CAMB NEW Y9 Maths LB ANS

Cambridge Lower Secondary Mathematics LEARNER’S BOOK 9 Cambridge Lower Secondary Mathematics Lynn Byrd, Greg Byrd & Chris Pearce LEARNER’S BOOK 9 Second edition Digital Access

Cambridge Lower Secondary Mathematics Greg Byrd, Lynn Byrd and Chris Pearce LEARNER’S BOOK 9

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 1 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 Learner’s Book answers Unit 1 Getting started 1 a 144 b 9 c 125 d 4 2 a 512 b 128 3 a 157 b 153 4 a 4 and 3000 and 225 b All of them. 5 106 Exercise 1.1 1 a integer 3 b irrational c irrational d integer 7 e irrational 2 a 1, 7 5 12, −38 and − 2 2. 5 are rational. b 200 is the only irrational number. 3 a integer b surd c surd d integer e integer f surd 4 a irrational because 2 is irrational b rational because it is equal to 4 2 = c irrational because 4 3 is irrational d 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 6 iii 10 iv 6 b They are all positive integers. c Learner’s own answer. d Learner’s own answer. 7 a 72=49 and 82=64 b 43=64 and 53=125 8 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 14 b 6 10 a i 1 ii 2 iii 3 b ( ) 5 1 + × ( ) 5 1− = 4, and so on c ( ) N N +1 1 × ( ) − = N −1 d 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 false b No. It might be a repeating pattern or it might not. Exercise 1.2 1 a 3×105 b 3.2×105 c 3.28×105 d 3.2871×105 2 a 6.3×107 b 4.88×108 c 3.04×106 d 5.2×1011 3 a 5400 b 1 410000 c 23370 000000 d 87 250000 4 Mercury 5.79×107km; Mars 2.279×108 ; Uranus 2.87×109 5 a Russia b Indonesia c The largest country is approximately 9 times larger than the smallest country. 6 a 7×10−6 b 8.12×10−4 c 6.691×10−5 d 2.05×10−7

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 2 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 7 a 0.0015 b 0.00001234 c 0.000000 079 d 0.000 9003 8 a 30 b 9.11×10−25kg 9 a z b y 10 a 65 is not between 1 and 10. b 6.5×105 c 4.83×107 11 a 1.5×10−2 b 2.73×10−3 c 5×10−8 12 a 6.1×106 b 6.17×105 c 1.75×105 13 a 7.6×10−6 b 8.02×10−5 c 1.6×10−7 14 a i 7×106 ii 3.4×107 iii 4.1×10−4 iv 1.37×10−3 b 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 1.3 1 a 1 4 b 1 8 c 1 81 d 1 216 e 1 10 000 f 1 32 2 3−3 , 2−4 and 4−2 are equal, 5−1 , 60 3 a 2−1 b 2−2 c 26 d 2−6 e 20 f 2−3 4 a 102 b 103 c 100 d 10−1 e 10−3 f 10−6 5 a 64−1 b 8−2 c 4−3 d 2−6 6 a 3−4 or 9−2 or 81−1 b The three ways in part a. 7 a 36 b 1 36 c 1 d 1 216 8 a 1 81 b 1 225 c 1 d 1 400 9 a i 2 ii 4 1 4 iii 9 1 9 b i x=5 ii x=10 10 a i 35 ii 39 iii 310 iv 36 b i 3 ii 3−1 iii 32 iv 3−2 v 3−3 c Learner’s own answers. d Learner’s own answers. 11 a 56 b 52 c 5−2 d 5−6 12 a 6−1 b 73 c 11−10 d 4−4 13 a x=4 b x=6 c x=−2 d x=5 14 a i 22 ii 43 iii 51 or 5 iv 23 b Learner’s own answers. c Learner’s own answers. 15 a 6−3 b 9−1 c 15−4 d 10−5 16 a 25 b 87 c 5−6 d 122 17 a 26 b 2−6 c 36 d 3−6 e 93 f 9−3 Check your progress 1 a rational b irrational c rational d irrational e rational 2 a rational because it is equal to 25 = 5 b irrational because it is 3 + 7 and 7 is a surd 3 n=3 4 a 8.6×1010 b 6.45×10−6 5 C, D, A, B 6 a 1 49 b 1 81 c 1 128

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 3 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 7 a 53 b 50 c 5−2 8 a 65 b 12−5 c 4−6 d 152 Unit 2 Getting started 1 x 3 +7 2 a 32×34=36 b 5 5 12 9 =53 c 72 5 ( ) =710 3 a x2+2x b 12y2−21yw 4 a 4(x+3) b 2x(2x+7) 5 a 17 12 or 1 5 12 b 6 5 or 11 5 6 a F=25 b a= F m c a=6 Exercise 2.1 1 a x y − = − × = − = − 2 3 2 5 3 10 7 b x xy 3 3 3 3 5 27 15 42 + = + × = + = c y x y 2 10 2 10 3 5 30 5 5 25 25 6 19 − = ( ) − = − = − = × 2 a 9 b 4 c 9 d 8 e 8 f 30 g 5 h 47 i −30 j −4 3 a Learner’s own answers. For example: i a=3, b=10, c=12, d=2 ii a=−3, b=−10, c=−12, d=−2 iii a=3, b=4, c=−36, d=3 b Learner’s own answers. c Learner’s own answers. 4 a Learner’s own answers. For example: Part a is incorrect as −32 should 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=6 b Learner’s own answer. For example: x=−4 and y=−1, x=−5 and y=−10, x=−6 and y=−21 c 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=−21 6 a 4 2 4 2 2 4 4 2 8 4 6 24 ( ) ( ) ( ) m p + = + × − = − = × − = − b p mp 3 3 3 4 3 2 4 64 24 40 − = − − × × − = − + = − ( ) c p m p ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ( ) = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ( ) − = − − = − − = − − 5 3 5 3 5 4 2 4 2 64 32 64 96 ( ) 7 a 21 b 36 c 16 d 64 e 68 f −18 g 14 h −25 i −7 j 82 Activity 2.1 Learner’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≠36 b For example: When y=2, (−y)4=(−2)4=16 and −y4=−24=−16, and 16≠−16 c 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≠10 9 a 26 b 49

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 4 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 10 5 9 7 5 2 9 1 2 7 2 1 549 1 2 2 2 2 1 5 5 a b a ab b − − + + = × − − − − − + + × − × − = × − × − ( ) ( ) ( ) ( ) + + = − − + = − 2 1 14 2092 14 23 − − − × − − − − − − − + = − − − − × − + 5 9 5 2 1 9 1 2 6 6 2 2 1 3 4 3 4 2 3 2 3 a b b a a a( ) b ( ) ( ) ( ) ( ) = − × − − + = − + − + = + = − + 10 1 9 1 8 9 9 6 8 2 10 48 16 22 1 23 4 ( ) Reflection: Learner’s own answers. Exercise 2.2 1 a n+5 b 5n−5 c n 5 +5 d 5(n+5) e n − 5 5 f 5−n 2 a 7x b 20−x c 2x+9 d x 6 −4 e x2 f 100 x g 5(x−7) h x i x3 j x3 k (3x)2+7 or 9x2+7 l (2x)3−100 or 8x3−100 3 a i 2x+2y ii xy b i 6x+2y ii 3xy c i 6x+4y ii 6xy d i 4x ii x2 e i 8x ii 4x2 f i 2x2+4x ii 2x3 4 a Perimeter=2(x+5)+2(2x)= 2x+10+4x=6x+10 b 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=48 d Perimeter=6x+10=6×3+10=28 Area=2x2+10x=2×32+10×3= 18+30=48 e Learner’s own answer. 5 a i P=2x+10 ii A=3x+6 iii When x=4, P=18 and A=18 b i P=2y−4 ii A=4y−24 iii When y=10, P=16 and A=16 c i P=4n+8 ii A=n2+4n iii When n=6, P=32 and A=60 d i P=2p2+8p ii A=4p3 iii When p=2, P=24 and A=32 6 a i 2 red+2 yellow=4 green; both=8x+4 ii 3 red+3 yellow=6 green; both=12x+6 iii 4 red+4 yellow=8 green; both=16x+8 b n red+n yellow=2n green (or similar explanation given in words) c i 6 red+2 yellow=12 blue; both=12x+12 ii 9 red+3 yellow=18 blue; both=18x+18 iii 12 red+4 yellow=24 blue; both=24x+24 d 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)=50 b 116 c (3w)2+2v(3v– 2w)+5w(w+v)= 9w2+6v2−4vw+5w2+5vw= 14w2+vw+6v2 d 116

