Lower Secondary Mathematics Learner's Book 9 Answers

StuDocu is not sponsored or endorsed by any college or universityLs maths 9 2ed tr learner book answersMasters in business administration (Cambridge Group of Institutions)Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 1 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 Learner’s Book answersUnit 1 Getting started1 a 144 b 9c 125 d 42 a 512 b 1283 a 157 b 1534 a 4 and 3000 and 225b All of them.5 106Exercise 1.11 a integer 3 b irrationalc irrational d integer 7e irrational2 a 1, 7512, −38 and − 2 25. are rational.b 200 is the only irrational number.3 a integer b surd c surdd integer e integer f surd4 a irrational because 2 is irrationalb rational because it is equal to 4 2 =c irrational because 43 is irrationald rational because it is equal to 8 2 3=5 a Learner’s own answer. For example: 2 and − 2 .b Learner’s own answer. For example: 2 and 2 2 −6 a i 4 ii 6iii 10 iv 6b They are all positive integers.c Learner’s own answer.d Learner’s own answer.7 a 7² = 49 and 8² = 64b 4³ = 64 and 5³ = 1258 a The square root of any integer between 16 and 25 is a possible answer.b The square root of any integer between 144 and 169 is a possible answer.9 a 14b 610 a i 1 ii 2 iii 3b ( )( ) 51 51 + × − = 4, and so onc ( ) N N + × 1 1 ( ) − = N −1d Learner’s own answer.11 a No. It is not a repeating pattern.b Learner’s own answer.Reflection:a i true ii true iii falseb No. It might be a repeating pattern or it might not.Exercise 1.21 a 3 × 105 b 3.2 × 105c 3.28 × 105 d 3.2871 × 1052 a 6.3 × 107 b 4.88 × 108c 3.04 × 106 d 5.2 × 10113 a 5400 b 1 410 000c 23 370 000 000 d 87 250 0004 Mercury 5.79 × 107 km; Mars 2.279 × 108; Uranus 2.87 × 1095 a Russia b Indonesiac The largest country is approximately 9 times larger than the smallest country.6 a 7 × 10−6 b 8.12 × 10−4c 6.691 × 10−5 d 2.05 × 10−7Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 2 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20217 a 0.0015 b 0.000 012 34c 0.000 000 079 d 0.000 900 38 a 30 b 9.11 × 10−25 kg9 a z b y10 a 65 is not between 1 and 10.b 6.5 × 105c 4.83 × 10711 a 1.5 × 10−2b 2.73 × 10−3c 5 × 10−812 a 6.1 × 106b 6.17 × 105c 1.75 × 10513 a 7.6 × 10−6b 8.02 × 10−5c 1.6 × 10−714 a i 7 × 106ii 3.4 × 107iii 4.1 × 10−4 iv 1.37 × 10−3b To multiply a number in standard form by 10, you add 1 to the index.c To multiply a number in standard form by 1000, you add 3 to the index. To divide a number in standard form by 1000, you subtract 3 from the index.Reflection: You can compare them easily. You can write the number without using a lot of zeros. You can enter them in a calculator.Exercise 1.31 a14b18c181d1216 e110 000 f 1322 3−3, 2−4 and 4−2 are equal, 5−1, 603 a 2−1 b 2−2 c 26d 2−6 e 20 f 2−34 a 102 b 103 c 100d 10−1 e 10−3 f 10−65 a 64−1 b 8−2c 4−3 d 2−66 a 3−4 or 9−2 or 81−1b The three ways in part a.7 a 36 b136 c 1 d12168 a181 b 1225 c 1 d 14009 a i 2 ii 414iii 919b i x = 5 ii x = 1010 a i 35ii 39iii 310 iv 36b i 3 ii 3−1 iii 32iv 3−2 v 3−3c Learner’s own answers.d Learner’s own answers.11 a 56 b 52 c 5−2 d 5−612 a 6−1 b 73c 11−10 d 4−413 a x = 4 b x = 6c x = −2 d x = 514 a i 22ii 43iii 51 or 5 iv 23b Learner’s own answers.c Learner’s own answers.15 a 6−3 b 9−1c 15−4 d 10−516 a 25 b 87c 5−6 d 12217 a 26 b 2−6 c 36d 3−6 e 93f 9−3Check your progress1 a rational b irrationalc rational d irrationale rational2 a rational because it is equal to 25 5 =b irrational because it is 3 7 + and 7 is a surd3 n = 34 a 8.6 × 1010 b 6.45 × 10−65 C, D, A, B6 a149b181c1128Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 3 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20217 a 53 b 50 c 5−28 a 65 b 12−5c 4−6 d 152Unit 2 Getting started1x3 + 72 a 32 × 34 = 36 b55129 = 53c 725( ) = 7103 a x2 + 2x b 12y2 − 21yw4 a 4(x + 3) b 2x(2x + 7)5 a1712 or 1 512b65 or 1156 a F = 25 b a = Fmc a = 6Exercise 2.11 a x y − = − ×= −= −2 3 2 53 107b x xy3 3 3 3 527 1542+ = + ×= +=c yxy2 10 10 3 2530552525 619− = ( ) −= −= −=×2 a 9 b 4 c 9d 8 e 8 f 30g 5 h 47 i −30j −43 a Learner’s own answers. For example:i a = 3, b = 10, c = 12, d = 2ii a = −3, b = −10, c = −12, d = −2iii a = 3, b = 4, c = −36, d = 3b Learner’s own answers.c Learner’s own answers.4 a Learner’s own answers. For example: Part a is incorrect as −32should be written as (−3)2, which is 9 and not −9; part b is incorrect as (−2)3 is −8 and not 8.b Learner’s own answer.5 a x = 1 and y = 14, x = 2 and y = 11, x = 3 and y = 6b Learner’s own answer. For example: x = −4 and y = −1, x = −5 and y = −10,x = −6 and y = −21c Learner’s own answer. For example: x = −1 and y = 14, x = −2 and y = 11,x = −3 and y = 6 or x = 4 and y = −1, x = 5 and y = −10, x = 6 and y = −216 a 4 2 4 2 2 44 2 84 624( ) ( )( )m p + = + × −= −= × −= −b p mp 3 3 3 4 3 2 464 2440− = − − × × −= − += −( )cpmp + ( ) = + −( )= − −= − −= −−535354242 6432 6496( )7 a 21 b 36 c 16d 64 e 68 f −18g 14 h −25 i −7j 82Activity 2.1Learner’s own answer.8 Learner’s own counter-examples.a For example: When x = 2, 3x2 = 3 × 22 = 3 × 4 = 12, and (3x)2 = (3 × 2)2 = 62 = 36, and 12 ≠ 36b For example: When y = 2, (−y)4 = (−2)4 = 16 and −y4 = −24 = −16, and 16 ≠ −16c For example: When x = 3 and y = 4, 2(x + y) = 2(3 + 4) = 2 × 7 = 14 and 2x + y = 2 × 3 + 4 = 10, and 14 ≠ 109 a 26b 49Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 4 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202110 5 9 7 5 2 9 1 27 2 1549 12 2 22155a b a abb− − + + = × − − − − − ++ × − × −= × − ×−( ) ( ) ( )( )+ += − − +=−21142092 1423−−− × −−− − −− − + = − − −− × − +5 9 5 2191 26 6 22 13 4 342 32 3ab b aa a( ) b ( )( )( ) ( )= − × − − += − + − += +=− +10191 8996 8 210 48 1622 1234( )Reflection: Learner’s own answers.Exercise 2.21 a n + 5 b 5n − 5cn5 + 5 d 5(n + 5)en − 55f 5 − n2 a 7x b 20 − xc 2x + 9 d x6 − 4e x2 f 100xg 5(x − 7) h xi x3j x3k (3x)2 + 7 or 9x2 + 7l (2x)3 − 100 or 8x3 − 100 3 a i 2x + 2y ii xyb i 6x + 2y ii 3xyc i 6x + 4y ii 6xyd i 4x ii x2e i 8x ii 4x2f i 2x2 +4x ii 2x34 a Perimeter = 2(x + 5) + 2(2x) = 2x + 10 + 4x = 6x + 10b Learner’s own answer.c Length of rectangle = x + 5 = 3 + 5 = 8 Width of rectangle = 2x = 2 × 3 = 6 Perimeter = 2 × length + 2 × width = 2 × 8 + 2 × 6 = 28 Area = length × width = 8 × 6 = 48d Perimeter = 6x + 10 = 6 × 3 + 10 = 28 Area = 2x2 + 10x = 2 × 32 + 10 × 3 = 18 + 30 = 48e Learner’s own answer.5 a i P = 2x + 10ii A = 3x + 6iii When x = 4, P = 18 and A = 18b i P = 2y − 4ii A = 4y − 24iii When y = 10, P = 16 and A = 16c i P = 4n + 8ii A = n2 + 4niii When n = 6, P = 32 and A = 60d i P = 2p2 + 8pii A = 4p3iii When p = 2, P = 24 and A = 326 a i 2 red + 2 yellow = 4 green; both = 8x + 4ii 3 red + 3 yellow = 6 green; both = 12x + 6iii 4 red + 4 yellow = 8 green; both = 16x + 8b n red + n yellow = 2n green (or similar explanation given in words)c i 6 red + 2 yellow = 12 blue; both = 12x + 12ii 9 red + 3 yellow = 18 blue; both = 18x + 18iii 12 red + 4 yellow = 24 blue; both = 24x + 24d 3n red + n yellow = 6n blue (or similar explanation given in words)e Learner’s own answer.7 a (3w)2 = 36, 2v(3v − 2w) = 30, 5w(w + v) = 50b 116c (3w)2 + 2v(3v – 2w) + 5w(w + v) = 9w2 + 6v2 − 4vw + 5w2 + 5vw = 14w2 + vw + 6v2d 116Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 5 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20218 a 3a2 − 7b = 61, 8b – 3a = 31, a2 + 6b = 37, 4(a + 3b) = 4b 133c 3a2 − 7b + 8b − 3a + a2 + 6b + 4(a + 3b) = 4a2 + 7b − 3a + 4a + 12b = 4a2 + a + 19bd 133e 11f Not valid because although the perimeter is positive, three of the side lengths are negative, which is not possible.9 a 2(3x2 + 4) + 2(5 − x2) or 3x2 + 4 + 3x2 + 4 + 5 − x2 + 5 − x2b 2(3x2 + 4) + 2(5 − x2) = 6x2 + 8 + 10 − 2x2 = 4x2 + 18 = 2(2x2 + 9) or 3x2 + 4 + 3x2 + 4 + 5 − x2 + 5 − x2 = 4x2 + 18 = 2(2x2 + 9)c Arun is correct. Learner’s own explanation. For example: The variable x only appears in the expression for the perimeter when it is squared. When you square 2 and −2 you get the same answer.or: 2(2(−2)2 + 9) = 2(2 × 4 + 9) = 2(8 + 9) = 34and 2(2(2)2 + 9) = 2(2 × 4 + 9) = 2(8 + 9) = 3410 a Side length = 25 5 = cm, Perimeter = 4 × 5 = 20 cmb Side length = 49 7 = cm, Perimeter = 4 × 7 = 28 cmc Perimeter = 4 × x or 4 x11 a Volume = x3b Side length = y3Exercise 2.31 a x x xx4 5 4 59× ==+ b y y yy2 4 2 46× ==+c u u uu8 6 8 62÷ ==− d w w ww5 5 14÷ ==−e g gg323 26( ) ==×f h hh512 5 1260( ) ==×g 5m3 + 3m3 = 8m3 h 8n2 − n2 = 7n22 a m14 b n12 c p7d q5 e r3f t5g x21 h y10 i z12j 5t7 k 5g2l −h93 a Sofia is correct. x2 ÷ x2 = x2−2 = x0 = 1b Learner’s own answer.c x2 ÷ x2 = 1d All the answers are 1. Learner’s own explanations. For example: When simplified, all the expressions have an index of 0, and anything to the power of 0 = 1. or Any expression divided by itself, always gives an answer of 1.4 a 6x5 b 12y9 c 30z7d 4m7 e 4n13 f 8p35 a Learner’s own answer.b Learner’s own answer.c Learner’s own answer. Sasha’s method would be easiest to use to simplify these expressions: 4x5 ÷ 6x3 = 2 53 324623xxx= , 12y7 ÷ 8y6 = 3 72 612832yyy= and 6z9 ÷ 36z4 = 636 696 45zzz= .6 a 3q4 b 3r4 c 3t6d 2u5 e 2v4f 5w7 a D 12x3 b A 25y6c C 53k d B 3138 a Arun is correct. Learner’s own explanation. For example: (3x2)3 = 33 × (x2)3 = 27 × x6 = 27x6 or (3x2)3 = 3x2 × 3x2 × 3x2 = 3 × 3 × 3 × x2 × x2 × x2 = 27 × x6 = 27x6 or (3x2)3 means everything inside the bracket must be cubed. That means the 3 must be cubed as well as the x2.b i 16x10 ii 125y12iii 16z28Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 6 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021Activity 2.3a Learner’s own spider diagram.b There are many possible expressions. For example: 3x2 × 12x10 4x8 × 9x4 36x14 ÷ x2 72x20 ÷ 2x8 (6x6)2 36(x3)4c Learner’s own answers.9 a q−3 = 13qb r−2 = 12rc t−5 = 15td v−1 = 1v10 a A and iii, B and iv, C and i, D and vii, E and vi, F and v.b Learner’s own answer. Any expression that simplifies to give 167y. For example: 53029yyReflection: Learner’s own answers.Exercise 2.41 a (x + 4)(x + 1) = x2 + 1x + 4x + 4 = x2 + 5x + 4b (x − 3)(x + 6) = x2 + 6x − 3x − 18 = x2 + 3x − 18c (x + 2)(x − 8) = x2 − 8x + 2x − 16 = x2 − 6x − 16d (x − 4)(x − 1) = x2 − x − 4x + 4 = x2 − 5x + 42 a x2 + 10x + 21 b x2 + 11x + 10c x2 + 2x − 15 d x2 + 4x − 32e x2 − 9x + 14 f x2 − 14x + 243 a Learner’s own answers and explanations.b Learner’s own answers and explanations.c Learner’s own answer.4 a y2 + 6y + 8 b z2 + 14z + 48c m2 + m − 12 d a2 − 7a − 18e p2 − 11p + 30 f n2 − 30n + 2005 a The plus at the end would change to a minus and the 9 changes to a 1. x2 + 1x− 20b The plus at the end would change to a minus and the 9 changes to a −1. x2− 1x− 20c The plus in the middle would change to a minus. x2− 9x + 20d i (x + A)(x + B) = x2+ Cx+ Dii (x + A)(x − B) = x2+ Cx− Diii (x − A)(x + B) = x2− Cx− Div (x − A)(x − B) = x2− Cx+ D6 a C w2 + 12w + 27 b A x2 + 2x − 35c B y2 − 2y − 48 d A z2 − 9z + 207 a (x + 2)2 = (x + 2)(x + 2)= x2 + 2x + 2x + 4= x2 + 4x + 4b (x − 3)2 = (x − 3)(x − 3)= x2 − 3x − 3x + 9= x2 − 6x + 98 a i y2 + 10y + 25ii z2 + 2z + 1iii m2 + 16m + 64iv a2 − 4a + 4v p2 − 8p + 16vi n2 − 18n + 81b (x + a)2 = x2 + 2ax + a29 a (x + 3)(x − 3) = x2 + 3x − 3x − 9 = x2 − 9b i x2 − 4ii x2 − 25iii x2 – 49c There is no term in x, and the number term is a square number.d x2 − 100e x2 − a2Activity 2.4a ① 33 × 29 = 957, ② 28 × 34 = 952, ③ 957 − 952 = 5Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 7 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021b ① 16 × 12 = 192, ② 11 × 17 = 187, ③ 192 − 187 = 5c The answer is always 5.d n n + 1n + 5 n + 6e ① (n + 5)(n + 1) = n2 + 6n + 5, ② n(n + 6) = n2 + 6n,③ n2 + 6n + 5 − (n2 + 6n) = n2 + 6n + 5 − n2 − 6n = 5 The answer is always 5. Learner’s own answer.Exercise 2.51 a25x b47xc8xd xe25xf4x2 a25310410310710yy yy y+ = + =b2512510251259yyyy 25y− = − =c34yd38ye119yf314y3 aa a a aa aa2 55102105 210710+ = +==+bb b b bb bb4 33124123 412712+ = +==+c57252535 3525353935c c c ccc+ = +==+1414d 5635 30 3025 1830730d d dd dd− = −==−25 18de78232124 2421 1624524e e ee ee− = −==−16ef 4 a A, D, F b B, C, Ec G; the answer is x35 a1226362656+ = + =b1 2232121+= =c56≠ 112d She cannot cancel the 3 with the 6, because the expression