CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Workbook answers 11 aThe number is 7.142… and there is no repeating pattern. bLearner’s own answer. For example: 2 and 5 − 2 . c Because the sum of two rational numbers must be rational. d No, because the product of two rational numbers is rational. 12 B 3 5.5 √25 A 25 Exercise 1.1 1 Number Rational 36 Irrational 48 64 3 100 2 a 27 , 500 3 a integer b surd c surd d integer e integer f surd a irrational because 3 is irrational i 20 + 2 = 6.4721… b rational because it is equal to 9 = 3 ii 20 − 2 = 2.4721… c rational because it is equal to 8 + 4 = 12 iii 16 d irrational because it is 2 + an irrational number a 2.25 b it is equal to 1.5 c yes, it is equal to 4.5 d yes, it is equal to 1.1 a 33 = 27 and 43 = 64 b 93 = 729 and 103 = 1000 c 1.12 = 1.21 and 1.22 = 1.44 4 5 6 7 b −36, − 3 8 √25 13 a b 5 519 She is correct. Substitute different values to see that ( n + 2)( n − 2) = n − 4 seems to be true. Exercise 1.2 1 2 Learner’s own answers. For example: 3 a 2.6 × 106 b 9.2 × 108 c 4.62 × 105 d 2.08 × 107 a 5.5 × 104 b 5.5 × 107 c 6.4 × 108 d 4.06 × 108 a 53 000 b 53 800 000 c 711 000 000 000 d 133 100 000 a 5 b a square root between 36 and 49 4 9.46 × 1012 km c 2 5 a 3 × 10−5 b 6.66 × 10−7 8 a 12 c 5.05 × 10−5 d 4.8 × 10−10 9 aNo. All fractions are rational. In fact, the repeating sequence is nine digits long. 6 a b c d 0.0015 0.000 012 34 0.000 000 079 0.000 900 3 7 a b c d 0.000 008 0.000 000 482 0.000 061 0.000 000 070 07 b 10 a b 1 84 b 7 4 It is rational. It is 1 . 9 The answer is 8. i 2 × 18 is a possible answer. ii 3 × 27 is a possible answer. iii 5 × 20 is a possible answer. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 8 4 × 10−7 m and 8 × 10−7 m 9 C, E, A, B, D 10 a Exercise 2.1 1 b 22 12 a 106 or one million. b 2.8 × 10 3 × 10 d 9.95 × 109 13 a 4.3 × 10−4 b 1.25 × 10−6 c 7 × 10−6 d 8 × 10−9 14 a 1.75 × 106 b 1.34 × 108 c 6.5 × 10−5 d 1.146 × 10−4 c 4.5 × 10 7 6 d a b 1 49 c e 1 225 f 1 125 1 400 4−1 b 4−3 c 40 d 44 e 4−4 f 4−2 3 a 5−1 b 5−3 e 50 4 a 1 8 b 1 27 c 1 125 d 1 or 0.001 1000 a 122 b 12−1 c 12−3 d 123 a 5 3 b 4 c 8 −5 d 150 or 1 e 5 −12 7 a 73 c 76 8 a 125 b 5−7 c 3−4 d 251 or 25 a 6 b −4 c −2 d 4 10 a −2 b 4 c 6 d 7 11 a 3 b 1 3 c 1 12 a 116 = 1 771 561 b 1 2 5 6 9 a c 11−3 = 52 c 5−2 d b 7−1 4 1 b x ÷ 2 − 4 = 10 ÷ 2 − 4 = 5−4 =1 c 4 × x 2 = 4 × 10 2 = 4 × 100 = 400 d 3 × ( x + 2 ) = 3 × (10 + 2 ) = 3 × 12 = 36 9 2 A and iii, B and v, C and i, D and vi, E and ii, F and iv 3 a x + y = 6 + −2 = 6 − 2 = 4 b x − y = 6 − −2 = 6 + 2 = 8 c x 2 + y 2 = 62 + (−2)2 = 36 + 4 = 40 d 3x + y = 3 × 6 + −2 = 18 − 2 = 16 e x + 4y = 6 + 4 × −2 = 6 − 8 = −2 f 3x + 4y = 3 × 6 + 4 × −2 = 18 − 8 = 10 a 2 b −14 c 35 d 13 e 7 f 100 a −4 b 5 c −8 d −26 e 94 f −4 g 12 h −11 Exercise 1.3 1 7 1 81 2 × x + 3 = 2 × 10 + 3 = 20 + 3 = 23 5.98 × 10 kg 11 aCopy and complete this sentence: 6.2 × 107 is 10 times larger than 6.2 × 106. b a 23 4 5 6 1 2 aIncorrect. He has worked out −12 and not (−1)2. Correct solution is −4 × (−1)2 −3 × −4 = −4 + 12 = 8 −6 b d Incorrect. He has worked out that (−4)3 = 64 and not −64. Correct solution is ( −4 ) − 3 7−1 −4 2 × −1 = −64 − −4 −2 = −64 − 2 = −66 7 Learner’s own values. For example: a x = 3 and y = 7, x = 4 and y = 44, x = 5 and y = 105 4 9 b x = −1 and y = −21, x = −2 and y = −28, x = −3 and y = −47 112 = 121 c x = 0 and y = −20, x = 1 and y = −19, x = 2 and y = −12 1331 13 7 2 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 8 9 a 15 b 20 c −20 d 11 e 8 f −64 g 2 h −7 i 8 j 2 k −25 l 4 10 a d Let y = 2, so (2y)3 = (2 × 2)3 = 43 = 64 and 2y3 = 2 × 23 = 2 × 8 = 16 e 1 −1 n f 64 ≠ 16, so (2y)3 ≠ 2y3. g 3(n + 20) h 3n Let x = 4 and y = 2, 3x − 3y = 3 × 4 − 3 × 2 = 12 − 6 = 6 and 3(y − x) = 3(2 − 4) = 3 × −2 = −6 i ( 4n)2 − 3 j 6 3 n + 10 6 ≠ −6, so 3x − 3y ≠ 3(y − x). k n −9 5 a 6x b 3x + 10 c 12x − 2 d 13x − 4 a xy b y2 c x3 d 16x2 a g2 = 25, g(8 − g) = 15, 2g(3g − 11) = 40 b 80 c g2 + g(8 − g) + 2g(3g − 11) = g2 + 8g − g2 + 6g2 − 22g = 6g2 − 14g. d 6g2 − 14g = 80 b 7 14 kg 4 5 Mass using expression ① 10.5 13 15.5 18 20.5 Mass using expression ② 10 12 14 16 18 Expression ②, 13.5 kg is closer to 14 kg than 15.5 kg. b 99 100 c2 100 5 − 3 × 5 × −3 − 5(5 − −3) 2 = 36 − 4 + 45 − 40 = 37 d + 8c ( c + d )2 = ( −3) + 3 + 9 2 8×5 ( 5 + − 3 )2 2 2 = −27 + 10 + 25 + 29 = 37 c 2a + 16 ii 5a + 15 i 2a + 16 = 22 ii 5a + 15 = 30 i 2b + 2 ii 5b − 20 i 2b + 2 = 26 ii 5b − 20 = 40 i 4c − 16 ii c2 − 8c when c = 10, d Exercise 2.2 i when b = 12, 2 3×5 −3 10 2n when a = 3, b ( ) − ( −4 − c ) +( ) − ( −4 − 5 ) 3c d 8 10 a − 3cd − c(c − d ) = 4 × ( −3)2 − 3 18 n 1000 n −5 4 3 3 12 4d 2 − 3 b 5 2n + 3 2 3 d c 1 2 a 5 b Age (A years) 1 c n − 10 c 11 a 6 a 18 kg d b 6 40 ≠ 400, so 10x 2 ≠ (10x)2. c 3 A and iii, B and vi, C and i, D and vii, E and viii, F and ii, G and iv, H and v Let x = 2, so 10x 2 = 10 × 22 = 10 × 4 = 40 and (10x)2 = (10 × 2)2 = 202 = 400 b a 5 Learner’s own counter-examples. For example: a 4 i 4c − 16 = 24 ii c2 − 8c = 20 i 2d 2 + 14d ii 7d 3 ii 7d 3 = 875 when d = 5, a 6 b 12 c x+2 d z+2 a 2 b 5 c y−3 d z−3 a 10 b 20 c 5a d 5b i 11 a 2d 2 + 14d = 120 i2(a + 3) + 2(3a + 1) = 8a + 8, 4(2a + 2) = 8a + 8 ii3(a + 3) + 3(3a + 1) = 12a + 12, 6(2a + 2) = 12a + 12 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE iii5(a + 3) + 5(3a + 1) = 20a + 20, 10(2a + 2) = 20a + 20 b c n black rods + n striped rods = 2n white rods (or similar explanation given in words) i4(a + 3) + 2(2a + 2) = 8a + 16, 8(a + 2) = 8a + 16 ii6(a + 3) + 3(2a + 2) = 12a + 24, 12(a + 2) = 12a + 24 iii8(a + 3) + 4(2a + 2) = 16a + 32, 16(a + 2) = 16a + 32 d 12 a 2n black rods + n white rods = 4n grey rods (or similar explanation given in words) i b $10 d 16 + 10d 13 a ii $26 c $46 $16 a2 42 When a = 4, + 3a = + 3 × 4 = 20 and 2 2 when b = 5, 2b (b2 − 4b − 3) = 2 × 5(52 − 4 × 5 − 3) = 10(25 − 20 − 3) = 20 15 aSide length = 3 27 = 3 cm, cube has 12 edges, so total length of edges = 12 × 3 = 36 cm b 80 c i 2a + 12a 1 A and ii, B and iv, C and i, D and iii 2 A and iii, B and iv, C and ii, D and i 3 a True b False y 5 × y 4 = y 9 c True d False y 9 ÷ y 3 = y 6 a g8 b h30 c i 21 d j 20 a 8x2 b 16x3 c 4y 4 d 11y 6 a a7 b b10 c c8 d d4 e e4 f f7 g g32 h y14 i i 72 j 13j 2 k k3 l −3l5 a 6a4 b 16b7 c 36c12 d 10e11 e 8g8 f 3h6 g 5x8 h 5x4 8 a B 9 aWhen the terms are simplified, one group has x6 terms and one group has x9 terms. 4 5 6 ii 8b − 32b − 24b 7 2 3 2 d When a = 4, 2a + 12a = 80 and when b = 5, 8b3 − 32b2 − 24b = 80 e Yes. Learner’s own explanations. 2 For example: When a is a positive integer, f g i −10 iii −18 ii −16 No, because the perimeter cannot be a negative number. For a < −6 the perimeter is positive, so is a valid measurement. 14 a2(4x2 + 3x) + 2(2x2 − 5x) = 8x2 + 6x + 4x2 − 10x = 12x2 − 4x 4 b 12x2 − 4x = 4x(3x − 1) c Arun is incorrect. When x = 3, perimeter = 96 and when x = −3 perimeter = 120. b A c A d D x6 terms: 3x3 × 2x3, 9x9 ÷ 3x3, 2x × 3x5 a2 a is positive, so is positive. Also 3a 2 2 is positive. When you add two positive numbers, you will get a positive answer, so the perimeter of the rectangle will always be positive. 12 3 x Exercise 2.3 As the side lengths are both 20, it must be a square. b c 48 cm x9 terms: x6 × 3x3, 12x12 ÷ 4x3, 6x6 × x3 b 10 a 9x12 ÷ x9 = 9x3: this is the only card, which when simplified, has an x3 term; all others have x6 terms or x9 terms. Zara is correct. (2x3)2 = 22 × x3×2 = 4x6 b i 9x14 ii 11 a C b 12 a c e 13 a c m−8 = A 64y27 iii 32z15 c B 4−4 = 4 1 4 b 5−3 = 3 8−5 = 5 1 8 d x−4 = 4 y−7 = 7 1 y f z−1 = 1 = x−3 = 3 1 x b y−4 = 4 1 m8 d n−5 = 5 d D 1 5 1 x 1 z 1 z 1 y 1 n Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 14 a A and v, B and iii, C and i, D and vii, E and ii, F and iv b Any expression that simplifies to give d 5 . 2 y7 For example: 10y3 ÷ 4y10 15 2 n × 3n ( 2 n 2 )3 2 16 Yes, 5 = 4 4 3x × x 2 × 3x 6n 3n = 8n6 4 a = 9 3 a 81x16 = 9x 4 9x12 23 × 34 b a +9 +9x +27 (x + 5)(x − 3) × x +5 x 2 x +5x −3 −3x −15 (x + 6)(x − 2) 20 3 × x +6 90 x 2 x +6x 4 80 12 −2 −2x −12 x2 + 6x − 2x − 12 = x2 + 4x − 12 18 × 42 c (x − 7)(x + 4) × 10 8 × x −7 40 400 320 x 2 x −7x 2 20 16 +4 +4x −28 x2 − 7x + 4x − 28 = x2 − 3x − 28 d (x + 2)(x + 3) (x − 8)(x + 2) +2 × x −8 x 2 x +2x x x2 −8x +3 +3x +6 +2 +2x −16 × x (x + 1)(x + 4) x2 − 8x + 2x − 16 = x2 − 6x − 16 4 a (x − 1)(x − 3) × x +1 × x −1 x 2 x +x x x2 −x +4 +4x +4 −3 −3x +3 (x + 5)(x + 6) x2 − x − 3x + 3 = x2 − 4x + 3 b (x – 4)(x – 8) × x +5 × x −4 x 2 x +5x x x2 −4x +6 +6x +30 −8 −8x +32 x2 + 5x + 6x + 30 = x2 + 11x + 30 5 +3x 600 x2 + x + 4x + 4 = x2 + 5x + 4 c x × x2 + 2x + 3x + 6 = x2 + 5x + 6 b x 30 400 + 320 + 20 + 16 = 756 2 +3 2 x2 + 5x − 3x − 15 = x2 + 2x − 15 600 + 90 + 80 + 12 = 782 b x x2 + 3x + 9x + 27 = x2 + 12x + 27 Exercise 2.4 1 × 7 6 x 2 × 3x 6 × 2 x 9 36 x17 = 13 = 9x 4 and 4 x13 4x (3x ) (x + 3)(x + 9) x2 − 4x − 8x + 32 = x2 − 12x + 32 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 5 Learner’s own answer. 6 a x2 + 7x + 10 b x2 + 2x − 8 c x2 − 3x − 18 d x2 − 6x + 9 e x2 + 15x + 50 f x2 − 13x + 40 g x2 + 5x − 50 h x2 − 3x − 40 7 a B b C d 8 1(x + 4)(x + 3) = x + 7x + 12 Rohan had the final term incorrect – he added 4 and 3 to get 7, not multiplied 4 by 3 to get 12. c A (x + 8)2 and (x + 7)(x + 9) giving x2 + 16x + 64 and x2 + 16x + 63. There is still a difference of 1. 15 a(2x + 1)(3x + 2) = 6x2 + 4x + 3x + 2 = 6x2 + 7x + 2 b C 12x2 + 19x + 5 ii 8y2 − 14y − 15 2 Exercise 2.5 a 1 1 2 + = 3 3 3 b 1 2 3 + = 5 5 5 c 2 3 5 + = 7 7 7 d 1 3 4 1 + = = 8 8 8 2 e 1 2 3 1 + = = 2 9 9 3 f 3 3 6 3 + = = 10 10 10 5 a x x 2x + = 3 3 3 b x 2 x 3x + = 5 5 5 c 2 y 3y 5 y + = 7 7 7 d y 3y 4 y y + = = 8 8 8 2 e m 2 m 3m m + = = 9 9 9 3 f 3n 3n 6 n 3n + = = 10 10 10 5 a 1 3 2 3 5 + = + = 4 8 8 8 8 b 1 2 3 2 5 + = + = 3 9 9 9 9 a2 − 16 c 2 1 4 1 3 1 − = − = = 3 6 6 6 6 2 b There is no term in a, and the number term is a square number. d 11 1 11 2 9 3 − = − = = 12 6 12 12 12 4 c a2 − 64 4 a 5x 8 b 5y 9 c p 2 d 3b 4 d a2 − b2 11 ( x + 4 )( x – 3) + x (5 – x ) = x 2 − 3x + 4 x − 12 + 5x − x 2 5 a x 2 b 4x 5 c 12 x d 6x 7 e 5 4x f y 6 g 2y 9 h y 18 i 5 16y j 17 24y 2(x + 5)(x − 9) = x2 − 4x − 45 Rohan simplified 5x − 9x to be 4x not −4x. 1 3(x − 3)(x − 2) = x2 − 5x + 6 Rohan had the final term incorrect – he multiplied −3 by −2 to get −6, and it should be +6. 9 a b c 10 a i a2 + 4a + 4 ii a2 − 4a + 4 iii b2 + 8b + 16 iv b2 − 8b + 16 v c2 + 2c + 1 vi c2 − 2c + 1 2 Learner’s own answer. For example: The first and last terms are the same, the middle terms have different signs. (x + y)2 = x2 + 2xy + y2 so (x − y)2 = x2 − 2xy + y2 i a2 − 1 iii a − 81 3 ii 2 = 6 x − 12 = 6 ( x – 2) 12 a b 13 a b i x2 + 12x + 36 ii x2 + 12x + 35 Learner’s own answer. For example: There is a difference of 1. i x2 + 14x + 49 ii x2 + 14x + 48 Learner’s own answer. For example: There is a difference of 1. 14 Learner’s own answer. For example: (x + 5)2 and (x + 4)(x + 6) giving x2 + 10x + 25 and x2 + 10x + 24 6 i 6 1 4 b 7 x and 4 1 x B, C both equal x or . 2 2 1 x E, which equals x or . 3 3 aA, D, F all equal x or a x+ y 2 b 2x + y 6 c 9x + y 12 d 15x − y 18 e 7x − 8 y 12 f 21a + 4b 28 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 8 g 10a + 15b 18 i 8ab − 45 36b a 17 c 17 ≠ 32. h ab − 35 7b 4 b b 5 32 Learner’s own explanation. For example: She has just crossed the 2s off and not cancelled properly. 9 8x + 2 2( 4 x + 1) 2 ( 4 x + 1) = = = 4x + 1 1 2 2 2 a 2x + 1 b 5x + 1 c 3x − 4 d 3x − 4 c 1 11 a b i 2x + 4 and 2(x + 2) ii 3x + 9 and 3(x + 3) iii 6x − 9 and 3(2x − 3) iv 4 − 6x and 2(2 − 3x) 12 a 2x + 3 2 b c 2x − 3 4 d y+x x+ y or xy xy b d +c c+d or cd cd c y−x xy d 2b + a a + 2b or ab ab e 5n − 2 m mn f 3h − 4 g gh 13 a Exercise 2.6 ii 45 A=b×h swap sides: A h b × h = A reverse the ×: b = F = bg F=b×g b×g=F reverse the ×: b = g T = mb T=m×b m×b=T reverse the ×: b = swap sides: F swap sides: T m i b S = , S = 20 a Polly’s age: d + 3, Max’s age: d − 2 b T = 3d + 1 c T = 25 d d= T −1 3 e d = 11 a F = 25 b F = 54 c I = 40 d e=5 e a=7 a 50% b 8% c 110% 10 a 450 m b 1303 m c 1078 m d 1615 m 11 a A b B c A d C 12 a n= p+8 3 b n = 7(q − k) n = 2pw − r d n= 9 c ii D = 150 D T c D = 180 D S T = , T = 5.5 hr 2 + 2 5 1 A and v, B and iv, C and ii, D and iii, E and vi, F and i 13 Arun is correct. 20 °C = 68 °F and 68 °F > 65 °F. 2 a 14 F = 120. Learner’s own explanation and working. For example: 3 i 24 ii 48 iii 72 iv 24d Use the formula a = b H = 24d = 24 × d = 24 × 10 = 240 a i 7 ii 14 iii 21 iv 7w b 7 19 a 6 8 2x + 3 5 5 − 7x 2 i M = b − kn swap sides: b − kn = M reverse the −: b = M + kn 7 2 8x + 24 8 ( x + 3) = = 2(x + 3) 1 4 4 7w + d 25 e 1 8x + 24 4 ( 2 x + 6 ) = = 2x + 6 and 1 4 4 iii ii X = b + rt swap sides: b + rt = X reverse the +: b = X − rt 1 x − 2 + 4x + 3 = 5x + 1 9 d 10 Evan is correct. 