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 5 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 8 a 3a2−7b=61, 8b – 3a=31, a2+6b=37, 4(a+3b)=4 b 133 c 3a2−7b+8b−3a+a2+6b+4(a+3b)= 4a2+7b−3a+4a+12b=4a2+a+19b d 133 e 11 f 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−x2 b 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)=34 and 2(2(2)2+9)=2(2×4+9)= 2(8+9)=34 10 a Side length= 25 = 5 cm, Perimeter=4×5=20cm b Side length= 49 = 7 cm, Perimeter=4×7=28cm c Perimeter=4 × x or 4 x 11 a Volume=x3 b Side length= y 3 Exercise 2.3 1 a x x x x 4 5 4 5 9 × = = + b y y y y 2 4 2 4 6 × = = + c u u u u 8 6 8 6 2 ÷ = = − d w w w w 5 5 1 4 ÷ = = − e g g g 3 2 3 2 6 ( ) = = × f h h h 5 12 5 12 60 ( ) = = × g 5m3+3m3=8m3 h 8n2−n2=7n2 2 a m14 b n12 c p7 d q5 e r3 f t5 g x21 h y10 i z12 j 5t7 k 5g2 l −h9 3 a Sofa is correct. x2÷x2=x2−2=x0=1 b Learner’s own answer. c x2÷x2=1 d All the answers are 1. Learner’s own explanations. For example: When simplifed, 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 30z7 d 4m7 e 4n13 f 8p3 5 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 5 3 3 2 4 6 2 3 x x x = , 12y7÷8y6= 3 7 2 6 12 8 3 2 y y y = and 6z9÷36z4= 6 36 6 9 6 4 5 z z z = . 6 a 3q4 b 3r4 c 3t6 d 2u5 e 2v4 f 5w 7 a D 1 2 x3 b A 2 5 y6 c C 5 3 k d B 31 3 8 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 125y12 iii 16z28

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 6 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 Activity 2.3 a 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 )4 c Learner’s own answers. 9 a q−3= 1 3 q b r−2= 1 2 r c t−5= 1 5 t d v−1=1 v 10 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 simplifes to give 1 6 7 y . For example: 5 30 2 9 y y Reflection: Learner’s own answers. Exercise 2.4 1 a (x+4)(x+1) =x2+1x+4x+4 =x2+5x+4 b (x−3)(x+6) =x2+6x−3x−18 =x2+3x−18 c (x+2)(x−8) =x2−8x+2x−16 =x2−6x−16 d (x−4)(x−1) =x2−x−4x+4 =x2−5x+4 2 a x2+10x+21 b x2+11x+10 c x2+2x−15 d x2+4x−32 e x2−9x+14 f x2−14x+24 3 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+48 c m2+m−12 d a2−7a−18 e p2−11p+30 f n2−30n+200 5 a The plus at the end would change to a minus and the 9 changes to a 1. x2+1x−20 b The plus at the end would change to a minus and the 9 changes to a −1. x2−1x−20 c The plus in the middle would change to a minus. x2−9x+20 d i (x+A)(x+B)= x2+Cx+D ii (x+A)(x−B)= x2+Cx−D iii (x−A)(x+B)= x2−Cx−D iv (x−A)(x−B)= x2−Cx+D 6 a C w2+12w+27 b A x2+2x−35 c B y2−2y−48 d A z2−9z +20 7 a (x+2)2=(x+2)(x+2) =x2+2x+2x+4 =x2+4x+4 b (x−3)2=(x−3)(x−3) =x2−3x−3x+9 =x2−6x+9 8 a i y2+10y+25 ii z2+2z+1 iii m2+16m+64 iv a2−4a+4 v p2−8p+16 vi n2−18n+81 b (x+a)2=x2+2ax+a2 9 a (x+3)(x−3)=x2+3x−3x−9=x2−9 b i x2−4 ii x2−25 iii x2 –49 c There is no term in x, and the number term is a square number. d x2−100 e x2−a2 Activity 2.4 a ① 33×29=957, ② 28×34=952, ③ 957−952=5

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 7 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 b ① 16×12=192, ② 11×17=187, ③ 192−187=5 c The answer is always 5. d n n+1 n+5 n+6 e 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 2.5 1 a 2 5 x b 4 7 x c 8 x d x e 2 5 x f 4 x 2 a 2 5 3 10 4 10 3 10 7 10 y y y y y + = + = b 2 5 1 25 10 25 1 25 9 y y y y 25y − = − = c 3 4 y d 3 8 y e 11 9y f 3 14 y 3 a a a a a a a a 2 5 5 10 2 10 5 2 10 7 10 + = + = = + b b b b b b b b 4 3 3 12 4 12 3 4 12 7 12 + = + = = + c 5 7 2 5 25 35 35 25 35 39 35 c c c c c c + = + = = + 14 14 d 5 6 3 5 30 30 25 18 30 7 30 d d d d d d − = − = = − 25 18d e 7 8 2 3 21 24 24 21 16 24 5 24 e e e e e e − = − = = − 16e f 4 a A, D, F b B, C, E c G; the answer is x 3 5 a 1 2 2 6 3 6 2 6 5 6 + = + = b 1 2 2 3 2 1 2 1 + = = c 5 6 ≠1 1 2 d She cannot cancel the 3 with the 6, because the expression is 3x+y, all divided by 6, not just 3x divided by 6. x y x y x y 2 6 3 6 6 3 6 + = + = + e Learner’s own answer. f i correct ii incorrect. Learners should show that the correct answer is 4 10 x y − iii correct iv incorrect. Learners should show that the correct answer is 9 8 20 x − 6 a i a b + 5 ii 5 9 12 a b + iii 2 9 15 a + iv ab b +12 4 v 3 40 10 ab b + vi 8 27 18 ab b + b Learner’s own checks. Activity 2.5 Learner’s own answers. 7 a 6 3 2 2 18 2 2 20 2 × + + = = =10 b 3×3+1=9+1=10 c 10=10 d 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+2 c 2x−3 d 2x−5 9 10 3 4 18 20 15 20 18 15 20 3 20 f f f f f f − = − = = −

