200 4 a C b B c A 5 a (x + 2)2 = (x + 2)(x + 2) = x2 + 2x + 2x + 4 = x2 + 4x + 4 6 a y2 + 10y + 25 d a2 – 4a + 4 7 a b c d d C b (x – 3)2 = (x – 3)(x – 3) = x2 – 3x – 3x + 9 = x2 – 6x + 9 b z2 + 2z + 1 e p2 – 8p + 16 c m2 + 16m + 64 f n2 – 18n + 81 ii x2 – 25 iii x2 – 49 i x2 – 4 There is no term in x, and the number term is a square number. x2 – 100 x2 – y2 8 a 33 × 29 = 957, 28 × 34 = 952, 957 – 952 = 5 b 16 × 12 = 192, 11 × 17 = 187, 192 – 187 = 5 c The answer is always 5. d n n+1 n+5 n+6 e (n + 5)(n + 1) = n2 + 6n + 5, n(n + 6) = n2 + 6n, n2 + 6n + 5 – (n2 + 6n) = 5 The answer is always 5. End-of-unit review 1 a x5 g q6 b y12 h r3 c z10 i t5 d 15m9 j 2u2 2 a 3a b 2b + 16 c 5c + 2d d 16z – 2 3 a ab b 40cd c w d 9e2 4 a 13 e 3 i 89 b 19 f 48 j 0 c 13 g –8 k 84 d 54 h 21 l – 42 5 a x = 19 b x=–4 c y = 65 6 a 2(x + 3) g x(5x + 1) b 4(y – 3) h a(3 – 5a) 7 a 2x b 3x 3 g x+y 4 5 h 12x − y 20 8 a x2 + 7x + 10 d x2 – 14x + 40 e 6n11 k 3v6 f l 6p7 7w d y = 60 e z=2 f z=6 c 3(a – 1) i 8(4y – x) d 10(2 – x) j 3y(2x – 1) e 6(4 + 5z) k 2m(9 + 4n) f l 10(5 – 3b) 3n(8 – 9n) c 3x d f i j 15 y 8 20a − 3b 24 y 6 12a − 14b 21 b x2 + x – 12 e x2 – 64 2 7 5a + 3b 15 2y 15 15a + 8b 20 c x2 – 3x – 54 f x2 – 12x + 36 e k l 9 4(2x + 5) + 3(8x – 4) = 8x + 20 + 24x – 12 = 32x + 8 = 8(4x + 1) Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 3 Answers to Coursebook exercises 10 Processing and presenting data F Exercise 10.1 1 Calculating statistics a i 21 minutes b i 21 minutes ii 21.125 minutes ii 23.375 minutes 2 a 51 b 14 3 a 53.5 b 13 4 a 54 b 12.52 5 a 9 years b 4 years iii 25 minutes iii 43 minutes 6 a $60–79 b Because the data is grouped. We do not know the exact amounts the workers earn. Please note that calculating the mean of grouped data is beyond the scope of the syllabus and will not be tested. But, you can inform students, if you wish, that it is only possible to find an estimate of the mean from grouped data. c Xavier could be correct. The range is between 40 (99.5 – 59.5) and 80 (119.5 – 39.5). 7 a The median is between 0 and 4 °C. b The range is between 10 and 20 °C. 8 52, 57 and 89 9 a 7 b 6.5 c Increases to 13.4 d Increases to 8 10 a Mean = 13.9 kg, median = 13.1 kg, range = 6.1 kg b Mean = 25.0 kg, median = 23.4 kg, range = 12.2 kg 11 a 30 b 1.50 m F Exercise 10.2 1 c Cannot say Using statistics The median is 5 minutes and the mean is 5.3 minutes. 2 T he median is 1 day. This is a better average, as there are as many values above as below. The mean is 1.63 days, but is affected by one high value. The mode is zero. 3 a 2.58 b The range for League One is 5 goals; for League two it is 4 goals. 4 The most useful average is the mode, which is 38 cm. 5 The median salary is $29 500. The mean is $33 000, which is affected by three very high values. 6 The modal class is 110–120 minutes 7 T he summer mean is 5.3 and the winter mean is 11.4. On average there are about twice as many breakdowns in the winter. The summer range is 7 and the winter range is 13. There is greater variation in the winter. 8 a The means are 87.7, 89.9 and 91.0 minutes. Andi is fastest, on average. Bart has the fastest single time. b The ranges are 8, 22 and 3 minutes. Chris is the most consistent. 9 O bi sent more messages (134 compared to 84) and on average they were longer than Darth’s (the modal class for Obi is 41–60 and for Darth it is 21–40). The range of lengths was greater for Obi. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 10 Answers to Coursebook exercises End-of-unit review 1 a 1 b 1.5 c 1.75 d 4 2 a 1 b 2 c 1.96 d 8 3 a 4 b 4.54 c 4 4 T he children are better. Use either the modal class or the median. The modal class for the children is 70–74 and for adults is 75–79. The median for the children is in the class 70–74 and that for adults is in the class 75–79. Also the range for children is smaller because there are no children’s estimates above 84. 5 F ind either the medians or the means, and the ranges. The medians are Coola 58 and Freezy 63. The means are 55.6 and 61.4. Freezy has a higher average. The ranges are Coola 37 and Freezy 64. Freezy sales are more varied. 6 aYou can use the median or the mean. The medians are 32 and 38.5 so Clancy has a better median. The means are about 38.5 and 39, which are about the same. b The ranges are 95 and 41 so Bristoe has more varied scores. 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 11 Percentages F Exercise 11.1 1 a 29.4 Using mental methods b 112.7 c 3388 d 7821 e 25 200 2 a 50% + 25% + 10% is a possible answer. There are others. b i 6120 g ii $54.4 iii 3.06 m iv 1530 ml 3 a 10 b 35 c 93 d 325 4 a $40.56 b 10.14 kg c 50.7 km d 81.12 million 5 a 24.32 b 24.32 c 97.28 d 48.64 6 a They are both 19.5. 