7 x − 14 8x + 6 7 ( x − 2 ) 2 ( 4 x + 3) + = + = 1 1 7 2 7 2 i a A = bh b 1 d a = 7w, 56 days So a = v−u to find the value of a. t v − u 32 − 12 = = 4. t 5 Then use the formula F = ma to work out the value of F. So F = 30 × 4 = 120. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 15 a r= 2A π b 4.8 cm 16 a bh A = a2 + 2 b A = 61 bh 2 d a = 12 c 17 a a= A− 6 side length of the larger cube = 2x c V 9 b V = 9x3 d Learner’s explanation and working. Example: Used the formula x = 3 x= 3 a 3.4 b 3.4 c 0.034 d 0.034 e 0.034 f 0.034 g 34 h 3.4 i 3400 j 30 400 k 30 l 340 7 POWERS OF TEN – EASY! 8 a V to work out the 9 i 5000 ii 500 iii 50 iv 5 v 0.5 vi 0.05 b larger a i 0.099 ii 0.99 Side length of larger cube is 2 × 4 = 8 cm iii 9.9 iv 99 Area of one face of larger cube = 8 × 8 = 64 cm2 v 990 vi 9900 576 value of x. x = 3 = 4 cm 9 9 Surface area of larger cube = 6 × 64 = 384 cm2 b smaller 10 a 0.004 × 103 Exercise 3.1 1 2 a 3.4 × 102 = 3.4 × 100 = 340 b 4.8 × 103 = 4.8 × 1000 = 4800 c 12.5 × 101 = 12.5 × 10 = 125 d 5 × 105 = 5 × 100 000 = 500 000 e 14 × 103 = 14 × 1000 = 14 000 A and ii, B and v, C and iv, D and i, E and iii 4 a 3.4 × 10–2 = 3.4 ÷ 100 = 0.034 b 8 × 10–3 = 8 ÷ 1000 = 0.008 c 15 × 10–4 = 15 ÷ 10 000 = 0.0015 d 12 × 10–1 = 12 ÷ 10 = 1.2 a 2800 b c 280 e 0.04 ÷ 10−2 =4 A and ii, B and vi, C and iv, D and i, E and iii, F and v 3 5 8 4 × 100 400 ÷ 102 b 0.4 × 101 40 ÷ 101 67 ÷ 102 670 × 10−3 670 ÷ 103 6.7 ÷ 101 = 0.67 67 × 10−2 6.7 × 10−1 11 a 45: A, D, H 4.5: B, E, J 0.45: C, G, I b 0.045: F is spare. Learner’s own answers. For example: 45 ×10−3, 4.5 ×10−2, etc. 12 a b c 28 000 d e f d 2880 13 a 270 b 280 000 f 0.2 g 28 h 0.2 14 a B i 0.028 j 0.28 Exercise 3.2 k 0.028 l 28.8 1 b 0.0048 c 125 000 c d A C B a 4 × 0.3 4 × 3 = 12 so 4 × 0.3 = 1.2 b 7 × 0.4 7 × 4 = 28 so 7 × 0.4 = 2.8 c 9 × −0.1 9 × −1 = −9 9 × −0.1 = −0.9 so Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE d −15 × 0.2 − 15 × 2 = −30 so −15 × 0.2 = −3 e 8 × 0.02 f −5 × −0.04 − 5 × −4 = 20 so −5 × −0.04 = 0.2 g 11 × 0.07 11 × 7 = 77 11 × 0.07 = 0.77 8 × 2 = 16 so 8 × 0.02 = 0.16 so 2 a6 ÷ 0.3 0.3 × 10 = 3 3 9 Denominator should be: 5 × 0.1 = 0.5, not 50. Answer = 1. 10 a 20 b 30 c 500 d 0.2 11 a 6 × 10 = 60 60 ÷ 3 = 20 b 8 ÷ 0.2 0.2 × 10 = 2 8 × 10 = 80 80 ÷ 2 = 40 c −9 ÷ 0.1 0.1 × 10 = 1 −9 × 10 = −90 −90 ÷ 1 = −90 d 12 ÷ 0.4 0.4 × 10 = 4 12 × 10 = 120 120 ÷ 4 = 30 e 6 ÷ −0.02 −0.02 × 100 = −2 6 × 100 = 600 600 ÷ −2 = −300 f 8 ÷ 0.04 0.04 × 100 = 4 8 × 100 = 800 800 ÷ 4 = 200 g −16 ÷ −0.08 −0.08 × 100 = −8 −16 × 100 = −1600 −1600 ÷ −8 = 200 a 1.2 b 3.6 d −8.1 e 3.3 f −0.24 g 0.28 h 0.45 i 1.4 j −5.55 i 1.1 ii 2.2 iii 3.3 iv 4.4 v 5.5 vi 6.6 b i larger ii smaller c i 80 ii 40 iii 20 iv 16 v 10 i larger ii larger d 12 a 158.4 b 158.4 c 0.015 84 d 352 e 0.352 f 3.52 13 a 2.6 c Hassan is incorrect. Numerator should be: 2.5 × 0.2 = 0.5, not 5. Estimate: 6 × 40 = 240 Accurate: 271.377 b Estimate: 200 ÷ 0.4 = 500 Accurate: 495 c Estimate: 80 × 5 0.2 = 400 0.2 = 2000 Accurate: 2400 14 a 5.4 m2 b 7.2 m2 c 0.48 m2 d 0.124 m2 True b True 15 4 m 4 A, C, E, I (0.024); D, G, J, L (0.24); B, F, H, K (2.4) 5 a 20 b 40 17 a c 30 d −40 c False, 0.0025 d False, 0.3 e 200 f −250 e True f True g 300 h 3000 i 200 j −400 6 a B c C 7 a 0.12 b 1.35 c 0.072 d 0.15 e 0.055 f 30 $200 increased by 20% 100% + 20% = 120% multiplier is 1.2 g 9 h 5 $200 × 1.2 = $240 i 7 j 40 a True b True c False d True 8 b B 16 0.35 m Exercise 3.3 d B 1 a$300 increased by 15% 100% + 15% = 115% multiplier is 1.15 $300 × 1.15 = $345 b c $400 increased by 32% 100% + 32% = 132% multiplier is 1.32 $400 × 1.32 = $528 2 a$300 decreased by 15% 100% − 15% = 85% multiplier is 0.85 $300 × 0.85 = $255 9 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE b $200 decreased by 20% 100% − 20% = 80% multiplier is 0.8 c The value of the scooter after 12 years. d 1800 × (0.88)4 = $1079.45 and 1800 × (0.88)5 = $949.92 e 1800 × (0.88)n $200 × 0.8 = $160 c $400 decreased by 32% 100% − 32% = 68% multiplier is 0.68 $400 × 0.68 = $272 3 A and vi, B and iii, C and i, D and iv, E and ii, F and v 4 a$800 increased by 10%, then increased by 20%. 800 × 1.1 = 880 b 15% decrease then 12% increase → multiplier = 0.952 → $800 × 0.952 = $761.60 c 45% increase then 24% increase → multiplier = 1.798 → $400 × 1.798 = $719.20 720 × 0.8 = $576 → 14 a a $2400 1.5 × 0.4 = 0.6 → 50% increase and 60% decrease 1.2 × 0.5 = 0.6 → 20% increase and 50% decrease 840 × 0.85 = $714 b 1.25 × 1.2 = 1.5 → 25% increase and 20% increase i b = c i = ii = a i 85.8 ii 362.5 b i 891 ii 48.72 7 a 1.071 8 a i 33.6 ii 120 a115, 116, 117, 118, 119, 120, 121, 122, 123, 124 b i 127.5 ii 76.95 b 115 a 0.63 c 124 a 65, 66, 67, 68, 69, 70, 71, 72, 73, 74 b 65 c 74 9 10 ii b 15 Learner’s own answers. For example: 1000 × 0.6 = $600 → 1.1016 a 6 198 b 198 $1.29 b Calculation: Amount: 1 4000 × 1.05 $4200.00 b 2 × 0.75 = 1.5 → 100% increase and 25% decrease 16 500 000 Exercise 3.4 1 $529.20 End of year: 11 a 10 b $800 increased by 5%, then decreased by 15%. 800 × 1.05 = 840 5 → E and 0.54 13 a60% increase then 45% decrease → multiplier = 0.88 → $600 × 0.88 = $528 880 × 1.2 = $1056 $800 increased by 25%, then decreased by 40%. 800 × 1.25 = 1000 d b $800 decreased by 10%, then decreased by 20%. 800 × 0.9 = 720 c → 12 a A and ii, B and v, C and i, D and iv, F and iii 2 4000 × (1.05) 2 $4410.00 3 3 4000 × (1.05) $4630.50 4 4000 × (1.05)4 $4862.03 5 4000 × (1.05)5 $5105.13 2 3 4 a24.5, 24.6, 24.7, 24.8, 24.9, 25.0, 25.1, 25.2, 25.3, 25.4 b 24.5 c 25.4 a 7.5, 7.6, 7.7, 7.8, 7.9, 8.0, 8.1, 8.2, 8.3, 8.4 i 1800 × 0.88 b 7.5 ii 1800 × (0.88)2 c 8.4 iii 1800 × (0.88)3 5 a 2.5 b 3.5 6 a 85 b 95 The value of the scooter after 7 years. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 7 8 9 a 6.5 ⩽ x < 7.5 Exercise 4.1 b 27.5 ⩽ x < 28.5 c 134.5 ⩽ x < 135.5 1 d 558.5 ⩽ x < 559.5 a 45 ⩽ x < 55 b 415 ⩽ x < 425 c 3735 ⩽ x < 3745 d 5205 ⩽ x < 5215 a 750 ⩽ x < 850 b 1150 ⩽ x < 1250 c 6650 ⩽ x < 6750 d 9050 ⩽ x < 9150 a y 2 b 19.5 m2 2 c 18.5 m2 ⩽ x < 19.5 m2 y b 2 y ii 55 12 A, ii and c; B, ii and a; C, i and e; D, iii and b; E, i and f; F, iii and d 13 a b 14 a b 2 a 495 g ⩽ x < 505 g i 2 × 495 g = 990 g ii 2 × 505 g = 1010 g i 145 cm ii 155 cm 3y iii 145 cm ⩽ x < 155 cm 4 18 = 9 x −2 18 x = −2 9 3y 4 ii 6.25 litres or 6250 mL iii5.75 litres ⩽ x < 6.25 litres or 5750 mL ⩽ x < 6250 mL +1 = 7 d =x 3(y + 5) = 2(20 − y) 3 y + 15 = 40 − 2 y 3 y + 2 y = 40 − 15 5 y = 25 = 7 −1 3y =6 4 3y = 6 × 4 3 y = 24 Carlos has worked out the correct answer as all pieces of wood can vary between 145 cm and 155 cm, so you must multiply the upper and lower bounds by 3. 5.75 litres or 5750 mL 6(3 − x) = 3x 18 − 6 x = 3 x 18 = 3 x + 6 x x=2 c i b 3 4 x= y= y= 25 5 y=5 24 3 y=8 3 a 30 x 4 =5 63 b y +1 =9 30 = 5 x 63 = 9( y + 1) 30 63 5 =x 9 = y +1 x=6 7 = y +1 7 −1 = y y=6 a g = 12 b g = −10 c p=7 d g=7 e y=5 f y = 12 g x = −3 h x = −2 iii1.15 litres ⩽ x < 1.25 litres or 1150 mL ⩽ x < 1250 mL b 5 − 2x = 9 −2 x = 9 − 5 −2 x = 4 iii 1.25 litres or 1250 mL 6 y = 4×2 y = 8 505 g ii x=2 d 5y + 3 = 9 + 2y 5y − 2y = 9 − 3 3y = 6 y=2 ii 1.15 litres or 1150 mL 12 y= 495 g i 24 x= =4 i Pepe is incorrect as he has multiplied the rounded number by three then worked out +/− 5 cm from that answer instead of +/− 15 cm from that answer (as there are three pieces of wood). 15 a = 1+ 3 2 65 55 ⩽ x < 65 12 x = 32 − 8 12 x = 24 2 −3=1 18.5 m i 4(3x + 2) = 32 12 x + 8 = 32 x=8 c b 16 x= 10 a 11 a 11 2x − 6 = 10 2 x = 10 + 6 2 x = 16 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 5 6 7 a 5x + 15 = 10x − 20 → x = 7 b x + 3 = 2x − 4 → x = 7 b y = 14 c Learner’s own answers. c 48 a 8x − 32 + 20 − 4x = 0 → 4x − 12 = 0 → x = 3 15 a b 2(x − 4) + 5 − x = 0 → 2x − 8 + 5 − x = 0 → x−3=0 → x=3 b c Learner’s own answers. 14 a a5(23 + 4) = 5 × 27 = 135 and 2(30 − 23) = 2 × 7 = 14, 135 ≠ 14 b 16 a Line 1: he added 5 and 4 instead of multiplying 5 and 4. Line 2: he subtracted 2x instead of adding 2x and added 9 instead of subtracting 9. c b b= c c=2 d 3 d =4 5 b n = 28 c 28 and 62 11 a b 1 C x= 5 c x = 32 s = 19 c 43 cm y 1 5 9 y 4 6 8 1 D x= 5 c 12 = −3 y=x+4 x 0 2 4 s + 2s + 2s + 5 = 100 → 5s + 5 = 100 b 4 y = 2x + 1 x 0 2 4 x + 50 and 2x + 80 2x + 80 = 144 −12 2 y = 6x + 3 = 6 × −3 + 3 = −18 + 3 = −15 4 x = −3 and y = −15 b A x = 6480 B x = 5 E x=5 b 13 a 6x + 3 = 2x − 9 6 x − 2 x = −9 − 3 4 x = −12 20 B and E x=2 3 y = 2x − 9 = 2 × −3 − 9 = −6 − 9 = −15 a 3 5(x − 8) = 2(x + 10) b 1 B and E give the correct answer of five grandchildren. 12 a = 60 9−x x= n + 2(n + 3) = 90 → 3n + 6 = 90 b 7 cm and 15 cm 420 2 y = 2x − 1 = 2 × 6 −1 = 12 − 1 = 11 3 Check values are correct. y = x + 5 = 6+5 = 11 4 x = 6 and y = 11 1 4 a 10 a ii 2 Learner’s checks. 9 3 cm i3(a − 2) + 3(a − 2) + a + a = 44 or 2a + 6(a − 2) = 44 or a + 3(a − 2) = 22 or 4a − 6 = 22 1 2x − 1 = x + 5 2 x − x = 5 + 1 x=6 5 7 a ii 3(a − 2) = a 1 5 25 4 4 5 5 5 + 4 = 5 × 9 = 45 + = 45 + 3 = 48 7 7 7 7 7 8 i Exercise 4.2 x=5 , 5 2 4 and 2 30 − 5 = 2 × 24 = 48 7 7 7 a = 21 y + 3y + y − 2 + 4(y − 2) = 116 y 10 9 8 7 6 5 4 3 2 1 0 y = 2x + 1 y=x+4 0 1 2 3 4x (3, 7); x = 3, y = 7 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE d 1 2x + 1 = x + 4 2x − x = 4 − 1 x=3 2 y = 2x + 1 = 2 × 3 +1 = 6 +1 =7 3 y = x+4 = 3+ 4 =7 4 9 4 x = 3 and y = 7 a x = 18, y = 2 b x = 9, y = 3 c x = 9, y = 6 d x = 12, y = 14 10 a x = 10, y = 20 b x = 3, y = 24 c x = 14, y = −9 d x = −2, y = 4 11 a b e Learner’s own answer. a x + y = 10 and x − y = 4 12 a b 1 x + y = 10 2 7 + y = 10 + x − y= 4 y = 10 − 7 2x + 0y = 14 =3 2x = 14, x = x + y = 37.74, x − y = 9.24 $23.49 and $14.25 14 a = 9, b = 3, c = 4, d = 10, e = 5, f = 11 14 =7 2 4 x = 7 and y = 3 1 x + 5y = 28 2 x + 5 × 5 = 28 − x + 3y = 18 x = 28 − 25 0x + 2y = 10 = 3 1 2 mean = b range = 11 − 3 = 8 x − 2y = 6 4x = 40, x = b x ⩾ −6 c x<0 d x ⩽ 10 e −8 ⩽ x < 0 f −3 < x ⩽ 3 0 1 a 13 x = 6, y = 18 6 x = 2, y = 5 7 x = 6, y = −3 8 a i, ii x = 2, y = 5 b Learner’s own check. c Learner’s own answers. –1 –4 –3 –2 –1 0 d 5 10 15 20 0 1 e 4 2 f 3 4 2 3 4 5 6 7 –5 – 4 –3 –2 –1 0 1 2 3 4 5 6 3 10 − 2 × 2 = 6 5 –2 0 0.5 1 1.5 2 2.5 3 3.5 2y = 4, y = = 2 4 x = 10 and y = 2 =7 x>2 2y = 34 − 30 40 = 10 4 6 c 2 3 × 10 + 2y = 34 4x + 0y = 40 42 a 3x + 2y = 34 and x − 2y = 6 + = b 4 x = 3 and y = 5 3x + 2y = 34 6 –3 3 3 + 3 × 5 = 18 1 9 + 3 + 4 + 10 + 5 + 11 a Exercise 4.3 x + 5y = 28 and x + 3y = 18 10 2y = 10, y = = 5 2 c cost of a cake, x = $1.50 and the cost of a coffee, y = $2 13 x = 13, y = 8, so 2x + 3y = 50 3 7−3=4 b 2x + 3y = 9, 2x + y = 5 a 9 b −6 c −3, −2, −1, 0, 1, 2 a Could be true. b Could be true. c Must be true. d Cannot be true. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 5 a b 6x > 18 2x − 3 < 19 2 x < 19 + 3 2 x < 22 18 x> 6 x>3 c x< 5x + 1 ⩽ −9 5x ⩽ −9 − 1 5x ⩽ −10 x⩽ −10 5 3x > 4x + 12 b 3 x − 4 x > 12 − x > 12 22 −x 2 −1 x < 11 3(x − 4) ⩾ 9 3x − 12 ⩾ 9 3x ⩾ 9 + 12 3x ⩾ 21 d 11 a c 7 a x ⩾ 0.5 b x<3 c x ⩽ 13 d x < 6.5 a 0 1 2 3 4 0 1 2 3 4 10 11 12 13 14 4 5 6 7 8 c d 8 a b 3(y − 4) + 7y ⩾ 8y − 5 3y − 12 + 7y ⩾ 8y − 5 10y − 8y ⩾ − 5 + 12 2y ⩾ 7 y ⩾ 3.5 i y = 33(3 − 4) + 7×3 ⩾ 8 × 3 − 5; 3 × −1 + 21 ⩾ 24 − 5; 18 ⩾ 19 false ii y = 3.53(3.5 − 4) + 7 × 3.5 ⩾ 8 × 3.5 − 5; 3 × −0.5 + 24.5 ⩾ 28 − 5; 23 ⩾ 23 true iii y = 43(4 − 4) + 7 × 4 ⩾ 8 × 4 − 5; 3 × 0 + 28 ⩾ 32 − 5; 28 ⩾ 27 true 9 a x ⩽ 10 b x>4 c x⩾2 d x < 20 Learner’s own checks. 10 a 14 −2 > −14 −2 x>7 −5x −18 ⩾ −5 −5 x⩾3 12 a 3 5 x + 2x + x + 30 < 360 or 4x + 30 < 360 b x < 82.5 ° c No, x cannot be 90 ° because it has to be less than 82.5 °. 13 a A + A + 5 + 2(A + 5) < 100 → 4A + 15 < 100 5 b −2 x −1 6 − 5x ⩽ −12 −5x ⩽ −12 − 6 −5x ⩽ −18 x⩾7 6 12 x < −12 21 x⩾ 3 x ⩽ −2 < 3x − 3 < 5x − 17 3 x − 5 x < −17 + 3 −2 x < −14 b A < 21.25 c No, because A < 21.25, so 2(A + 5) < 52.5. 14 a x + 2x + 3(x − 10) < 360 → 6x − 30 < 360 b x < 65 c Yes. 2x = 3(x − 10) → x = 30 and this is in the solution set. 15 a 2z + 9 > 13 b 3(z − 4) > −6 c 4 + 2z > 8 d 5(3z − 2) > 20 16 a 0<x⩽7 5 6 0 1 2 b 2 ⩽ y ⩽ 10 0 c 2 4 7 8 4 6 8 10 12 0 1 2 3 4 2 3 −1 < n < 3 –2 d 3 –1 0 < m < 4.5 0 1 4 5 5x − 14 > 2x + 1 b x>5 c Learner’s own checks. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 5.1 ii 1 a = 80, b = 80, c = 55, d = 125 2 B and C 3 B = 48, C = 66, A = 180 − (48 + 66) = 66 so A and C are equal. 