is 3x + y, all divided by 6, not just 3x divided by 6.xy xy x y2 636 636+= +=+e Learner’s own answer.f i correctii incorrect. Learners should show that the correct answer is 410x y −iii correctiv incorrect. Learners should show that the correct answer is 9 820x −6 a ia b +5ii 5 912a b +iii 2 915a +iv abb+124v3 4010abb+vi 8 2718abb+b Learner’s own checks.Activity 2.5Learner’s own answers.7 a6 3 2218 22202× + += = = 10b 3 × 3 + 1 = 9 + 1 = 10c 10 = 10d Learner’s own explanation. For example: He factorises the bracket to give 2 × bracket, which is then divided by 2. The × 2 and ÷ 2 cancel each other out, leaving just the bracket.e When x = 3, 6 × 3 + 1 = 18 + 1 = 19, 19 ≠ 10, so the answer is wrong. Learner’s own explanation. For example: The expression shows that 6x + 2 must all be divided by 2. Arun has only divided the 2 in the numerator by 2, and not the 6x by 2 as well.f Learner’s own answer.8 a 2x + 1 b x + 2c 2x − 3 d 2x − 5910341820152018 1520320f f f fff− = −==−Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 8 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202196 4220 2552 3 225 4 55x x − + x x − + + = + = ( ) ( ) 3x − 2 + 4x + 5 = 7x + 310 a 2(x + 3) = 2 × x + 2 × 3 = 2x + 6b Learner’s own choice and explanation.c i 2(x + 3) or 2x + 6ii 2(x + 2) or 2x + 4iii 4(x − 3) or 4x – 12iv 3(1 − 3x) or 3 − 9xReflection: Learner’s own answers.Exercise 2.61 a S = 60M b S = 900c M = S60d M = 22.52 a i F = 60 ii F = −78b m = Fa, m = 12c a = Fm, a = −1.753 a3D Shape Number of facesNumber of vertices Number of edgesCube 6 8 12Cuboid 6 8 12Triangular prism 5 6 9Triangularbased pyramid4 4 6Square-based pyramid 5 5 8b E = F + V − 2, or any equivalent versionc V = E − F + 2i V = 6 ii V = 7d c i is a pentagonal-based pyramid and c ii is a hexagonal-based pyramide F = E − V + 2, F = 0, it is not possible to have a shape with five edges and seven vertices.f Learner’s own answer.4 a Ben’s age is x + 2, Alice’s age is x − 6b T = 3x − 4 c T = 53d x = T + 43e x = 225 a v = 87 b v = 125c u = 27 d u = 46e t = 10 f a = 26 a 20% b 60%c 125%7 a 65 kg b 49.1 kg (1 d.p.)c 95.9 kg (1 d.p.) d 57.3 kg (1 d.p.)8 a i B x = y z −2ii C x = 2 35( ) y h +iii A x = 7k(y − 6)iv C x = 3ny + mv A x = w y −7b Learner’s own answer.9 a t = m − 97b t = 5(k + m)c t = pv − h d t = 95q w+10 a A = a2 + bcb A = 49.5c A = a2 + bc, A − bc = a2, a = − A bcd a = 811 a 78.5 cm b r = Aπ c 6.25 cm12 a l = 3V b 2 cm13 Sasha is correct as 30 °C = 86 °F and 86 °F > 82 °F (or 82 °F = 27.8 °C and 27.8 °C < 30 °C).14 a She is not underweight as her BMI is 20.05, which is greater than 18.5.b 3.7 kgCheck your progress1 a 39 b 161c 122 perimeter = 16x + 8, area = 5x(3x + 4) = 15x2 + 20xDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 9 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20213 a x5 b q6 c h10d 15m9 e 2u2f 3p24 a x2 + 7x + 10 b x2 + x − 12c x2 − 3x − 54 d x2 – 14x + 40e x2 − 64 f x2 − 12x + 365 a23xb215yc1220x y −d 3x − 56 a x = 31 b z = x y −25, z = 6c y = ± x z −5 , y = ±6Unit 3 Getting started1 a 8 b 32.5 c 6 d 0.85e 90 f 625 g 700 h 322 B3 a 15.4 b 6404 a $345 b $2405 63.6 cm2 (3 s.f.)Exercise 3.11 a, D and ii; b, A and v; c, E and iv; d, C and i; e, B and iii2 a 3.2 × 103 = 3.2 × 1000 = 3200b 3.2 × 102 = 3.2 × 100 = 320c 3.2 × 101 = 3.2 × 10 = 32d 3.2 × 100 = 3.2 × 1 = 3.2e 3.2 × 10−1 = 3.2 ÷ 10 = 0.32f 3.2 × 10−2 = 3.2 ÷ 100 = 0.032g 3.2 × 10−3 = 3.2 ÷ 1000 = 0.0032h 3.2 × 10−4 = 3.2 ÷ 10 000 = 0.000 323 a Yes. Learner’s own explanation.b i smaller ii the sameiii greater4 a 1300 b 7800 c 240d 85 500 e 65 f 8000g 17 h 0.8 i 0.085j 0.45 k 0.032 l 1.255 a 320 ÷ 103 = 320 ÷ 1000 = 0.32b 320 ÷ 102 = 320 ÷ 100 = 3.2c 320 ÷ 101 = 320 ÷ 10 = 32d 320 ÷ 100 = 320 ÷ 1 = 3206 a 2.7 b 0.45c 0.36 d 0.017e 0.08 f 0.0248g 9 h 0.00257 a Learner’s own answer.b i 6.8 ÷ 10−3 = 6800ii 0.07 ÷ 10−4 = 700c Learner’s own answer.d Learner’s own answer. For example: An alternative method is to realise that ÷ by 10−x and × by 10x are the same. So, in this case 2.6 ÷ 10−2 = 2.6 × 102e Learner’s own answer.8 a 3.2 ÷ 103 = 3.2 ÷ 1000 = 0.0032b 3.2 ÷ 102 = 3.2 ÷ 100 = 0.032c 3.2 ÷ 101 = 3.2 ÷ 10 = 0.32d 3.2 ÷ 100 = 3.2 ÷ 1 = 3.2e 3.2 ÷ 10−1 = 3.2 × 10 = 32f 3.2 ÷ 10−2 = 3.2 × 100 = 320g 3.2 ÷ 10−3 = 3.2 × 1000 = 3200h 3.2 ÷ 10−4 = 3.2 × 10 000 = 32 0009 a Yes. Learner’s own explanation.b i greater ii the sameiii smaller10 a 2.5 b 47 600c 70 d 8.511 Do not tell anyone the secret!12 a i 400 ii 40iii 4 iv 0.4v 0.04 vi 0.004b Smallerc Smallerd i 0.12 ii 1.2iii 12 iv 120v 1200 vi 12 000e Largerf Largerg Learner’s own answer.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 10 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202113 a= 80.8 × 10180 × 10–10.008 × 1030.08 ÷ 10–2800 ÷ 1028 ÷ 100b= 0.3232 ÷ 1023.2 ÷ 10132 × 10–2 3.2 × 10–10.32 × 100320 ÷ 103Activity 3.1Learner’s own answers.Reflection: Learner’s own answers.Exercise 3.21 a 1.6 b −5.6 c −5.4d 6 e 0.3 f −0.66g 3.6 h −0.442 a 0.08 × 0.2 8 × 2 = 16 8 × 0.2 = 1.6 0.08 × 0.2 = 0.016b 0.4 × 0.007 4 × 7 = 28 4 × 0.007 = 0.028 0.4 × 0.007 = 0.00283 C, D, I, K (0.015); A, F, H, J (0.15); B, G, L (1.5); E (15)4 a 20 b −50c −30 d 600e 40 f −400g 200 h −3005 a0 81 1000 09 1008199..××= =b6 4 10000 004 1000640041600 ..××= =6 a D b B c C d D7 a i 0.8 ii 2.4 iii 4iv 5.6 v 7.2 vi 8.8b i Larger ii Smallerc i 60 ii 30 iii 20iv 15 v 12 vi 10d i Smaller ii Largere Learner’s own answer.8 a False b Truec False d True9 He has made a mistake. The denominator is 0.12, not 1.2; he wrote the answer with only one decimal place. Answer = 50.10 a 200 b 120c 300 d 4011 a A and iv, B and v, C and vi, D and vii, E and iii, F and ib Learner’s own answer. Any question that gives an answer of 0.024. For example: 0.03 × 400 × 0.002c Learner’s own answer.12 Learner’s own answers and discussions. For example: 28 × 0.057 = 1.596, 2.8 × 0.57 = 1.596, 28 × 5.7 = 159.6, 2.8 × 5.7 = 15.96 15.96 ÷ 0.57 = 28, 159.6 ÷ 0.57 = 280, 15.96 ÷ 28 = 0.57, 15.96 ÷ 280 = 0.05713 a 123 × 57 = 7011b i 701.1 ii 701.1 iii 70.11iv 7.011 v 7.011 vi 0.070 1114 a Learner’s own answer.b Learner’s own answer.c i Estimate: 4 × 30 = 120 Accurate: 119.625ii Estimate: 10 ÷ 0.2 = 50 Accurate: 62iii Estimate: 60 40 01×. = 24 000 Accurate: 19 20015 a 0.2 ÷ 0.4 = 0.5 mb 0.45 mc Learner’s own answer.Exercise 3.31 a 200 × 1.1 = $220 220 × 1.15 = $253b 200 × 0.9 = $180 180 × 0.85 = $153c 200 × 1.2 = $240 240 × 0.95 = $228Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 11 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20212 a Learner’s choice of who they think is correct, with reason. b Sofia is correct. Learner’s explanation. For example: 10% of $800 is $80, so the value goes up to $880. 10% of $880 is $88, so the value goes down to $792. The 10% decrease is greater than the 10% increase. It is not the same value.c The coin is now worth less than $800. Learner’s explanation. For example: The 10% decrease will be $80, but the 10% increase will be less than $80 as it is 10% of a smaller amount than $800. $800 − $80 = $720, $720 + $72 = $792.d Learner’s own answer.3 a i 57.6 ii 57.6b =c i = ii =4 a–e Learner’s own answers.5 a i 195 ii 64.4b i 630 ii 108.8646 a 1.1235 b $67.417 a i 72 ii 52.8b i 285 ii 48.4128 a 0.7216 b $4618.249 a A and iii, B and iv, C and i, E and ii, F and vb D and 0.8110 a Zara is correct. 1.04 × 1.04 is the same as (1.04)2, so 5000 × 1.04 × 1.04 = 5000 × (1.04)2b 5000 × (1.04)3c 5000 × (1.04)4d 8. The power on the 1.04 is the number of years.e i 5000 × (1.04)12ii 5000 × (1.04)20iii 5000 × (1.04)nf 15 years11 a i 10 000 × 0.9ii 10 000 × 0.92iii 10 000 × 0.93b The population after 5 years.c The population after 10 years.d Five years. 10 000 × 0.94 = 6561, 10 000 × 0.95 = 5904.9e 10 000 × 0.9nActivity 3.3 Learner’s own answers.Exercise 3.41 a i 25, 26, 27, 28, 29, 30, 31, 32, 33, 34ii 25iii 34b i 85, 86, 87, 88, 89, 90, 91, 92, 93, 94ii 85iii 94c i 265, 266, 267, 268, 269, 270, 271, 272, 273, 274ii 265iii 274d i 845, 846, 847, 848, 849, 850, 851, 852, 853, 854ii 845iii 8542 a 11.5, 11.6, 11.7, 11.8, 11.9, 12.0, 12.1, 12.2, 12.3, 12.4b 11.5c 12.43 a i 54.5, 54.6, 54.7, 54.8, 54.9, 55.0, 55.1, 55.2, 55.3, 55.4ii 54.5iii 55.4b 42 × 1.3 = 54.6 = $554 a–c Learner’s own answers.5 a–c Learner’s own answers and discussions.6 a 3.5 ⩽x < 4.5b 11.5 ⩽x < 12.5c 355.5 ⩽x < 356.5d 669.5 ⩽x < 670.57 a 15 ⩽x < 25 b 335 ⩽x < 345c 4745 ⩽x < 4755 d 6295 ⩽x < 63058 a 250 ⩽x < 350 b 1850 ⩽x < 1950c 4650 ⩽x < 4750 d 7950 ⩽x < 8050Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 12 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20219 Learner’s own answers and discussions.a i 0.5 ii 5 iii 50b The lower and upper bounds of a rounded number will always be +/− half of the degree of accuracy.10 a i 1555 cm ii 1565 cmb 1555 cm ⩽x < 1565 cm11 a i 171.5 cm ii 172.5 cmb 171.5 cm ⩽x < 172.5 cm 12 A, i and e; B, i and f; C, ii and b; D, iii and a; E, ii and c; F, iii and dCheck your progress1 a 74 500 b 12c 0.046 d 59e 0.0728 f 5g 37 h 182 a −1.6 b 3.6c −0.0028 d 600e 300 f 9g 7.5 h 0.113 $265.204 a i 20 000 × 1.08ii 20 000 × (1.08)2iii 20 000 × (1.08)3b The value of the painting after 5 years.c The value of the painting after 20 years.d 6 years. 20 000 × (1.08)5 = 29 386.561 54, 20 000 × (1.08)6 = 31 737.486 46e 20 000 × (1.08)n5 a i 7150 m2ii 7250 m2b 7150 m2⩽x < 7250 m2Unit 4 Getting started1 a x = 5 b x = 9c y = 25 d y = 252 a 5 b 7 c 5, 6, 73 a 2x > 10 b 4x < 36c y + 5 ⩾ 13 d y − 5 ⩽ −11Exercise 4.11 a 8 30 148 162168xxxx= − += −== −−bc232348211 5162 16 32 4824yyyyy= +== ×== =d 6 3 22 79 1511595323y yyyyy+=−====2 a x = −11 b x = −3c y = 4 d y = 8e a = −6 f a = −1g x = 2 h z = 43 a, b x = 15c Learner’s own answers.4 Learner’s own answers and explanations. For example:a Substitute x = 26 back into the original equation and check that left hand side = right hand side.b When he expanded the bracket on the lefthand side he didn’t multiply the 8 by 2. When he brought the −3x to the left-hand side he forgot to make it +3x. When he brought +8 to the right-hand side he forgot to make it −8.c 2 16 18 35 185 20 4 25x xxxx+=−+ === =16. Check: When x = 0.4, 2(0.4 + 8) = 2 × 8.4 = 16.8 and 3(6 − 0.4) = 3 × 5.6 = 16.8d Learner’s own answer.5 a, b x = 13c Learner’s own answers.15 10 910 9 1510 661035− =− = −− = −=−−=xxxxDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 13 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20216 a4242427776cccc====c2127e +=21 7 223 23 21217= += += +− ==( ) eeeee7 a a = 27 b b = 37c c = 3 d d = 118 a, b, c and e Learner’s own answers and explanations.d i x = 1414 ii x = 635iii x = −159 a i A + 10 ii A − 6b A + 10 = 2(A − 6)c A = 2210 a 2(x + 3) + 7x − 5 + 5(7 − x) = 48 OR 4x + 36 = 48b x = 3c 12 cm, 16 cm, 20 cm11 a 9a = 4a + 20b a = 4c Triangle sides 12 cm, rectangle sides 7 cm and 11 cm12 a B and Db A x =115; B x = 15; C x = 8640; D x = 15; E x =115 There are 15 sectors in the pie chart.13 a85 5y= b15228y +=c85 8555 17y= → y = = and 152215288 2 19 2 17yy y y+= → = + → = + → =d Learner’s own answer.Activity 4.1 i, ii and iii Learner’s answers and discussions.a 10x − 8 = 5x + 12, x = 4b 12(x − 5) = 4(x + 1), x = 8c 5x − 4 = 2x + 20, x = 8d 5757=x +, x = 8e 91262=x, x = 714 a 54 2704=x −b x = 9c 54 °, 54 °, 72 °15 a Learner’s own problem. For example:i A quadrilateral has sides of length x cm, 2(x + 1) cm, 3(x + 2) cm, and 4(x + 3) cm. The perimeter is 80 cm. Work out the value of x.ii The two shorter sides of a rectangle have side lengths of 6(3a − 4) and 3(4a − 3). Work out the value of a.iii There are x sweets in bag A. There are five fewer sweets in bag B than bag A. The sweets in bag B are shared between 180 people. Each person gets 15 sweets. How many sweets are in bag A?b i x = 6ii a = 2.5iii x = 17Exercise 4.21 1 Work out x. 5 3 2 155 2 15 33 186183x xx xxx− = +− = +== =2 Work out y. y x = −= × −= −= 5 35 6 330 3273 Check values are correct. y x = += × += += 2 152 6 1512 15274 Write the answers: x = 6 and y = 272 x = 5, y = 9b12 15d=12 151215121545=== =dddDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 14 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20213 x = 4, y = 134 x = 7, y = −55 ay = 3x + 1 x 0 3 6y 1 10 19y = x + 9 x 0 3 6y 9 12 15b024681012141618202 3 4 5 6y = 3x + 1 y = x + 9 0 1xyc (4, 13)d The coordinates give the solution of the equations; x = 4 and y = 13e Learner’s own answer. For example: The solution of simultaneous equations is the point of intersection of the straight-line graphs.6 a i x = 2, y = 6ii x = 2, y = 6b x = 2, y = 6c Learner’s own answers and explanations.7 a i x = 2, y = 7ii x = 6, y = 2b Learner’s own answers.8 a i x = 9, y = 4ii x = 10, y = 8b i x = 2, y = 3ii x = 4, y = 89 a x = 5, y = 2 b x = 16, y = 3c x = 7, y = 4 d x = 3, y = 610 Sofia is correct, x = −3 and y = 6. Zara got the signs round the wrong way.11 a1 Add the two equations.2 Substitute x = 18 into first equation2x + y = 50 2 × 18 + y = 50+ x − y = 4 y = 50 − 36 3x + 0y = 54 = 143x = 54, x = =54318 3 Check in second equation18 − 14 = 44 x = 18 and y = 14b1 Subtract the two equations.2 Substitute y = 9 into first equation x + 4y = 41 x + 4 × 9 = 41− x + 2y = 23 x = 41 − 360x + 2y = 18 = 52y = 18, y = =18293 Check in second equation5 + 2 × 9 = 234 x = 5 and y = 9c1 Subtract the two equations.2 Substitute y = 4 into first equation3x + 2y = 38 3x + 2 × 4 = 38− 3x − y = 26 3x = 38 − 80x + 3y = 12 3x = 30, x = =303103y = 12, y = =12343 Check in second equation3 × 10 − 4 = 264 x = 10 and y = 412 a Learner’s own answer.i You can add or subtract. If you add, you eliminate the ys, if you subtract you eliminate the xs.ii Subtract to eliminate the xs.iii Add to eliminate the ys.iv Subtract to eliminate the ys.b Learner’s own answer.