7 1 e 39.6 b They are both 9. c A% of B is the same as B% of A. Percentage 5% 20% 40% 60% 80% 120% Amount ($) 9 36 72 108 144 216 F Exercise 11.2 v 72.25 seconds Comparing different quantities a Cinema 62%, theatre 51% b Cinema 38%, theatre 49% 2 a 7% (or 7.3%) b 13% (or 12.6%) 3 a Alphatown 37% (or 36.7%); Betatown 30% (or 30.1%) c 10% (or 9.8%) b Alphatown 4 a 16% (or 15.8%) bThe percentage of women who are non-smokers (78%) is less than the percentage of non-smoking men (84%). 5 3 9% of the engineers own a car less than five years old but only 29% of accountants do. (Or any equivalent statement.) F Exercise 11.3 1 Percentage changes Game 25%, phone 8%, computer 2% 2 a 6.6 kg, 15.4 kg and 27.5 kg 3 a 17% (or 16.7%) b 39 cm (or 39.2 cm) 4 7% (or 7.1%) 5 a 33.3% b i 37.5% increase ii 27.3% increase iii 42.9% decrease 6 a The sale price was $19 200. So the second percentage should be based on $19 200, not $20 000. b $19 968 7 a 57% b 92% c 200% Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 11 Answers to Coursebook exercises F Exercise 11.4 1 Practical examples 25% 2 40% (or 40.3%) 3 a $287 b $1107 4 $3.90 5 a i $48.53 b i $0.84 ii $22.42 or $22.43 ii $0.39 6 a $360 b $4860 7 a $25.20 b 3.8% iii $74.52 iii $1.30 8 $1408 9 a $8293.50 b 15.3% c $1382.25 10 15% discount is best. 15% is $1046.25 and 1 off is $871.88. 8 11 53% (or 53.1%) profit 12 a Clock 71% profit, necklace 126% profit, 19% loss b 3% (or 2.6%) loss End-of-unit review 1 a 50.4 b 288 c 6.39 d 195.3 2 a 700 mm b 384 kg c 912 hours 3 a 232.2 b 11.61 c 116.1 4 a 27% b 68% c No. 32% of the women cycle and only 27% of the men. d 580.5 5 School A 67%, school B 91%. The percentage from school B is greater. 6 a The population is 9600. b The price is $40 420. c The mass is 5.76 kg. 7 A price can increase by any percentage. It cannot decrease by more than 100%. 8 $123.76 9 The percentage profit is 53% (or 53.3%). 10 $590.75 11 a $11 172 2 b 21.2% Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 12 Tessellations, transformations and loci F Exercise 12.1 1 Tessellating shapes Chek students’ diagrams showing tessellations of each shape. 2 Exterior angle = 360° ÷ 6 = 60° Interior angle = 180° − 60° = 120° Angles around a point = 360°, 360 ÷ 120 = 3 hexagons Three hexagons will fit around a point, leaving no gaps. 3 a Exterior angle = 360° ÷ 8 = 45° Interior angle = 180° − 45° = 135° Angles around a point = 360°, 360 ÷ 135 = 2.6... hexagons Only two octagons will fit around a point, leaving a gap of 360° − 270° = 90°. b Same working as for part a. Only two octagons will fit around a point, leaving a gap of 360° − 270° = 90°. A square tile has an interior angle of 90°, so will fit around a point with two octagons, leaving no space. F Exercise 12.2 1 Solving transformation problems y 3 2 1 c a A 0 –4 –3 –22 –11 –11 d 1 2 3 –2 4 x b –3 2 a y 6 5 4 3 2 1 b iv ii i B iii 0 0 3 y 6 5 4 3 2 1 1 2 3 4 5 6 7 iii iv i B 0 x b y 6 5 4 3 ii 2 1 0 1 2 3 4 5 6 7 x c a 0 0 1 2 3 4 5 6 7 8 Copyright Cambridge University Press 2013 9 10 x Cambridge Checkpoint Mathematics 9 1 Unit 12 Answers to Coursebook exercises 4 a b (1, −2) y X 4 3 2 1 0 –4 –3 –2 –1 –1 Y –2 –3 –4 1 2 3 4 x Z 5 a A (2, 5), B (6, 5), C (5, 1), D (1, 3) b y=x y 6 5 4 3 2 1 B′ C′ A B D A′ C D′ 0 0 1 2 3 4 5 6 7 1 2 3 x c A'(5, 2), B'(5, 6), C'(1, 5), D'(3, 1) d The x and y coordinates have changed places. F Exercise 12.3 1 a Transforming shapes A –4 –3 –2 –1 –1 –2 2 y A 0 1 2 3 4 x 4 3 2 1 –4 –3 –2 –1 –1 –2 0 4 x y b 6 5 4 c 3 2 1 –6 –5 –4 –3 –2 –1 –1 –2 –3 B –4 –5 –6 2 b y 4 3 2 1 a 0 1 2 3 4 5 6 x d Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 3 a Reflection in the y-axis c Reflection in the line y = 1 b Reflection in the x-axis d Reflection in the line x = 1 4 a Translation 4 b Translation 3 −1 −4 e Translation −3 d Translation −2 1 Translation −5 f 3 5 a Rotation 90° anticlockwise about (−1, 2) c Rotation 180° about (−1, 1) e Rotation 90° anticlockwise about (0, −4) 6 c Translation 0 −2 2 Unit 12 b Rotation 90° clockwise about (−3, 2) d Rotation 180° about (2, −1) y 6 5 4 3 2 1 aii Y ci X 0 –6 –5 –4 –3 –2 –1 1 –1 bii –2 ai –3 –4 –5 cii –6 2 3 Z 4 5 bi 6 x d i The positions of the shapes are different, even though the elements of the transformations are the same. ii Yes, as a different order often results in a different finishing position. iii For example: Reflect in line y = −2, then reflect in line x = 3 7 a i Rotation 90° clockwise, centre (1, 2) ii Translation 2 −4 iii Reflection in the line x = 4.5 b i For example: Translation 0 then reflection in the line x = 1 2 2 ii For example: Rotation 90° anticlockwise, centre (−3, −6) followed by a translation 8 F Exercise 12.4 1 Enlarging shapes y 4 3 2 1 –4 –3 –2 –1 –1 –2 0 1 2 3 4 x Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 3 Unit 12 Answers to Coursebook exercises 2 a b y 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4 c y 4 3 2 1 0 1 2 3 4 x –4 –33 –22 –11 –1 –1 –2 –2 –33 –4 3 a Enlargement scale factor 2, centre (−5, 2) y 4 3 2 1 0 1 2 3 4 x –4 –3 –22 –11 –1 –1 –22 –3 –4 0 1 2 3 4 x b Enlargement scale factor 4, centre (−6, −2) 4 Enlargement scale factor 3, centre (4, −5) 5 a y 10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 x b For example, triangle with vertices at (1, 1), (2, 1) and (1, 3); enlargement scale factor 2, centre (0, 1). If Ahmad is correct, the coordinates of the vertices of the enlargement should be at (2, 2), (4, 2) and (2, 6). y 6 5 4 3 2 1 0 0 1 2 3 4 5 6 x Vertices are at (2, 1), (4, 1) and (2, 5), not (2, 2), (4, 2) and (2, 6). F Exercise 12.5 Drawing a locus 1 P 5 cm 4 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 2 3 2 cm 2 cm 2 cm 2 cm 3 cm 6 cm 3 cm 3 cm A Unit 12 B 3 cm 4 The circle needs to have a radius of 4 cm. D 4 cm 5 C 1.5 cm 10 cm 12 cm 6 a b c 7 W X Z Y 8 P 70 km Q End-of-unit review 1 Exterior angle = 360° ÷ 5° = 72°. Interior angle = 180° − 72° = 108° Angles around a point = 360°, 360 ÷ 108 = 3.3... pentagons. Only three pentagons will fit around a point, leaving a gap of 360° − 324° = 36°. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 5 Unit 12 2 y 6 5 4 3 2 1 Answers to Coursebook exercises C D A B 0 0 1 2 3 4 5 6 7 x 3 a Reflection in the line y = −1 4 Translation −4 c b Rotation 180°, centre (1, 0) y 4 a b 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4 0 e Translation −2 d Rotation 90° anticlockwise, centre (−3, 2) 1 2 3 4 x 5 y 4 3 2 1 –7 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 0 1 2 3 4 x 5 Enlargement scale factor 3, centre (1, 2) 6 Q 4 cm 7 2 cm X 6 2 cm 8 cm 2 cm Y 2 cm Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 13 Equations and inequalities F Exercise 13.1 1 a x = 1.5 Solving linear equations b x = −1.5 c x = −5.5 d x = −1.5 2 a a = −11 b a=5 c a=6 d a = −6 3 a y=8 b y = 16 c y=4 d y=8 4 a d=5 b d=7 c d = −1 d d = −2 5 a 2x + 24 = 4x − 6 → x = 15 b x + 12 = 2x − 3 → x = 15 6 a x=1 b x = −0.25 or − 1 4 c x = −1 7 a p = 0.5 b p = 3.5 c p = 1.5 c z = 3 13 8 a x = 1 17 b y= 7 9 d p = −1.5 9 2x + 16 = 18 − 3x → 5x = 2 → x = 0.4 10 a 5x = 65 → x = 13 b 2(x − 4) = x + 5 → 2x − 8 = x + 5 → x = 13 b −19 = 5x → x = −3.8 d −5x = 19 → x = −3.8 11 a Add 2x to both sides. c Subtract 6 from both sides. 122(x + 3) + x = 0 x + 2(x − 3) = 0 3x − 2(x + 3) = 0 −(x + 2) + 2(x − 3) = 0 x − (2 − x) = 0 13 a m = 11 b m = −5 14 a x = 2 b x=4 F Exercise 13.2 x=8 x=6 x=2 x=1 x = −2 c m = 11 Solving problems a 4 b 3a + 6 = 100 c a = 31 13 2 a x−2 b 4x − 4 = 84 c x = 22 3 a 6N + 3 = 141 b 23 c 69 and 72 1 4 a i A + 10 ii A − 6 b A + 10 = 2(A − 6) 5 a 5x + 2 = 57 b 11 cm, 19 cm and 27 cm 6 a 9a = 4a + 20 b a=4 F Exercise 13.3 1 c Both are correct. c d 31 13 , 33 13 , 35 13 d 440 cm2 c 22 Triangle sides 12 cm, rectangle sides 7 cm by 11 cm Simultaneous equations 1 x = 5, y = 9 2 x = −5, y = −14 3 x = 7, y = −5 4 a x = 3, y = −2 b x = 3, y = 16 c x = 4, y = −6 5 x = 8, y = 24 6 x = 3, y = 6 7 x = −3, y = 6 8 x = 6, y = 2 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 13 Answers to Coursebook exercises F Exercise 13.4 1 a x = 9, y = 6 Simultaneous equations 2 b x = 15.5, y = 14.5 2 a x=8 b y=3 3 a y=2 b x = −3 4 a 40 b x = 10, y = 4 5 a x = 9, y = 4 b x = 5, y = −2 F Exercise 13.5 1 a x=5 c x = 8, y = −6 c x = −4, y = 8 Trial and improvement b x = 10 c x=4 d x=7 2 a x = 1.5 b x = 5.5 c x = 4.5 d x = 10 3 a 18 b x = 5.8 4 a w=7 b w = 7.2 c w = 7.4 5 a = 4 gives 36; solution is a = 3.6. 6 x = 4.3 7 a 8² + 10 × 8 = 144 < 150; 9² + 10 × 9 = 171 > 150 b x = 8.2 8 x = 1.5 9 y = 4.3 10 a x = 2.8 b x = 7.2 F Exercise 13.6 1 a x≤2 Inequalities b x > −2 c x ≥ 10 d x < −20 2 a 0 3 b –3 0 c 0 d –20 0 3 a 7 b −4 c –2, –1, 0 or 1 4 a x > 1.4 b x ≤ 3.5 c x < −2 13 d x ≥ −3 5 a 0 1.4 0 3.5 b c –2 13 0 –3 0 d 6 a 2z + 9 > 13 b 3(z − 4) > −6 c 4 + 2z > 8 d 5(3z − 2) > 20 7 a a < 3.5 b b ≥ 11 c c≤6 d d > −27 8 a 5n + 5 ≤ 30 b n≤5 c 5, 12 and 13 9 a 4x + 30 < 360 2 b x < 82.5 Cambridge Checkpoint Mathematics 9 c 90 > 82.5 Copyright Cambridge University Press 2013 Answers to Coursebook exercises Unit 13 End-of-unit review 1 a x=9 b x = 12 c x = −4.5 2 a m = 8.5 b m = 22 c m = 4.25 3 a 8x = 2x + 36 or 4x = x + 18 b x=6 d x = −9 c 14 cm 4 a 2(N + 10) = 4(N − 10) b N = 30 5 a x = 8, y = 16 b x = 30, y = 40 c x = 15, y = 11 6 Solve the equations x + y = 100 and x – y = 95 to get the numbers 2.5 and 97.5. 