4 a = 100, b = 105, c = 25 5 a = 49, b = 49, c = 48 6 x = 30 7 a, bLearner’s own sketch showing the exterior angles 135 °, 120 °, 105 ° 8 x = 95, y = 39, z = 124 Exercise 5.3 9 x = 50, y = 30, z = 80 1 a = 123; b = 109 2 a 3 Two exterior angles are 74 ° and two are 106 °. 4 60 5 72 ° 6 a 7 20, 30 and 40 are factors of 360. 50 is not a factor of 360. 8 a 9 12 10 a b 11 a b Using the exterior angle property, A = 122 − 59 = 63 and E = 122 − 63 = 59. The third angle is 180 − 122 = 58, so both triangles have angles of the same size. 13 It is correct. Substitute values of n. To show it algebraically requires factorising. Exercise 5.2 b 1 a 2 aLearner’s own answer; divide pentagon into three triangles. c 1080 ° 1800 ° b 60 ° c All the angles are 108 °. The fifth angle is also 108 degrees. It is a regular polygon if all the sides are the same length but this may not be the case. 3 a 1260 ° 4 a 7 sides b The sum must be a multiple of 180. b 140 ° 5 One of the angles marked is not inside the hexagon. The angle there is 360 − 90 = 270 °. 6 84 7 a 8 1800 ÷ 180 = 10 so that is 12 sides. However, 180 is not a factor of 2000. 9 a Angles are 135 ° + 135 ° + 90 ° = 360 ° b i2 × 120 ° + 2 × 60 ° = 360 ° OR 120 ° + 4 × 60 ° = 360 ° 900 ° Third angle = 360 − 108 − 108 = 144 ° 12 36 ° Angles B and D are not equal. 360 ° 1440 ° 11 30 ° 10 Angles of the quadrilateral are 118 °, 127 °, 75 °; a = 360 − 320 = 40 ° 15 There are three ways: b 50 ° b 90 ° 6 b 10 ° 8 c b 120 ° 9 d 10 36 10 36 11 12 sides; the interior angle is 150 °, the exterior angle is 30 °, 360 ÷ 30 = 12. 12 55 13 142 ° Exercise 5.4 All questions except questions 7, 9 and 10 have the answer included for self-assessment. 1–6 Learner’s own diagrams and checks. 7 Learner’s own pattern. Assess by looking. 8 Learner’s own diagrams and checks. 9 Learner’s own pattern. Assess by looking. 10 a b Learner’s own diagram. 10 cm Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 5.5 e 1 a = 10 cm; b = 13 cm; c = 34 cm 2 a = 8.6 cm; b = 14.4 cm; c = 16.7 cm aLearner’s own answer. For example: Time of day, size of car, reason for travel, day of the week. 3 a = 12 cm; b = 7.5 cm; c = 14 cm b 4 a = 35.3 cm; b = 17.9 cm; c = 16.2 cm 5 a 10.1 cm b Learner’s own diagram. Learner’s own questions. For example: Do larger cars have more passengers? Are there more passengers in cars early in the morning? Are cars likely to have more passengers at the weekend? a i 4.2 cm ii 7.1 cm c iii 11.3 cm iv 14.1 cm Learner’s own predictions. For example: During the rush hour cars are more likely to have only one passenger. Cars on Sundays will have more passengers than cars on Mondays. d Learner’s own answer. For example: Observing cars at different times of day or different days of the week. e Learner’s own answer and explanation. 6 b Learner’s own checks. Check with answers in part a and try other side lengths. 7 3.7 m OR 3.71 m 8 14.7 m OR 14.72 m 9 a 10 a 2 17.3 cm b 173 cm² 12 cm b 15 cm 3 11 The calculator answer is 36.37… so 36.3 cm or 36.4 cm are acceptable answers. 12 a b 24 cm 720 cm² There are alternative answers to many questions in this unit. 1 aLearner’s own answer. For example: Gender, other interests, availability of equipment. b c d aLearner’s own questions. For example: Are young people faster using a keyboard than older people? Is there a difference between the speed of young people and old people writing on paper? Is there a difference between boys’ speed and girls’ speed? b Exercise 6.1 16 Learner’s own answer and explanation. Learner’s own questions. For example: Do girls spend the same amount of time playing computer games as boys? Do young children play on computer games less than older children? Does playing computer games affect the time spent doing homework? Learner’s own predictions. For example: Girls spend less time playing computer games than boys. Learners who play sports spend less time playing computer games than learners who don’t play sports. Learner’s own predictions. For example: Girls can type more quickly than boys. Older people can write more quickly on paper than younger people. c, d, e Learner’s own answers. These will depend on the predictions in part b. Exercise 6.2 1 a 34 2 a Teacher might not choose at random. b Learner’s own answer. For example: Using random numbers or names from a hat or particular positions in the register. 3 b 26 aPeople might be more likely to phone if they have a complaint. b Learner’s own answer. Learner’s own answer. For example: Use random numbers or names from a hat or a number of learners from different year groups. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 4 a Using social media Advantage Disadvantage Easy to do Some people do not use social media b Sending Can select letters to who to ask people c Asking people in the street 5 Age Can choose a Can be representative expensive sample and take a lot of time c 31 2 diameter, d = 2 × 4 = 8 cm C = π × d = π × 8 = 25.14 cm 3 C = π × d = π × 12 = 37.704 cm 1 of the circumference = 37.704 ÷ 2 = 18.852 cm 2 Perimeter = 12 + 18.852 = 30.85 cm 4 a b 42 9 Learner’s own answer. Any method should take account of the fact that parents might have more than one child in the school and you do not want to choose any parent twice. Exercise 7.1 d = 6 cm b = 50.24 cm 2 r = 1 cm A = π × r 2 = π × 12 = π ×1 c = 3.14 cm 2 r = 6 cm A = π × r 2 = π × 62 = π × 36 Sample size. Q1: How many portions of fruit or vegetables did you eat yesterday? Q2: How often do you eat meat in the main meal of the day? Q3: Why do you think people are overweight? This could be a multiplechoice question. r = 4 cm A = π × r 2 = π × 42 = π × 16 Learner’s own questions. For example: How were the adults chosen? What age were the adults? What questions were the adults asked? How did the adults measure their energy levels? 8 a radius, r = 4 cm 12 aQ1: People might say yes because they think they should. Q2: This question will encourage people to say no. Q3: This question asks people to say something they might feel uncomfortable about because it is being rude about people. b 1 = 28.26 cm aNo. Adults from a small sample said that vitamins gave them more energy but that is not the same thing as proving that they work. b 7 7 d = 9 cm C = π×d=π×9 Under 18 18 to 55 Over 55 Sample 6 People might not reply c 5 = 113.04 cm 2 diameter, d = 6 cm radius, r = 6 ÷ 2 = 3 cm A = π × r2 = π × 32 = π × 9 = 28.28 cm2 6 A = π × r2 = π × 52 = π × 25 = 78.55 cm2 Area of semicircle = 78.55 ÷ 2 = 39.3 cm2 7 8 a 12.6 cm2 b 44.2 m2 c 616 cm2 d 8.04 m2 a Learner’s own answers and explanations. For example: Dipti has the incorrect answer. She has not halved the diameter to get the radius. For example: Gabir has the incorrect answer. He has used the formula for the circumference not for the area. d = 5 cm C = π × d = π × 6 C = π × d = π × 5 = 18.84 cm 17 = 15.7 cm Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE b c Area = πr2 d = 2.4, so r = 2.4 ÷ 2 = 1.2 9 r2 = 1.22 = 1.44 Circle A : Circle B Radius Circumference Area (cm) (cm) (cm2) Area = π × 1.44 = 4.5238… Ratio 4:8 8π : 16π 16π : 64π Area = 4.52 cm2 (3 s.f.) Ratio in its simplest form 1:2 1:2 1:4 a i A = 22.9 cm2 ii C = 17.0 cm b i A = 1590.4 mm2 ii C = 141.4 mm 10 a i A = 113.5 cm b i c d ii P = 43.7 cm A = 904.8 mm ii P = 123.4 mm i A = 402.1 cm ii P = 82.3 cm i A = 88.4 m2 ii P = 38.6 m 2 2 2 11 Sofia is correct. Learner’s own explanation and working. For example: d Learner’s own answers. For example: The ratios of the radius and circumference are the same. e Learner’s own answers. For example: The ratio of the areas is the square of the ratio of the radius. f i1 : 3 (the ratios of the radius and circumference are the same) ii12 : 32 = 1 : 9 (the ratio of the areas is the square of the ratio of the radii) Area of semicircle 1 1 2 2 = × π × r2 = × π × 22 = 6.28 m2 Area of quarter-circle = 1 4 × π × r2 = 1 4 12.57 m > 6.28 m 2 Perimeter of semicircle = ×π×d +d = 1 2 1 2 12 Marcus is incorrect. Learner’s own explanation and working. For example: 2 Perimeter of semicircle = × π × 42 = 12.57 m2 2 1 19 Zara is incorrect. 4 3 4 1 a b × π × 6 + 3 + 3 = 20.14 m Area = base × height =6×4 Area = 1 2 = 1 2 13 a, b A and v, B and ii, C and i, D and vi, E and iv, F and iii c Area = 1 2 15 7.59 m = 759 cm = Circumference = 6π cm b Area = 49π m2 Circumference = 14π m c Area = 100π mm2 Circumference = 20π mm 18 aArea = πr2 = π × 42 = 16π cm2 Circumference = πd = π × 8 = 8π cm b 18 1 2 2 Area = 9π cm2 Area = πr2 = π × 82 = 64π cm2 Circumference = πd = π × 16 = 16π cm × base × height ×6×4 = 12 cm 2 14 1.4 cm = 14 mm 17 a × π × 6 + 6 = 3π + 6 m = 24 cm 2 20.57 m > 20.14 m 16 27 cm 2 Exercise 7.2 × π × 8 + 8 = 20.57 m ×π×d +r+r = 1 not 6π + 6 cm Perimeter of three-quarter circle = 3 ×π×d +d = × π × radius 2 × π × 32 = 14.14 cm 2 2 a Area = rectangle + triangle = 24 + 12 = 36 cm2 b Area = rectangle + semicircle = 24 + 14.14 = 38.14 cm2 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE c 3 = 14.14 + 12 = 26.14 cm a The two semicircles make one circle, so: Area = semicircle + triangle Area of circle = πr2 = π × 5.52 = 95.033… 2 Total area = 110 + 95.033… = 205 cm2 (3 s.f.) Area A = l × w = 8 × 10 = 80 9 Area B = l × w = 12 × 1 = 12 b Total area = 80 + 12 = 92 cm2 Area of large semicircle Area A = l × w = 6 × 6 = 36 = πr 2 = 1 1 Area B = × b × h = × 6 × 4 = 12 2 2 Total area = 36 + 12 = 48 mm2 c 1 1 2 2 × π × 52 = 39.27 Total area = 30 + 39.27 = 69.27 cm2 1 2 1 2 Area triangle = × b × h = × 2 × 6 = 6 Area circle = πr2 = π × 42 = 50.26 Shaded area = 50.26 − 6 = 44.26 cm2 4 a 5 cm and 3 cm b Area A = base × height = 3 × 7 = 21 cm2 7 cm ii 135 cm2 b i 3 cm, 6 cm ii 90 cm2 6 a 104 cm2 7 a i Estimate = 42 cm2 ii A = 39.44 cm2 i Estimate = 49.5 m2 ii A = 47.7 m2 i Estimate = 108 cm2 ii A = 120.7 cm2 i Estimate = 3600 mm2 ii A = 4156.3 mm2 d 8 Area of small semicircle 1 1 2 2 × π × 3.42 = 18.16 1 1 2 2 Area of triangle = bh = × 6.8 × 9.2 = 31.28 Total area = 33.24 + 18.16 + 31.28 = 82.68 = 83 cm2 (2 s.f.) 10 a 60 m2 b 54.54 cm2 c 59.69 m2 Shape B, shaded area = π × 42 − 5.662 = 18.23 cm2 i c × π × 4.62 = 33.24 Shape A, shaded area = 82 − π × 42 = 13.73 cm2 a b 2 12 Learner’s own answers and explanations. For example: Arun is incorrect. The shaded area in Shape A is less than, not greater than, the shaded area in Shape B. Total area = Area A + Area B = 21 + 15 = 36 cm2 b 1 2 11 338 cm2 Area B = base × height = 5 × 3 = 15 cm2 5 1 = πr 2 = Area A = l × w = 10 × 3 = 30 Area B = π r 2 = d Chatri is not correct as the area of this compound shape is 83 cm2 not 82 cm2 (2 s.f.) 13 aWhen radius = 2 cm, height of rectangle = 4 cm and length of rectangle = 3 × 4 = 12 cm. 152.55 cm2 Seb’s method is incorrect. Learner’s own explanation and working. For example: When he works out the area of the circle he doesn’t use the correct radius. He actually uses the diameter of 11 cm rather than the radius of 5.5 cm. The answer should be: Shaded area = 4 × 12 − π × 22 = 48 − 4π = 4(12 − π) cm2. b i 12 − π cm2 ii 9(12 − π) cm2 iii 25(12 − π) cm2 iv 100(12 − π) cm2 c Learner’s own answer and explanation. For example: The number outside the bracket is the radius squared, and inside the bracket is always (12 − π). d r2(12 − π) cm2 14 a 400 m c 46.56 m e 14 669 m 2 b 10 465 m2 d 461 m f 4204 m2 15 8.8 cm Area of rectangle = 10 × 11 = 110 19 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 7.3 1 2 3 a 1 hectometre = 100 metres b 1 kilogram = 1000 grams c 1 megatonne = 1 000 000 tonnes a1 km = 1000 m, so 17.2 km = 17.2 × 1000 = 17 200 m d 1 gigalitre = 1 000 000 000 litres b 1 hL = 100 L, so 0.9 hL = 0.9 × 100 = 90 L a1 centimetre = 0.01 metres OR 1 metre = 100 centimetres c 1 Gg = 1 000 000 000 g, so 1.5 Gg = 1.5 × 1 000 000 000 = 1 500 000 000 g 1 microlitre = 0.000 001 litres OR 1 litre = 1 000 000 microlitres b 1000 mL = 1 L, so 43 000 mL = 43 000 ÷ 1000 = 43L d 1 nanometre = 0.000 000 001 metres OR 1 metre = 1 000 000 000 nanometres c 1 000 000 μg = 1 g, so 900 000 μg = 900 000 ÷ 1 000 000 = 0.9 g a3 nanolitres, 3 millilitres, 3 centilitres, 3 litres, 3 teralitres 9 3 nL, 3 mL, 3 cL, 3 L, 3 TL a9 micrograms, 9 milligrams, 9 grams, 9 kilograms, 9 gigagrams 9 μg, 9 mg, 9 g, 9 kg, 9 Gg aA millimetre is a very small measure of length. It is represented by the letters mm. You can also say that there are one thousand millimetres in a metre or that 1 millimetre is one thousandth of a metre. b A microgram is a very small measure of mass. It is represented by the letters μg. From the Sun to: Distance in … Venus 47.9 Mm Earth 108 Mm Jupiter 0.228 Gm Uranus 1.4 Gm Neptune 2.9 Gm 10 A and iii, B and v, C and i, D and ii, E and iv 11 aMarcus is incorrect. 