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 15 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021c Learner’s own answer. For example: Subtract to eliminate one of the letters when the coefficients of that letter are the same number and both positive or both negative. Add to eliminate one of the letters when the coefficients of that letter are the same number and one positive and one negative.d i x = 9, y = 6 ii x = −3, y = 2iii x = 8, y = 3 iv x = 9, y = 5Activity 4.2 All answers should be x = 6, y = 1813 a x = 9, y = 4 b x = 5, y = −2c x = 2, y = 4 d x = 7, y = 114 a x = 2, y = 2b 3 × 2 + 2 = 6 + 2 = 8 and 4 × 2 + 2 × 2 = 8 + 4 = 12Reflection: Learner’s own answers.Exercise 4.31 a x⩽ 2 b x > −2c x⩾ 10 d x < −20e −2 ⩽x < 2 f −10 < x⩽ 152 a0 1 2 3 4b–5 –4 –3 –2 –1 0c–2 –1 0 1d–20 –15 –10 –5 0ef–4 –3 –2 –1 0 1 2 3 4 5 63 a 7 b −4c −2, −1, 0 or 14 a x > 2 b x⩽ 4c x < −3 d x⩾ −35 a0 1 2 3 4b–1 0 1 2 3 4 5c–5 –4 –3 –2 –1 0d–4 –3 –2 –1 06 a x < 3b, c Learner’s own answers.7 a He has multiplied out the bracket incorrectly. 3(x + 2) ⩽ 2x − 5 3x + 6 ⩽ 2x − 5 3x − 2x ⩽ −5 − 6 x ⩽ −11b i x = −12 3(−12 + 2) ⩽ 2 × −12 − 5 −30 ⩽ −29 Trueii x = −11 3(−11 + 2) ⩽ 2 × −11 − 5 −27 ⩽ −27 Trueiii x = −10 3(−10 + 2) ⩽ 2 × −10 − 5 −24 ⩽ −25 False For x⩽ −11 the substitutions give values that are true and when x > −11 it gives a false value.8 a 4 2 3 5 188 12 5 188 5 18 124 61 5y y yy y yy y yyy+ − < −+ − < −− + < −<<( ).b i y = 1 4(2 × 1 + 3) − 5 × 1 < 18 − 1 15 < 17 True–1 0 1 2 3 4 5 6 7 8 9Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 16 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021ii y = 1.5 4(2 × 1.5 + 3) − 5 × 1.5 < 18 − 1.5 16.5 < 16.5 Falseiii y = 2 4(2 × 2 + 3) − 5 × 2 < 18 − 2 18 < 16 False9 a a < 3.5 b b⩾ 11c c⩽ 6 d d > −27 Learner’s checks for each solution.10 a 5n + 5 ⩽ 30 b n⩽ 5c 5, 12 and 1311 a Learner’s own answer. For example: To make the x positive, Sergey adds x to both sides and subtracts six from both sides. He then rewrites the final inequality with the x on the left and so he has to change the < to >. To make the x positive, Natalia divides both sides by −1, but this has the effect of changing the < to >.b Learner’s own answers.c Learner’s own answer. For example: 2(x − 8) ⩾ 4x − 26 2x − 16 ⩾ 4x − 26 2x − 4x ⩾ −26 + 16 − 2x ⩾ −10 10 ⩾ 2x 5 ⩾ x x ⩽ 512 a x > −4 or −4 < xb x⩾ 5 or 5 ⩽xc x > 6 or 6 < xd x⩽ −13 or −13 ⩾xe x < 4 or 4 > xf x⩾ −2 or −2 ⩽x13 a 3x − 7 < 4x − 11 b For example: 3 7 4 117 11 4 344x xx xxx− < −− + < −<>c When x = 5, 3 × 5 − 7 < 4 × 5 − 11 8 < 9 True When x = 4, 3 × 4 − 7 < 4 × 4 − 11 5 < 5 False14 a 2 < x⩽ 5 0 1 2 3 4 5 6b 5 ⩽y⩽ 200 5 10 15 20 25 30c 3 < n < 92 3 4 5 6 7 8 9 10 11d −3 < m < 6–4 –3 –2 –1 0 1 2 3 4 5 6 7 8Check your progress1 a x = −4 b a = −2.5 c x = 2.4d y = 9 e m = 16 f n = 10Learner’s own checks for each solution.2 x = 5, y = 193 x = 19, y = 74 a a < 2 b b⩾ 5c c > −1 d d⩾ −5 Learner’s own checks for each solution.5 a −1 < x⩽ 2–2 –1 0 1 2 3 4b −4 < n < 1–5 –4 –2 –3 –1 0 1 2Unit 5 Getting started1 140 °2 62 ° 3 a a and d OR b and e OR c and fb c and dc a and c OR d and fDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 17 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20214 The angle next to a = c (alternate angles); the third angle at the same point is b(corresponding angles); the 3 angles on a line have a sum of 180 °.5 a Learner’s own diagram.b Each angle should be 37.5 °.c Learner’s own check.Exercise 5.11 60 °, 25 °, 95 °2 a x = 36, y = 50 b 122 °c A + B + C + D = 116 ° + 72 ° + 122 ° + 50 ° = 360 °3 a = 40 °, b = 30 °, c = 70 °, d = 120 °4 755 a Trapezium. One pair of parallel sides.b A = 60 °, B = 120 °, C = 135 °, D = 45 °6 C = 40 °, B = D = 100 °, A = 120 °7 a 54 ° (angle of isosceles triangle AOB)b 36 ° (angle BOC is 108 ° and triangle OBCis isosceles)c 90 ° = 54 ° + 36 °8 x = 65 ° (angles on a straight line); y = 45 ° = 115 ° (corresponding angles) − 70 ° (alternate angles)9 105 °Reflection: Learner’s own answer10 a 45 ° + 51 ° = 96 °b A + B + C + D = 96 ° + 65 ° + 127 ° + 72 ° = 360 °Exercise 5.21 110 °2 40 °3 136 °4 a 103 ° b 128 °5 a 88 ° b 128 °6 a, b Learner’s own diagram of a hexagon split into four triangles.c 4 × 180 ° = 720 ° d 120 °7 a 109 ° b 1008 a Six triangles; 6 × 180 ° = 1080 °b Eight triangles; 8 × 180 ° = 1440 °9 a b The sum of the angles = (n − 2) × 180 °c 7 × 180 ° = 1260 °; correct because there are seven triangles.10 a 100 ° b 135 °11 144 °12 a, b There are two ways: The second way could be drawn in a reflected form.c There is no other way. Either the two squares are adjacent or they have one triangle between them on one side and two triangles between them on the other side. This way will look different if it is reflected, but it is still the same arrangement.13 a Learner’s own diagram of a regular arrangement of triangles.b Learner’s own diagram of a regular arrangement of hexagons.c Because 108 ° is not a factor of 360 °.d Learner’s tessellations based on the two drawings in Question 12.e Learner’s own diagram: two octagons (135 ° angle) and one square (90 °) at every point.f Learner’s own answer.Polygon Number of sidesSum of interior anglestriangle 3 180 °quadrilateral 4 360 °pentagon 5 540 °hexagon 6 720 °octagon 8 1080 °decagon 10 1440 °Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 18 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021Reflection: In this case, subtract the 360 ° at the centre. 5 × 180 − 360 = 540 gives the same answer.Exercise 5.31 a–c Learner’s own diagram and explanation. The explanation is the same as for a pentagon. Walking round the hexagon you turn through each angle in turn and the total is 360 °.2 a = 99 °; b = 112 °; c = 125 °3 a Yes, vertically opposite angles.b Yes. They are not all on the same side, but the vertically opposite angles will be the same as you walk round the quadrilateral.4 a 120 ° b 90 ° c 72 °5 a 360 ° b 360 ÷ 8 = 45 °6 aRegular polygon Sides Exterior angleEquilateral triangle 3 120 °Square 4 90 °Regular pentagon 5 72 °Regular hexagon 6 60 °Regular octagon 8 45 °Regular decagon 10 36 °b The exterior angle = 360 ÷ n degreesc i 30 ° ii 18 °7 a 9 b 140 °8 a i 150 ° ii 160 ° iii 170 °b i 12 ii 18 iii 369 15 sides10 a 8 b 12 c 20 d 2411 a 360 − 2 × 135 = 90b Learner’s own diagram.12 (360 − 60) ÷ 2 = 150 ° is the interior angle. The exterior angle is 180 − 150 = 30 °. The number of sides is 360 ÷ 30 = 12.13 Interior angle 168 ° means exterior angle 12 ° and 360 ÷ 12 = 30 so it has 30 sides. Interior angle 170 ° means exterior angle 10 ° and 360 ÷ 10 = 36 so it has 36 sides. But interior angle 169 ° means exterior angle 11 ° and 11 is not a factor of 360 so that is not possible.Reflection: Yes they do. Check with some values for n. It is easier to see if you write (n − 2) × 180 ÷ nas (180n − 360) ÷ nExercise 5.4The answers to all the questions in this exercise are diagrams. Each question asks the learner to check their accuracy either by measuring themselves or by asking a partner to measure.Question 12 asks learners to think about whether there are different ways to complete the construction. They should be able to decide which method is easier or more likely to give an accurate drawing.Exercise 5.51 a 10 cm b 13 cm c 17 cm2 a 4.3 cm b 12.1 cm c 14.2 cm3 a 12 cm b 4.8 m c 75 mm4 a 6.6 cm b 5.0 cm c 13.5 m5 a 2 b 3 c 4 2 =d Learner’s own diagram. A continuation of the spiral pattern.e The 3rd hypotenuse is 2, the 8th hypotenuse is 3 and the 15th hypotenuse is 4.6 a 39 70 2 2 + = 80 cm to the nearest cm.b 105 58 2 2 + = 120 cm to the nearest cm.7 3 50 0 91 2 2. . − = 3.38 m to the nearest cm.8 a Learner’s drawing.b 5.12 + 6.82 = 8.52, so it is a right-angled triangle.c 5.12 + 6.82 = 72.25 = 8.52. The triangle satisfies Pythagoras’ theorem, and so is right-angled.9 Either 15 20 25 2 2 + = cm or 20 15 13 2 2 2− = . cm to 1 d.p.10 a 90 + 40 = 130 mb 130 90 40 31 5 2 2 − + ( ) = . m to 1 d.p.11 a Square perimeter = 4 × 25 = 100 mm, rectangle perimeter = (2 × 20) + (2 × 30) = 40 + 60 = 100 mmb Diagonal of square = 35.4 mm; diagonal of rectangle = 36.1 mmDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 19 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021c Learner’s diagram and value.d The values so far support Sofia’s conjecture and any further values should too. The square has the minimum diagonal for a given perimeter. All the examples here are for a perimeter of 100 mm, but it is true for any given perimeter.12 There are two possible answers. Either the two shorter sides are 1 and 4 OR the hypotenuse is 9 and one of the other sides is 8.13 a 7.52 + 5.52 = 86.5 and so length of diagonal = 86 5. .b x2 + 5.52 = x2 + 30.25 and so length of diagonal = x2+ 30 25 . .c d = + x y 2 214 a i 7 7 98 2 2 + =ii 98 49 2 7 2 7 22= × = × =b x x x x 2 2 2+ = 2 2 =Check your progress1 a = 65. The reason could use corresponding angles and the exterior angle of a triangle.2 116 ° (x = 106)3 10 sides4 a Learner’s own diagram.b Each side should be 8.5 cm.5 35 m or 35.3 m or 35.36 m are possible answers.6 x = 10 and y = 24Unit 6 Getting startedIn many questions these are suggested answers and there are many other possibilities. It is not possible to give a complete list of answers.1 Learner’s own answers.a For example: length or width.b For example: number of doors or passenger seats.c For example: colour or manufacturer.2 Learner’s own answer. For example: Using random numbers of position on the register. It could include a specific number from each year group.3 A number is assigned to each person. 50 numbers between 1 and 632 are generated. Any number that is a repeat is ignored.Exercise 6.1These are suggested answers but there are many other possibilities. It is not possible to give a complete list of answers.1 Learner’s own answers.a For example: Can boys estimate more accurately than girls? Can learners estimate acute angles more accurately than obtuse angles? Can learners accurately estimate how long one minute is?b For example: Girls can estimate the length of a short line more accurately than boys. Older learners can estimate an obtuse angle more accurately than younger learners. Learners tend to underestimate one minute of time.c Learner’s own answers. This will depend on the predictions. For example: Methods could take names from a hat or use random numbers. The method could take learners from different groups in the school.d Learner’s own answer and explanation.e Learner’s own answer.f Learner’s own generalisation, depending on their data.2 Learner’s own answers.a For example: Are lessons too long? Are there too many lessons in a day? Should school start earlier in the day?b For example: Learners want longer lessons. Learners want fewer lessons in a day. Learners would prefer to start school one hour later.c Learner’s own answers. This will depend on the predictions. For example: The method could take learners from different groups in the school.d Learner’s own answer and explanation.e Learner’s own answer.f Learner’s own generalisation, depending on their data.