7 x = 6.4. Here are some possible values. x 6 7 6.5 6.4 6.3 3x + x² 54 70 61.75 60.16 58.59 8 a x≥7 b x ≤ −4 c x > 14.5 10 a 6x + 3 < 50 b x < 75 c 7 11 a True b True c False 9 a b 0 7 –4 c 0 0 14.5 6 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 3 Answers to Coursebook exercises 14 Ratio and proportion F Exercise 14.1 1 Comparing and using ratios a Sky blue 1 : 1.67, Sea blue 1 : 1.75 b Sky blue 2 a Angelica 1 : 2.5, Shani 1 : 2.2 3 a 1 : 1.38 b Angelica’s b 1 : 1.35 c The Seals 4 200 g 5 a 7.5 kg cement and 30 kg gravel 6 Age of children Child : staff ratios Number of children Number of staff up to 18 months 3:1 10 4 18 months up to 3 years 4:1 18 5 3 years up to 5 years 8:1 15 2 5 years up to 7 years 14 : 1 24 2 Total: 13 7 a 13 g b 16 g F Exercise 14.2 1 b 52.5 kg c 9g Solving problems a Yes, as the number of cartons bought increases, so does the total cost (the ratio stays the same). b No, the ratio does not stay the same. c Yes, as the number of tickets bought increases, so does the total cost (the ratio stays the same). d Yes, as the distance increases, so does the number of litres of petrol (the ratio, on average, stays the same). e No, the ratio does not stay the same. f No, the ratio does not stay the same. 2 a $6 b $18 3 a $28.50 b $13.30 4 a $0.65 b $0.62 5 a Box of 150 paper towels c $4.50 c 750 ml b 400 g pack of cheese 6 £160 7 a €383.50 b £68 8 $449 = €342.75; €359 = $470.29. She should buy the camera in America. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 14 Answers to Coursebook exercises End-of-unit review 1 a Sea green 1 : 1.4, Fern green 1 : 1.375 2 a 1 : 1.5 b 1 : 1.75 b Fern green c The Dales 3 160 ml 4 a 42 g b Check students’ checking methods. 5 a Yes, as the number of packets bought increases, so does the total cost (the ratio stays the same). b No, the ratio does not stay the same as the numbers are not related. c Yes, as the amount of fuel bought increases, so does the total cost (the ratio stays the same). d No, the ratio does not stay the same as the numbers are not related. 6 a $44 b $13.75 7 a $3.28 b $3.16 8 a €522 b £55 c 300 g jar 9 $695 = £439.87, £479 = €756.82. He should buy the laptop in America. 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 15 Area, perimeter and volume F Exercise 15.1 1 a d g j 40 000 cm2 800 mm2 5 m2 9 cm2 Converting units of area and volume b e h k 5000 cm2 80 mm2 4.2 m2 7.6 cm2 c f i l 16 500 cm2 1240 mm2 0.8 m2 0.2 cm2 2 a 7 000 000 cm3 e 400 mm3 i 12.3 m3 b 750 000 cm3 f 6350 mm3 j 4 cm3 c 1 200 000 cm3 g 6 m3 k 0.54 cm3 3 a 60 ml d 8 litres g 3000 cm3 b 125 ml e 2.4 litres h 4200 cm3 c 4700 ml f 0.85 litres i 750 cm3 d 3000 mm3 h 0.35 m3 l 62.5 cm3 4 a 44.8625 m2 (Check students’ explanations of estimation.) b $2520 (Check students’ explanations of inverse operations.) 5 2 tins (Check students’ explanations.) 6 67.5 (Check students’ explanations.) 7 a 196 cm3 or 196 ml b 1 jar holds about 200 ml, so 10 of these would be 2 litres, so Eloise is wrong as she only has 1.2 litres. c 5 large + 3 small = 1205 ml (1 jar with a 5 ml gap) or 3 large + 8 small = 1188 ml (wasting 12 ml, but all jars full) F Exercise 15.2 1 a 30 000 m2 d 124 000 m2 Using hectares b 46 000 m2 e 7500 m2 c 8000 m2 f 250 m2 2 a 5 ha d 0.15 ha b 8.9 ha e 0.09 ha c 24 ha f 126.5 ha 3 a 429 000 m2 b 42.9 ha 4 a 47.3 ha b $186 835 c 50 × 4000 = $200 000 5 3567 ha × $5120 = $18 263 040 They cannot afford to buy the land as $18.26 million > $16 million. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 15 Answers to Coursebook exercises F Exercise 15.3 1 Solving circle problems a A = 201.1 cm2, C = 50.3 cm c A = 38.5 m2, C = 22.0 m e A = 63.6 m2, C = 28.3 m b A = 706.9 cm2, C = 94.2 cm d A = 113.1 cm2, C = 37.7 cm f A = 490.9 mm2, C = 78.5 mm 2 a A = 56.55 cm2, P = 30.85 cm c A = 31.81 m2, P = 23.14 m e A = 226.19 mm2, P = 61.70 mm b A = 157.08 cm2, P = 51.42 cm d A = 127.23 cm2, P = 46.27 cm f A = 5.09 m2, P = 9.25 m 3 a d = 18 cm e d = 90 mm b d = 25 mm f d = 64 cm c d = 13 m d d = 2m 4 a r = 8.7 cm e r = 9.0 mm b r = 6.1 cm f r = 12.4 cm c r = 2.5 m d r = 1.4 m b 36.15 m2 c 1539.38 mm2 5 0.9 cm (9 mm) 6 5.23 m 7 121 cm2 8 a 104.55 cm2 F Exercise 15.4 1 b 130 cm2 c 134.4 cm2 Area of cross-section Length of prism Volume of prism a 2 12 cm 10 cm 120 cm3 b 24 cm2 8.5 cm 204 cm3 c 18.5 m 6.2 m 114.7 m3 2 3 a V = 490 cm3, SA = 406 cm2 c V = 462 cm3, SA = 522 cm2 b V = 456 cm3, SA = 438 cm2 4 a V = 942.5 cm3, SA = 534.1 cm2 c V = 17 592.9 mm3, SA = 4272.6 mm2 b V = 353.4 cm3, SA = 322.0 cm2 5 Radius of circle Area of circle Height of cylinder Volume of cylinder a 2.5 m 19.63 m 4.2 m 82.47 m3 b 6 cm 113.10 cm2 4.48 cm 507 cm3 c 2.52 m 20 m2 2.5 m 50 m3 d 4.56 mm 65.25 mm 16 mm 1044 mm3 6 a 5.5 cm 2 Calculating with prisms and cylinders a 120 cm2 2 d 20.26 cm2 2 2 b 4.2 cm Cambridge Checkpoint Mathematics 9 c 2.1 cm Copyright Cambridge