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 8 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 9 6 4 2 20 25 5 2 3 2 2 5 4 5 5 x x − + x x − + + = + = ( ) ( ) 3x−2+4x+5=7x+3 10 a 2(x+3)=2×x+2×3=2x+6 b Learner’s own choice and explanation. c i 2(x+3) or 2x+6 ii 2(x+2) or 2x+4 iii 4(x−3) or 4x–12 iv 3(1−3x) or 3−9x Reflection: Learner’s own answers. Exercise 2.6 1 a S=60M b S=900 c M= S 60 d M=22.5 2 a i F=60 ii F=−78 b m= F a , m=12 c a= F m, a=−1.75 3 a 3D Shape Number of faces Number of vertices Number of edges Cube 6 8 12 Cuboid 6 8 12 Triangular prism 5 6 9 Triangularbased pyramid 4 4 6 Square-based pyramid 5 5 8 b E=F+V−2, or any equivalent version c V=E−F+2 i V=6 ii V=7 d c i is a pentagonal-based pyramid and c ii is a hexagonal-based pyramid e F=E−V+2, F=0, it is not possible to have a shape with fve edges and seven vertices. f Learner’s own answer. 4 a Ben’s age is x+2, Alice’s age is x−6 b T=3x−4 c T=53 d x=T + 4 3 e x=22 5 a v=87 b v=125 c u=27 d u=46 e t=10 f a=2 6 a 20% b 60% c 125% 7 a 65kg b 49.1kg (1 d.p.) c 95.9kg (1 d.p.) d 57.3kg (1 d.p.) 8 a i B x= y z − 2 ii C x=2 3 5 ( ) y h + iii A x=7k(y−6) iv C x=3ny+m v A x=w y − 7 b Learner’s own answer. 9 a t= m − 9 7 b t=5(k+m) c t=pv−h d t=9 5 q w+ 10 a A=a2+bc b A=49.5 c A=a2+bc, A−bc=a2 , a = A b − c d a=8 11 a 78.5 cm b r= A π c 6.25cm 12 a l= 3 V b 2 cm 13 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 3.7kg Check your progress 1 a 39 b 161 c 12 2 perimeter=16x+8, area=5x(3x+4)=15x2+20x

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 9 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 3 a x5 b q6 c h10 d 15m9 e 2u2 f 3p2 4 a x2+7x+10 b x2+x−12 c x2−3x−54 d x2 –14x+40 e x2−64 f x2−12x+36 5 a 2 3 x b 2 15 y c 12 20 x y − d 3x−5 6 a x=31 b z= x y − 2 5 , z=6 c y=± x z −5 , y=±6 Unit 3 Getting started 1 a 8 b 32.5 c 6 d 0.85 e 90 f 625 g 700 h 32 2 B 3 a 15.4 b 640 4 a $345 b $240 5 63.6cm2 (3 s.f.) Exercise 3.1 1 a, D and ii; b, A and v; c, E and iv; d, C and i; e, B and iii 2 a 3.2×103=3.2×1000=3200 b 3.2×102=3.2×100=320 c 3.2×101=3.2×10=32 d 3.2×100=3.2×1=3.2 e 3.2×10−1=3.2÷10=0.32 f 3.2×10−2=3.2÷100=0.032 g 3.2×10−3=3.2÷1000=0.0032 h 3.2×10−4=3.2÷10000=0.00032 3 a Yes. Learner’s own explanation. b i smaller ii the same iii greater 4 a 1300 b 7800 c 240 d 85500 e 65 f 8000 g 17 h 0.8 i 0.085 j 0.45 k 0.032 l 1.25 5 a 320÷103=320÷1000=0.32 b 320÷102=320÷100=3.2 c 320÷101=320÷10=32 d 320÷100=320÷1=320 6 a 2.7 b 0.45 c 0.36 d 0.017 e 0.08 f 0.0248 g 9 h 0.0025 7 a Learner’s own answer. b i 6.8÷10−3=6800 ii 0.07÷10−4=700 c 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=2.6×102 e Learner’s own answer. 8 a 3.2÷103=3.2÷1000=0.0032 b 3.2÷102=3.2÷100=0.032 c 3.2÷101=3.2÷10=0.32 d 3.2÷100=3.2÷1=3.2 e 3.2÷10−1=3.2×10=32 f 3.2÷10−2=3.2×100=320 g 3.2÷10−3=3.2×1000=3200 h 3.2÷10−4=3.2×10000=32000 9 a Yes. Learner’s own explanation. b i greater ii the same iii smaller 10 a 2.5 b 47 600 c 70 d 8.5 11 Do not tell anyone the secret! 12 a i 400 ii 40 iii 4 iv 0.4 v 0.04 vi 0.004 b Smaller c Smaller d i 0.12 ii 1.2 iii 12 iv 120 v 1200 vi 12 000 e Larger f Larger g Learner’s own answer.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 10 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 13 a = 8 0.8 × 101 80 × 10–1 0.008 × 103 0.08 ÷ 10–2 800 ÷ 102 8 ÷ 100 b = 0.32 32 ÷ 102 3.2 ÷ 101 32 × 10–2 3.2 × 10–1 0.32 × 100 320 ÷ 103 Activity 3.1 Learner’s own answers. Reflection: Learner’s own answers. Exercise 3.2 1 a 1.6 b −5.6 c −5.4 d 6 e 0.3 f −0.66 g 3.6 h −0.44 2 a 0.08×0.2 8×2=16 8×0.2=1.6 0.08×0.2=0.016 b 0.4×0.007 4×7=28 4×0.007=0.028 0.4×0.007=0.0028 3 C, D, I, K (0.015); A, F, H, J (0.15); B, G, L (1.5); E (15) 4 a 20 b −50 c −30 d 600 e 40 f −400 g 200 h −300 5 a 0 81 100 0 09 100 81 9 9 . . × × = = b 6 4 1000 0 004 1000 6400 4 1600 . . × × = = 6 a D b B c C d D 7 a i 0.8 ii 2.4 iii 4 iv 5.6 v 7.2 vi 8.8 b i Larger ii Smaller c i 60 ii 30 iii 20 iv 15 v 12 vi 10 d i Smaller ii Larger e Learner’s own answer. 8 a False b True c False d True 9 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 120 c 300 d 40 11 a A and iv, B and v, C and vi, D and vii, E and iii, F and i b Learner’s own answer. Any question that gives an answer of 0.024. For example: 0.03×400×0.002 c 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=0.057 13 a 123×57=7011 b i 701.1 ii 701.1 iii 70.11 iv 7.011 v 7.011 vi 0.07011 14 a Learner’s own answer. b Learner’s own answer. c i Estimate: 4×30=120 Accurate: 119.625 ii Estimate: 10÷0.2=50 Accurate: 62 iii Estimate: 60 4 0 01 × . =24000 Accurate: 19 200 15 a 0.2÷0.4=0.5 m b 0.45 m c Learner’s own answer. Exercise 3.3 1 a 200×1.1=$220 220×1.15=$253 b 200×0.9=$180 180×0.85=$153 c 200×1.2=$240 240×0.95=$228

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 11 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 2 a Learner’s choice of who they think is correct, with reason. b Sofa 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 57.6 b = c i = ii = 4 a–e Learner’s own answers. 5 a i 195 ii 64.4 b i 630 ii 108.864 6 a 1.1235 b $67.41 7 a i 72 ii 52.8 b i 285 ii 48.412 8 a 0.7216 b $4618.24 9 a A and iii, B and iv, C and i, E and ii, F and v b D and 0.81 10 a Zara is correct. 1.04×1.04 is the same as (1.04)2 , so 5000×1.04×1.04=5000×(1.04)2 b 5000×(1.04)3 c 5000×(1.04)4 d 8. The power on the 1.04 is the number of years. e i 5000×(1.04)12 ii 5000×(1.04)20 iii 5000×(1.04)n f 15 years 11 a i 10 000×0.9 ii 10 000×0.92 iii 10 000×0.93 b The population after 5 years. c The population after 10 years. d Five years. 10 000×0.94=6561, 10000×0.95=5904.9 e 10000×0.9n Activity 3.3 Learner’s own answers. Exercise 3.4 1 a i 25, 26, 27, 28, 29, 30, 31, 32, 33, 34 ii 25 iii 34 b i 85, 86, 87, 88, 89, 90, 91, 92, 93, 94 ii 85 iii 94 c i 265, 266, 267, 268, 269, 270, 271, 272, 273, 274 ii 265 iii 274 d i 845, 846, 847, 848, 849, 850, 851, 852, 853, 854 ii 845 iii 854 2 a 11.5, 11.6, 11.7, 11.8, 11.9, 12.0, 12.1, 12.2, 12.3, 12.4 b 11.5 c 12.4 3 a i 54.5, 54.6, 54.7, 54.8, 54.9, 55.0, 55.1, 55.2, 55.3, 55.4 ii 54.5 iii 55.4 b 42×1.3=54.6=$55 4 a–c Learner’s own answers. 5 a–c Learner’s own answers and discussions. 6 a 3.5⩽x<4.5 b 11.5⩽x<12.5 c 355.5⩽x<356.5 d 669.5⩽x<670.5 7 a 15⩽x<25 b 335⩽x<345 c 4745⩽x<4755 d 6295⩽x<6305 8 a 250⩽x<350 b 1850⩽x<1950 c 4650⩽x<4750 d 7950⩽x<8050