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 20 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20213 Learner’s own answers. a, b For example: Questions and predictions could be about lengths of words, lengths of sentences, lengths of articles or vocabulary used. c, d, e Learner’s own answers. This will depend on the predictions.f Learner’s own generalisation, depending on their data.Reflection: Learner’s own suggestions about making predictions and choosing a sample to test them.Exercise 6.21 17 girls and 13 boys2 a To encourage people to buy Supremo Shampoo.b For example: Sample choice, asking a question suggesting a particular answer, people giving an answer they think the questioner wants.3 a For example: It is cheap. It is quick. It gives a large sample.b For example: Many people do not use social media. Many people will not reply. People who reply might only do so because they have a strong opinion.4 a 8 b 26c Learner’s own explanation. For example: The vertical axis starts at 30 and not at zero.d Learner’s own diagram. The vertical axis should start at 0, and they should use a uniform scale.5 a 30%b The people who reply might all have a similar opinion and not be representative.6 a The questioner is suggesting the answer they want, i.e. ‘yes’.b For example: Do not let the person know which drink is the new recipe. Ask ‘Which drink do you prefer?’. Arrange for half the people to have the original drink first and for half of the people to have the original drink second.7 a i If you ask people to agree with you, they might do so just to avoid conflict.ii What do you think is the cause of global warming?b i People are likely to say ‘yes’.ii What is a fair price for entry to this exhibition?c i People will not want to admit they are overweight.ii The question is too personal. A better question would be, for example, ‘Do you weigh less than …’ and give a particular value.d i People might not know what ‘enough exercise’ is. They might say they do enough exercise when they do not.ii How many times a week do you take exercise, such as walking for 30 minutes, cycling or going to a gym?8 People are more likely to reply if they have a complaint.9 A good survey would choose men and women of different ages in the correct proportions questioned at different times of the day. These are the numbers required:Men WomenUnder 30 15 1530 or more 45 45Ask the first question about age. When the required number has been reached, do not ask any more people in that particular category.10 a No. Learner’s own explanation. For example: The sample is too small to make a valid conclusion.b Learner’s own explanation. For example: The scale does not start at zero, which makes the proportional differences between men and women look greater than they really are.c Learner’s own diagram. The vertical axis should start at 0, and they should use a uniform scale.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 21 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021Check your progress1 a Which cake do you think tastes best? Which cake looks most attractive? Do you dislike any of the cakes?b People will prefer type A. Type A looks most attractive. Most people dislike Type A.2 Learner’s own answer. For example: Including random numbers or using registers and a particular number from each year.3 a It will be biased towards people travelling to work.b Choose people on trains on different days and at different times of day.Unit 7 Getting started1 a 37.70 cm b 21.99 m2 4.8 cm or 48 mm3 a 34 cm2 b 44 m24 Group 1: A, D, G, H; Group 2: B, F; Group 3: C, E5 a 320 000 b 560 000 000c 6.82 d 4.5Exercise 7.11 a radius = 2 cm A r == ×= ×=π2223 14 23 14 412 6... cm (1 d.p.)b radius = 9 cm A r == ×= ×=π2223 14 93 14 81254 3... cm (1 d.p.)c radius = 4.2 m A r == ×= ×=π2223 14 4 23 14 17 6455 4. .. .. m (1 d.p.)2 a diameter = 16 cm r dA r= ÷= ÷=== ×= ×=216 283 142 83 142 64201 09222 cm cm (2 d.p.)π...b diameter = 9 cm r dA r= ÷= ÷=== ×= ×=29 24 53 142 4 53 142 20 2563 63222.. .. .. cm cm (2πd.p.)c diameter = 2.6 m r dA r= ÷= ÷=== ×= ×=22 6 21 33 142 1 33 142 1 695 31222... .. .. m m (2 d.πp.)3 a 153.938 cm2b i 153.86 cm2 ii 153.958 cm2iii 154 cm2c i 0.05% ii 0.01%iii 0.04%d π = 3.142e Learner’s own answers and explanations. For example: It is best to use the π button for the most accurate answer, but if you have to use an approximation, then π = 3.142 is the best to choose as it gives an approximate answer closest to the accurate answer.4 a 113 cm2 b 56.7 m2c 415 cm2 d 18.1 m25 a Learner’s own answers and explanations. For example: Ellie has made the mistake of multiplying the radius by pi and then squaring, rather than squaring the radius and then multiplying by pi. Hans has made the mistake of multiplying the radius by 2, rather than squaring the radius.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 22 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021b 3 14 17 314 2 899 07469 07 322. . . ...× = ×=Area = m s.f. ( )6 Learner’s own answers.Ad=π 22 or Ad=π247 a i A = 98.5 cm2ii C = 35.2 cmb i A = 804.2 mm2ii C = 100.5 mm8 a Estimate: Accurate: A = × × =126 2 60 38 2 2 π . . cmb Estimate: Accurate: A = × × =1214 85 346 40 2 2 π . . mc r = 7.35 cm; Estimate: Accurate: A = × × =127 35 84 862 2 π . . cmd r = 9.64 m; Estimate: Accurate: A = × × =129 64 145 97 2 2 π . . m9 a i A = 245.4 m2 ii P = 64.3 mb i A = 831.0 mm2 ii P = 118.3 mmActivity 7.1 Learner’s own answers.10 Marcus is correct. Area of semicircle = 10.618 cm2, Area of quarter-circle = 9.0792 cm2 and 10.618 > 9.0792.11 a Learner’s own answers and explanations.b Learner’s own answers and explanations.c i 3.3 cm ii 2.4 m iii 9.0 mm12 a, b A and v, B and i, C and vi, D and iii, E and iv, F and ii13 16.44 m14 84 m215 a Learner’s own answers and explanations.b i 25π mm ii 144π mm2iii 45π cm iv 400π cm2c iiiReflection: Learner’s own answers.Exercise 7.21 a Area A = l × w = 5 × 4 = 20 Area B = l × w = 11 × 2 = 22 Total area = 20 + 22 = 42 cm2b Area A = × × = × × =1212b h 12 6 36 Area B = l × w = 12 × 3 = 36 Total area = 36 + 36 = 72 cm2c Area A = l × w = 5 × 12 = 60 Area B = = × × =12122 2 π π r 6 56 5. 5 Total area = 60 + 56.55 = 116.55 cm2d Area rectangle = l × w = 4 × 1.5 = 6 Area circle = πr2 = π ×32 = 28.27 Shaded area = 28.27 − 6 = 22.27 cm22 a i 3 cm ii 68 cm2b i 7 cm, 8 cm ii 98 cm2c i 7 cm ii 138 cm23 a i 7 × 4 + 0.5 × 7 × 5 = 45.5 cm2ii 48.1 cm2b i 3 × 3 + 0.5 × 3 × 1.52 = 12.375 m2ii 10 m2c i 0.5 × 4 × 10 + 0.5 × 3 × 52 = 57.5 cm2ii 50.5 cm2d i 0.5 × 3 × 302 + 0.5 × 3 × 152 = 1687.5 mm2. The following could be accepted as an alternative: 0.5 × 3 × 302 + 0.5 × 3 × 152 = 1687.5ii 1539.4 mm24 a Learner’s own answer.b Learner’s own answers and explanations.c Learner’s own discussions.5 a 34 cm2 b 34.365 cm2 c 187.56 mm2A ≈ × × = × × =12123 6 3 36 542 2 cm ;A ≈ × × = × × =12123 15 3 225 337.5 m ; 2 2A ≈ × × = × × =12123 7 3 49 73 5 2 2 . ; cmA ≈ × × = × × =12123 10 3 100 150 2 2 m ;A r = = × × = × ×=1212122 2212 14472π π ππ mP = + d d = × × + = +1212π π 24 24 12 24 π mDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 23 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20216 Sofia is correct, the two shaded areas are approximately the same size. Area of 1st shape = 86.31 cm2, Area of 2nd shape = 87.96 cm2Activity 7.2 Learner’s own answers.7 a i 18(π − 2) cm2 ii 50(π − 2) cm2iii 72(π − 2) cm2 iv 4.5(π − 2) cm2b Learner’s own answer. For example: The answer is always a number times the bracket π − 2. The number outside the bracket is always half of the square of the radius.c122r ( ) π − 2d Learner’s own discussions.8 Learner’s own answers and explanations. For example: The shaded areas are the same as they are both ‘Area of square of side length 10 cm − Area of circle of radius 5 cm’. The areas of both are 21.46 cm2.9 a When radius = 4, Area of circle = π × 42 = 16π. When radius = 4, side length of square = 4 × 2 = 8 cm. Area of square = 8 × 8 = 64. Shaded area = 64 − 16π = 16(4 − π) cm2.b i 25(4 − π) cm2 ii 9(4 − π) cm2iii 36(4 − π) cm2 iv 100(4 − π) cm2c Learner’s own answers. For example: The answer is always a number times the bracket 4 − π. The number outside the bracket is always the radius squared.d r2(4 − π)Exercise 7.31 a A milligram is a very small measure of mass. It is represented by the letters mg. 1 milligram = 0.001 grams which is the same as 1 mg = 1 × 10−3 g. You can also say that there are one thousand milligrams in a gram or 1 milligram is one thousandth of a gram.b A nanometre is a very small measure of length. It is represented by the letters nm. 1 nanometre = 0.000 000 001 metres which is the same as 1 nm = 1 × 10−9 m. You can also say that there are one billion nanometres in a metre or 1 nanometre is one billionth of a metre.2 a A kilolitre is a very large measure of capacity. It is represented by the letters kL. 1 kilolitre = 1000 litres which is the same as 1 kL = 1 × 103 L. You can also say that there are one thousand litres in a kilolitre or 1 litre is one thousandth of a kilolitre.b A gigametre is a very large measure of length. It is represented by the letters Gm. 1 gigametre = 1 000 000 000 metres which is the same as 1 Gm = 1 × 109 metres. You can also say that there are one billion metres in a gigametre or that 1 metre is one billionth of a gigametre.3 a 8 micrometres, 8 millimetres, 8 centimetres, 8 metres, 8 kilometres, 8 gigametresb 8 μm, 8 mm, 8 cm, 8 m, 8 km, 8 Gm4 a Learner’s own answers and explanations. For example: Marcus is correct. 1 tonne = 1000 kg. Also 1 kg = 1000 g and 1 Mg = 1 000 000 g = 1000 kg = 1 t. Arun is incorrect. 1 litre = 1000 mL and 1 litre = 100 cL, so 1000 mL = 100 cL →10 mL = 1 cL, not 100 mL = 1 cLb Learner’s own discussions.c Learner’s own answers and explanations.d Learner’s own discussions.5 a 2.5 Mm to m → 1 Mm = 1 000 000 m, so 2.5 Mm = 2.5 × 1 000 000 = 2 500 000 mb 0.75 GL to L →1 GL = 1 000 000 000 L, so 0.75 GL = 0.75 × 1 000 000 000 = 750 000 000 Lc 13.2 hg to g → 1 hg = 100 g, so 13.2 hg = 13.2 × 100 = 1320 g6 a 364 cL to L → 100 cL = 1 L, so 364 cL = 364 ÷ 100 = 3.64 Lb 12 000 mg to g → 1000 mg = 1 g, so 12 000 mg = 12 000 ÷ 1000 = 12 gc 620 000 μm to m → 1 000 000 μm = 1 m, so 620 000 μm = 620 000 ÷ 1 000 000 = 0.62 mDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 24 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20217 From Earth to: Distance in …Mars 78.34 GmJupiter 628.7 GmSaturn 1.28 TmUranus 2.724 TmNeptune 4.35 Tm8 A and v, B and iv, C and i, D and iii, E and ii9 a Learner’s own answers and explanations. For example: Sofia is correct. 300 000 000 × 60 × 60 × 24 × 365.25 = 9.467 28 × 1015, which rounds to 9.47×1015.b 299 792 458 × 60 × 60 × 24 × 365.25 = 9.460 730 473 × 1015c 9 460 000 000 000 000d 6 × 9 460 000 000 000 000 = 56 760 000 000 000 000 = 5.676 × 1016e Learners own discussions.10 a D, B, C, Ab 2 147 483 648 bytesc 10 880 photosd 1864 films11 Learner’s own answers and explanations. For example: Magnar is incorrect. The fastest is model B because 10 ns is quicker than 40 ns and 60 ns.Reflection: Learner’s own answers.Check your progress1 a 39.27 cm b 21.36 m2 a 123 cm2 b 36.3 m23 49.1 cm24 170 cm25 a 5 nanograms, 5 micrograms, 5 milligrams, 5 grams, 5 kilograms, 5 tonnesb 5 ng, 5 μg, 5 mg, 5 g, 5 kg, 5 tUnit 8 Getting started1 a58= 0.625 terminatingb56 = 0.83. recurring2 a 513b 612c 65123 a 68 b 104 a13b 17155 a12b 720c 10 d12Exercise 8.11 a14= 0 2. 5 which is a terminating decimal2414=×=× 2 2 0 2. . 5 0 = 5 which is a terminating decimal3414=×=× 3 3 0 2. . 