University Press 2013 Answers to Coursebook exercises Unit 15 End-of-unit review 1 a d g j 50 000 cm2 8.2 cm2 7000 mm3 450 cm3 b e h k 4 m2 90 00 000 cm3 0.27 cm3 9 litres 2 a 18.6875 m2 (Check: 6 m × 3 m = 18 m2) 3 a 30 000 m d 2 ha 2 4 a 153.9 cm2 b 46 000 m e 9.4 ha 2 c f i l 900 mm2 24.5 m3 80 ml 3600 cm3 b $988 c 8000 m2 f 0.56 ha b 44.0 cm 5 7 cm 6 3.4 cm 7 a 152.5 cm2 b 46.8 cm 8 a 192 cm 3 b 180 cm3 9 b 222 cm2 c 492 cm2 c 444 cm3 10 V = 3619 mm3, SA = 1508 mm2 11 5.2 cm Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 3 Answers to Coursebook exercises 16 Probability F Exercise 16.1 1 Calculating probabilities a 5% b 65% 2 a 11 c 95% b 25 36 b 1 4 c 1 4 a 0.04 b i 0.90 ii 0.84 5 a 30% b 60% c 10% 3 a 36 1 2 F Exercise 16.2 1 a, b T H 8 Sample space diagrams c i 1 + + + + + + + + + + + + 2 ii 1 iii 1 7 ii 15 iii 4 4 4 1 2 3 4 5 6 2 a 1 3 a b 1 9 1 4 c 16 = 4 4 b 1 4 36 2 4 a 1, 2, 3 on one; 1, 2, 3, 4, 5 on the other b i 5 a 6 + + + 4 + + + 2 + + + 1 2 3 ii 0 3 4 b 2 3 7 a 9 8 7 6 5 4 3 2 1 0 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 2 5 15 The axes can be the other way round. b i 13 6 a 9 c 1 + + + + + + + + + + + + + + + + + + + + iii 2 9 11 c 12 + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + iv 5 9 1 b i 100 ii 19 100 3 b i 20 ii 1 81 iii 100 iv 1 4 0 1 2 3 4 5 6 7 8 9 8 a T H + + + + + + + + + + + + + + + + + + + + 5 1 2 3 4 5 6 7 8 9 10 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 16 Answers to Coursebook exercises 9 a D + + + + A + + + + O + + + + R + + + + D E A R b i 1 16 1 7 16 ii F Exercise 16.3 The letters can be in any order. 3 16 iii iv 13 16 Using relative frequency a 0.45 b 0.75 2 a 0.21 b 0.47 3 a 0.13 b 0.31 4 a 0.016 (to 3 d.p.) b 0.984 (to 3 d.p.) c 0.63 all to 2 d.p. 5 a i A 0.26, B 0.18 ii A 0.39, B 0.58 b A; it is more likely to last a long time and less likely to last a short time. 6 a Solong 0.07, HQ 0.18, Tooloo 0.05. b The probability it is faulty is 0.11. Better than HQ, not as good as the others. 7 T he probabilities for germination are A 0.71, B 0.55, C 0.84; C is most likely, as it has the greatest relative frequency. 8 a i 0.65 ii 0.73 b Yes. The probability of 9 or 10 increased. End-of-unit review 1 a 0.98 b 0.89 c 0.66 2 a 0.60 b 0.20 c 0.04 3 a 5 4 3 2 1 + + + + + + + + + + + + + + + + + + + + + + + + + 1 2 3 4 5 b i 1 5 4 a T H ii 8 25 iii 6 + + + + + + + + + + + + + + + + c i 2 25 b i 1 4 1 iii 25 ii 14 5 25 ii 1 4 iii 7 16 1 2 3 4 5 6 7 8 5 a i 0.4 ii 0.72 b 0.875 c Hassan because his probability of scoring two baskets is greater. 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 17 Bearings and scale drawing F Exercise 17.1 1 Using bearings a 070° b 150° 2 a 045° b 225° 3 a 050° b 165° c 230° d 300° c 260° d 335° e 120° 4 a i The bearing of Y from X is 058°, the bearing of X from Y is 238°. ii The bearing of Y from X is 142°, the bearing of X from Y is 322°. iii The bearing of Y from X is 033°, the bearing of X from Y is 213°. b Check students’ own diagrams and answers. c The bearing of Y from X + 180° = the bearing of X from Y. d When the bearing of Y from X is m°, the bearing of X from Y is m° + 180°. 5 a i 077° b i 118° c i 016° ii 257° ii 298° ii 196° 6 a i 244° b i 348° c i 204° ii 064° ii 168° ii 024° F Exercise 17.2 1 Making scale drawings a Check students’ scale drawings. b 128 km c 249° 2 a Check students’ scale drawings. b 12.4 km c 143° 3 8.2 km, 137° 4 No, they will not collide – students should draw a diagram to show this. 5 a 67 km b 49 km 6 a 4.5 km b 36 cm 7 a 20 km b €1120 End-of-unit review 1 a 085° 2 a b c d i i i i 065° 124° 308° 236° b 138° ii ii ii ii c 245° d 330° 245° 304° 128° 056° 3 a Check students’ scale drawings. b 150 km c 267° 4 10.7 km, 075° 5 a 5.5 km b 56 cm 6 $1920 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Answers to Coursebook exercises 18 Graphs F Exercise 18.1 Gradient of a graph a 3 b 1 2 a 1 b 4 c 1 3 a −4 b −1 c −1 1 4 4 p 1 2 3 2 1 r − 3 q −2 5 a Gradient of d = 52 = 2.5 b Gradient of e is 5, of f is −10. 6 a 1 c − 1 b 2 7 a 1 20 50 b 1 y 6 2 5 4 3 2 1 –4 –3 –2 –1 0 –1 1 2 3 x 4 8 a 2 b 1 c −4 d 0 9 a 6 b −4 c 12 d 0.1 F Exercise 18.2 1 a, b, c The graph of y = mx + c d The gradient of every line is 1.5. y 7 b 6 a 5 4 c 3 2 1 –3 –2 –1 0 –1 1 2 3 4 x –2 –3 –4 –5 –6 2 a 2 b −2 c 3 d −3 3 a A and C b A c B and C d D 4 a y = 6x b y = 6x + 8 5 a 5 b 2 c −5 d −2 6 A and C are parallel; B, D and E are parallel. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 18 Answers to Coursebook exercises F Exercise 18.3 1 Drawing graphs a y = −x + 10 b y = −2x + 10 c y = − 1x + 5 d y = − 1x + 21 2 2 2 2 a y = 2x − 5; this is of the form y = mx + c, which is the equation of a straight line. b 2 c y 2 1 –1 0 –1 1 3 2 4 x –2 –3 –4 –5 –6 3 a y = − 1 x − 2; this is of the form y = mx + c, which is the equation of a straight line. b −1 2 2 c y 3 2 1 –4 –3 –2 –1 0 –1 1 x 2 –2 –3 4 a y = − 2 x + 8; this passes through (0, 8) and (12, 0). 