1 MW = 1000 kW not 100 kW. 1 MW = 1 000 000 W and 1 kW = 1000 W, 1 000 000 ÷ 1000 = 1000, so 1 MW = 1000 kW b 12 630 MW = 630 000 kW = 630 000 000 W Name of star Distance in ly Distance in m Wolf 359 7.78 7.36 × 1016 You can also say that there are one million micrograms in a gram or that 1 microgram is one millionth of a gram. Ross 154 9.68 9.16 × 1016 YZ Ceti 12.13 1.15 × 1017 aA kilometre is a very large measure of length. It is represented by the letters km. Gliese 832 16.08 1.52 × 1017 1 microgram = 0.000 001 grams which is the same as 1 μg = 1 × 10−6 g. 1 kilometre = 1000 metres which is the same as 1 km = 1 × 103 metres. You can also say that there are one thousand metres in a kilometre or that 1 metre is one thousandth of a kilometre. b A megatonne is a very large measure of mass. It is represented by the letters Mt. 1 megatonne = 1 000 000 tonnes which is the same as 1 Mt = 1 × 106 t. 20 a100 cm = 1 m, so 760 cm = 760 ÷ 100 = 7.6 m c 1 millimetre = 0.001 metres which is the same as 1 mm = 1 × 10−3 m. 6 8 1 milligram = 0.001 grams OR 1 gram = 1000 milligrams b 5 7 b b 4 You can also say that there are one million tonnes in a megatonne or that 1 tonne is one millionth of a megatonne. 13 a 12 KB, 936 KB, 42.5 MB, 1.14 GB, 6.3 TB b i1 GB can store 178 photos, 16 GB can store 16 × 178 = 2848 photos. ii 28 288 photos iii238 photos. Working: 1.8 MB can store 476 photos. 1.8 MB × 2 = 3.6 MB, so double the file size means half as many photos. So, 1 GB can store 476 ÷ 2 = 238 photos. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE ivLearner’s own working and answer. For example: 32 GB = 32 000 MB and 32 000 MB ÷ 13 000 photos = 2.46… MB per photo. d 4 Suggest Sue uses a 2.4 MB file size for each photo as this will keep her just below the 32 GB limit. 1 = 0.125 which is a terminating decimal. 8 2 1 c Yes. Learner’s own explanations. For example: When a unit fraction is a terminating decimal then any multiple of a terminating decimal is also a terminating decimal. 8 terminating decimal. 3 8 1 = 3 × = 3 × 0.125 = 0.375 which is a 8 terminating decimal. 5 8 5 aSometimes true. Learner’s own explanations. For example: Apart from 1 = 5 × = 5 × 0.125 = 0.625 which is a 7 8 terminating decimal. b 14 b 1 = 0.05 which is a terminating decimal. 20 3 1 20 = 3× 20 20 = 5× 1 20 = 3 × 0.05 = 0.15 which is a 20 2 a b c = 9× = 9 × 0.05 = 0.45 which is a recurring decimal . 2 i = 0.3 recurring decimal iii iv 6 3 6 4 6 5 6 . = 0.6 recurring decimal b terminating decimal ii iii 2 25 5 25 11 25 = 0.44 terminating decimal Sometimes true. Learner’s own 1 1 1 , , are 5 10 20 1 1 1 , , are recurring. 15 35 45 d Always true. Learner’s own explanations. For example: Even if the fraction can be simplified, the denominator will be a multiple of 3, so will be recurring. aRecurring decimals. Learner’s own explanations. For example: When they are all written in their simplest form, the denominators are multiples of 3, so recurring. b B and D can be simplified. Learner’s own explanations. For example: They can both 1 be simplified to 6 . It doesn’t change the answer to part a, because the denominators are still multiples of 3, so recurring. c Any fraction which has a denominator which is a multiple of 9, when it is written in its simplest form, is a recurring decimal. a 1 terminating 4 b 4 terminating 5 c 8 recurring 15 d 1 terminating 5 = 0.08 terminating decimal = 0.2 terminating decimal = 0.5. Never true. Learner's explanations. For example: A denominator which is a multiple of 15 is also a multiple of 3, which is a recurring, not terminating, decimal. . = 0.83 recurring decimal 0.04 i 6 = 0.5 terminating decimal a c 21 20 2 c = 5 × 0.05 = 0.25 which is a terminating decimal. . 0.16 ii 3 1 1 terminating, but terminating decimal. 9 = explanations. For example: terminating decimal. 5 Learner’s own answers. Yes. Learner’s own explanations. For example: When a unit fraction is a terminating decimal then any multiple of a terminating decimal is also a terminating decimal. = 2 × = 2 × 0.125 = 0.25 which is a 8 = 0.8 terminating decimal b Exercise 8.1 a 25 aTerminating decimals. Learner’s own explanations. For example: All the denominators will divide exactly into 10, 100, 1000 or 10 000. 14 140 kg 1 20 iv 7 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 8 9 7 terminating 10 e 1 recurring 6 a recurring b terminating c recurring d terminating f 3 A and ii, B and iii, C and i 4 a–d i Learner’s own estimates. a, b Number of days off work due to illness Abi 8 4 = 30 15 Dave 6 1 = 30 5 Bim 5 1 = 30 6 Enid 2 1 = 30 15 Caz 3 1 = 30 10 Fin 9 3 = 30 10 5 ii 6 c ii 40 10 a b c 9 m. Learner’s own answer and 10 explanation. iv 2 1 13 + = 5 4 20 8 i 5 2 17 + = 18 3 18 ii 3 2 37 + = 5 9 45 9 iii 1 1 1 + = 6 3 2 iv 2 1 13 + = 5 4 20 10 a 5 8 b 40 11 a 50 25 1 2 4 2 3 3+ × 1 2 2 4 3 22 ÷ − 1 1 1 2 3 8 4 8 8 8 2 3 5 5 6 8 15 =3 2 a c 22 3 4 3 2 4 4 1 a 4 b c 4 2×4 8 = = 5 3 × 5 15 d 8 Indices: e 22 = 4 3 5 20 5 3 3 20 −1 = 5 40 6 6 b 2 d 5 3 1 15 Division: 4 ÷ = 4 × = Subtraction: 4 1 Addition: 3 + c 1 4 Multiplication: × 4 5 91 ×7 = 910 4 5 1 + 1 × 3 or equivalent. 6 6 3 17 2 m 36 5 8 cm2 9 9 b 12 cm Exercise 8.3 1 1 1 2 + 1 − Brackets: 1 − = 1 − = 1 8 2 3 2 7 14 ÷ = × = 9 3 7 3 3 1 13 2 2 14 91 28 819 1 Addition: + = + = 47 2 18 18 18 9 29 16 m2 36 3 7 a 15 b 35 c 4 16 2 5 1 5 1 1 1 + = 6 3 2 Addition: 2 + 1 = 2 + 1 = 3 b Division: iii Exercise 8.2 a 7 kg 10 7 3 2 37 + = 5 9 45 10 1 3 56 ii 4 13 c 1 2 5 + = 6 3 6 4 15 Learner’s own answer and explanation. For example: He cannot be correct because if you round both sides up and add them to 6 you get 6 + 5 + 13 = 24. This is nearly 2 m less than the perimeter, so the third side must be at least 2 m more than 6 m. i 2 ii 7 2 − 4 + 12 or equivalent. 5 25 2 (terminating), 3 + 4 = 23 (terminating). d 14 Multiplication: 6 × 7 = No. Learner’s own examples. For example: 1 + 1 = 3 (terminating), 2 + 3 = 31 11 b 50 6 For example: ii 25 OR A, B and E are recurring decimals; C, D and F are terminating decimals. 49 5 8 7 18 b a Learner’s own decisions on how to group the students. For example: A and F are not unit fractions; B, C, D and E are unit fractions. 1 4 a − f 11 2 5 6 = 29 6 =4 g 5 2 3 3 5 5 6 4 9 3 4 5 8 4 7 6 2 3 a 6 a 1 8 × 12 = 2 3 3 ×3× 4 = 2× 4 = 8 × 20 = × 5 × 4 = 3 × 4 = 12 5 × 18 = × 27 = × 32 = 5 6 4 9 3 4 5 × 6 × 3 = 5 × 3 = 15 × 9 × 3 = 4 × 3 = 12 × 4 × 8 = 3 × 8 = 24 × 48 = × 8 × 6 = 5 × 6 = 30 × 35 = 8 4 7 × 7 × 5 = 4 × 5 = 20 b × 20 = 1 4×2 c 18 ×4×5 = 1× 5 2 = 5 2 =2 28 1 2 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 3 b × 25 = 10 c 3 d 5 4 9 × 14 = × 24 = 3 2×5 3 2×2 5 3× 3 ×5 ×5= ×2×7 = ×3×8= 3×5 2 3× 7 2 5×8 3 = 15 = 21 = 2 2 40 3 4 a 5 a 9 b 12 c 15 d 33 e 30 f 6 g 9 1 3 h i 13 1 5 j 1 3 1 11 5 a 18 35 b 7 15 6 3 3 4 d e 21 40 f 9 13 g 1 9 7 11 h 1 5 1 9 b 2 1 3 1 e.g. × 3 9 b 1 3 2 13 + 2 × = 2 3 3 5 45 2 A and ii, B and v, C and i, D and iii, E and iv 2 a 2 12 ÷ b 18 ÷ 3 3 c 20 ÷ 8 Sarah is incorrect. 21 cm is the smallest whole number value for d so that the circumference is greater than 64 cm. d 30 ÷ Learner’s own working. For example: e 24 ÷ 1 69 > 64 7 × 22 = 7 1 = 69 cm, 7 7 3 a 15 ÷ 6 22 When d = 21 cm, C = × 21 = 22 × 3 = 66 cm, 7 66 > 64 When d = 20 cm, C = 6 62 < 64 22 7 × 20 = 440 7 6 = 62 cm, a–f i a c ii e ii 10 a 1 c 3 3 4 2 4 2 4 3 7 3 7 4 3 4 3 2 5 2 5 4 4 = 20 × = 5 × 4 × = 5 × 7 = 35 = 30 × = 15 × 2 × = 15 × 3 = 45 5 = 24 × = 6 × 4 × = 6 × 5 = 30 7 7 6 3× 2 8 9 9 9 8 4×2 7 = 15 × = 5 × 3 × = 5×7 2 = 35 = 27 2 = 17 1 = 13 1 2 12 ÷ = 12 × = 3 × 4 × = 3×9 2 2 2 c 3 20 9 4 14 16 2 25 b 1 5 4 2 11 3 1 22 2 ii d ii f ii 11 2 m 25 b 4 m2 1 5 d 1 5 m2 7 2 3 = 18 × = 6 × 3 × = 6 × 4 = 24 7 Learner’s own estimates. ii 4 4 3 = 12 × = 6 × 2 × = 6 × 3 = 18 b 7 9 14 1 1 15 484 4 Exercise 8.4 a When d = 22 cm, C = 23 1 7 22 2 2 2 3 6 = , × = 5 15 7 11 77 Learner’s own answer and explanation. For example: When the four numbers in the fractions are all different and are all prime numbers or 1, then you will not be able to cancel. When one of the numerators and denominators are the same, then you will be able to cancel. When one of the numerators and denominators are even, then you will be able to cancel. When one of the numerators and denominators are a multiple of each other, then you will be able to cancel. 3 c j 12 aLearner’s own examples of two proper fractions that when multiplied do not cancel. 1 3 4 19 24 1 16 2 c 35 48 i 11 24 m3 1 = 10 = 13 1 4 2 b =7 1 6 43 2 m 56 6 15 ÷ 10 4 A and iii, B and v, C and ii, D and iv, E and i 5 a 25 b d 35 e 5 5 5 6 2 ×3 10 × 5 20 ÷ = 20 × = 10 × 2 × = 3 = 50 = 39 = 16 2 = 19 1 3 3 d 13 = 15 × 1 5 13 10 = 3×5× 13 5× 2 = 3 × 13 2 2 3 4 88 1 2 20 15 c 16 f 2 1 4 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 6 15 16 a 8 9 4 c 4 9 1 5 1 11 1 D 1, A 1 , C 1 , B 1 3 16 14 d 7 1 6 b 2 e f 4 1 b 19 26 3 c Estimates are given first, then the accurate answers: a 1, 25 26 b 4, 3 c 3, 3 5 5 d e 5, 4 f 1 5 4 =1 ,1 3 3 21 7 1 1 =3 ,3 2 9 2 1 8 4 5 9 5 3 3 1 5 × = = , check 12 5 12 4 12 10 a 3 4 5 ⇒ 25 − 5 = 20 2 a 3 a 4 , check c 13 13 12 12 4 , check × = = 21 13 21 7 21 6 3 1 5 1 7 2 4 1.75 × 2 × 32 ⇒ 4.7 × 35 ⇒ 4 7 10 × 35 = 175 cm3 13 2 m 6 a 14 aLearner’s own explanation. For example: 1 3 He rounded 3 to 3 and he rounded 9 to 0.44 × 52 ⇒ 0.44 = 7 15 1 c 1 12 25 3 4 4 8 Completing the working gives 1 1: 15 8 75 4 75 23 × = 1: = 1: 2 13 26 26 8 2 23 Yes, 1: 2 ≈1 : 3 26 2 1: 3 2 1:1 5 47 10 b 45 44 100 47 × 35 ⇒ 4 2 7 × 35 2 10 329 = 164 1 2 c 234 , 52 = 25 ⇒ 44 49 1 2 1 4 100 3 2 3 0.9 × 6 ⇒ 0.9 = × 25 = 11 2 9 2 20 9 20 , 6 = ⇒ × 10 3 3 31 1 10 =6 c ( ) 2.4 × 33 − 7 ⇒ 2.4 = 24 10 , 33 − 7 = 20 ⇒ 24 1 10 × 20 2 = 48 4 3 2 1 2 1 ÷ = 2 , 1 +3 =5 , 7 14 3 2 3 6 2 1 3 2 × 1 = 3 , 8 3 − 2 5 = 5 11 5 2 5 4 6 12 7 a 8 4 2 m 5 9 a b 1 1 2 38 cm2 Exercise 8.5 10 1 m 1 11 8 c 12 b 57 5 8 19 m2 5 a 2 18 × 32 b 9. So 3 : 9 = 1 : 3. e 2 35 ⇒ = 5 d 35 9 12 50 kg c 5 × = c 18 b 3 5 = 35 × 4 = 140 a 7 20 4 3 7 3 21 21 × = ⇒ × 12 = 21 × 3 = 63 2 2 4 4 1 4 b c 81 1 , check 5 × 4 = 20 = 10 11 a 2 1 2 1 1 + 1.5 + 9 ⇒ + 1 = ( 2) = 4 ⇒ 4 + 9 = 13 2 2 2 24 b 80 b 24 1 4 × = 5 6 5 b d 1 1 3 52 − 4 + 0.75 ⇒ 4 + = 5 ⇒ 52 = 25 4 4 4 3.5 × 1.5 × 12 ⇒ Sofia is incorrect. Learner’s own examples. e.g. 1 1 2 3 10 ÷ = 2, ÷ = 2 3 3 3 3 3 2 − 0.6 − 3 ⇒ 2 − = ( 2) = 8 ⇒ 8 − 3 = 5 5 5 5 10 11 3 4 12 Terms are 12, 15, 18. nth term rule is 3n + 9, so 50th term = 159. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 13 a b c V = 154 cm3 c −6, −5.6, −5.2, −4.8, −4.4, −4, −3.6 22 22 × 12 3V = 3×154, πh = × 12 = 7 7 1 7 3V 3 × 154 3 × 154 × 7 49 = 22 × 12 = = πh 4 and 1 22 × 12 4 7 49 7 r= = = 3.5 4 2 d 7.5, 6.25, 5, 3.75, 2.5, 1.25, 0 a B b The 6th term, which is 2390. (Sequence is 3, 10, 38, 150, 598, 2390, …) 3 × 27 = 22 22 × 154 17 r= 81 = 222 92 9 = 222 22 Check: 1 1 3 V = π r 2h = 3 71 1 22 9 9 × 1 × × × 154 = 3 × 9 = 27 3 22 1 22 1 7 1 2 3 4 5 25 a 5, 7, 9, 11, 13 b 0, 3, 6, 9, 12 c 11, 9, 7, 5, 3 d 2, 2.5, 3, 3.5, 4 e 210, 190, 170, 150, 130 a 4, 5, 7, 10, 14 b 5, 7, 11, 17, 25 c 20, 17, 13, 8, 2 a non-linear b linear c linear d non-linear a linear b non-linear c linear d linear e non-linear f linear a 9, 5, 1, −3, … b 12, 17, 22, 27, … c 3, 4, 6, 9, … d 10, 9, 7, 4, … e 64, 40, 28, 22, … f 8, 10, 14, 22, … c 17, 15, 11, 5, … d 32, 24, 12, −4, … 11 a A 343; B 64; C 179 b 64, 179, 343 or B, C, A If ✱ = 6, then the sequence is −2, −2, −2, −2, … 5, 6, 17, … b −1, 2, 5, … c 4, 1, 4, … d −2, 0, 4, … a 3, 4 , 5 , 7, 8 , 9 , 11 b 10, 9 , 8 2, 7 , 6 , 6, 5 2 3 1 3 3 5 2 3 4 5 If ✱ = 5, then the sequence is −2, −3, −22, −10 643, … So, as long as ✱ is greater than 6, there will be positive numbers in the sequence. 13 Timo’s method is incorrect. Learner’s own explanation. For example: He has reversed the order of the operations, but he hasn’t used inverse operations to reverse the actual operations. Correct solution is: 4th term = 72 + 8 = 80, 80 ÷ 2 = 40, 3rd term = 40 + 8 = 48, 48 ÷ 2 = 24. 14 16 a 5 15, 19, 26, 36, … If ✱ = 7, then the sequence is −2, −1, 6, 223, … 7 1 5 b If ✱ = 8, then the sequence is −2, 0, 8, 520, … A and iv, B and ii, C and i, D and iii 1 3 7, 8, 11, 16, … If ✱= 9, then the sequence is −2, 1, 10, 1009, … 6 8 10 a 12 Sofia is incorrect. Learner’s own explanation. For example: Exercise 9.1 1 9 1 5 15 Two of the terms in the sequence are negative. Learner’s own working. For example: Sequence is −4, 2, −10, 86, … 16 First term = −10. Learner’s own working. For example: 4th term = 512, 3rd term = 0, reverse the function so rule is cube root and subtract ?. 