5 0 = 75 which is a terminating decimalb15= 0 2. which is a terminating decimal2515=×=× 2 2 0 2. . = 0 4 which is a terminating decimal4515=×=× 4 4 0 2. . = 0 8 which is a terminating decimal2 a19 = 0.1.b Recurring decimal.c All recurring decimals.i29 = 0.2.ii 39 = 0.3.iii 49 = 0.4.iv 59 = 0.5.v69 = 0.6.vi 79 = 0.7.vii 89 = 0.8.3 a3913= = 0.3. and 6923= = 0.6.b Learner’s own discussions. Their answers are not different because 99 = 0.9. = 1.4 a18= 0.125b Terminating decimal.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 25 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021c They are all terminating decimals.i28= 0 2. 5 ii 38= 0.375iii 48= 0 5. iv 58= 0.625v68= 0 7. 5 vi 78= 0.875d Learner’s own answers. The following three fractions can be simplified. c i 2814= = 0 2. 5, iii 4812= = 0 5. and v6834= = 0 7. 55 a No, 3612= = 0 5. which is not a recurring decimal.b Yes.c Learner’s own explanations. For example: 6 is even, so it can be halved. So 3612= . However, 7 is odd and so it cannot be halved, so there is not an equivalent fraction such that ?712= .d Learner’s own investigations and answers. For example: If the denominator is even, then there will be a fraction such that ??=12 which will not be a recurring decimal. If the denominator is odd and the unit fraction is a recurring decimal, then it’s possible that all the fractions with the same denominator will be recurring decimals as well. However, there are exceptions such as: 115 is recurring, but31515= = 0 2. which is terminating.6 a Recurring decimals. Learner’s own explanations. For example: The denominators are multiples of 3. The numerators are all 1.b They are still recurring decimals. Learner’s own explanations. For example: The fractions that can be cancelled down still have a denominator with a multiple of 3, and once cancelled are not even.c Learner’s own explanations. For example: B is now 3612= , D is now 31214= , E is now 31515= . These are all terminating decimals.d No. Learner’s own discussions.7 a Always true: 7 is odd and a prime number, so all fractions with a denominator of 7 cannot be simplified. 17 is a recurring decimal, so all fractions with a denominator of 7 are recurring.b Sometimes true: For example: 16, the denominator is a multiple of 2, and the fraction is a recurring decimal. However, it is not always true because they can also be terminating decimals, e.g. 14, the denominator is a multiple of 2, and the fraction is a terminating decimal.c Sometimes true: For example: 120, the denominator is a multiple of 10, and the fraction is a terminating decimal. However, it is not always true because they can also be recurring decimals e.g. 130, the denominator is a multiple of 10, and the fraction is a recurring decimal.d Never true: A fraction with a denominator which is a power of 2 is a terminating decimal. 121212= = 05 02 . , 2 3 . , 5 0 = . , 12512124 5 = = 0 0625 0 03125 . , . , etc. Each decimal can be divided by 2 to get the next decimal in the sequence, so they will all be terminating.8 a Learner’s own answers and explanations. For example: Recurring decimals. All the denominators are multiples of 7 and they are all written in their simplest form (apart from E).b Learner’s own answers and explanations. For example: E is not written in its simplest form, but when it is, it is equivalent to 114 which is recurring. So it doesn’t change the answer to part a.c Learner’s own answers. For example: She must add ‘when it is written in its simplest form’ so her statement now is: Any fraction which has a denominator that is a multiple of 7, when it is written in its simplest form, is a recurring decimal.9 a206013= recurringb366035= terminatingc456034= terminatingDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 26 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021d55601112= recurringe 18601215= recurringf 3 32160720= terminating10 a recurring b terminatingc recurring d terminating11 The fractions written in their simplest form are:Abi 2116818= Bim 2816816= Caz 32168421=Dave 35168524= Enid 40168521= Fin 4216814=a, b Learner’s own decisions on how to sort the friends into two groups. For example: Abi and Fin – the fractions they work are terminating decimals. Bim, Caz, Dave and Enid – the fractions they work are recurring decimals. OR Abi, Bim and Fin – the fractions they work are unit fractions. Caz, Dave and Enid – the fractions they work are not unit fractions. OR Abi, Bim, Dave and Fin – the denominators of the fractions they work are even numbers. Caz and Enid – the denominators of the fractions they work are odd numbers. etc.Activity 8.1 Learner’s own answers.Reflection: Learner’s own answers.Exercise 8.21 a 5233512+ −  Brackets: 3512610510110− = − = Addition: 5 5 5 2311020303302330+ = + =b 1056710− × Multiplication: 567105 76 103560712× = = =×× Rewrite 10: 10 91212= Subtraction: 9 91212712512− =c 534232÷ +  Brackets: 2323232 23 3492××= × = = Division: 5 5 3443203÷ = × = Addition: 203496094964919+= += = 72 a 2516b 314c 2 d 3343 a Learner’s own answers. For example: 7 + 3 − (6 − 3) = 10 − 3 = 7b 7112c Learner’s own answers and explanations.d Learner’s own discussions.4 a i 9 − (2 + 4) = 9 − 6 = 3 ii 3340b i 8 + (2 − 1) = 8 + 1 = 9 ii 9524c i 5 + 2 × 16 = 5 + 32 = 37 ii 42 49d i 16 16 1512121434− × = − = ii 1511245 a Learner’s own answers.b Learner’s own answers. For example: It might be easier to work with the whole numbers and fractions separately and not convert into improper fractions.6 a 25 5 8 19715− +  or 25 5 8 19715− −b Learner’s own answer and explanation. For example: Her estimate is too long as the length of her third side is more than the sum of the other two sides, which is not possible in a triangle.c 111945. Learner’s own answer and explanation. For example: Yes, the third side is less than the total of the other two sides.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 27 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20217 314kg8 Division: 6 6 4554304÷ = × = Multiplication: 3 5 5 14134654×= ×= Addition: 30465495434+ = = 239 a 1782 m b 8182 cm c 13112 m10 a Learner’s own answer and explanation. For example: They get different answers. Marcus is correct. His method does multiply 112 by 112. Arun’s method multiplies 1 by 1 and 12 by 12, which does not give the same answer. This can be shown using a multiplication box.× 1 121 1121212141 1 1 2121212121414× = +++ = . Marcus’s method gets this answer. Arun’s method only gets the 1 and the 14, it doesn’t get the other two 12s.b Learner’s own discussions. General rule: change the mixed number to an improper fraction. Square the numerator, square the denominator. Change the answer back to a mixed number.11 a 334b 29 59c 183512 a 2 2 51313122 + × or 2 2 5131312×+ b 18518 m2Exercise 8.31 a343412 12 3 3 913× = × = × =b575728 28 5 4 2014× = × = × =c45445 45 4 9 36195× = × = × =d38372 72 3 9 27198× = × = × =2 a383 322721236 36 9 13298× = × = × = =b494 435231339 39 13 173139× = × = × = =c565 53203238 8 4 6346× = × = × = =d7107 726321245 45 9 312910× = × = × = =3 a Learner’s own discussions. For example: She cancelled using a common factor of 4, but she should have cancelled using the highest common factor of 8.b 16 16 2 7112411 113223132324× = × = × = =c highest common factord Learner’s own discussions.4 a 84 b 140c 212d 22 125 a1021 b516c839d23e635 f386 a18b147 Lewis is correct, he travels 18313 km which is more than 180 km.8 Estimate Accuratea 1 31235× 1 4 6 12× = 525b 2 3 1423× 2 3 712× = 814c 1 31816× 1 3 3 1212× = 3916d 3 123522× 3 1 3 1212× = 412e 3 43435× 4 4 18 12× = 1714f 4 2 47516× 4 2 10 12× = 10479 a Learner’s own working. For example: 8 4 12× = and 4 < 8, 4 3 1223× = and 3 < 412, 5931016× = and 1659<Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 28 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021b i When you multiply any number by an improper fraction, the answer will always be greater than the original number.ii When you multiply any number by a mixed number, the answer will always be greater than the original number.c Learner’s own discussions.10 a smaller, 214b bigger, 813c bigger, 11211 a A 1733, B 1112, C 913, D 13, E 1516, F 29b, c Learner’s own decisions on how to sort the cards into two groups. For example:A, D and F are proper fractions; B, C and E are improper fractions. ORB and E have an even number for the denominator; A, C, D and F have an odd number for the denominator. ORA, B, C, D and F have a denominator which is a multiple of 3; E does not have a denominator which is a multiple of 3, etc.Reflection: Learner’s own answers.Exercise 8.41 a 16 16 4 7 2847744÷ = = × 1 × =b 21 21 7 5 35 355371÷= =× × =c 14 14 7 9 63299271÷ = × = × =d 8 8 2 11 22 41111421÷ = × = × =2 A and iii, B and i, C and iv, D and v, E and ii3 a894789742 79 11495921÷ = × = = = 1××b792579527 59 235181718÷ = × = = = 1××c67314671432 21 1212÷ = × = 1 = 4××d5615245624155 246 1543131 41 3÷ = × = = = 1××4 C910, D2021, B 119, A19315Estimate Accuratea 1 11245÷ 2 ÷ 2 = 1 56b 2 1 1423÷ 2 ÷ 2 = 1 1720c 4 5 1816÷ 4 5 45÷ =99124d 2 3 2314÷ 3 ÷ 3 = 1 3239e 5 2 1234÷ 6 ÷ 3 = 2 2f 4 2 4523÷ 5 3 123÷ = 145g 1141011÷ 1 ÷ 1 = 1 138h35110÷2 1 212÷ =276 a Learner’s own working. For example: 3 612÷ = and 6 > 3, 1 2122314÷ = and 2 1 1412> , 581634÷ = 3 and 33458>b i When you divide any number by an improper fraction, the answer will always be smaller than the original number.ii When you divide any number by a mixed number, the answer will always be smaller than the original number.c Learner’s own discussions.7 a bigger, 913b smaller, 4c smaller, 22218 Learner’s own answer and explanation. For example: His conjecture is not true. If you divide a mixed number by a larger mixed number the answer will be a proper fraction, not a mixed number, e.g. 2 3 12141013÷ =9 a1415 b 267c 117d 119e1127 f 1111Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 29 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202110 a Learner’s answer and explanation. For example: π ≈ 3, and diameter = circumference ÷ π. 15 cm is slightly more than the circumference, 15 ÷ 3 = 5, so the diameter will be just under 5 cm.b 14 4 172279977229212÷ = × = =11 a 123b 112c 11312 9212 km/h13 23561221 5  + ÷ is greater212453415342736− ÷ = = and 23561279283621 5  + ÷ = =Exercise 8.51 a1212125 5 1 5 6 36 1 352 22+ − ⇒ +.  = ( ) = ⇒ 36 − =b3 0 2 23 3 3 2723 501515153 33− + ⇒ = ( ) =⇒ + =. − 27c6 3 0 7 3 4 364 322 2 310310710− + ⇒ = ⇒ =⇒ − =. + 6 362 a 48 b 49 c 123 a1.5 2.5 40 × × ⇒ × = ⇒ × =×= 3252154151 410 40 15 10 150b1.25 3 × × ⇒ × = ⇒ × = × = 12547235835856 56 35 7 24517c2.75 × ⇒ 18 18 2 8 × = × ⇒ 1 1 × = 8 4 = 93411411499212 294 a 126 b 108 c 1055 a Learner’s own answers. For example: Akeno’s method – advantage: shorter, disadvantage: involves changing decimals to improper fractions and cancelling before multiplying (which learners might not like). Dae’s method – advantage: can work on one step at a time and could easily do this method mentally, disadvantage: method is longer (which learners might not like).b Learner’s own answers and explanations.c Learner’s own answers and explanations.i For example: Dae’s method because when 14 is multiplied by 2.5 it gives a whole number.ii For example: Akeno’s method because 15 cannot be divided by 2 exactly, so it is easier to use improper fractions and to work out the answer as a mixed number.6 a0 28 5 5 25 25 7 2 2428 110028100. , × ⇒ 0.28 = = ⇒ × =b 0 7 4 4 3277102730771030711 13. × ⇒ 0.7 = = , ⇒ = ×c1 3 4 4 4 4 60 60 783 3113 6101310. × − ( ) ⇒ − 1.3 = , = ⇒ × =7 a 1 b910c 418 2 m29 a Learner’s own answers and explanations. For example: Write the decimal as a fraction, square the fraction then multiply by the mixed number which has been written as an improper fraction.b Learner’s own answers and explanations. For example: Fraction, because 23 and 49 are both recurring decimals so it is easier to write them as fractions.c Learner’s own discussions.d 0 8 7 4 2 1245. × = ; example strategy:0 8 74228531124515216251528 35 124545. × = ×= ×===××10 213mDownloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 30 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202111 a Learner’s own answers and explanations. For