3 3 ii y = 2 x + 2 5 a i y=x+6 3 y 8 b b −2 y=x+6 7 6 5 4 2 y=3x+2 3 2 1 –3 –2 –1 0 –1 c i 1 6 a C 7 1 2 3 4 x ii 2 3 b B d A b B c A is x + 4y = 80 or y = − 1 x + 20; C is x + 4y = 0 or y = − 1 x. 4 2 c D a y = − 1 x + 10 4 Cambridge Checkpoint Mathematics 9 4 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 8 a i y = 10x − 7.5 c i y ii y = 0.05x − 3 ii 20 15 10 y 2 5 1 –1 0 –5 1 2 b i 10 3 x –20 0 –1 –10 –2 –15 –3 20 40 60 80 Unit 18 ii 0.05 Other scales are possible. x –20 9 a 1 or 0.05 20 10 a y 60 50 b (40, 2) c y = 0.05x + 4 or an equivalent equation 5x + 2y = 100 40 30 20 2x + 5y = 100 10 –20 –10 0 –10 10 20 30 40 50 60 x –20 b i −2.5 ii −0.4 F Exercise 18.4 1 c At approximately (14, 14) Simultaneous equations a x = −2 and y = −6 b x = 2 and y = 2 c x = 4 and y = 0 2 a x = −2 and y = −3 b x = 2 and y = 5 c x = 8 and y = 2 3 a y 8 6 i ii iii 4 2 –4 –2 0 –2 2 4 6 8 10 x –4 b i x = 5 and y = 2 ii x = 8 and y = 5 4 Answers from the graph should be approximately these. a x = 17.5 and y = 3.75 b x = 21.4 and y = 5.7 5 y 8 iii x = 4 and y = 3 c x = 12.0 and y = 11.9 x = 2.2 or 2.3; y = 4.7 or 4.8 6 4 2 –6 –4 –2 0 –2 2 4 6 x –4 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 3 Unit 18 6 Answers to Coursebook exercises y 8 x = 3.0; y = −1.6 6 4 2 –4 –2 0 –2 2 6 4 x 8 –4 –6 b y = − 1x − 1 7 a y = −1.5x + 6 F Exercise 18.5 1 a c x = 6 and y = −3 3 Direct proportion C 70 Cost ($) 60 50 40 30 20 10 0 8 4 6 Length (metres) 2 b 6.2 c C = 6.2M b d i $77.50 ii 32 or 32.3 p 100 Pages 2 a p = 16m M 10 80 60 40 20 0 c 16 Mass (grams) 3 a 5g c d i 120 2 3 Minutes 4 5 m ii 19.5 minutes b m = 5n d 77 m 2500 2000 1500 1000 500 0 4 1 100 200 300 Sheets 400 Cambridge Checkpoint Mathematics 9 500 n Copyright Cambridge University Press 2013 Answers to Coursebook exercises a $4.17 y 400 b Cost (dollars) 4 c 4.17 Unit 18 d 120 or 119.9 300 200 100 0 F Exercise 18.6 1 a y = 0.5x + 6 40 60 Fuel (litres) 20 80 100 x Practical graphs b y 10 c i 8.5 m ii 8 years c i 22 cm ii 7.5 hours c i $32.50 ii 11 km 8 6 4 2 0 2 a h = 30 − 2t b 2 4 6 8 10 x h 30 25 20 15 10 5 0 3 a c = 5d b 2 4 6 8 10 t c 60 50 40 30 20 10 0 4 a y = 2t + 6 b 2 4 6 8 10 12 d y 28 24 d 20 16 12 8 4 0 c i 16 2 ii 9 minutes Copyright Cambridge University Press 2013 4 6 8 10 12 t d The graph becomes horizontal. Cambridge Checkpoint Mathematics 9 5 Unit 18 Answers to Coursebook exercises 5 a c = 20 − 0.5t b c $13.50 c 20 16 12 8 4 0 6 a P = 8000 − 500Y b 4 8 12 16 20 t P 8000 c i 6000 ii 8 years 7000 6000 5000 4000 3000 2000 1000 0 1 2 3 4 5 6 7 8 9 y 7 a 12 million b 15 million c 0.1 d P = 0.1t + 12 8 a A = 2000 + 50t b c i $2250 ii 12 years A 2800 2600 2400 2200 2000 1800 1600 0 2 4 6 8 10 12 y End-of-unit review b 7 c 1 2 a 0.2 b −2 c 1 3 a 4 b −5 c 3 d −1 4 a y = −2x + 4 b y=− 1x+ 1 2 4 1 b 2 c y = 1x − 2 2 d y=x− 2 1 a −1 3 5 a −2 2 2 3 c 2 6 A and C are parallel; B and E are parallel 7 a x = 4.2 and y = 0.4 6 b x = 6.4 and y = −0.7 Cambridge Checkpoint Mathematics 9 c x = 5.5 and y = −3.5 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 8 a Unit 18 b x = −1.7 and y = 1.8 or 1.9 y 3 2 1 –4 –2 2 x 4 b c = 30d + 40 c 300 c Cost in dollars 9 a $100 0 –1 d 6 days 250 200 150 100 50 0 10 a b about 41 HK$ y 1500 Pakistani rupees 1 2 3 4 5 Days 6 7 8 d c 40.98 HK$ 1000 500 0 20 40 60 80 100 Hong Kong dollars Copyright Cambridge University Press 2013 x Cambridge Checkpoint Mathematics 9 7 Answers to Coursebook exercises 19 Interpreting and discussing results F Exercise 19.1 1 a Interpreting and drawing frequency diagrams Mass, m (kg) Frequency Midpoint 40 ≤ m < 50 4 45 50 ≤ m < 60 12 55 60 ≤ m < 70 8 65 b Masses of students in 9T y 14 Frequency 12 10 8 6 4 2 0 40 45 50 55 60 65 70 Mass (kg) c 24 d x 2 3 2 a 50 at each surgery b Oaklands Surgery Birchfields Surgery Time, t (minutes) Frequency Midpoint Time, t (minutes) Frequency Midpoint 0 ≤ t < 10 25 5 0 ≤ t < 10 8 5 10 ≤ t < 20 10 15 10 ≤ t < 20 14 15 20 ≤ t < 30 12 25 20 ≤ t < 30 17 25 30 ≤ t < 40 3 35 30 ≤ t < 40 11 35 c Waiting times at two doctors’ surgeries Oaklands Birchfields y 30 Frequency 25 20 15 10 5 0 0 30 10 20 Waiting time (minutes) 40 x d Over three times as many people had to wait less than 10 minutes in Oaklands surgery compared to Birchfields. More people had to wait over 10 minutes in Birchfields surgery compared to Oaklands. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 1 Unit 19 3 a Answers to Coursebook exercises Number of hours year 9 students spend doing homework each week y Girls Boys 20 16 12 8 4 0 0 4 8 12 20 x 16 b More girls spend between 0 and 4, and 12 and 20 hours doing homework each week, whereas more boys spend between 4 and 12 hours doing homework each week. c 40 boys and 50 girls d No, as there were 10 more girls than boys surveyed. There should have been the same number of boys and girls in order to make a fair comparison. F Exercise 19.2 a y 5 Rainfall (mm) 1 Interpreting and drawing line graphs Monthly rainfall in Lima, Peru 4 3 2 1 0 J F M A M J J A Month S O N D x b Example: Rainfall decreases steadily from January to April, then increases between April and July. Apart from an increase between August and Spetember there is a steady decrease between July and November. The year finishes with a small increase between November and December. c June and July Number of tourists (millions) 2 a y 1000 Number of tourists worldwide 950 900 850 800 750 700 650 2002 2004 2006 2008 Year 2010 2012 x b Number of tourists is increasing each year. The number increased at a similar rate between 2002 and 2008, then from 2008 to 2010 it increased at a smaller rate. c 880 million d Answer between 950 and 970 million 2 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises 3 a Unit 19 Daily temperatures recorded in Athens in one week Maximum temperature (°C) Minimum temperature (°C) Temperature (°C) y 24 22 20 18 16 14 12 10 Mon Tues Wed Thur Day of week Fri Sat Sun x b For example: The maximum temperatures increased gradually during the week, peaked on Thursday, then gradually decreased for the rest of the week. The minimum temperatures started increasing from Tuesday, peaked on Thursday, dropped back down on Friday and Sunday, with a slight increase on Saturday. c Friday 4 a 43 million b 1998 and 2000 c 2000 and 2002 d No because the graph is increasing and decreasing by different amounts. There is no real pattern to the figures. 5 a Mass is increasing every year. b Age 10 and Age 12 c 50 kg d No because by age 18 a girl should be almost fully grown. You cannot tell whether she will put on more mass or stay the same. F Exercise 19.3 a Hours watching TV 1 Interpreting and drawing scatter graphs Hours spent by students doing y homework and watching TV 20 15 10 5 0 0 5 10 15 Hours doing homework 20 x b Negative correlation. The more homework the students does, the less TV they watch. 2 a History and music test results of 15 students y 80 Music result (%) 70 60 50 40 30 20 10 0 0 20 40 60 80 History result (%) 100 x b No correlation. Getting a good result in one subject doesn’t mean a student will get a good, or bad, result in the other. Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 3 Unit 19 Number of cold drinks sold 3 a Answers to Coursebook exercises Temperature and number of cold drinks sold y over a 14 day period 32 28 24 20 25 30 35 Maximum daytime temperture (°C) x b Positive correlation. The higher the temperature the more cold drinks were sold. 4 a Positive correlation. The greater the distance, the longer the journey took. b 6 km in 16 minutes. It should have taken less time, so the taxi may have been delayed in traffic. F Exercise 19.4 1 a Interpreting and drawing stem-and-leaf diagrams Beach car park 7 2 City car park 3 0 4 9 6 6 6 5 4 2 5 5 9 7 7 6 5 4 6 9 2 1 0 0 6 8 8 9 5 7 Key: For the Beach car park, 5 | 4 means 45 ice-creams For the City car park, 3 | 0 means 30 ice-creams b i Mode ii Median iii Range Beach car park 46 57 17 City car park 45 46 39 c For example: On average Antonino had better sales at the Beach car park. His median was higher. This shows that 50% of his daily sales were 57 ice-creams or more, compared to only 46 for the City car park. His mode was also higher. The range was smaller, showing that his sales were more consistent, however it was at the City car park where he had his highest daily sale of 69 ice-creams. d For example: No, Antonino’s sales were better at the City car park as he had a higher median and mode and sales were more consistent. 2 a i Mode ii Median iii Range iv Mean Boys’ times 17.4 s 16.3 s 2.9 s 16.56 s Girls’ times 16.8 s 17.5 s 4s 17.72 s b For example: On average the boys ran faster than the girls, as their mean and median were lower. The girls had the fastest modal time, but they had a larger range showing that their times were more varied than the boys. c For example: No, as the girls mean and median are both slower. This shows that on average the boys are faster. 