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 12 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 9 Learner’s own answers and discussions. a i 0.5 ii 5 iii 50 b The lower and upper bounds of a rounded number will always be +/− half of the degree of accuracy. 10 a i 1555 cm ii 1565cm b 1555 cm⩽x<1565 cm 11 a i 171.5cm ii 172.5cm b 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 d Check your progress 1 a 74500 b 12 c 0.046 d 59 e 0.0728 f 5 g 37 h 18 2 a −1.6 b 3.6 c −0.0028 d 600 e 300 f 9 g 7.5 h 0.11 3 $265.20 4 a i 20 000×1.08 ii 20 000×(1.08)2 iii 20 000×(1.08)3 b 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=29 386.56154, 20000×(1.08)6=31737.48646 e 20000×(1.08)n 5 a i 7150m2 ii 7250m2 b 7150 m2⩽x<7250 m2 Unit 4 Getting started 1 a x=5 b x=9 c y=25 d y=25 2 a 5 b 7 c 5, 6, 7 3 a 2x>10 b 4x<36 c y+5⩾13 d y − 5⩽−11 Exercise 4.1 1 a 8 30 14 8 16 2 16 8 x x x x = − + = − = = − − b c 2 3 2 3 48 2 11 5 16 2 16 3 2 48 24 y y y y y = + = = × = = = d 6 3 22 7 9 15 1 15 9 5 3 2 3 y y y y y y + = − = = = = 2 a x=−11 b x=−3 c y=4 d y=8 e a=−6 f a=−1 g x=2 h z=4 3 a, b x=15 c 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 3 5 18 5 2 0 4 2 5 x x x x x + = − + = = = = 16 . Check: When x=0.4, 2(0.4+8)=2×8.4=16.8 and 3(6−0.4)=3×5.6=16.8 d Learner’s own answer. 5 a, b x=13 c Learner’s own answers. 15 10 9 10 9 15 10 6 6 10 3 5 − = − = − − = − = − − = x x x x

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 13 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 6 a 42 42 42 7 7 7 6 c c c c = = = = c 21 2 7 e + = 21 7 2 2 3 2 3 2 1 21 7 = + = + = + − = = ( ) e e e e e 7 a a=27 b b= 3 7 c c=3 d d=11 8 a, b, c and e Learner’s own answers and explanations. d i x=14 1 4 ii x=6 3 5 iii x=− 1 5 9 a i A+10 ii A−6 b A+10=2(A−6) c A=22 10 a 2(x+3)+7x−5+5(7−x)=48 OR 4x+36=48 b x=3 c 12cm, 16 cm, 20cm 11 a 9a=4a+20 b a=4 c Triangle sides 12 cm, rectangle sides 7cm and 11 cm 12 a B and D b A x = 1 15 ; B x=15; C x=8640; D x=15; E x = 1 15 There are 15 sectors in the pie chart. 13 a 85 5 y = b 152 2 8 y + = c 85 85 5 5 17 y = → y = = and 152 2 152 8 8 2 19 2 17 y y 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=4 b 12(x−5)=4(x+1), x=8 c 5x−4=2x+20, x=8 d 5 75 7 = x + , x=8 e 9 126 2 = x , x=7 14 a 54 270 4 = x − b x=9 c 54°, 54 °, 72° 15 a Learner’s own problem. For example: i A quadrilateral has sides of length xcm, 2(x+1) cm, 3(x+2) cm, and 4(x+3) cm. The perimeter is 80cm. 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 fve 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=6 ii a=2.5 iii x=17 Exercise 4.2 1 1 Work out x. 5 3 2 15 5 2 15 3 3 18 6 18 3 x x x x x x − = + − = + = = = 2 Work out y. y = x − = × − = − = 5 3 5 6 3 30 3 27 3 Check values are correct. y = +x = × + = + = 2 15 2 6 15 12 15 27 4 Write the answers: x=6 and y=27 2 x=5, y=9 b 12 15 d = 12 15 12 15 12 15 4 5 = = = = d d d

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 14 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 3 x=4, y=13 4 x=7, y=−5 5 a y=3x+1 x 0 3 6 y 1 10 19 y=x+9 x 0 3 6 y 9 12 15 b 0 2 4 6 8 10 12 14 16 18 20 2 3 4 5 6 y = 3x + 1 y = x + 9 0 1 x y c (4, 13) d The coordinates give the solution of the equations; x=4 and y=13 e 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=6 ii x=2, y=6 b x=2, y=6 c Learner’s own answers and explanations. 7 a i x=2, y=7 ii x=6, y=2 b Learner’s own answers. 8 a i x=9, y=4 ii x=10, y=8 b i x=2, y=3 ii x=4, y=8 9 a x=5, y=2 b x=16, y=3 c x=7, y=4 d x=3, y=6 10 Sofa is correct, x=−3 and y=6. Zara got the signs round the wrong way. 11 a 1 Add the two equations. 2 Substitute x=18 into frst equation 2x+ y=50 2×18+y=50 + x− y= 4 y=50−36 3x+0y=54 =14 3x=54, x = = 54 3 18 3 Check in second equation 18−14=4 4 x=18 and y=14 b 1 Subtract the two equations. 2 Substitute y=9 into frst equation x+4y=41 x+4×9=41 − x+2y=23 x=41−36 0x+2y=18 =5 2y=18, y = = 18 2 9 3 Check in second equation 5+2×9=23 4 x=5 and y=9 c 1 Subtract the two equations. 2 Substitute y=4 into frst equation 3x+2y=38 3x+2×4=38 − 3x− y=26 3x=38−8 0x+3y=12 3x=30, x = = 30 3 10 3y=12, y = = 12 3 4 3 Check in second equation 3×10−4=26 4 x=10 and y=4 12 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.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 15 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 c Learner’s own answer. For example: Subtract to eliminate one of the letters when the coeffcients of that letter are the same number and both positive or both negative. Add to eliminate one of the letters when the coeffcients of that letter are the same number and one positive and one negative. d i x=9, y=6 ii x=−3, y=2 iii x=8, y=3 iv x=9, y=5 Activity 4.2 All answers should be x=6, y=18 13 a x=9, y=4 b x=5, y=−2 c x=2, y=4 d x=7, y=1 14 a x=2, y=2 b 3×2+2=6+2=8 and 4×2+2×2=8+4=12 Reflection: Learner’s own answers. Exercise 4.3 1 a x⩽2 b x>−2 c x⩾10 d x<−20 e −2⩽x<2 f −10<x⩽15 2 a 0 1 2 3 4 b –5 –4 –3 –2 –1 0 c –2 –1 0 1 d –20 –15 –10 –5 0 e f –4 –3 –2 –1 0 1 2 3 4 5 6 3 a 7 b −4 c −2, −1, 0 or 1 4 a x>2 b x⩽4 c x<−3 d x⩾−3 5 a 0 1 2 3 4 b –1 0 1 2 3 4 5 c –5 –4 –3 –2 –1 0 d –4 –3 –2 –1 0 6 a x<3 b, 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 ⩽ −11 b i x=−12 3(−12+2) ⩽ 2×−12−5 −30⩽−29 True ii x=−11 3(−11+2) ⩽ 2×−11−5 −27⩽−27 True iii 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 18 8 12 5 18 8 5 18 12 4 6 1 5 y y y y y y y y y y y + − < − + − < − − + < − < < ( ) . 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 9