example: Write the decimal as a fraction, square root the fraction then complete the calculation using fractions.b Learner’s own answers and explanations. For example: Fraction, because the square roots are easier to work out if the decimals are changed to fractions.c Learner’s own discussions.d 4 25 1 57923. × = ; example strategy:4 25 1 457914169174431732311. × = ×= ×==12 a K = 2b v = = = =2 2 ×2144214221489, but v = 11389≠c vKm=2d vKm= = = = =2 2 ×1825362565151 andK = = mv × × = × =12126525236252225 18Activity 8.5 Learner’s own answers.Reflection: Learner’s own answers.Check your progress1 a recurring b terminatingc recurring d terminating2 a 514b 2712c 429303 a 1212b4154 a 48 b345 a 80 b 50 c13Unit 9 Getting started1 a i add 25ii 5 6 35, b i subtract 0.3 ii 7.3, 72 a 5, 7, 9, …b add 2c Pattern 4dPosition number 1 2 3 4term 5 7 9 112 × position number 2 4 6 82 × position number + 3 5 7 9 11 Position-to-term rule is: term = 2 × position number + 33 a 12 13 131212, , , …, 17b 0.5, 4.5, 8.5, …, 36.54 a 3n + 5 b 24 − 5n5 a i x 6 11 18 25y 6 81212 1512ii x −2 1128 1112y −15 21235 5212b i yx= +23 ii y = 5(x − 1)Exercise 9.11 a linear b linearc non-linear d non-lineare linear f non-linearg linear h non-lineari linear Learner’s own explanations. For example: The linear sequences go up/down by the same amount each time. The non-linear sequences do not go up/down by the same amount each time.2 a 3.5, 4.2, 4.9, … b 2, 5, 11, …c 4 3 31323, , , ... d 40, 18, 7, …e 1.25, 3.25, 7.25, … f 1 2 7 12, , , …Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 31 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20213 A and iii, B and i, C and iv, D and ii4 a 4, 5, 14, … b 2, 7, 52, …c 5, 9, 49, … d 0, 9, 144, …5 a 3, 3, 3, … All the terms of the sequence are the same.b Learner’s own two sequences. For example: First term is 5, term-to-term rule is square and subtract 20. first term is 16, term-to-term rule is subtract 12 and square.c Learner’s own answers. For example: It is not possible if the numbers are positive integers because if you square then add or add then square, you will have sequences where the terms are getting bigger every time. However, if you use fractions, it is possible – e.g. first term is 12, term-to-term rule is ‘square and add 14’, or first term is 19, term-to-term rule is ‘add 29 and square’. It is also possible if you add negative numbers – e.g. first term is 2, term-to-term rule is ‘square and add −2’, or first term is 9, term-to-term rule is ‘add −6 and square’.d Learner’s own discussions.6 a 2, 3 27, 4 47, 5 67, 7 17 , 8 37, 9 57b 90, 8434, 79 12, 74 14, 69, 6334, 5812c −4, −3.7, −3.4, −3.1, −2.8, −2.5, −2.2d 31, 24.8, 18.6, 12.4, 6.2, 0, −6.27 a Cb The fifth term, which is 126 382 570 (fourth term = 11 242 which is less than one million)8 a 3, 4, 6, 9, …b 6, 8, 12, 18, …c 20, 19, 16, 11, …d 100, 90, 75, 55, …Activity 9.1Learner’s own questions and discussions.9 Zara is correct. Learner’s own explanation. For example: The first term is 3 and when you cube 3 you get 27. Then: If you subtract 24, you get a second term which is also 3, so all the terms of the sequence are 3 and so you don’t get a negative number. If you subtract a number less than 24, the second term is greater than 3, so all further terms get bigger so you don’t get a negative number – e.g. if you subtract 23, the sequence will be 3, 4, 41, 68 898, … If you subtract a number greater than 24, the second term is smaller than 3, so all further terms get smaller so you do get a negative number – e.g if you subtract 25, the sequence will be 3, 2, −17, −4938, …10 a 4, 8, 216, … b −6, −8, −64, …c 2, 4, 244, …11 Tania’s method is incorrect. Learner’s own explanation. For example: She needs to reverse the term-to-term rule to find the previous terms in the sequence, not just halve the 6th term to get the 3rd term. Correct answer is: 5th term = 486 ÷ 3 = 162, 4th term = 162 ÷ 3 = 54, 3rd term = 54 ÷ 3 = 18.12 4th term = (11.5 − 6) × 2 = 11, 3rd term = (11 − 6) × 2 = 10, 2nd term = (10 − 6) × 2 = 813 3Reflection: Learner’s own answers.Exercise 9.21 a 1st term = 4 × 1 − 5 = −1 2nd term = 4 × 2 − 5 = 3 3rd term = 4 × 3 − 5 = 7 4th term = 4 × 4 − 5 = 11b 1st term = 12 + 1 = 2 2nd term = 22 + 1 = 5 3rd term = 32 + 1 = 10 4th term = 42 + 1 = 17c 1st term = 13 2nd term = 23 3rd term = 33=1 4th term = 4313=1d 1st term = 13 = 1 2nd term = 23 = 8 3rd term = 33 = 27 4th term = 43 = 64Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 32 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20212 a 7, 11, 15, …, 43 b −3, −1, 1, …, 15c 3 4 4 1212, , , …, 8 d152535, , , …, 2e 0, 3, 8, …, 993 A and iv, B and iii, C and i, D and ii4 a Learner’s own answer and reason.b Card A has the greater value. A: 8th term = 82 − 14 = 50, B: 20th term = 45× + 20 33 = 49c Learner’s own answer.5 a 7, 8, 11, 16, 23, 32, …b 7, 8, 11, 16, 23, 32, … + 1 + 3 + 5 + 7 + 9 + 2 + 2 + 2 + 2c The second differences are all the same (+2).d i 5, 7, 11, 17, 25, 35, … + 2 + 4 + 6 + 8 + 10 + 2 + 2 + 2 + 2ii 2, 7, 13, 20, 28, 37, … + 5 + 6 + 7 + 8 + 9 + 1 + 1 + 1 + 1iii 3, 5, 11, 21, 35, 53, … + 2 + 6 + 10 + 14 + 18 + 4 + 4 + 4 + 4 In each sequence the second differences are all the same.e i quadratic ii lineariii neither iv linearv neither vi quadraticf Learner’s own discussions.6 a n2 + 3 b n2 + 10c n2 − 1 d n2 − 97 Learner’s own explanation. For example: When you square a number you get a positive answer and then once you add 5 you know that all the terms in the sequence will be positive. You cannot have a first term of −1 as this is a negative number not a positive number, so it cannot be in the sequence.8 a in7ii n8iii n6b Learner’s own discussions.9 a A 121545= , B 141879= , C 91234=b C 34, B 79, A 4510 a, b, c Learner’s own answers.d i Yes, when n = 13, 132 − 76 = 93, so 93 is the 13th term.ii No, 4896 16 98 3= . …, so not a whole number. OR No, when n = 16, 163 = 4096, when n = 17, 173 = 4913 and 4896 lies between 4096 and 4913, so cannot be in the sequence.11 Marcus is correct. Learner’s own explanations. For example: when n = 1, 4 1 41212− × = , when n = 2, 4 2 3121212− × = , when n = 3, 4 3 31212− × = , etc.12 a 91214− n b 20.2 − 0.2nc − −112nd −3.5 − 1.5nExercise 9.31 a i x 0 1 2 3y 0 1 4 9ii x 0 1 2y 0 1 8b i 0 1 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 10yx1ii 0 1 2 3 4 5 6 7 8 9 100 2 3 4 5 6 7 8 9 10yx1c i y = x2ii y = x3Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 33 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20212 a i x 2 5 9 11y 7 28 84 124 ii x 1 3 5 10y −2 24 122 997b i y = x2 + 3 ii y = x3 − 33 a i x −3 1312y 18 2912ii x −5 1412y 100 141iii x −4 0 3y −8 8 125b Learner’s own discussions.c i y = 2x2ii y = (2x)2iii y = (x + 2)3d Learner’s discussions.4 a i x −8 −4 15y 9 1 400 ii x −2 1312y 12 1334 iii x −2 1412y −712336458b i y = (x + 5)2ii y = 3x2iii y = + x3 125 a×4 yx2 b x1413121y14491 46 a i x −4 −3 3 4y 16 9 9 16ii Learner’s own answer. For example: x = −4 and 4 have the same y-value. x = −3 and 3 have the same y-value.iii Learner’s own discussions. For example: Yes, when you square +xand −x, you get x2.b ix 1 or −1 2 or −2 4 or −4 10 or −10y 5 20 80 500ii Learner’s own answer. For example: There are two possibilities for x for each y-value.iii Learner’s own discussions. For example: You could say that either all the x-values are positive or that all the x-values are negative.7 a i x 2 4 5 12y 8 32 50 288 ii x 7 10 11 13y 16 49 64 100b i y = 2x2ii y = (x − 3)2Activity 9.3Learner’s own answers.8 a i y = x2ii x y = ±iii Learner’s own check. b i y = x3ii x y =3iii Learner’s own check. c i yx=22ii x y = ±2iii Learner’s own check. d i y = x2 + 3 ii x y = − ± 3iii Learner’s own check.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 34 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 e i y = (x − 4)2ii x y = + ± 4iii Learner’s own check. f i y = (2x)3ii xy=32iii Learner’s own check.9 A and iii, B and i, C and v, D and vi, E and iv, F and ii10 They are both correct. Learner’s own explanations. For example: The x-values match the y-values for both function equations and y = (2x)2 = 2x × 2x = 4x2.11×8 yx2x14123 or –3y122 72y = 8x212 Arun is incorrect. Learner’s own explanations. For example: He is correct for the function y = 2x4 because any positive or negative number to the power of four gives a positive answer. This is then multiplied by two to still give a positive answer. He is incorrect for the function y x =123because when a negative number is cubed, the answer will be negative. When this is multiplied by 12, the answer will still be negative.Reflection: Learner’s own answers.Check your progress1 a 3, 4, 11, 116, …b −3, 1, 9, 121, …c 5, 6, 9, 14, …d 40, 38, 34, 28, …2 a1232, , 1 , ..., 5 b 8, 11, 16, …, 1073 a n2 b n2 − 2 cn94 The number 178 is not a term in this sequence, because when you solve the equation n2 + 32 = 178 to find the value of n you do not get a whole number.nnnn22232 178178 32146146 12 08+ == −== = . ...5 a i x −2 4 5 9y 2 8 12 1240 12ii x −15 − 8121 4y 49 1481 144b i yx=22ii y = (x + 8)2Unit 10 Getting started1 a $155 b c = 20d + 352 a x −2 −1 0 1 2 3y −5 −3 −1 1 3 5b Learner’s own graph; A straight line through (0, −1), (0.5, 0) and (3, 5).c 2d −13 a 40 °Cb 20 °Cc At the startExercise 10.11 a $31b The number of days multiplied by 3 plus 10 for the fixed charge.2 a 3 days b t = 10n +153 a 27 kg b b = 2g − 34 a s + f = 50b s + f = 52c s + f = 60Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 35 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20215 a 6 × 5 + 6 × 10 = 90b 5f + 10t = 90c 86 a 12 × 6 + 12 × 4 = 72 + 48 = 120b 6l + 4s = 120c s = 2l7 a The total value is 80 cents.b 48 a 3x + 2y = 50b If 3x = 21 then 2y = 19 and that is impossible if y is a whole number.9 a–d Learner’s own answers.10 a 32 b 3r + 4q = 100c 10 d 32e No. Each pair would have a total of 7 edges and 7 is not a factor of 100.f q = 3r − 5Reflection: Two possible ways are x = 20 − 2yand 2y = 20 − x.Exercise 10.21 a x −1 0 1 2 3y 5 15 25 35 45b When x = 5, then y = 10 × 5 + 15 = 652 a x −10 0 10 20 30y −30 −10 10 30 50b At (0, −10)c When x = 23, then y = 2 × 23 − 10 = 36, so (23, 36) is on the graph.3 a x 0 1 3 5 6y 20 16 8 0 −4b At (0, 20) and (5, 0)4 a x 0 10 20 30 40y 12 8 4 0 −4b 2 × 15 + 5 × 6 = 605 a x −2 0 2 4 6y 10 6 10 22 42b When x = 5, then y = 52 + 6 = 316 a x 0 2 6 10 16y 8 7 5 3 0b i (16, 0) ii (0, 8)7 a x 0 1 2 4 6y 9 7.5 6 3 0b i (6, 0) ii (0, 9)c Learner’s own graph. A straight line through (6, 0) and (0, 9).8 a x 15 10 5 0y 0 1 2 3b Learner’s own graph. A straight line through (15, 0) and (0, 3).9 a x 0 2 4 6 8 10y 10 8 6 4 2 0b Learner’s own graph. A straight line through 10 on each axis.c Learner’s own graph. A straight line through 7 on each axis.d x −1 0 1 2 3 4 5y 5 4 3 2 1 0 −1e Learner’s own graph. A straight line through 4 on each axis.f A straight line through c on each axis.g Learner’s own graph. A line parallel to the others through the origin.10 a Learner’s own graph. A straight line through 12 on each axis.b x 0 1 2 3 4 5 6y 12 10 8 6 4 2 0c Learner’s own graph. A straight line through 12 on the y-axis and 6 on the x-axis.d Learner’s own graph. A straight line through 12 on the y-axis and 4 on the x-axis.e Learner’s own graph. A straight line through 12 on the y-axis and 3 on the x-axis.f A straight line through 12 on the y-axis and 12k on the x-axis.