3 a Website A 4 8 Website B 3 0 0 12 8 9 13 4 6 8 7 6 5 5 5 2 1 14 5 5 5 6 6 8 9 8 5 3 3 2 2 15 4 5 6 7 7 8 1 0 16 6 7 8 9 Key: For Website A, 0 | 13 means 130 hits For Website B, 12 | 8 means 128 hits 4 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013 Answers to Coursebook exercises b i Mode ii Median iii Range iv Mean Website A 145 147 31 147.1 Website B 145 148 41 149.9 Unit 19 For example: Website A and Website B both had the same mode and almost the same median. The median for Website B was only one more than Website A, so this average is almost the same. The mean was also very similar with only a difference of 2.8 hits per day. So on average Website B had slightly more hits than Website A. Website B’s range is a lot higher than Website A, showing that the number of hits it had per day varied a lot more. c Example: They could use either website. Website A was more consistent. Website B was only slightly better on average than Website A. F Exercise 19.5 1 Comparing distributions and drawing conclusions or example: The heights of the Stage 7 students are more varied. There were no Stage 8 students between 120 cm F and 130 cm tall, compared with three Stage 7 students. The greatest difference was between 160 cm and 170 cm tall where there were nine more Stage 8 students than Stage 7 students. 2 F or example: The number of goals scored at home matches was less varied than at away matches i.e. they were more consistent at home. The greatest number of goals they scored at a home match was 5 (twice), compared to 4 (once) at an away match. They never scored less than 2 goals at a home match, whereas at 9 away matches they scored less than 2 goals. 3 aFor example: The scatter graph showing monthly milk production and average daytime temperature has positive correlation. The scatter graph showing monthly milk production and average rainfall has negative correlation. b Yes, because the graphs show that in warmer months more milk is produced, and in wetter months less milk is produced. c Mode Median Range Mean 2010 59 000 59 000 20 000 63 000 2011 52 000 61 000 17 000 60 700 For example: Claude is correct because the mean milk production was higher in 2010 than 2011, so on average his cows produced more milk per month. Although the median was higher in 2011, there were a few months when milk production was low and so the overall mean was less than for 2010. The range in 2011 was less than 2010, which means that milk production in 2011 was more consistent. 4 aFor example: In 1960 there were more people aged under 40 years in the village, compared to 2010. In 2010 there were 8 people in the village over the age of 80, compared with none in 1960. There were nine times as many people aged 60 to 80 in the village in 2010 as there were in 1960. b In 1960, 37 were over the age of 40 out of 158 altogether, 37 × 100 = 23.4% (approx 25%) 158 In 2010, 96 were over the age of 40 out of 162 altogether, 96 × 100 = 59.3% (approx 60%) 162 Yes they are correct. c For example: People are living longer, so there are more people in the older age ranges. The number of young people may be declining as they move to towns or cities to look for work. End-of-unit review 1 a 60 b Andersons Supermarket Chattersals Supermarket Time, t (minutes) Frequency Midpoint Time, t (minutes) Frequency Midpoint 0 ≤ t < 15 5 7.5 0 ≤ t < 15 32 7.5 15 ≤ t < 30 8 22.5 15 ≤ t < 30 13 22.5 30 ≤ t < 45 38 37.5 30 ≤ t < 45 10 37.5 45 ≤ t < 60 9 52.5 45 ≤ t < 60 5 52.5 Copyright Cambridge University Press 2013 Cambridge Checkpoint Mathematics 9 5 Unit 19 Answers to Coursebook exercises c Frequency y 40 Time it takes employees to travel to work Andersons supermarket Chattersals supermarket 30 20 10 0 0 15 30 45 Time (minutes) 60 x 2 a Number of visitors (millions) d For example: More than 6 times as many employees took less than 15 minutes to travel to work to Chattersals than Andersons, whereas nearly 4 times as many took between 30 and 45 minutes to travel to Andersons than Chattersals. Only 5 employees (8%) from Chattersals took longer than 45 minutes to travel to work, compared with 9 employees (15%) from Andersons. Number of visitors to a theme park y 3 2.5 2 1.5 1 2002 2004 2006 2008 Year 2010 2012 x b Visitor numbers are steadily increasing. c 1.65 million d Answer between 2.5 and 2.6 million (inclusive) 3 a i Mode ii Median iii Range iv Mean Boys times 67 s 69 s 32 s 69.1 s Girls times 56 s 63 s 32 s 64.5 s b The range is the same for the boys and the girls so they are both as varied as each other. The median and the mean for the boys and girls are all over 60 seconds. The boys’ mean and median are higher than the girls’. The girls’ mean and median are closer to 60 seconds. The girls’ mode is only 4 seconds under 60 seconds, whereas the boys’ mode is 7 seconds over 60 seconds. c No, the boys’ median is higher, but is further away from 60 seconds, as is their mean, so the boys are worse at estimating 60 seconds. 6 Cambridge Checkpoint Mathematics 9 Copyright Cambridge University Press 2013