3 512 − ? = 0 8−? = 0 ?=8 The reverse function is cube root and subtract 8. 3 3 2nd term = 0 − 8 = −8, 1st term = −8 − 8 = −10 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 9.2 1 a1st term = 3 × 1 = 3 2nd term = 3 × 2 = 6 3rd term = 3 × 3 = 9 4th term = 3 × 4 = 12 b 2 8 1 1 1st term = × 1 = 2nd term = 22 = 4 4th term = 42 = 16 d 1st term = 13 = 1 3rd term = 33 = 27 2nd term = 23 = 8 4th term = 43 = 64 b 1st term = 12 − 1 = 0 2nd term = 22 − 1 = 3 3rd term = 32 − 1 = 8 10th term = 102 − 1 = 99 c 1st term = × 1 = 2nd term = × 2 = 1 1 1 2 2 2 2 2 d 10 2 2 =1 They are the same. b i =5 1 2 2 1 n The sequence n is the same as . 3 3 1 n The sequence n is the same as . 5 5 3 3n The sequence n is the same as . 4 4 iv 4 A and iv, B and i, C and v, D and vi, E and iii, F and ii 5 a 15, 23, 31, …, 87 c −3 , −3, −2 , …, 1 e 1 1 2 2 b −1, 6, 13, …, 62 d 1 2 3 , , , …, 1 10 10 10 21, 24, 29, …, 120 6 A and iv, B and iii, C and i, D and ii 7 a Learner’s own answers. b B has the smaller value. A 11th term = 121 − 33 = 88, B 120th term = 2 × 120 + 7 = 87. c quadratic e neither f linear a n2 b n2 + 20 c n2 − 2 d n2 + 7 n 13 12 a A 3, B 5, C b 5 8 n 9 n 12 c 4 7 b 4 7 3 5 C ,A ,B 5 8 13 aNo. n2 + 34 = 292, n2 = 258, n = 258 = 16.06…, which is not a whole number. b 14 a Yes. 5832 = 18 , which is a whole number. 5832 is the 18th term in the sequence. 3 1 n+8 1 3 1 4 2 b 12.8 − 0.3n −2 − n d 3.5 − 3.5n 15 a 2, 7, 14, … b 119 16 a 9, 17, 27, … 153 b 1, 10, 25, …, 298 c 9, 12, 19, …, 180 b 64 and 125 c n The sequence n is the same as . iii d 11 a 2 1 2 2 2 3 1 10 3rd term = =1 10th term = = 5 2 2 2 a neither 1 1st term = 2nd term = = 1 ii 26 =1 1 10th term = × 10 = 3 4 =1 a1st term = 6 × 1 + 1 = 7 2nd term = 6 × 2 + 1 = 13 3rd term = 6 × 3 + 1 = 19 10th term = 6 × 10 + 1 = 61 2 c 10 Learner’s own explanations. For example: For the sequence n2 − 10, the first term will be negative, so 7 cannot be the first term. 1st term = 12 = 1 3rd term = 32 = 9 2 quadratic 4 c 3rd term = × 3 = b 9 4th term = × 4 = 3 linear 1 2 1 ×2= = 4 4 2 1 4 2nd term = 4 4 3 1 3rd term = × 3 = 4 4 1 a 3 3 Exercise 9.3 1 a 16 and 25 2 a i ii x 2 y 4 16 25 x 2 4 y 8 64 125 4 b They are the same. c i y = x2 ii y = x3 5 5 d i x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 y 0 2 4 6 8 10 12 14 16 18 20 22 24 26 3 Learner’s own answers. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE b ii x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 y 0 10 20 30 40 50 60 70 80 90 100 110 120 130 3 a i ii x 3 6 8 y 10 37 65 x 1 10 3 7 a x 1 8 1 4 1 2 2 y 1 16 1 4 1 16 i x 2 −4 −6 −10 y 2 8 18 50 x 5 −3 −15 −7 y 64 0 144 16 i y= x2 2 ii y = (x + 3)2 ii y −1 25 998 4 b i a i ii x −2 1 3 y 16 4 36 x −1 3 4 y iii 5 a x −1 2 y = 4x2 ii y = (3x)2 iii y = (x + 1)3 x −2 y ii iii 6 27 8 9 12 36 16 64 x −3 2 y 14 9 x −3 1 2 i y = (x − 4)2 ii y=x +5 iii y = x3 − x ×2 Sofia is correct. Learner’s own explanation. For example: When you square the positive and negative of the same number you get the same answer, e.g. 22 = (−2)2 = 4 and 52 = (−5)2 = 25. Zara is incorrect. Learner’s own explanation. For example: When you add 1 to the positive and negative of the same number you get different answers, so when you square these answers, your final answers will be different, e.g. (2 + 1)2 = 32 = 9 and (−2 + 1)2 = (−1)2 = 1. 4 0 y −27 b y=x −2 0 27 125 i i b 3 9 81 144 y b ii y=x +1 2 a b 1 2 5 − 1 2 − 5 8 c 1 4 1 2 − d 3 8 e 2 1 2 a 2 y f i y = x4 ii x=± 4 y iii Learner’s own check. i y = x5 ii x= 5 y iii Learner’s own check. i y = (3x)2 ii x=± iii Learner’s own check. i y = x3 − 10 ii x = 3 y +10 iii Learner’s own check. i y = ii x=±4 y iii Learner’s own check. i y= ii x = 3 2y iii Learner’s own check. y 3 x 4 2 x3 2 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 10 a, b Learner’s own answers for grouping the functions: For example: •One step functions: B, F, G, L Two step functions: A, C, D, E, H, I, J, K •Contains a power: A, C, D, F, H, J Contains a root: B, E, I, K Contains no powers or roots: G, L •Contains the number 4: A, D, E, L Contains the number 9: G, H, I, K Contains the number 2: C, J Contains no numbers: B, F •Contains fractions: C, D, E, I Contains no fractions: A, B, F, G, H, J, K, L 11 Sofia and Zara are both correct. The table of values works for both equations. This is because x = ± 4y = ± 4 × y = ± 2 × y = ± 2 y 12 3 x ×10 y Exercise 10.1 1 a 20 + 15 × 4 = $80 c y = 15w + 20 2 a 3 4 5 6 1 4 −3 1 4 5 32 −270 1 y Learner’s own explanation. For example: Start by working out the missing number in the function machine using the first pair of values in the table. 20 kg b 2x + 4y = 22 a a + b = 36 b 23 c b = 3a a 24 b i a 84 minutes c 36 minutes a 85 c 15 (with two pentagons) His conjecture is incorrect for the second function y = 5 − x5. If x5 is greater than 5, the y-value will be negative. For example: when x = 2, y = 5 − 25 = 5 − 32 = −27. 28 b t = 2g − 20 b 5p + 6h = 100 y = 10 + x because all the other functions are equivalent. 9 a r + b = 18 c 15 red and 3 blue b r + 4b = 27 Exercise 10.2 1 2 1 1 5 5 1 × ? = 1 , × ? = , ? = × 8 = 10, so the 2 4 8 4 4 missing number in the function machine is 10. His conjecture is correct for the first function y = (x − 5)4. When you work out x − 5, if the answer is positive or negative, once you have raised it to the power of 4, the answer is always positive. For example, 34 = (−3)4 = 81. 2 8 a 3 5 −1 0 1 2 3 4 y 7 12 17 22 27 32 at (0, 12) c 5 × 5 + 12 = 37, but 5 × 10 + 12 = 62 a x −10 0 10 20 30 40 y 8 10 12 14 16 18 b at (0, 10) c 11.4 a x 0 5 10 15 20 25 y 20 15 10 5 0 −5 b 4 x b 3 13 Marcus is incorrect. Learner’s own explanations. For example: ii 4s + 6l = 40 a, b Learner’s own answers. y = 10x 1 2 $170 7 3 x b a at (0, 20) and (20, 0) x 0 1 2 3 4 5 6 y 10 8 6 4 2 0 −2 b Learner’s own graph; A straight line through (0, 10) and (5, 0). c 3.5 a x 0 3 6 9 12 15 y 4 3 2 1 0 −1 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 6 7 8 b Learner’s own graph; A straight line through (0, 4) and (12, 0). c at (4.5, 2.5) a x 0 6 2 5 y 9 0 6 1.5 b Learner’s own graph; A straight line through (0, 9) and (6, 0). c at (0, 9) and (6, 0) a 11 aLearner’s own graph; A straight line through (0, 7) and (14, 0). i ii c 5 and −5 49 1 2 3 Exercise 10.3 y 6 −2 −3 −2 1 6 1 1 d Yes; (−9) − 3 = 78 a gradient 10 and y-intercept 20 b gradient −20 and y-intercept 10 c gradient 0.5 and y-intercept −2.5 a gradient b gradient 0 and y-intercept 12 c gradient −30 and y-intercept −45 3 a 1 2 4 a y =15 − x b gradient −1 and y-intercept 15 c (15, 0) a y = 4− x b gradient − and y-intercept 4 2 2 x −3 −2 −1 0 1 2 3 x2 + 1 10 5 2 1 2 5 10 y 8 y = x2 + 1 5 4 y=x −1 2 c 2 0 −2 2 x 0 12 4 8 y 9 0 6 3 Learner’s own graph; A straight line through (0, 9) and (12, 0). c Learner’s own graph; A straight line through (0, 6) and (8, 0). d Learner’s own graph; A straight line through (0, 3) and (4, 0). 6 × 5 + 5 × 6 = 60 b at (0, 12) and (10, 0) c Learner’s own graph; A straight line through (0, 12) and (10, 0). 1 10 and y-intercept 3 3 b − 1 3 c −2 1 3 1 3 x 0 12 6 3 y 4 0 2 3 d Learner’s own graph; A straight line through (0, 4) and (12, 0). e Learner’s own checks. a 4 × 2.5 = 10 and 20 − 10 = 10 b y= c gradient and y-intercept − 2 7 a C− 8 A and iv, B and ii, C and iii, D and i 9 a gradient −0.2 and y-intercept 2 b gradient −2.5 and y-intercept 5 c gradient −1 and y-intercept 0.4 It is on y = x2 − 1. x 45 a 6 29 b 0 52 − 3 = 22 10 a A is y = x2; B is y = x2 − 4 −1 c b (2, 6) −3 −2 Learner’s own graph; A parabola with the bottom at (0, −3). a c x b and c 9 Learner’s own graph; A straight line through (0, 12) and (4, 0). 12 a b d b 6 10 a 1 4 x−2 1 2 1 4 3 5 b 1 2 C2 4 × 0 − 2 × (−6) + 8 = 0 + 12 + 8 = 20 b 4 × 5 − 2 × 4 + 8 = 20 − 8 + 8 = 20 c y = 0.5x + 3 d gradient 0.5 and y-intercept 3 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 10.4 1 2 3 4 5 6 30 9 aLearner’s own graph; A line from the origin through (50, 875). a 400 m b gradient = 8; the speed is 8 m/s c y = 8t d 560 m a 15 dollars b 6 dollars/metre b 75 m c d = 6x c 2.5 m/s d 51 dollars d e 5 metres y = 300 − 2.5x (learners could use other letters). a i b 0.35 c y = 0.35x d 32.55 dinars e 2000 dollars a Time (hours) 0 5 10 Temperature (°C) 20 17 14 b y = 17.5x c You can exchange 1 Franc for 17.5 Rand. d 2275 Rand 10 aLearner’s own graph; A straight line from (0, 300) to (120, 0). ii 14 dinars 7 dinars Exercise 11.1 1 aFlour:2 parts = 250 g, 1 part = 250 ÷ 2 = 125 g Butter: 1 part = 125 g b 2 Total = 250 + 125 = 375 g aPeach juice: 3 parts = 450 mL, 1 part = 450 ÷ 3 = 150 mL b −0.6; the temperature decreases at a rate of 0.6 °C/hour c y = 20 − 0.6t d 12.8 °C a i 15 000 ii 23 000 Kim: 2 parts = 2 × 13 = $26 iii 27 000 Total they share = 65 + 26 = $91 Pineapple juice: 4 parts = 4 × 150 = 600 mL b 3 b 400/year or 0.4 thousand/year c p = 0.4t + 15 4 Total = 450 + 600 = 1050 mL Tina: 5 parts = $65 → 1 part = 65 ÷ 5 = $13 a Benji: 2 parts = $24, 1 part = 24 ÷ 2 = $12 Abdul: 1 part = $12 aLearner’s own graph; A line from the origin through (25, 42). b about 30 dollars c The gradient is 42 ÷ 25 = 1.68, so the equation is d = 1.68l. d 30.24 is the exact value e 100.8 dollars f 40 litres 7 He is not correct. Priya’s speed is 50 km/h and Mei’s is 60 km/h. 8 aLearner’s own graph; A straight line from (0, 12) through (20, 40). b 33 litres c 1.4 litres/second d y = 1.4x + 12 Caen: 3 parts = 3 × 12 = $36 b Total = 12 + 24 + 36 = $72 5 a 21 b 35 6 a 180 g b 480 g 7 a 6, 15, 24 b 45 8 aInstead of using $40 = 5 parts (for travel) he has used $40 = 4 parts (for food). He has also added up the total number of parts incorrectly. The total is 16 not 15. b 5 parts = $40, so 1 part = 40 ÷ 5 = $8 Total number of parts = 4 + 7 + 5 = 16 Total spent = 16 × 8 = $128 9 650 mL Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 10 a b $135 Zosia gets $60, Abie gets $75 11 12 : 16 → divide both numbers by 4 → 3 : 4 9 : 12 → divide both numbers by 3 → 3 : 4 12 550 mL vanilla ice cream, 2200 mL grape juice, 2750 mL ginger ale. Learner’s own method. For example: Grape juice: 2250 ÷ 4 = 562.5 mL per part, Ginger ale: 2750 ÷ 5 = 550 mL per part. Use 550 mL per part as smallest amount. Ice cream: 1 × 550 mL = 550 mL, Grape juice: 4 × 550 mL = 2200 mL, Ginger ale: 5 × 550 mL = 2750 mL 13 0.03 and 0.025 or 0.036 and 0.03 Activity Child : staff ratio Number of children Horseriding 4:1 20 20 ÷ 4 = 5 so 5 Sailing 5:1 15 15 ÷ 5 = 3 so 3 Rockclimbing 8:1 32 32 ÷ 8 = 4 so 4 Canoeing 10 : 1 28 28 ÷ 10 = 2.8 so 3 Total number of staff = 5 + 3 + 4 + 3 = 15 Exercise 11.2 1 14 22.5 cm 15 a b c 90 °, 35 ° and 55 ° or 90 °, 70 ° and 20 ° Two solutions. Learner’s own explanation. For example: The 20 ° difference could be between the right angle and one of the other angles, or it could be between the two other angles (not the right angle). 2 3 35 ° : 55 ° : 90 ° → 7 : 11 : 18 or 20 ° : 70 ° : 90 ° → 2 : 7 : 9 a i b direct proportion a i $4.40 iii $8.80 Number of children Horseriding 4:1 22 Number of staff a less time c i ×2 2 people = 6 days 4 people = 3 days a less than 60 seconds 22 ÷ 4 = 5.5 so 6 b more than 60 seconds c i ×2 Sailing 5:1 17 Rockclimbing 8:1 30 30 ÷ 8 = 3.75 so 4 ÷2 Canoeing 10 : 1 26 26 ÷ 10 = 2.6 so 3 d Learner’s own answer. For example: Move two children from horse riding to rock climbing and move two children from sailing to canoeing. New table is: b c ×2 ÷2 normal speed = 60 seconds ÷2 2 × speed = 30 seconds ii a more time 1 person = 12 days inverse proportion 5 160 $6.60 2 people = 6 days d Total number of staff = 6 + 4 + 4 + 3 = 17 31 b ÷2 17 ÷ 5 = 3.4 so 4 b ii direct proportion ii 4 iii 120 b Child : staff ratio ii 80 16 a Activity Number of staff normal speed = 60 seconds 1 speed = 120 seconds 2 ×2 inverse proportion ÷2 ×2 ×7 4 people = 7 hours 2 people = 14 hours 4 people = 7 hours 8 people = 3.5 hours 4 people = 7 hours 28 people = 1 hour ×2 ÷2 ÷7 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 7 a 100 g b x × y = 180, yes b 225 g c c 400 mL a i 3 hours i and iii Number of days it takes sheep to eat a bag of feed ii 12 hours iii 4 hours b Number of days 6 60 km/h 8 Number of students 5 2 3 4 6 7 40 35 30 25 20 15 10 5 0 8 0 1 hour 20 minutes 11 Zara is incorrect. It will take the same amount of time as 20 minutes is the time the journey takes. It doesn’t matter how many people are on the train. 