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 16 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 ii y=1.5 4(2×1.5+3)−5×1.5 < 18−1.5 16.5<16.5 False iii y=2 4(2×2+3)−5×2 < 18−2 18<16 False 9 a a<3.5 b b⩾11 c c⩽6 d d>−27 Learner’s checks for each solution. 10 a 5n+5⩽30 b n⩽5 c 5, 12 and 13 11 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 fnal 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 ⩽ 5 12 a x>−4 or −4<x b x⩾5 or 5⩽x c x >6 or 6<x d x⩽−13 or −13⩾x e x <4 or 4>x f x⩾−2 or −2⩽x 13 a 3x−7<4x−11 b For example: 3 7 4 11 7 11 4 3 4 4 x x x x x x − < − − + < − < > c When x=5, 3×5−7<4×5−11 8<9 True When x=4, 3×4−7<4×4−11 5<5 False 14 a 2<x⩽5 0 1 2 3 4 5 6 b 5⩽y⩽20 0 5 10 15 20 25 30 c 3<n<9 2 3 4 5 6 7 8 9 10 11 d −3<m<6 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 Check your progress 1 a x=−4 b a=−2.5 c x=2.4 d y=9 e m=16 f n=10 Learner’s own checks for each solution. 2 x=5, y=19 3 x=19, y=7 4 a a<2 b b⩾5 c c>−1 d d⩾−5 Learner’s own checks for each solution. 5 a −1<x⩽2 –2 –1 0 1 2 3 4 b −4<n<1 –5 –4 –3 –2 –1 0 1 2 Unit 5 Getting started 1 140° 2 62° 3 a a and d OR b and e OR c and f b c and d c a and c OR d and f

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 17 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 4 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 5.1 1 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 75 5 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 OBC is 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 answer 10 a 45°+51 °=96° b A+B+C+D=96°+65°+127°+72°=360° Exercise 5.2 1 110° 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 100 8 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 refected 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 refected, 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 sides Sum of interior angles triangle 3 180° quadrilateral 4 360° pentagon 5 540° hexagon 6 720° octagon 8 1080° decagon 10 1440°

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 18 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 Reflection: In this case, subtract the 360 ° at the centre. 5×180−360=540 gives the same answer. Exercise 5.3 1 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 a Regular polygon Sides Exterior angle Equilateral 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 degrees c i 30 ° ii 18° 7 a 9 b 140 ° 8 a i 150 ° ii 160 ° iii 170 ° b i 12 ii 18 iii 36 9 15 sides 10 a 8 b 12 c 20 d 24 11 a 360−2×135=90 b 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÷n as (180n−360)÷n Exercise 5.4 The 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 5.5 1 a 10cm b 13 cm c 17cm 2 a 4.3 cm b 12.1cm c 14.2cm 3 a 12cm b 4.8m c 75mm 4 a 6.6 cm b 5.0 cm c 13.5m 5 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 + = 80cm to the nearest cm. b 105 58 2 2 + = 120cm to the nearest cm. 7 3 50 0 91 2 2 . . − = 3.38m 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 satisfes 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=130m b 130 90 40 31 5 2 2 − ( ) + = . m to 1 d.p. 11 a Square perimeter=4×25=100mm, rectangle perimeter=(2×20)+(2×30)= 40+60=100 mm b Diagonal of square=35.4mm; diagonal of rectangle=36.1mm

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 19 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 c Learner’s diagram and value. d The values so far support Sofa’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 100mm, 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 2 14 a i 7 7 98 2 2 + = ii 98 49 2 7 2 7 2 2 = × = × = b x x x x 2 2 2 + = 2 2 = Check your progress 1 a=65. The reason could use corresponding angles and the exterior angle of a triangle. 2 116° (x=106) 3 10 sides 4 a Learner’s own diagram. b Each side should be 8.5 cm. 5 35m or 35.3m or 35.36m are possible answers. 6 x=10 and y=24 Unit 6 Getting started In 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 specifc 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 6.1 These 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.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 20 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 3 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 6.2 1 17 girls and 13 boys 2 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 26 c 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 frst 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 confict. 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 Women Under 30 15 15 30 or more 45 45 Ask the frst 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.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 21 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 Check your progress 1 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 started 1 a 37.70cm b 21.99m 2 4.8cm or 48mm 3 a 34cm2 b 44m2 4 Group 1: A, D, G, H; Group 2: B, F; Group 3: C, E 5 a 320000 b 560000000 c 6.82 d 4.5 Exercise 7.1 1 a radius=2cm A r = = × = × = π 2 2 2 3 14 2 3 14 4 12 6 . . . cm (1 d.p.) b radius=9cm A r = = × = × = π 2 2 2 3 14 9 3 14 81 254 3 . . . cm (1 d.p.) c radius=4.2m A r = = × = × = π 2 2 2 3 14 4 2 3 14 17 64 55 4 . . . . . m (1 d.p.) 2 a diameter=16cm r d A r = ÷ = ÷ = = = × = × = 2 16 2 8 3 142 8 3 142 64 201 09 2 2 2 cm cm (2 d.p.) π . . . b diameter=9cm r d A r = ÷ = ÷ = = = × = × = 2 9 2 4 5 3 142 4 5 3 142 20 25 63 63 2 2 2 . . . . . . cm cm (2 π d.p.) c diameter=2.6m r d A r = ÷ = ÷ = = = × = × = 2 2 6 2 1 3 3 142 1 3 3 142 1 69 5 31 2 2 2 . . . . . . . m m (2 d. π p.) 3 a 153.938 cm2 b i 153.86cm2 ii 153.958cm2 iii 154cm2 c i 0.05% ii 0.01% iii 0.04% d π=3.142 e 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 113cm2 b 56.7 m2 c 415cm2 d 18.1m2 5 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.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 22 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 b 3 14 1 7 3 14 2 89 9 0746 9 07 3 2 2 . . . . . . × = × = Area = m ( ) s.f. 6 Learner’s own answers. A d = ⎛ ⎝ ⎜ ⎞ ⎠ π ⎟ 2 2 or A d = π 2 4 7 a i A=98.5cm2 ii C=35.2cm b i A= 804.2mm2 ii C=100.5mm 8 a Estimate: Accurate: A = × × = 1 2 6 2 60 38 2 2 π . . cm b Estimate: Accurate: A = × × = 1 2 14 85 346 40 2 2 π . . m c r=7.35cm; Estimate: Accurate: A = × × = 1 2 7 35 84 86 2 2 π . . cm d r=9.64m; Estimate: Accurate: A = × × = 1 2 9 64 145 97 2 2 π . . m 9 a i A=245.4 m2 ii P=64.3m b i A=831.0mm2 ii P=118.3mm Activity 7.1 Learner’s own answers. 10 Marcus is correct. Area of semicircle=10.618cm2 , Area of quarter-circle=9.0792cm2 and 10.618>9.0792. 11 a Learner’s own answers and explanations. b Learner’s own answers and explanations. c i 3.3cm ii 2.4m iii 9.0mm 12 a, b A and v, B and i, C and vi, D and iii, E and iv, F and ii 13 16.44m 14 84m2 15 a Learner’s own answers and explanations. b i 25πmm ii 144πmm2 iii 45πcm iv 400πcm2 c i ii Reflection: Learner’s own answers. Exercise 7.2 1 a Area A=l×w=5×4=20 Area B=l×w=11×2=22 Total area=20+22=42cm2 b Area A = × × = × × = 1 2 1 2 b h 12 6 36 Area B=l×w=12×3=36 Total area=36+36=72cm2 c Area A=l×w=5×12=60 Area B= = × × = 1 2 1 2 2 2 π π r 6 56 5. 5 Total area=60+56.55=116.55 cm2 d Area rectangle=l×w=4×1.5=6 Area circle= πr2=π×32=28.27 Shaded area=28.27−6=22.27cm2 2 a i 3cm ii 68 cm2 b i 7cm, 8cm ii 98 cm2 c i 7cm ii 138 cm2 3 a i 7×4+0.5×7×5=45.5cm2 ii 48.1cm2 b i 3×3+0.5×3×1.52=12.375m2 ii 10m2 c i 0.5×4×10+0.5×3×52=57.5cm2 ii 50.5cm2 d i 0.5×3×302+0.5×3×152=1687.5mm2 . The following could be accepted as an alternative: 0.5×3×302+0.5×3×152=1687.5 ii 1539.4mm2 4 a Learner’s own answer. b Learner’s own answers and explanations. c Learner’s own discussions. 5 a 34cm2 b 34.365cm2 c 187.56mm2 A ≈ × × = × × = 1 2 1 2 3 6 3 36 54 2 2 cm ; A ≈ × × = × × = 1 2 1 2 3 15 3 225 337.5 m ; 2 2 A ≈ × × = × × = 1 2 1 2 3 7 3 49 73 5 2 2 . ; cm A ≈ × × = × × = 1 2 1 2 3 10 3 100 150 2 2 m ; A r = = × × = × × = 1 2 1 2 1 2 2 2 2 12 144 72 π π π π m P = + d d = × × + = + 1 2 1 2 π π 24 24 12π 24 m