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 36 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202111 a Learner’s own graph. A straight line through (14, 0) and (0, 7).b A straight line through (n, 0) and ( , 0 )12n .12 a x −3 −2 −1 0 1 2 3y 9 4 1 0 1 4 9b Learner’s own graph. A parabola with the base at the origin. c x −3 −2 −1 0 1 2 3y 11 6 3 2 3 6 11d Learner’s own graph. A parabola with the base at (0, 2). e x −3 −2 −1 0 1 2 3y 5 0 −3 −4 −3 0 5f Learner’s own graph. A parabola with the base at (0, −4).g A curve with the y-axis as a line of symmetry and the lowest point at (0, c). (Learners are not expected to know the word parabola.)13 A and iii, B and iv, C and i, D and ii14 ax −5 −4 −3 −2 −1 0 1 2 3 4 5y 16 7 0 −5 −8 −9 −8 −5 0 7 16b Learner’s own graph. A parabola with the bottom at (0, −9).c i (−10, 91) ii (8, 55)iii (20, 391)iv (−3, 0) or (3, 0)v (6, 27) or (−6, 27)Exercise 10.31 a gradient 4 and y-intercept −6b gradient 6 and y-intercept 4c gradient −6 and y-intercept 42 a gradient 0.5 and y-intercept 3b gradient −1 and y-intercept 8c gradient 14 and y-intercept 03 a 3 b 1 c134 a −12b −1 c −45 a x 0 2 4 6 8 10y 5 4 3 2 1 0b Learner’s own graph. A straight line through (10, 0) and (0, 5).c y x = −512d gradient −12 and y-intercept 5e Learner’s own check.6 a y = 15 − 3xb gradient −3 and y-intercept 15 c x 0 5 2 4y 15 0 9 3d Learner’s own graph. A straight line through (0, 15) and (5, 0).e Learner’s own check.7 a y x = −634b gradient −34 and y-intercept 6 c x 0 8 4y 6 0 3 Learner’s own (correct) values in the last column.d Learner’s own graph. A straight line through (0, 6) and (8, 0).e Learner’s own check.8 a i y = 18 − 2xii y x = −912iii y = 9 − 2xiv y = −3 x12 b Line Gradient y-intercept2x + y = 18 −2 18x + 2y = 18 −1294x + 2y = 18 −2 93x + 6y = 18 −123c The gradient is −ab and the y-intercept is 18b.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 37 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20219 a y = 4x + 8b straight line, gradient 4, y-intercept 8, passes through (0, 8) and (−2, 0)10 a i 3 ii 12 iii y = +x123b i 3 ii −12 iii y x = − +123Exercise 10.41 a 250 m b 16 sc 12.5 m/s d d =12.5te 625 m2 a 400 b 20c i 8ii There are 8 HK dollars to 1 US dollar.d y = 8xe 920 HK dollars3 a 28 dollars b 15c 1.4 dollars d y = 1.4xe 64.12 dollars f 36.5 litres4 a 1 m b Weeks 0 1 2 3 4 5Height (m) 1 1.2 1.4 1.6 1.8 2c 0.2 md y = 0.2t + 1e 3.2 m5 a 1500 m b 750 mc 50 m/minute d y = 1500 − 50xe 350 m f 30 minutes6 a Learner’s own graph. A straight line from (0, 0) through (50, 45). b Dollars 50 20 30 15Euros 45 18 27 13.5c i 0.9ii 1 dollar buys 0.9 eurosd y = 0.9xe 252f 1707 a Learner’s own graph. A straight line from the origin through (25, 40).b 24 dollars c 1.6d 1.6 dollars e y = 1.6xf 152 dollars g 62.5 kg8 a Learner’s own graph. A straight line from (0, 20) going through (30, 32).b 24 °Cc 0.4d y = 0.4t + 20e 44 °Cf 200 secondsg Learner’s own answers.9 a Learner’s own graph. A straight line from (0, 100) going through (8, 72).b 79 litresc 3.5 litres/hourd y = 100 − 3.5h10 a 4800 b 33 00011 a The y-intercept is 24. Gradient = =−−32 2410 00 8. , so the equation of the line is p = 0.8t + 24.b 36 = 0.8t + 24 so t = =36 24 −0 815.; it takes 15 years.12 a 8 m/sb Marcus. Arun’s speed is 5 m/s.13 The rate for A is 2 cm/minute and the rate for B is 5 cm/minute.14 a12010=12 m/sb280 1201016 −= m/sc400 280524 −= m/s15 a Decreasing at a rate of 2 litres/hour.b y = 18 − 2tc 9 hours16 Learner’s own answers.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 38 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021Check your progress1 a 5x + 10y = 100b 10y = 100 − 5x, then divide both sides by 10.c −122 a x 0 1 2 3 4 5y 15 12 9 6 3 0b Learner’s own graph. A straight line through (0, 15) and (5, 0).c −33 a Learner’s own graph. The usual parabola shape with the bottom at (0, 5).b 5 and −54 a 4.5 m b 0.3 m/yearc y = 0.3x + 3 d 5.7 mUnit 11 Getting started1 a 20 : 1 b 1 : 4 c 1 : 52 a 90 b 108 c 723 a47 b 324 a Sky blue: 34, Ocean blue: 57b Sky blue is lighter. Learner’s own method. For example: Sky blue 1 : 3 = 2 : 6 = 2 parts blue and 6 parts white Ocean blue 2 : 5 = 2 parts blue and 5 parts white There is more white in sky blue, so this shade is lighter.5 $6.80Exercise 11.11 a Cherries: 2 parts = 80 g, 1 part = 80 ÷ 2 = 40 g Sultanas: 5 parts = 5 × 40 = 200 gb Total = 80 + 200 = 280 g2 a Strawberries: 2 parts = 400 g, 1 part = 400 ÷ 2 = 200 g Raspberries: 1 part = 200 gb Total = 400 + 200 = 600 g3 a $125 b $2004 a $18 b $425 a Sand: 2 parts = 15 kg, 1 part = 15 ÷ 2 = 7.5 kg Cement: 1 part = 7.5 kg Gravel: 4 parts = 4 × 7.5 = 30 kgb Total = 15 + 7.5 + 30 = 52.5 kg6 a 24 and 42 b 1207 a Learner’s own answers.b Learner’s own answers.c Learner’s own discussions.8 a 750 mL b 1.5 L9 1. Difference in number of parts = 4 − 1 = 3 2. 3 parts = 39 g 3. 1 part = 39 ÷ 3 = 13 g 4. 4 parts = 13 × 4 = 52 g 5. Total mass = 13 + 52 = 65 g10 a $70b Moira gets $21 and Non gets $49.11 a There are two possible solutions. The numbers are either 6 and 9 or 4 and 6.b i Learner’s own answers.ii There are two possible solutions. Either the first number is 6 or the second number is 6.iii 6 : 9 → dividing both numbers by 3 gives 2 : 3 4 : 6 → dividing both numbers by 2 gives 2 : 3c Learner’s own discussions.12 0.18 or 1.28; Check: 0.48 : 0.18 = 8 : 3 or 1.28 : 0.48 = 8 : 313 440 g of oats, 220 g of butter and 110 g of syrup. Learner’s own method. For example: Butter: 250 ÷ 2 = 125 g per part, Oats: 440 ÷ 4 = 110 g per part. Use 110 g per part as smallest amount. Syrup: 1 × 110 g = 110 g, Butter: 2 × 110 g = 220 g, Oats: 4 × 110 g = 440 g14 12 g15 3 : 4 : 5Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 39 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202116 Learner’s own working. For example: When working out the number of members of staff the number must be rounded up to make sure there are enough members of staff.Age of childrenChild : staff ratiosNumber of childrenNumber of members of staffup to 18 months 3:1 10 10 ÷ 3 = 3.3… = 418 months up to 3 years4:1 18 18 ÷ 4 = 4.5 = 53 years up to 5 years 8:1 15 15 ÷ 8 = 1.875 = 25 years up to 7 years 14:1 24 24 ÷ 14 = 1.7…= 2 Total number of members of staff needed = 4 + 5 + 2 + 2 = 13Reflection: Learner’s own answers. Exercise 11.21 a direct proportionb neitherc inverse proportiond direct proportione neitherf inverse proportiong neither2 a $7 b $17.50c $1.75 d $8.753 a 50 g b 150 g c 1.875 L4 a 4 horses = 2 days 1 horse = 8 daysb 4 horses = 2 days 8 horses = 1 day5 a normal speed = 36 seconds 12 speed = 72 secondsb normal speed = 36 seconds 3 × speed = 12 seconds6 a 20 minutes b 30 km/h7Number of people 4 12 2 1 6 10 5Cost per person (€) 300 100 600 1200 200 120 2408 a–d Learner’s own answers and discussions.9 2 hours 24 minutes10 a Learner’s own answers. Marcus is correct because the length of the ride is 4 minutes and it doesn’t matter how many people are on the roller coaster.b Learner’s own discussions.Activity 11.2 Learner’s own questions and answers.11 a Yes. Learner’s own explanations. For example: The height of bounce is 0.8 × the height it is dropped from.b 96 cmc i005010015020025050 100 150 200 250 300Height when dropped (cm)Height of bounce (cm)Height of ball before and after bounceii They are in a straight line.iii Yesiv 225 cm12 a Direct proportion. Learner’s own explanation. For example: The mass : length increase ratio is the same as 5 g : 3 mm for all pairs of values÷4÷ 2× 2× 3× 4× 2÷ 2÷ 3Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 40 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021bMass (g)20101530252025 30 35 40 45 50 Length of increase (mm)Length of increase of string when differentmasses are addedc Use your graph to work outi 27 mmii 33 g −34 g (accurate answer is 3313 g)d True. Learner’s own explanation. For example: Because one set of values is a multiple of the other, so the gradient of the line is constant.Check your progress1 a 750 g b 1050 g or 1.05 kg2 a 24, 30 and 42 b 1143 Sugar = 50 g, Butter = 100 g, Flour = 400 g4 a $6 b $18 c $4.505 a 12 days b 3 days c 6 peopleUnit 12 Getting started1 0.852 a H 1 H 2 H 3 H 4 H 5 H 6T 1 T 2 T 3 T 4 T 5 T 6b i112ii 31214=3 a1350= 0 2. 6 b15 or 0.24 a45b325c58d332Exercise 12.11 25%2 a16 b3612= c4623=3 a41025= b310 c710d3104 a 0.3 b 0.45 c 0.7 d 0.255 a 0.084 b 0.9166 a 0.85 b 0.7 c 0.057 a 0.4 b 0.52c 0.6 d 0.488 a18 b2814=c2814= d58e389 a 0.45 b 0.710 P(A) = 47; P(B) = 27; P(C) = 1711 a 0.2 b 0.95 c 0.412 a 0.1 b 0.09 c 0.19 d 0.8113 a112b Learner’s own answers. For example: The smallest possible numbers are black 3, white 8, yellow 1. Or learners could have any multiples of these.Exercise 12.21 P(S) is always 12 whether the first spin is a head or a tail.2 If A happens, the number is 2, 3 or 4 and then P(1 or 2) = 13; if A does not happen, the number is 1, 5 or 6 and then P(1 or 2) = 13; as these are the same, the events are independent.3 No. If the first two spins are tails then the probability that all three are = P(Y) = 12. If the first two spins are not both tails then Y is impossible and P(Y) = 0.4 They are independent. The coin is fair and so the probability is always 12. The coin has no memory of the previous throws!5 Fog will decrease the probability that the flight will leave on time because the flight could be cancelled.6 a If R happens, the number is 1, 2, 3, 4, 5 or 6 and P(even) = 3612= . If R does not happen, the number is 1, 2, 3 or 4 and P(even) = 2412= . The probabilities are the same and so the events are independent.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 41 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021b If B happens, the number is 1, 2, 3 or 4 and P(2) = 14. If B does not happen, the number is 1, 2, 3, 4, 5 or 6 and P(2) = 16. The probabilities are not the same, so the events are not independent.7 a They are independent. If the first ball is replaced then the situation is exactly the same both times.b They are not independent. If the first ball is black, the probability that the second ball is black is smaller than if the first ball is white.8 a Learner’s own explanation. For example: Arun and Sofia are not friends and do not travel together and there are no external factors such as weather or traffic.b Learner’s own explanation. For example: Arun and Sofia are brother and sister and travel to school together.9 If X happens then one of the cards must be A, C or D. Of these, 2 out of 3 are in the word CODE, so the probability of Y is 23. If X does not happen the card must be B or E. Then 1 out of 2 is in the word CODE, so the probability is 12. These probabilities are different, so the events are not independent.Exercise 12.31 a14b14c142 a136b112c1123 a16b19c25364 a 0.09 b 0.49 c 0.21 d 0.215 a 0.48 b 0.32 c 0.12 d 0.086 a i 0.015 ii 0.085 iii 0.135 iv 0.765b Learner’s own explanation. For example: They are mutually exclusive and one of them must happen.7 a First5not 5Second5not 55not 5Outcome5, 55, not 5not 5, 5not 5, not 5191989891989× = 1811919× = 8811989× = 8818919× = 64818989b i181ii 6481iii 881iv 881c Not getting a 5 either time.