1 3 b 2 3 c 1 4 d 5 12 e 7 12 2 a 0.35 b 0.2 3 a 85% b 35% c 20% b 0.7 c 0.52 169 cm b 14 a 2 houses b 12 people 4 a 0.34 c d 15 days 5 a i Houses b 0.85 Days 165 cm 6 20 4 6 1 120 4 1 30 1 7 a 0.4 b 0.9 6 60 12 8 a 0.01 b 0.98 4 60 8 c 0.5 d 0.95 e 0.89 f 0.81 15 a 32 ii 0.45 5 12 9 10 15 20 30 40 60 Number of days (y) 36 x×y 180 180 180 180 180 180 180 18 70 a 1 13 a 5 60 ivAnswer between 25 and 26 days (accurate answer is 25.7 to 1 d.p.) 12 A = 14, B = 15, C = 49, D = 7.5 Number of sheep (x) 30 40 50 Number of sheep Exercise 12.1 10 4 days People 20 iiNo, the points do not form a straight line. Cost per student 240 600 400 300 200 171.43 150 ($) 9 10 12 9 6 4.5 3 0.65 aT is 3, 6, 9 or 12 and F is 5 or 10 and these have no common element. b i 19 36 ii 2 3 iii 17 36 10 a i 13 25 ii 12 25 iii 22 25 b 15 is a multiple of both 3 and 5, so the events are not mutually exclusive. Adding the probabilities will not give the correct answer. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 12.2 1 2 3 7 aLearner’s own explanation. For example: The score on one dice does not affect the score on the other dice. b Yes. For example: The score on one dice does not affect the score on the other dice. c Learner’s own explanation. For example: If you get an even number you cannot get an odd number on the same dice and vice versa. aLearner’s own explanation. For example: The thunderstorm increases the probability that Zara will be late. b Learner’s own explanation. For example: They are not independent. If Arun is late on Monday he will probably make more effort not to be late on Tuesday. a i b 1 4 d Learner’s own explanation. For example: 1 8 ii 1 2 c iii 8 1 4 a 0.5 c Learner’s own explanation. For example: The probability of the letter being in the word HEAD is the same, whether it is in the word FACE or not. d Learner’s own explanation. For example: If the letter is in the word FACE, then P(in EACH) = 0.75. If the letter is not in the word FACE it is B, D, G or H and P(in EACH) = 0.25. The probabilities are different, so the events are not independent. b 1 4 ii 2 3 ii 2 3 1 6 of 12 is impossible and P(Z) = 0. The probabilities are different, so the events are not independent. i 1 2 iii 1 3 b i 1 3 iii 1 3 c Learner’s own explanation. For example: 9 6 aLearner’s own explanation. For example: 2 If A happens then P(B) = ; if A does not 5 3 happen then P(B) = . These are not the 5 same so the events are not independent. b Yes. Learner’s own explanation. For 1 example: This time P(B) = 2 if A happens and if A does not happen. 33 ii 4 11 b For all 3 to be the same colour the third must be red. The probability of this is iii 2 11 i 1 C is correct. Learner’s own explanation. For example: The results of the first three rolls and of the fourth roll are independent. 5 11 a P(multiple of 3) = 3 both if the number is even and if it is not. 5 Learner’s own explanation. For example: If blue is 6 then P(Z) = P(total of 12) = P(yellow is 6) = . If blue is not 6 a total 1 a 0.5 aLearner’s own explanation. For example: If the blue dice is 6 then P(Y) = P(yellow 1 is 6) = 6 ; If the blue dice is not 6 it will be another number and then P(Y) is 1 again . The same probability implies 6 independent events. Whether or not R happens, P(F) = 4 . 4 b 2 4 = . 10 5 Exercise 12.3 1 a 0.32 b 0.6 c 0.2 d 0.12 a b c 1 6 5 6 1 36 25 36 3 a 0.42 b 0.3 c 0.18 4 a 1 64 b 1 32 c 1 32 d 5 32 e 25 64 2 d Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 5 a First late 0.25 0.75 b 6 i not late Outcome 0.4 late late, late 0.25 × 0.4 = 0.1 0.6 not late late, not late 0.25 × 0.6 = 0.15 0.4 late not late, late 0.75 × 0.4 = 0.3 0.6 not late not late, not late 0.75 × 0.6 = 0.45 ii 0.1 0.45 iii 0.9 iii 0.14 a First Second red 0.6 blue red, red 0.6 × 0.35 = 0.21 blue red, blue 0.6 × 0.65 = 0.39 0.35 red blue, red 0.4 × 0.35 = 0.14 blue blue, blue 0.4 × 0.65 = 0.26 0.65 b i Outcome 0.35 red 0.65 0.4 7 Second ii 0.21 0.26 a First Second Outcome 3 4 odd odd, odd 2 3 × 34 = 6 12 = 1 2 1 4 even odd, even 2 3 × 14 = 2 12 = 1 6 3 4 odd even, odd 1 3 × 34 = 3 12 = 1 4 1 4 even even, even 1 3 × 4 1 = 1 12 odd 2 3 1 3 even b 34 i 1 2 ii 1 12 iii 11 12 iv 5 12 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 8 a Learner’s own tree diagram. For example: Outcome First Second 3 5 T TT 3 5 × 35 9 = 25 or 0.36 2 5 F TF 3 5 × 25 6 or 0.24 = 25 3 5 T FT 2 5 × 35 6 = 25 or 0.24 2 5 F FF 2 5 × 5 2 = 25 or 0.16 T 3 5 2 5 F 4 The branches for F and T could be reversed. b 9 9 or 0.36 25 i 21 or 0.84 25 ii 0.23. Learner’s own method. For example: This could be done with a tree diagram. 0.8 × 0.05 + 0.2 × 0.95 = 0.23 Exercise 12.4 1 a 0.46 2 a 0.45 b c Yes. Learner’s own explanation. For example: 0.325 is quite close to 0.35, so there is no reason to reject the conjecture. 3 a 0.58 4 a 17 = 0.68 25 b i c 8 is the best estimate. For example: The relative frequency is tending to 0.8 (1 d.p.) and 0.8 × 10 = 8 a Total 10 20 30 40 50 60 70 80 Silver cars 2 7 11 16 19 23 27 31 Relative frequency 0.2 0.35 0.367 0.4 0.38 0.383 0.386 0.388 5 6 0.36 0.325 b 0.7 ii c 0.18 0.785 iii 0.14 0.782 b Learner’s own graph. Check that the points from the table in part a have been plotted correctly. c Learner’s own estimate. 0.38 or 0.39 or 0.4 would be a sensible estimate from the data. a b c 35 b Flips 20 40 60 80 100 Frequency of 2 heads 5 9 11 17 19 Relative frequency 0.25 0.225 0.183 0.2125 0.19 Learner’s own graph. Check that the points from the table in part a have been plotted correctly. Flips 20 40 60 80 100 Frequency of 2 heads 4 11 16 20 24 Relative frequency 0.2 0.275 0.267 0.25 0.24 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE d 7 8 Learner’s own graph. Check that the points from the table in part c have been plotted correctly. 5 a N a, b and c Learner’s own results. 8 cm (80 m) d Learner’s own graphs. 85 ° e Learner’s own comparison of relative frequencies with 0.5. 145 ° 6 cm (60 m) a and b Learner’s own results. c Learner’s own comparison of relative 1 6 Keri frequencies with 0.1666 … (that is ). b Exercise 13.1 1 2 a 4.5 × 8 = 36 km b 18 ÷ 8 = 2.25 cm a6 cm. Learner’s own diagram. Check that the length of the line is 6 cm and that the angle between the N arrow and the line is 120 °. b 3 Dom 8.5 cm. Learner’s own diagram. Check that the length of the line is 6 cm and that the angle between the N arrow and the line is 35 °. 6 No. Learner’s own diagram and explanation. For example: The diagram shows that the jeeps are moving in different directions and they will not meet at all. 7 a, b Yes. Learner’s own diagram and explanation. For example: The diagram shows that the ship is much closer to C than B. 8 a N N Learner’s own diagram. Check that the length of the line is 4 cm and that the angle between the N arrow and the line is 95 °. 4 Learner’s own measurement. In the range 70–75 m. 80 ° N 12.5 cm (125 km) 145 ° 8.5 cm (85 km) Harbour Ship Fabia 3 cm (300 m) 50 ° Luca b Learner’s own measurement and conversion. In the range 175–180 km. c Learner’s own measurement. In the range 283 °–289 °. Gate 230 ° 2 cm (200 m) 9 N a 190 ° 6 cm (120 km) 12 cm (240 km) Airport Aeroplane b 36 N 300 ° Learner’s own measurement and conversion. In the range 225–230 km. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE c Learner’s own measurement. In the range 088 °–093 °. 10 a Greg N 95 ° Car 14 In the range 72–78 km and 338 °– 342 °. 15 a, b Learner’s own answers. Allow ± 2 ° on the bearings and ± 2 mm on the distances on the map. For example: N 35 ° 7 cm (14 km) 8 cm (16 km) Distance in real life (m) Start A 080 ° 3.5 700 A B 100 ° 4.6 920 B E 140 ° 2.4 480 b Learner’s own measurement and conversion. In the range 25.8–26.2 km. E F 227 ° 4.5 900 c Learner’s own measurement. In the range 245 °–250 °. F C 345 ° 3.5 700 C D 255 ° 3.6 720 D Finish 328 ° 3.3 660 11 In the range 7.3–7.5 km and 215 °–220 °. 12 a In the range 45–48 km. Exercise 13.2 b In the range 52–55 km. 1 13 a160 °. Learner’s own explanation. For example: Triangle is equilateral so angle ABC = 60 °. Line BD is parallel to the north arrow so angle ABD = 40 ° and so angle DBC = 60 − 40 = 20 °. Bearing of C from B = 180 − 20 = 160 °. b c 37 Distance Bearing on map (cm) From To 280 °. Learner’s own explanation. For example: Triangle is equilateral so angle ACB = 60 °. Line CE is parallel to line BD so angle ECB = angle DBC = 20 °. Bearing of C from B = 360 − 20 − 60 = 280 °. 2 3 a 1 1 × 3, × 6 = (3 ÷ 3, 6 ÷ 3) = (1, 2) 3 3 b 2 2 × 3, × 6 = (3 ÷ 3 × 2, 6 ÷ 3 × 2) = ( 2, 4) 3 3 a 1 1 × 4, × 12 = ( 4 ÷ 4, 12 ÷ 4) = (1, 3) 4 4 b 3 3 × 4, × 12 = ( 4 ÷ 4 × 3, 12 ÷ 4 × 3) = (3, 9) 4 4 H and iii, I and ii, J and vi, K and i, L and v, M and iv Learner’s own accurate sketch and checks. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 4 a (9, 12) b Letter A B C D E F K R W Position in alphabet 1st 2nd 3rd 4th 5th 6th 11th 18th 23rd xcoordinate 1 × 3 = 3 2 × 3 = 6 3 × 3 = 9 4 × 3 = 12 5 × 3 = 15 6 × 3 = 18 11 × 3 = 33 18 × 3 = 54 23 × 3 = 69 ycoordinate 1 × 4 = 4 2 × 4 = 8 3 × 4 = 12 4 × 4 = 16 5 × 4 = 20 6 × 4 = 24 11 × 4 = 44 18 × 4 = 72 23 × 4 = 92 Coordinate pair (3, 4) c (6, 8) (9, 12) (12, 16) (15, 20) (18, 24) (33, 44) (54, 72) (69, 92) iThe x-coordinates are the numbers in the 3 times table. To work out the x-coordinate of any letter, multiply the position number of the letter in the alphabet by 3. iiThe y-coordinates are the numbers in the 4 times table. To work out the y-coordinate of any letter, multiply the position number of the letter in the alphabet by 4. 5 6 a B (4, 5) b C (12, 15) c A (3, 2) d C (15, 10) a (10, 16) b (15, 24) c Learner’s own explanation. For example: E is the 5th letter of the alphabet, so has coordinates (5 × 5, 5 × 8) = (25, 40). d T is the 20th letter of the alphabet, so has coordinates (20 × 5, 20 × 8) = (100, 160). e nth letter of the alphabet has coordinates (5n, 8n). 7 H (28, 36) 8 Difference in x-coordinates = 10 − 1 = 9 Difference in y-coordinates = 13 − 1 = 12 1 3 1 3 ×9 = 3 × 12 = 4 E = C(1, 1) + (3, 4) = (1 + 3, 1 + 4) = (4, 5) 9 aDifference in x-coordinates = 7 − 2 = 5 Difference in y-coordinates = 18 − 3 = 15 2 ×5= 2 5 2 × 15 = 6 5 H = F(2, 3) + (2, 6) = (2 + 2, 3 + 6) = (4, 9) b Learner’s own check by drawing a diagram. 10 aYes. Learner’s own justification. For example: 3 2 1 of the way = 4 × 3 = 12. 3 3 3 1 3 2 y-coordinates: of the way = 10, so of the way = 10 ÷ 2 = 5, so of the way = 5 × 3 = 15. 3 3 3 x-coordinates: of the way = 8, so of the way = 8 ÷ 2 = 4, so So, B is the point (12, 15). b 38 No. Learner’s own justification. For example: OA is 2 2 1 of OB so ratio OA : AB = : = 2 : 1. 3 3 3 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 13.3 11 aLearner’s own diagram. Coordinate grid with points A (2, 7), B (10, 3), C (5, 3) and D (11, 5). Line segments AB and CD drawn. Point where AB crosses CD labelled ‘E’. b E (8, 4) c Midpoint of CD = 5 + 11 3 + 5 16 8 , = , = (8, 4) = E 2 2 2 2 d 1 a, b Learner’s own diagram. Coordinate grid. Triangle A with vertices (1, 4), (2, 3) and (2, 5). Triangle B with vertices (4, 3), (5, 4) and (4, 5). Triangle C with vertices (6, 2), (7, 3) and (6, 4). 2 a, b Learner’s own diagram. Coordinate grid. Rectangle A with vertices (1, 4), (3, 4), (3, 5) and (1, 5). Rectangle B with vertices (3, 4), (4, 4), (4, 6) and (3, 6). Rectangle C with vertices (3, 0), (4, 0), (4, 2) and (3, 2). 3 a and ii, b and iii, c and i 4 aA to B is a reflection in the mirror line x = 3. 3 of the way along AB: 4 Difference in x-coordinates = 10 − 2 = 8 3 ×8 = 6 4 Difference in y-coordinates = 3 − 7 = −4 3 × −4 = −3 4 So, A(2, 7) + (6, −3) = (2 + 6, 7 − 3) = (8, 4) = E. 12 Difference in x-coordinates = −4 − −9 = −4 + 9 = 5 3 ×5= 3 5 Difference in y-coordinates = −2 − −12 = −2 + 12 = 10 3 × 10 = 6 5 b Learner’s own working. For example: d B to E is a reflection in the mirror line y = 5. e E to C is a rotation 180 °, centre (6, 6). f C to F is a translation − . 1 −3 a–d Learner’s own diagram. Coordinate grid. Triangle A with vertices (−1, −4), (−3, −2) and (−4, −4). Triangle a with vertices (3, −3), (6, −3) and (5, −1). Triangle b with vertices (0, 0), (−3, 0) and (−1, 2). Triangle c with vertices (0, 2), (−2, 3) and (0, 5). Triangle d with vertices (3, 3), (6, 3) and (5, 5). 7 Difference in y-coordinates is 5 − −7 = 12. Points: P S T U V She has only got part b correct. Learner’s own explanations. For example: The object and its image are always congruent, so answers are: x-coordinates −4 −3 −2 −1 0 1 2 a Corresponding lengths are the same. y-coordinates −7 −5 −3 −1 1 3 5 b Corresponding angles are the same. c The object and the image are congruent. a Translation b Translation 0 c Translation − Q R 8 39 A to D is a rotation 90 °, anticlockwise, centre (1, 4). 