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 23 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 6 Sofa is correct, the two shaded areas are approximately the same size. Area of 1st shape=86.31cm2 , Area of 2nd shape=87.96 cm2 Activity 7.2 Learner’s own answers. 7 a i 18(π − 2) cm2 ii 50(π − 2) cm2 iii 72(π − 2) cm2 iv 4.5(π − 2) cm2 b 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. c 1 2 2 r ( ) π − 2 d 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 10cm−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=8cm. Area of square=8×8=64. Shaded area=64−16π=16(4−π) cm2 . b i 25(4−π) cm2 ii 9(4−π) cm2 iii 36(4−π) cm2 iv 100(4−π) cm2 c 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 7.3 1 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 1mg=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=0.000000 001 metres which is the same as 1nm=1×10−9m. 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 1kL=1×103L. 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=1000000000 metres which is the same as 1Gm=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 gigametres b 8μm, 8mm, 8 cm, 8m, 8km, 8Gm 4 a Learner’s own answers and explanations. For example: Marcus is correct. 1 tonne=1000kg. Also 1kg=1000g and 1Mg=1 000000 g=1000kg=1 t. Arun is incorrect. 1 litre=1000mL and 1 litre=100cL, so 1000mL=100 cL →10mL=1cL, not 100 mL=1cL b Learner’s own discussions. c Learner’s own answers and explanations. d Learner’s own discussions. 5 a 2.5 Mm to m → 1Mm=1 000000m, so 2.5Mm=2.5×1000000=2 500000m b 0.75GL to L →1GL=1000 000000L, so 0.75GL=0.75×1000000000 =750000000L c 13.2hg to g → 1hg=100 g, so 13.2hg =13.2×100=1320g 6 a 364 cL to L → 100cL=1L, so 364cL=364÷100=3.64L b 12000mg to g → 1000mg=1g, so 12000mg=12000÷1000=12g c 620000μm to m → 1000000μm=1m, so 620 000μm=620000÷1000000 =0.62m

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 24 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 7 From Earth to: Distance in … Mars 78.34Gm Jupiter 628.7Gm Saturn 1.28Tm Uranus 2.724Tm Neptune 4.35Tm 8 A and v, B and iv, C and i, D and iii, E and ii 9 a Learner’s own answers and explanations. For example: Sofa is correct. 300000 000×60×60×24×365.25 =9.46728×1015, which rounds to 9.47×1015. b 299792458×60×60×24×365.25 =9.460730 473×1015 c 9460000000000 000 d 6×9460000000000000 =56760 000000000000 =5.676×1016 e Learners own discussions. 10 a D, B, C, A b 2147483648 bytes c 10880 photos d 1864 flms 11 Learner’s own answers and explanations. For example: Magnar is incorrect. The fastest is model B because 10ns is quicker than 40ns and 60ns. Reflection: Learner’s own answers. Check your progress 1 a 39.27cm b 21.36m 2 a 123cm2 b 36.3m2 3 49.1cm2 4 170cm2 5 a 5 nanograms, 5 micrograms, 5 milligrams, 5 grams, 5 kilograms, 5 tonnes b 5ng, 5μg, 5mg, 5 g, 5kg, 5t Unit 8 Getting started 1 a 5 8 = 0.625 terminating b 5 6 =0.83. recurring 2 a 51 3 b 6 1 2 c 6 5 12 3 a 68 b 10 4 a 1 3 b 1 7 15 5 a 1 2 b 7 20 c 10 d 1 2 Exercise 8.1 1 a 1 4 = 0 2. 5 which is a terminating decimal 2 4 1 4 = 2 2 × = × 0 2. . 5 0 = 5 which is a terminating decimal 3 4 1 4 = 3 3 × = × 0 2. . 5 0 = 75 which is a terminating decimal b 1 5 = 0 2. which is a terminating decimal 2 5 1 5 = 2 2 × = × 0 2. . = 0 4 which is a terminating decimal 4 5 1 5 = 4 4 × = × 0 2. . = 0 8 which is a terminating decimal 2 a 1 9 =0.1. b Recurring decimal. c All recurring decimals. i 2 9 =0.2. ii 3 9 =0.3. iii 4 9 =0.4. iv 5 9 =0.5. v 6 9 =0.6. vi 7 9 =0.7. vii 8 9 =0.8. 3 a 3 9 1 3 = =0.3. and 6 9 2 3 = =0.6. b Learner’s own discussions. Their answers are not different because 9 9 =0.9. =1. 4 a 1 8 = 0.125 b Terminating decimal.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 25 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 c They are all terminating decimals. i 2 8 = 0 2. 5 ii 3 8 = 0.375 iii 4 8 = 0 5. iv 5 8 = 0.625 v 6 8 = 0 7. 5 vi 7 8 = 0.875 d Learner’s own answers. The following three fractions can be simplifed. c i 2 8 1 4 = = 0 2. 5, iii 4 8 1 2 = = 0 5. and v 6 8 3 4 = = 0 7. 5 5 a No, 3 6 1 2 = = 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 3 6 1 2 = . However, 7 is odd and so it cannot be halved, so there is not an equivalent fraction such that ? 7 1 2 = . d Learner’s own investigations and answers. For example: If the denominator is even, then there will be a fraction such that ? ? = 1 2 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: 1 15 is recurring, but 3 15 1 5 = = 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 3 6 1 2 = , D is now 3 12 1 4 = , E is now 3 15 1 5 = . 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 simplifed. 1 7 is a recurring decimal, so all fractions with a denominator of 7 are recurring. b Sometimes true: For example: 1 6 , 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. 1 4 , the denominator is a multiple of 2, and the fraction is a terminating decimal. c Sometimes true: For example: 1 20, 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. 1 30, 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. 1 2 1 2 1 2 = = 0 5. , 2 3 0 2. , 5 0 = . , 125 1 2 1 2 4 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 1 14 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 a 20 60 1 3 = recurring b 36 60 3 5 = terminating c 45 60 3 4 = terminating