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 42 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20218 a Firstrednot redSecondrednot redrednot redOutcomered, redred, not red 0.3 × 0.4 = 0.120.3 × 0.6 = 0.18not red, red0.7 × 0.4 = 0.28not red, not red0.7 × 0.6 = 0.420.30.70.60.40.60.4b i 0.18 ii 0.28 iii 0.12 iv 0.42c Learner’s own explanation. For example: They are mutually exclusive and one of them must happen.9 a BlackbirdYesNoRobinYesNoYesNoOutcomeYes, YesYes, NoNo, YesNo, No0.90.10.80.20.80.20.9 × 0.8 = 0.720.9 × 0.2 = 0.180.1 × 0.8 = 0.080.1 × 0.2 = 0.02b i 0.72 ii 0.02c 0.9810 a FirstBlueYellowSecondBlueYellowBlueYellowOutcomeBlue, BlueBlue, YellowYellow, BlueYellow, Yellow× = = 212231416× = = 612233412× = 1121314× = = 312133414143423131434b i16ii 14iii 12iv 34v56Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 43 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202111 a Learner’s own diagram. For example: The best way to do this is with a tree diagram.FirstYesNoSecond OutcomeYesNoYesNoYes, YesYes, NoNo, YesNo, No0.40.60.90.10.90.10.4 × 0.9 = 0.360.4 × 0.1 = 0.040.6 × 0.9 = 0.540.6 × 0.1 = 0.06b Miss the first time, and get a basket the second time.c 0.94Exercise 12.41 a325 or 0.12 b725 or 0.28 c15 or 0.22 a Red 0.39; white 0.27; blue 0.34b The probability of each colour is 0.333. Blue is closest to this, white is furthest from this.3 a Rolls 10 20 30 40 50 60 70 80 90 100Total frequency 2 4 5 8 9 10 11 16 17 18Relative frequency 0.2 0.2 0.167 0.2 0.18 0.167 0.157 0.2 0.189 0.18b Learner’s own graph. Check that the relative frequency values from the table in part b have been plotted correctly.c Line through 0.167 on vertical axis.4 a Flips 20 40 60 80 100Frequency of heads 8 19 30 38 44Relative frequency 0.4 0.475 0.5 0.475 0.44b Learner’s own graph. Check that the relative frequency values from the table in part a have been plotted correctly.c The probability is 0.5. The relative frequency values are close to this. The values are below or equal to this.5 a Learner’s own table. Check that they have calculated the relative frequencies correctly.b Learner’s own graph. Check that the relative frequency values from the table in part a have been plotted correctly.c Learner’s own estimate. The probability is 0.583 and the estimate could be close to this.d Learner’s own discussions.6 a Draws 20 40 60 80 100 120 140 160 180 200Frequency 10 14 27 36 42 50 55 62 70 79Relative frequency 0.5 0.35 0.45 0.45 0.42 0.417 0.393 0.388 0.389 0.395Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 44 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021b Learner’s own graph. Check that the relative frequency values from the table in part a have been plotted correctly.c Learner’s own estimate. For example: 8 black and 12 white.7 aDigits 20 40 60 80 100Frequency of 0 2 5 7 7 8Relative frequency 0.1 0.125 0.117 0.088 0.08b Learner’s own graph. Check that the relative frequency values from the table in part a have been plotted correctly.cDigits 20 40 60 80 100Frequency of 0 2 6 8 9 15Relative frequency 0.1 0.15 0.133 0.113 0.15d Learner’s own graph. Check that the relative frequency values from the table in part c have been plotted correctly.eDigits 100 200 300 400 500Frequency of 0 11 27 40 52 60Relative frequency 0.11 0.135 0.133 0.13 0.12f Learner’s own graph. Check that the relative frequency values from the table in part e have been plotted correctly.g The probability is 0.1. The probabilities vary around this value. Sofia has the closest final value. You might expect her final value to be close because she has the largest sample size.8 Learner’s own answers and experiments.Check your progress1 Learner’s own answers. There are many possible answers. For example:a Roll a 2 and roll an odd number.b Roll a 2 and roll an even number.2 a If X happens then the number is 2, 4 or 6 and P(Y) = 13. If X does not happen then the number is 1, 3 or 5 and again P(Y) = 13.b If X happens the numbers are 2, 4 or 6 then P(Z) = 13. If X does not happen the numbers are 1, 3 or 5 then P(Z) = 23. Different probabilities so they are not independent.3 a 0.36 b 0.164 a 0.2 b 0.22c The probability is 0.2. The relative frequencies are the same or similar.Unit 13 Getting started1 a b c NBA 220° d NBA305°2 a 16 km b 30 cm3 a (8, 8) b (5, 8)4 a−−56b56NBA155°NBA60°Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 45 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 20215 axy420 3 7 4 862 10y = 451 93156bxy420 3 7 4 862 10x = 651 931566 Rotation, 90 ° clockwise, centre (−1, 0).7Exercise 13.11 Distance on scale drawing = 800 ÷ 100 = 8 cmN8cm50°2 a N5cmJR7cm140°230°b Learner’s own measurement. Answer in range 85 m–88 m.3 Yes they could meet. Learner’s own answers and discussions. Learner’s own explanation. For example: In a sketch of the situation, the two lines cross, showing the point where the yacht and the speedboat could meet. You don’t know if the yacht and the speedboat will meet because you don’t know their speeds, but if they do meet it will be at this point.4 NShip8cmAB42°152°5 a Teshi’s sketch is incorrect. He has drawn Yue south of Jun instead of Jun south of Yue.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 46 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021b N(8km)4 cm(6km)3cmYueJun4.1cm = 8.2km70°137°6 a NN4 cm(80km)5cm(100km)30°120°b Learner’s own measurement. Answer in range 125 km–130 km (Accurate answer is 128 km to 3 s.f.)c Learner’s own measurement. Answer in range 246 °–252 ° (Accurate answer is 249 ° to 3 s.f.)d Learner’s own discussions.7 a NN6 cm(12km) 45°275°8cm(16km)b Learner’s own measurement. Answer in range 12 km–13 km (Accurate answer is 12.4 km to 3 s.f.)c Learner’s own measurement. Answer in range 140 °–145 ° (Accurate answer is 143 ° to 3 s.f.)Activity 13.1 Learner’s own question and discussions.8 a, b i, c i, d i NPQNShopFarmhousecaféb ii Learner’s own measurements. In the range 14 km–14.5 km and 275 °–280 °.c ii Learner’s own measurements. In the range 6.5 km–7 km, 140 °–145 °.d ii Learner’s own measurements. In the ranges: Distance from P = 11.5 km–12 km, Distance from Q = 1.2 km–1.6 km.9 aP LNShip7.5cm (75km)Nb Learner’s own measurements. In the range 46 km–47 km.c Learner’s own measurements. In the range 53 km–54 km.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 47 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 202110 a i 120 ° ± 2 ° ii 247 ° ± 2 °iii 351 ° ± 2 °b 20 km c $1120Reflection: Learner’s own answers.Exercise 13.21 Learner’s own diagram. Check that all of the points are plotted and labelled correctly.a A (0, 2) and B (3, 2)b (1, 2)c (2, 2)d C (4, 0) and D (4, 8)e (4, 2)f (4, 6)2 A and v, B and iii, C and vi, D and ii, E and iv, F and i3 a–d Learner’s own answers.e Learner’s own answer. For example: Chesa’s method will work as she takes into account the position of S. When S moves she will add her distances on to the coordinates of S. Tefo’s method will not work as he is just finding the fraction of the coordinates of T. When S moves this will give the incorrect answer.f Learner’s own discussions.4 a B (4, 3) b A (12, 9)c C (2, 3) d B (8, 12)5 a B (4, 6)b C (6, 9)c J (2 × 10, 3 × 10) = (20, 30)d P (2 × 16, 3 × 16) = (32, 48)e (2 × 20, 3 × 20) = (40, 60)f Coordinates of the point labelled with the nth letter are (2n, 3n).6 a Yes. Learner’s own explanation. For example: E is at (4 × 3, 4 × 7) = (12, 28).b No. Learner’s own explanation. For example: OD lies 14 of the distance OE and so DE lies 34 of the distance OE. This means the ratio OD : DE is 1434: : = 1 3 and not 1 : 4.7 R (12, 15)8 a Learner’s own explanation. For example: The point (1, 2) is not on the line. It is two units to the left of where the line starts at point A.b Learner’s own explanation. For example: She needs to add (1, 2) on to the coordinates of A (3, 2).c The coordinates of C are (3 + 1, 2 + 2) = (4, 4). Learner’s own check.d i 1 : 5 ii 1 : 4e Learner’s own discussions.9 Difference in x-coordinates = 9 − 3 = 6, 23× = 6 4 Difference in y-coordinates = 13 − 4 = 9, 23× = 9 6 H = F(3, 4) + (4, 6) = (3 + 4, 4 + 6) = (7, 10)10 a L (10, 11)b Learner’s own check using a diagram.11 a, b Learner’s own diagram. Check that the points and diagonals are drawn accurately.c E 3 3 1212,d2 525 22727212123 3  + += = , , ,e Difference in x-coordinates of AC = 5 − 1 = 4 5812× = 4 2 Difference in y-coordinates of AC = 5 − 1 = 4 5812× = 4 2 E = A (1, 1) + 2 2 1212, = 1 2 1 2 1212 + + , = 3 3 1212,12 J (13, 13). Learner’s own working. For example:Difference in x-coordinates is 17 − 5 = 12Difference in y-coordinates is 19 − 1 = 18 There are six points after F, so the x-coordinates increase by 12 ÷ 6 = 2 for each point, and the y-coordinates increase by 18 ÷ 6 = 3 for each point.Points: F G H I J K Lx-coordinates 5 7 9 11 13 15 17y-coordinates 1 4 7 10 13 16 19Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 48 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021Exercise 13.31 a and iii, b and i, c and ii2 a Learner’s own diagram. The image should have vertices (2, 0), (4, 0), (4, 1) and (3, 1).b Learner’s own diagram. The image should have vertices (3, 0), (4, 0), (4, 1) and (3, 2).c Congruent. Learner’s own explanation. For example: In both parts the object and the image are identical in shape and size.3 a Learner’s own diagram. The image should have vertices (2, 1), (4, 3) and (1, 3).b Learner’s own diagram. The image should have vertices (−2, 0), (−5, 0) and (−3, 2).c Learner’s own diagram. The image should have vertices (−2, 1), (−2, 4) and (0, 3).d Learner’s own diagram. The image should have vertices (2, −1), (4, −3) and (2, −4).4 a i Learner’s own diagram. The image should have vertices (−1, −3), (1, −3), (0, −2) and (0, −3).ii Learner’s own diagram. The image should have vertices (−3, 5), (−5, 5), (−4, 4) and (−4, 5).b i Learner’s own diagram. The image should have vertices (3, −3), (5, −2), (5, −1) and (4, −1).ii Learner’s own diagram. The image should have vertices (−1, −3), (1, −2), (1, −1) and (0, −1).c i Learner’s own diagram. The image should have vertices (−2, 2), (−2, 4), (−3, 3), (−4, 3) and (−4, 2).ii Learner’s own diagram. The image should have vertices (−2, −4), (−2, −6), (−4, −6), (−4, −5) and (−3, −5).d i The positions of the shapes are different, even though the elements of the transformations are the same.ii Yes. Learner’s own explanation. For example: A different order often results in a different finishing position.iii Learner’s own discussions.iv Learner’s own transformations. For example: Reflection in line y = −2, then reflection in line x = 3.v Learner’s own checks.5 a Reflection in the y-axis.b Reflection in the x-axis.c Reflection in the line y = 1.d Reflection in the line x = 1.6 a 2−1 b51c−−64d 5−1e −4 4f137 a 90 ° clockwise, centre (3, 3)b 90 ° anticlockwise, centre (3, 0)c 180 °, centre (3, 0)d 90 ° clockwise, centre (−1, 0)e 90 ° anticlockwise, centre (−1, −1)8 a i Rotation 180 °, centre (−2, 1) OR reflection in the line y = 1 OR translation 0−3.ii Translation 2−4 OR rotation 180 °, centre (2.5, 3)iii Reflection in the line x = 4.5 OR rotation 180 °, centre (4.5, 1) OR translation 20.b Learner’s own discussions. For example: Yes, for all of them there is more than one transformation. Because each object and image are in the same orientation, they can all just be translated from one shape to the other shape. The shapes can all also be rotated 180 °. For the two pairs where the translation is either horizontal only or vertical only, it is also possible to reflect the shapes in a vertical or horizontal mirror line.c i For example: Rotation 180 °, centre (3, 5) followed by a translation 1−4.ii For example: Rotation 90 ° anticlockwise, centre (1, 4) followed by a translation −−25.iii For example: Rotation 90 ° anticlockwise, centre (3, 0) followed by a translation  −42.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 49 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021d Learner’s own discussions. For example: Yes, for all of them there is more than one combined transformation. Each object can be rotated about any point to get it in the same orientation as the image, and then you can use a translation to move it into the correct position. In part i, you could also use a reflection in any vertical or horizontal line and then you can use a translation to move it into the correct position.9 a They are both correct. When you start with triangle G and follow Sofia’s instructions, the final image is triangle H. When you start with triangle G and follow Zara’s instructions, the final image is triangle H.b For example: Reflection in the line x = 3 then translation −−62. For example: Translation −−82 then reflection in the line x = −4.c There are an infinite number of combined transformations. Learner’s own explanation. For example: G can be reflected in any line x = ‘a number’ then translated to H.10 a i Learner’s own diagram. Shape B with vertices (6, 4), (8, 5), (8, 2) and (6, 2). Shape C with vertices (2, 5), (4, 6), (4, 8) and (2, 8).ii Reflection in the line y = 5.b i Learner’s own diagram. Shape D with vertices (5, 8), (8, 8), (8, 10) and (6, 10). Shape E with vertices (2, 5), (5, 5), (5, 7) and (3, 7).ii Rotation 90 ° anticlockwise, centre (2, 5).Activity 13.3Learner’s own answers and discussions.11 a A to B b A to Cc B to D d C to EReflection:a It is the same shape and size.b • corresponding lengths are equal• corresponding angles are equal• the object and the image are congruentExercise 13.412 aScale factor 2bScale factor 3cScale factor 43 a Learner’s own explanation. For example: She hasn’t enlarged the shape correctly from the centre of enlargement. She has incorrectly used the centre as one of the vertices of the triangle.Downloaded by Tanishka Doshi ([email protected])lOMoARcPSD|16792922