6 Difference in x-coordinates is 2 − −4 = 6. There are 6 points after P, so the x-coordinates increase by 6 ÷ 6 = 1 for each point, and the y-coordinates increase by 12 ÷ 6 = 2 for each point. c a–c Learner’s own diagram. Coordinate grid. Triangle A with vertices (6, 4), (7, 6) and (5, 6). Triangle B with vertices (1, 1), (3, 1) and (2, 3). Triangle C with vertices (3, 1), (5, 2) and (3, 3). Triangle D with vertices (5, 0), (7, 0) and (6, 2). S (−1, −1) T (0, 1) A to C is a translation 2 . 5 So, K(−9, −12) + (3, 6) = (−9 + 3, −12 + 6) = (−6, −6). 13 a 6 b 4 6 −6 4 −6 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 9 d Reflection in line y = 1. e Reflection in line x = 0 (or y-axis). f Reflection in line y = −2. g Reflection in line x = −2. h Rotation 90 ° clockwise, centre (0, −4). i Rotation 90 ° anticlockwise, centre (0, 0). j Rotation 180 °, centre (0, 2). a iRotation 90 ° anticlockwise about (−1, 3). b 3 Learner’s own diagram. The enlarged rightangled triangle should have the corresponding vertex on the cross, base length eight squares and height four squares. 4 4 −4 ii Translation iii Reflection in the line x = −1. iv Reflection in the line x = −3.5. iCheck learners’ own combinations of at least two transformations. For example: 5 Rotation 90 ° anticlockwise about (−2, 3) followed by translation 0 −5 iiCheck learners’ combinations of at least two transformations. For example: aLearner’s own explanation. For example: The top vertex of the kite is only two squares from the centre of enlargement, not three. All the other vertices are in the correct position. b Reflection in line x = 3, followed by translation 1 3 10 Zara is not correct. Arun is correct. Learner’s own diagrams. 11 a A to C b A to B c A to D d B to E 12 aLearner’s own diagram. Check that each of the transformed shapes are drawn correctly. The final image should have vertices (4, 0), (5, 0), (5, 1), (5, −1) and (4, 1). b 8 −3 A translation 6 Exercise 13.4 1 Learner’s own diagram. Enlarged triangle with vertices (1, 0), (5, 2) and (1, 2). aLearner’s own diagram. Enlarged square with vertices (0, 1), (3, 1), (3, 4) and (0, 4). b 40 scale factor 2 aLearner’s own diagram. Enlarged triangle with vertices (3, 1), (5, 1) and (3, 5). b 2 a Learner’s own diagram. Enlarged triangle with vertices (0, 2), (3, 2) and (0, 5). Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE b scale factor 3 f A = 22 × 12.566 = 50.265 cm2 (3 d.p.) g A = πr2 = π × 42 = 16π = 50.265 cm2 (3 d.p.) 15 aArun is incorrect. He has multiplied the perimeter of H by 3 instead of dividing by 3. Perimeter = 36 ÷ 3 = 12 cm b c Area = 54 ÷ 32 = 6 cm2 Exercise 14.1 scale factor 2 1 a Volume = area of cross-section × length = 20 × 8 = 160 cm3 b Volume = area of cross-section × length = 15 × 6 = 90 cm3 c Volume = area of cross-section × length = 12 × 9 = 108 cm3 d 7 aLearner’s own diagram. Shape B with vertices (6, 0), (10, 0), (10, 2), (8, 4) and (6, 4). b c 8 9 = 30 × 12 = 360 cm3 2 a Learner’s own diagram. Shape C with vertices (0, 1), (6, 1), (6, 4), (3, 7) and (0, 7). Volume = area of cross-section × length = 32 × 10 = 320 cm3 b Learner’s own diagram. Shape D with vertices (1, 0), (9, 0), (9, 4), (5, 8) and (1, 8). 1 = 15 × 7 = 105 cm3 c Area of cross-section = area of circle = π × r 2 = π × 42 = π × 16 = 50.265… cm2 Area of N = 9 × 4 = 144 cm 2 Volume = area of cross-section × length = 50.265… × 11 = 552.92 cm3 10 Perimeter of Z = 36 cm 3 a 150 cm3 11 a Enlargement scale factor 3, centre (6, 2). b 129.6 cm3 b Enlargement scale factor 2, centre (3, 5). c 427.5 cm3 4 Area of Length cross-section of prism Volume of prism a Enlargement scale factor 2, centre (2, 4). 8.4 cm2 20 cm 168 cm3 b 24 cm2 6.5 cm 156 cm3 b C = πd = 4π = 12.566 cm (3 d.p.) c 58 m2 5.7 m 330.6 m3 c C = 2 × 12.56637… = 25.133 cm (3 d.p.) d 56.85 mm2 62 mm 3524.7 mm3 d C = πd = 8π = 25.133 cm (3 d.p.) e A = πr2 = π × 22 = 4π = 12.566 cm2 (3 d.p.) 12 Enlargement scale factor 3, centre (6, 1). 13 Enlargement scale factor 2, centre (3, 4). 14 a 41 1 Volume = area of cross-section × length Perimeter of N = 12 × 4 = 48 cm Area of Z = 72 cm2 Area of cross-section = area of triangle = 2 × base × height = 2 × 6 × 5 = 15 cm2 (1, 1), (1, 5), (7, 5), (3, 1) 2 Area of cross-section = area of rectangle = base × height = 8 × 4 = 32 cm2 aLearner’s own diagram. Enlarged shape with vertices as given in part b. b Volume = area of cross-section × length Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 5 6 7 a 480 cm3 c 675 cm b 480 cm3 b 3 Learner’s own answer and explanation. For example: Timo has used the diameter instead of the radius in the volume formula. Correct answer is 226 cm3 (3 s.f.) a 754.0 cm c 42 411.5 mm b 3 Base: 24 × 8 = 192 Total = 120 + 208 + 192 = 520 cm2 c 492.6 cm 3 3 1 Triangle ends: 2 × × 12 × 9 = 108 2 Sloping face: 15 × 12.5 = 187.5 Back face: 9 × 12.5 = 112.5 8 Radius of circle Area of circle a 7 cm 153.94 cm2 12 cm 1847.26 cm3 b 1.5 m 7.07 m2 2.4 m 16.96 m3 3 c 9 cm 254.47 cm2 7.51 cm 1910 cm3 = π × 32 d 2.19 m 15 m2 3.8 m 57 m3 e 4.55 mm 65 mm 22 mm 1430 mm 9 a x = 4.3 b x = 3.3 c x = 2.1 10 a 729 cm3 b 13 851 g c $96 957 Height of Volume of cylinder cylinder 2 a b 2 a Total = 108 + 187.5 + 112.5 + 150 = 558 cm2 Area of circle = πr2 = 28.27 cm2 (2 d.p.) Circumference of circle = πd =π×6 = 18.85 cm (2 d.p.) Area of rectangle = circumference of circle × 8 Exercise 14.2 1 Base: 12 × 12.5 = 150 3 11 The smallest. Learner’s own explanation. For example: In the smallest tin you get 233 cm3 per dollar, compared with 201 cm3 per dollar for the medium tin and 225 cm3 per dollar for the large tin. = 18.85 × 8 = 150.80 cm2 (2 d.p.) Total area = 2 × area of circle + area of rectangle = 2 × 28.27 + 150.80 = 207 cm2 (3 s.f.) 4 a 477.5 cm2 Square base: 10 × 10 = 100 b 401.1 cm2 1 Triangular faces: 4 × × 10 × 8 = 160 2 2 Total = 100 + 160 = 260 cm c 8482.3 mm2 5 The tetrahedron has the greater surface area. Triangular face: × 20 × 17 = 170 Tetrahedron: SA = 448 cm2, Cylinder: SA = 427.25… cm2 Total = 4 × 170 = 680 cm2 448 cm2 > 427.25… cm2 1 2 Front and back: 2 × (4 × 12) = 96 6 151 cm2 (3 s.f.) Sides: 2 × (4 × 10) = 80 7 a 3840 mm2 b 1548 cm2 Top and base: 2 × (12 × 10) = 240 Total = 96 + 80 + 240 = 416 cm2 42 1 Triangle ends: 2 × × 24 × 5 = 120 2 Sides: 2 × (13 × 8) = 208 8 344 cm2 (3 s.f.) Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 9 Carlos is incorrect. Learner’s own working and explanations. For example: 5 b The surface area of the polytunnel = 1 2 ( ( ) ) aThere are two vertical and one horizontal planes of symmetry. × 2 × π × 4.52 + π × 9 × 27 = 445.32 ... m 2 Carlos will need more than 445 m2 of plastic to make the polytunnel as the total surface area is more than 445 m2 and he will need to allow extra plastic for overlaps at the edges, wastage, etc. 6 a 2D regular polygon Number of lines of symmetry 3D pyramid Number of planes of symmetry Learner’s own answers. triangle 3 triangular 3 b Prism B has a greater surface area than prism A. square 4 square 4 pentagon 5 pentagonal 5 c Prism A: hexagon 6 hexagonal 6 octagon 8 octagonal 8 10 a 1 1 2 2 SA = 2 × × π × 52 + × π × 10 × 15 + 10 × 15 = 464 cm2 (3 s.f.) b The number of lines of symmetry of a regular 2D polygon is the same as the number of planes of symmetry of its matching 3D pyramid. Learner’s own explanation. For example: This is because all the lines of symmetry become vertical planes of symmetry, but the triangular sides of the pyramid meet at a point so there are no horizontal lines of symmetry. c i Prism B: 1 1 4 4 SA = 2 × × π × 82 + × π × 16 × 13 + 2 × 8 × 13 = 472 cm2 (3 s.f.) Difference = 472 − 464 = 8 cm2 Exercise 14.3 1 a and d 2 a A and F c C 3 Learner’s own diagram. Two vertical and one horizontal planes of symmetry, each plane splitting the shape into two congruent shapes. b B and D 7 aLearner’s own diagram. One vertical plane of symmetry, splitting the shape into two congruent shapes. b ii 12 aLearner’s own diagram. Check that the drawn part is a reflection of the shape drawn in the question. b One c Learner’s own diagram. One plane of symmetry splitting the shape into two copies of the shape in the question. The other plane is perpendicular to the first, splitting the shape into two congruent shapes. Learner’s own diagram. One vertical plane of symmetry, splitting the shape into two congruent shapes. 4 8 10 aLearner’s own diagram. Check that the drawn part is a reflection of the shape drawn in the question. b Cube c 9 Two like this Two like this 43 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 15.1 1 2 a 44 Frequency Midpoint 1 6 8 11 14 5 a Time, t (seconds) Tally Frequency Midpoint 5 15 25 0 < t ⩽ 10 ll 2 5 10 < t ⩽ 20 llll 5 15 35 45 55 20 < t ⩽ 30 llll lll 8 25 30 < t ⩽ 40 llll 4 35 40 < t ⩽ 50 l 1 45 b Learner’s own frequency polygon. Same axes and labels as in the diagram in the question. Points (5, 1), (15, 6), (25, 8), (35, 11), (45, 14) and (55, 5) plotted and joined with straight lines. a Height, h (cm) Frequency Midpoint b 3 Time, t (seconds) 0 < t ⩽ 10 10 < t ⩽ 20 20 < t ⩽ 30 30 < t ⩽ 40 40 < t ⩽ 50 50 < t ⩽ 60 4 260 ⩽ h < 280 3 270 280 ⩽ h < 300 7 290 300 ⩽ h < 320 9 310 320 ⩽ h < 340 1 330 Learner’s own frequency polygon. Same axes and labels as in the diagram in the question. Points (270, 3), (290, 7), (310, 9) and (330, 1) plotted and joined with straight lines. a 32 b Height, t (cm) Frequency Midpoint 10 ⩽ t < 12 4 11 12 ⩽ t < 14 16 13 14 ⩽ t < 16 7 15 16 ⩽ t < 18 5 17 c Learner’s own frequency polygon. Make sure that they use suitable axes labels and scales. Points (11, 4), (13, 16), (15, 7) and (17, 5) plotted and joined with straight lines. d 20 5 = 32 8 e Learner’s own answer and explanation. For example: Zara is incorrect. You don’t know from the frequency polygon what the fastest time is. All you can say is that the fastest time is between 10 and 12 minutes. 5 b Learner’s own frequency polygon. Make sure that they use suitable axes labels and scales. Points (5, 2), (15, 5), (25, 8), (35, 4) and (45, 1) plotted and joined with straight lines. a 50 b Wednesday Height, h (cm) Frequency Midpoint 120 ⩽ h < 140 4 130 140 ⩽ h < 160 6 150 160 ⩽ h < 180 22 170 180 ⩽ h < 200 18 190 Saturday Height, h (cm) Frequency Midpoint 120 ⩽ h < 140 25 130 140 ⩽ h < 160 16 150 160 ⩽ h < 180 7 170 180 ⩽ h < 200 2 190 c Learner’s own diagram showing two frequency polygons on one set of axes. Make sure that they use suitable axes labels and scales. Make sure that each polygon is labelled clearly. Wednesday points (130, 4), (150, 6), (170, 22) and (190, 18) plotted and drawn with straight lines. Saturday points (130, 25), (150, 16), (170, 7) and (190, 2) plotted and joined with straight lines. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE d 6 aLearner’s own diagram showing two frequency polygons on one set of axes. Make sure that they use suitable axes labels and scales. Make sure that each polygon is labelled clearly. Falcons Club points plotted at (2.5, 4), (7.5, 24), (12.5, 18), (17.5, 12) and (22.5, 10) and joined with straight lines. Harriers Club points plotted at (2.5, 10), (7.5, 8), (12.5, 10), (17.5, 26) and (22.5, 16) and joined with straight lines. b 7 d i 12 iiNo, Sienna cannot fill in the correct frequencies in her table. Learner’s own explanation. For example: From the first table Sienna knows that there are two girls with a mass between 7.0 and 7.1 kg. However, this does not tell her how many girls had masses between 7.0 and 7.05 kg and how many girls had masses between 7.05 and 7.1 kg, so it is impossible for her to complete her table. She would have to find the original data, before it was grouped, in order to group it the way she wants to. 8 Learner’s own answers. For example: Learner’s own comments. For example: The most popular training time for the Falcons Club was between 5 and 10 hours, whereas for the Harriers Club it was between 15 and 20 hours. In the Falcons Club only 22 athletes trained for more than 15 hours a week compared with 42 athletes from the Harriers Club. Time to solve maths problem, t (seconds) Tally 20 ⩽ t < 30 llll llll 10 25 c Falcons Club 68, Harriers Club 70. 