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 26 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 d 55 60 11 12 = recurring e 1 8 60 1 2 15 = recurring f 3 3 21 60 7 20 = terminating 10 a recurring b terminating c recurring d terminating 11 The fractions written in their simplest form are: Abi 21 168 1 8 = Bim 28 168 1 6 = Caz 32 168 4 21 = Dave 35 168 5 24 = Enid 40 168 5 21 = Fin 42 168 1 4 = 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 8.2 1 a 5 2 3 3 5 1 2 + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Brackets: 3 5 1 2 6 10 5 10 1 10 − = − = Addition: 5 5 5 2 3 1 10 20 30 3 30 23 30 + = + = b 10 5 6 7 10 − × Multiplication: 5 6 7 10 5 7 6 10 35 60 7 12 × = = = × × Rewrite 10: 10 912 12 = Subtraction: 9 9 12 12 7 12 5 12 − = c 5 3 4 2 3 2 ÷ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Brackets: 2 3 2 3 2 3 2 2 3 3 4 9 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × = × = = Division: 5 5 3 4 4 3 20 3 ÷ = × = Addition: 20 3 4 9 60 9 4 9 64 9 1 9 + = + = = 7 2 a 2 5 16 b 31 4 c 2 d 33 4 3 a Learner’s own answers. For example: 7+3−(6−3)=10−3=7 b 7 1 12 c Learner’s own answers and explanations. d Learner’s own discussions. 4 a i 9−(2+4)=9−6=3 ii 3 3 40 b i 8+(2−1)=8+1=9 ii 9 5 24 c i 5+2×16=5+32=37 ii 42 4 9 d i 16 16 15 1 2 1 2 1 4 3 4 − × = − = ii 15 11 24 5 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 1 9 7 15 − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ or 25 5 8 1 9 7 15 − − 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 1119 45 . Learner’s own answer and explanation. For example: Yes, the third side is less than the total of the other two sides.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 27 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 7 31 4 kg 8 Division: 6 6 4 5 5 4 30 4 ÷ = × = Multiplication: 3 5 5 1 4 13 4 65 4 × = × = Addition: 30 4 65 4 95 4 3 4 + = = 23 9 a 1 7 8 2 m b 81 8 2 cm c 1 3 11 2 m 10 a Learner’s own answer and explanation. For example: They get different answers. Marcus is correct. His method does multiply 11 2 by 11 2 . Arun’s method multiplies 1 by 1 and 1 2 by 1 2 , which does not give the same answer. This can be shown using a multiplication box. × 1 1 2 1 1 1 2 1 2 1 2 1 4 1 1 1 2 1 2 1 2 1 2 1 2 1 4 1 4 × = +++ = . Marcus’s method gets this answer. Arun’s method only gets the 1 and the 1 4 , it doesn’t get the other two 1 2 s. 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 33 4 b 29 5 9 c 183 5 12 a 2 2 5 1 3 1 3 1 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + × or 2 2 5 1 3 1 3 1 2 × ⎛ ⎝ ⎜ ⎞ ⎠ + ⎟ b 18 5 18 m2 Exercise 8.3 1 a 3 4 3 4 12 12 3 3 9 1 3 × = × = × = b 5 7 5 7 28 28 5 4 20 1 4 × = × = × = c 4 5 4 45 45 4 9 36 1 9 5 × = × = × = d 3 8 3 72 72 3 9 27 1 9 8 × = × = × = 2 a 3 8 3 3 2 27 2 1 2 36 36 9 13 2 9 8 × = × = × = = b 4 9 4 4 3 52 3 1 3 39 39 13 17 3 13 9 × = × = × = = c 5 6 5 5 3 20 3 2 3 8 8 4 6 3 4 6 × = × = × = = d 7 10 7 7 2 63 2 1 2 45 45 9 31 2 9 10 × = × = × = = 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 7 11 24 11 11 3 22 3 1 3 2 3 24 × = × = × = = c highest common factor d Learner’s own discussions. 4 a 84 b 140 c 2 1 2 d 22 1 2 5 a 10 21 b 5 16 c 8 39 d 2 3 e 6 35 f 3 8 6 a 1 8 b 1 4 7 Lewis is correct, he travels 1831 3 km which is more than 180km. 8 Estimate Accurate a 1 3 1 2 3 5 × 1 4 6 1 2 × = 5 2 5 b 2 3 1 4 2 3 × 2 3 7 1 2 × = 81 4 c 1 3 1 8 1 6 × 1 3 3 1 2 1 2 × = 3 9 16 d 3 1 2 3 5 22 × 3 1 3 1 2 1 2 × = 4 1 2 e 3 4 3 4 3 5 × 4 4 18 1 2 × = 17 1 4 f 4 2 4 7 5 16 × 4 2 10 1 2 × = 10 4 7 9 a Learner’s own working. For example: 8 4 1 2 × = and 4<8, 4 3 1 2 2 3 × = and 3<4 1 2 , 5 9 3 10 1 6 × = and 1 6 5 9 <

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 28 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 b 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, 2 1 4 b bigger, 81 3 c bigger, 11 2 11 a A 17 33, B 1 1 12, C 91 3 , D 1 3 , E 1 5 16, F 2 9 b, 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. OR B and E have an even number for the denominator; A, C, D and F have an odd number for the denominator. OR A, 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 8.4 1 a 16 16 4 7 28 4 7 7 4 4 ÷ = = × 1 × = b 21 21 7 5 35 3 5 5 3 7 1 ÷ = × = × = c 14 14 7 9 63 2 9 9 2 7 1 ÷ = × = × = d 8 8 2 11 22 4 11 11 4 2 1 ÷ = × = × = 2 A and iii, B and i, C and iv, D and v, E and ii 3 a 8 9 4 7 8 9 7 4 2 7 9 1 14 9 5 9 2 1 ÷ = × = = = 1 × × b 7 9 2 5 7 9 5 2 7 5 9 2 35 18 17 18 ÷ = × = = = 1 × × c 6 7 3 14 6 7 14 3 2 2 1 1 2 1 2 ÷ = × 1 = = 4 × × d 5 6 15 24 5 6 24 15 5 24 6 15 4 3 1 3 1 4 1 3 ÷ = × = = = 1 × × 4 C 9 10, D 20 21, B 11 9 , A1 9 31 5 Estimate Accurate a 1 1 1 2 4 5 ÷ 2÷2=1 5 6 b 2 1 1 4 2 3 ÷ 2÷2=1 1 7 20 c 4 5 1 8 1 6 ÷ 4 5 4 5 ÷ = 99 124 d 2 3 2 3 1 4 ÷ 3÷3=1 32 39 e 5 2 1 2 3 4 ÷ 6÷3=2 2 f 4 2 4 5 2 3 ÷ 5 3 12 3 ÷ = 14 5 g 11 4 10 11 ÷ 1÷1=1 13 8 h 3 5 1 10 ÷2 1 2 1 2 ÷ = 2 7 6 a Learner’s own working. For example: 3 6 1 2 ÷ = and 6>3, 1 2 1 2 2 3 1 4 ÷ = and 2 1 1 4 1 2 > , 5 8 1 6 3 4 ÷ = 3 and 33 4 5 8 > 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, 91 3 b smaller, 4 c smaller, 2 2 21 8 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 1 2 1 4 10 13 ÷ = 9 a 14 15 b 2 6 7 c 11 7 d 11 9 e 11 27 f 1 1 11