30 ⩽ t < 40 llll llll llll llll llll 25 35 d Learner’s own answer and explanation. For example: Yes, because the number of athletes surveyed at each club was nearly the same. 40 ⩽ t < 50 llll llll llll lll 18 45 50 ⩽ t < 60 llll ll 7 55 a Frequency Midpoint aLearner’s own frequency polygon. Make sure that they use suitable axes labels and scales. Points (7.05, 2), (7.15, 12), (7.25, 14), (7.35, 9), (7.45, 7) and (7.55, 6) plotted and joined with straight lines. b Learner’s own frequency polygon. Make sure that they use suitable axes labels and scales. Points (25, 10), (35, 25), (45, 18) and (55, 7) plotted and joined with straight lines. b c Learner’s own comments. For example: Most students took less than 40 seconds to solve the puzzle. i Mass, m (kg) Frequency 7.0 ⩽ m < 7.2 14 7.2 ⩽ m < 7.4 23 7.4 ⩽ m < 7.6 13 iiLearner’s own frequency polygon. Make sure that they use suitable axes labels and scales. Points (7.1, 14), (7.3, 23) and (7.5, 13) plotted and joined with straight lines. c 45 are more groups so it shows you more information on the mass of the girls. The second frequency polygon only has three groups so less information can be taken from the graph. Learner’s own answer and explanation. For example: On Saturday there were fewer taller people and more shorter people. There were only two people with a height between 180 cm and 200 cm on Saturday compared with 18 on Wednesday. There were 25 people with a height between 120 cm and 140 cm on Saturday compared with four on Wednesday. Learner’s own answers and explanations. For example: The first frequency polygon gives you better information because there 9 Mass, m (kg) Frequency Midpoint 3.6 ⩽ m < 3.8 8 3.7 3.8 ⩽ m < 4.0 12 3.9 4.0 ⩽ m < 4.2 9 4.1 4.2 ⩽ m < 4.4 11 4.3 4.4 ⩽ m < 4.6 5 4.5 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE Exercise 15.2 1 aLearner’s own scatter graph with axes labelled as in the graph in the question. Points (3, 5), (11, 10), (18, 18), (19, 20), (5, 6), (20, 18), (14, 16), (8, 9), (9, 11), (7, 6), (5, 5), (16, 15), (10, 11), (9, 7) and (16, 16) marked with crosses. b 2 4 46 b 5 A. Learner’s own explanation. For example: The scatter graph is showing a positive correlation. This happens when as one value increases, the other value also increases. In this case, as the French results increase, the Spanish results also increase. aLearner’s own scatter graph with axes labelled as in the graph in the question. Points (3, 19), (10, 11), (15, 7), (8, 11), (10, 10), (13, 9), (4, 17), (16, 5), (12, 10), (8, 14), (17, 2), (11, 9), (5, 15), (20, 4) and (7, 13) marked with crosses. b 3 (8, 28), (25, 27), (16, 25), (14, 18), (9, 17) and (28, 25) plotted. B. Learner’s own explanation. For example: The scatter graph is showing a negative correlation. This happens when as one value increases, the other value decreases. In this case, as the art results increase, the science results decrease. aLearner’s own scatter graph. Horizontal axis labelled ‘Hours reading’. Vertical axis labelled ‘Spelling test score’. Both axes shown from 0 to 25. Points (4, 6), (13, 12), (20, 20), (9, 8), (18, 17), (1, 2), (11, 13), (8, 10), (18, 19), (2, 3), (15, 16), (10, 12), (4, 5), (14, 12) and (7, 7) plotted. b Positive correlation. The more hours reading a student does, the better their spelling test score. c Learner’s own line of best fit. Strong positive correlation. d Learner’s own estimate from their line of best fit. e No. It is not a good idea to use the line of best fit to make predictions outside the range of the data, because you do not know what happens beyond the data you are given. aLearner’s own scatter graph. Horizontal axis labelled ‘Number of packets of cookies sold’ and shown from 0 to 30. Vertical axis labelled ‘Number of packets of oranges sold’ and shown from 0 to 30. Points (15, 12), (12, 22), (26, 14), (22, 7), No correlation. The number of packets of cookies sold has no relationship to the number of packets of oranges sold. aLearner’s own answers. For example: Negative correlation because learners are often good at maths and science or languages and drama, but are not good at all these subjects. b Learner’s own scatter graph. Horizontal axis labelled ‘Maths result’ and shown from 0 to 100. Vertical axis labelled ‘Drama result’ and shown from 0 to 100. Points (72, 27), (34, 62), (81, 19), (57, 41), (32, 66), (78, 25), (65, 37), (67, 32), (53, 59), (61, 48), (35, 63), (42, 59), (55, 40), (79, 35) and (31, 77) plotted. c Strong negative correlation. The better the students’ result in maths, the worse their drama result. d Learner’s own answer. e Learner’s own line of best fit. f Learner’s own estimate from their line of best fit. g No, because 10% for drama lies outside the range of the data we are given, so we cannot predict what will happen. 6 Learner’s own explanation. For example: It is a coincidence that the graph shows a negative correlation. While it might be true that if you have no hair or short hair you need a hat to keep your head warm or protect it from the sun, it does not mean that you are going to buy lots of hats. In this study, the people with longer hair might have big families, and so they bought lots of hats for their family members. The number of hats you need does not depend on the length of your hair. It depends on whether you like to wear hats or not. 7 aPositive correlation. The better the score in algebra, the better the score in geometry. b 4 c e Learner’s own diagram. Scatter graph from the question with the point (10, 13) plotted. f Learner’s own line of best fit. Make sure the line of best fit passes through the point (10, 13). 10 d 13 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE g iLearner’s own estimate using their line of best fit. For example: 8. 4 a June iiLearner’s own estimate using their line of best fit. For example: 11. 8 aNegative correlation. The further the house is from the railway station, the lower its value is. b c a 9 7 3 6 7 8 6 6 6 5 3 3 2 1 0 4 0 3 7 5 0 2 4 5 6 8 8 6 2 8 0 Key: For June, 0 2 means 20 customers For August, 3 6 means 36 customers b i Mode ii Median iii Range iv Mean June 46 43 48 44 August 58 51 26 49 Key: 0 5 means 5 c Learner’s own answers. For example: In August the mode, median and mean are all greater than in June, showing that on average there are more customers. The range, however, is smaller in August than in June, showing that there is more variation in the numbers of customers riding in June. d Learner’s own answers. For example: Yes, because the mode, median and mean are all greater in August than in June. a i–iv 5 7 9 9 5 4 6 0 8 7 9 6 2 5 1 2 7 3 0 4 Key: 0 5 means 5 0 1 2 2 2 y = 6.4 km 0 1 2 b 0 The house that doesn’t fit the trend is worth $146 000 and is 6 km from the railway station. Learner’s own explanation. For example: The house might not be in a very good state of repair, which is why it isn’t worth as much as it should be. Exercise 15.3 1 5 7 9 9 0 4 5 6 7 8 9 0 1 2 2 3 4 5 6 7 Unordered: Key: 0 6 means 6 0 6 8 4 6 1 8 2 1 4 7 3 9 0 2 2 1 8 1 0 8 1 2 Ordered: Key: 0 4 means 4 0 4 6 6 8 1 0 1 2 2 3 4 7 8 9 2 0 1 1 1 2 8 8 5 i Mode ii Median iii Range iv Mean Girls’ times 27.3 26.05 2.6 26.1 Boys’ times 26.5 27.4 3.6 27.3 b Learner’s own answer. For example: The range is larger for the boys, showing that their times are more varied. The girls have a lower median and mean which shows that using these averages they were faster at solving the puzzle. c iThe mode, as the boys’ mode is lower than the girls’, which makes them appear faster. 3 Spanish test results for class 9R 9 9 7 5 9 8 7 6 5 4 0 7 6 5 4 3 2 2 1 0 Key for class 9R: 5 0 means 5 47 August Spanish test results for class 9T 0 1 2 4 6 6 8 0 1 2 2 3 4 7 8 9 0 1 1 1 2 8 8 Key for class 9T: 0 4 means 4 iiThe median or the mean, as the girls’ median and mean are lower than the boys’, so the girls were faster. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE d 6 Location A e 7 On average, using the median and mean, class 9R were taller than class 9T. Class 9T had more variation in heights, and their modal height was taller than class 9R. Learner’s own answer. For example: The girls, as their median and mean are lower, therefore they were faster than the boys. Location B a 1 4 2 3 b 33 % 1 3 0% c Range = 305 g (most variation) d Mean = 792.5 g, Median = 790 g Exercise 15.4 1 Range = 295 g i b c 8 minutes 1 3 Median = 652.5 g 2 Key: For the top shelf, 4 10 means 104 boxes of cereal 112 123 26 120.5 Middle shelf 139 137 32 134.5 Learner’s own answers. For example: The sales of cereal were better on the middle shelf as on average more boxes were sold (the mean, median and mode were all greater on the middle shelf than the top shelf). The sales on the middle shelf were more varied, but included the largest number of boxes sold on one day. The smallest number of boxes sold on one day were on the top shelf. 8 aMissing numbers from the diagram: top row 0, second top row 5, third top row 2 and 8, bottom row 5. Missing numbers from the table: Class 9T row 144 and 145, Class 9R row 149. b 48 13 12 13 × 12 = 156 15 10 15 × 10 = 150 17 2 17 × 2 = 34 Totals: 31 417 417 3 = 13 minutes Learner’s own explanation. For example: The answers for the range and mean are only estimates because the data is grouped and you do not have the individual values of the data. a i 290 ⩽ h < 310 ii 290 ⩽ h < 310 b Learner’s own explanation. For example: you can only give the modal class and class where the median lies, because the data is grouped and you do not know the individual values. c i a 150 men, 140 women b Top shelf 11 × 7 = 77 d For the middle shelf, 11 5 means 115 boxes of cereal Mode Median Range Mean Midpoint × frequency 7 31 Middle shelf 5 0 7 9 0 2 6 8 9 9 2 4 5 7 12 ⩽ t < 14 11 Estimate of mean = a 10 11 12 13 14 ii Midpoint Frequency Mean = 658 g, Learner’s own answers. For example: Location A because the mean and median mass of potatoes was greater than location B. The range was very similar showing that the variation in the mass of potatoes was similar at both locations. Top shelf 9 4 9 2 2 9 8 7 6 5 4 2 0 0 12 ⩽ t < 14 a ii 80 cm 290 cm b Modal class interval Class interval Estimate where the of mean median lies 20 ⩽ a < 30 40 ⩽ a < 50 39.6 Women 50 ⩽ a < 60 40 ⩽ a < 50 42.5 Men c Learner’s own answer. For example: The mean age of the women is approximately three years more than the men. The median age lies in the same interval for both men and women. The modal class interval for the men is a lot younger than for the women. Learner’s own comments. For example: Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE d 4 a b Learner’s own answer. For example: On average the men are the younger competitors because their mean and modal group are both younger than the women. c mean = 15.6, median = 15.5 and mode = 8 Table A Number of texts Tally Frequency 5–11 llll l 6 12–18 llll ll 7 19–25 llll 5 26–32 ll 2 Table B Number of texts Tally Frequency 5–13 llll lll 8 14–22 llll llll 9 23–31 lll 3 Class interval Estimate where the of mean median lies Table A 12–18 12–18 16.05 Table B 14–22 14–22 15.75 d iThe estimate of the mean from table B is closer to the accurate mean. iiThe accurate median is within both the class intervals for the medians. However, the midpoint of 12–18 is 15 and 14–22 is 18, and so the accurate median of 15.5 is closer to the midpoint of the median class for table A than table B. iiiThe accurate mode is 8. This isn’t in either of the modal class intervals. 5 a b 49 i 40 < a ⩽ 50 ii 30 < a ⩽ 40 i 35.25 years ii 80 years Frequency 0 < a ⩽ 20 13 20 < a ⩽ 40 8 40 < a ⩽ 60 12 60 < a ⩽ 80 7 d i 0 < a ⩽ 20 ii 20 < a ⩽ 40 e i 36.5 years ii 80 years f iLearner’s own answers and explanations. For example: I think the answers in parts a and b are the more accurate answers because the groups are smaller in size so the individual values are more likely to be nearer the midpoints in the smaller groups than in the bigger groups. The range is the same for both sets of answers because the smallest and greatest possible values are the same. iiLearner’s own answers and explanations. For example: The answers in parts d and e were quicker to work out because there were fewer groups, so there were fewer calculations to do for the median and mean. c Modal class interval Age, a (years) 6 Karim is correct. Estimate of the mean = 5 × 4 + 15 × 14 + 25 × 12 + 35 × 20 + 45 × 10 = 28 emails. 60 7 Mean mass of chicks after hatching = 30 × 4 + 34 × 8 + 38 × 6 + 42 × 2 = 35.2 g. 20 Mean mass of chicks at eight weeks old = 0.8 × 3 + 1 × 6 + 1.2 × 7 + 1.4 × 4 = 1.12 kg. 20 1120 ÷ 35.2 = 31.8 > 30 So, the mean mass of the chicks at eight weeks old is more than 30 times the mean mass of the chicks after hatching. Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021