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 29 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 10 a Learner’s answer and explanation. For example: π ≈ 3, and diameter=circumference÷ π. 15cm is slightly more than the circumference, 15 ÷ 3=5, so the diameter will be just under 5cm. b 14 4 1 7 22 7 99 7 7 22 9 2 1 2 ÷ = × = = 11 a 12 3 b 11 2 c 11 3 12 92 1 2 km/h 13 2 3 5 6 1 2 2 1 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ÷ is greater 2 1 2 4 5 34 15 3 4 27 36 − ÷ = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ and 2 3 5 6 1 2 7 9 28 36 2 1 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ÷ = = Exercise 8.5 1 a 1 2 1 2 1 2 5 5 1 5 6 36 1 35 2 2 2 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⇒ + ⎛ ⎝ ⎜ ⎞ ⎠ . ⎟ = ( ) = ⇒ 36 − = b 3 0 2 23 3 3 27 23 50 1 5 1 5 1 5 3 3 3 − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⇒ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( ) = ⇒ + = . − 27 c 6 3 0 7 3 4 36 4 32 2 2 3 10 3 10 7 10 − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⇒ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⇒ = ⇒ − = . + 6 36 2 a 48 b 49 c 12 3 a 1.5 × × 2.5 40 ⇒ × = ⇒ × = × = 3 2 5 2 15 4 15 1 4 10 40 15 10 150 b 1.25 × × 3 ⇒ × = ⇒ × = × = 1 2 5 4 7 2 35 8 35 8 56 56 35 7 245 1 7 c 2.75 × ⇒ 18 2 8 × 18 = × ⇒ 1 1 × 8 4 = = 9 3 4 11 4 11 4 99 2 1 2 2 9 4 a 126 b 108 c 105 5 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 a 0 28 5 5 25 25 7 2 2 4 28 1 100 28 100 . , × ⇒ 0.28 = = ⇒ × = b 0 7 4 4 3 2 7 7 10 2 7 30 7 7 10 30 7 1 1 1 3 . × ⇒ 0.7 = = , ⇒ × = c 1 3 4 4 4 4 60 60 78 3 3 1 13 6 10 13 10 . × − ( ) ⇒ − 1.3 = , = ⇒ × = 7 a 1 b 9 10 c 41 8 2m2 9 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 2 3 and 4 9 are both recurring decimals so it is easier to write them as fractions. c Learner’s own discussions. d 0 8 7 4 2 1 2 4 5 . × = ; example strategy: 0 8 7 4 2 2 8 5 3 1 1 2 4 5 15 2 16 25 15 2 8 3 5 1 24 5 4 5 . × = × = × = = = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × × 10 21 3 m

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 30 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 11 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 5 7 9 2 3 . × = ; example strategy: 4 25 1 4 5 7 9 1 4 16 9 17 4 4 3 17 3 2 3 1 1 . × = × = × = = 12 a K=2 b v = = = = 2 2 × 2 1 4 4 2 1 4 2 2 1 4 8 9 , but v=11 3 8 9 ≠ c v K m = 2 d v K m = = = = = 2 2 ×18 25 36 25 6 5 1 5 1 and K = = mv × × = × = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 1 2 6 5 25 2 36 25 2 2 25 18 Activity 8.5 Learner’s own answers. Reflection: Learner’s own answers. Check your progress 1 a recurring b terminating c recurring d terminating 2 a 5 1 4 b 2 7 12 c 4 29 30 3 a 12 1 2 b 4 15 4 a 48 b 3 4 5 a 80 b 50 c 1 3 Unit 9 Getting started 1 a i add 2 5 ii 5 6 3 5 , b i subtract 0.3 ii 7.3, 7 2 a 5, 7, 9, … b add 2 c Pattern 4 d Position number 1 2 3 4 term 5 7 9 11 2×position number 2 4 6 8 2×position number+3 5 7 9 11 Position-to-term rule is: term=2×position number+3 3 a 12 13 13 1 2 1 2 , , , …, 17 b 0.5, 4.5, 8.5, …, 36.5 4 a 3n+5 b 24−5n 5 a i x 6 11 18 25 y 6 8 1 2 12 15 1 2 ii x −2 11 2 8 111 2 y −15 2 1 2 35 52 1 2 b i y x = + 2 3 ii y=5(x − 1) Exercise 9.1 1 a linear b linear c non-linear d non-linear e linear f non-linear g linear h non-linear i 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 3 1 3 2 3 , , , ... d 40, 18, 7, … e 1.25, 3.25, 7.25, … f 1 2 7 1 2 , , , …

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 31 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 3 A and iii, B and i, C and iv, D and ii 4 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. frst 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. frst term is 1 2 , term-to-term rule is ‘square and add 1 4 ’, or frst term is 1 9 , term-to-term rule is ‘add 2 9 and square’. It is also possible if you add negative numbers – e.g. frst term is 2, term-to-term rule is ‘square and add −2’, or frst term is 9, term-to-term rule is ‘add −6 and square’. d Learner’s own discussions. 6 a 2, 32 7 , 44 7 , 56 7 , 71 7 , 83 7 , 95 7 b 90, 84 3 4 , 79 1 2 , 74 1 4 , 69, 633 4 , 581 2 c −4, −3.7, −3.4, −3.1, −2.8, −2.5, −2.2 d 31, 24.8, 18.6, 12.4, 6.2, 0, −6.2 7 a C b The ffth term, which is 126 382570 (fourth term=11242 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 9.1 Learner’s own questions and discussions. 9 Zara is correct. Learner’s own explanation. For example: The frst term is 3 and when you cube 3 you get 27. Then: c 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. d 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, 68898, … e 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 fnd 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=8 13 3 Reflection: Learner’s own answers. Exercise 9.2 1 a 1st term=4×1−5=−1 2nd term=4×2−5=3 3rd term=4×3−5=7 4th term=4×4−5=11 b 1st term=12+1=2 2nd term=22+1=5 3rd term=32+1=10 4th term=42+1=17 c 1st term=1 3 2nd term= 2 3 3rd term=3 3 =1 4th term= 4 3 1 3 =1 d 1st term=13=1 2nd term=23=8 3rd term=33=27 4th term=43=64

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 32 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 2 a 7, 11, 15, …, 43 b −3, −1, 1, …, 15 c 3 4 4 1 2 1 2 , , , …, 8 d 1 5 2 5 3 5 , , , …, 2 e 0, 3, 8, …, 99 3 A and iv, B and iii, C and i, D and ii 4 a Learner’s own answer and reason. b Card A has the greater value. A: 8th term=82−14=50, B: 20th term= 4 5 × 20 + 33 = 49 c 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 + 2 c The second differences are all the same (+2). d i 5, 7, 11, 17, 25, 35, … +2 +4 +6 +8 +10 +2 +2 +2 +2 ii 2, 7, 13, 20, 28, 37, … +5 +6 +7 +8 +9 +1 +1 +1 +1 iii 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 linear iii neither iv linear v neither vi quadratic f Learner’s own discussions. 6 a n2+3 b n2+10 c n2−1 d n2−9 7 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 frst term of −1 as this is a negative number not a positive number, so it cannot be in the sequence. 8 a i n 7 ii n 8 iii n 6 b Learner’s own discussions. 9 a A 12 15 4 5 = , B 14 18 7 9 = , C 9 12 3 4 = b C 3 4 , B 7 9 , A 4 5 10 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 4 1 2 1 2 − × = , when n=2, 4 2 3 1 2 1 2 1 2 − × = , when n=3, 4 3 3 1 2 1 2 − × = , etc. 12 a 9 1 2 1 4 − n b 20.2−0.2n c − −1 1 2 n d −3.5−1.5n Exercise 9.3 1 a i x 0 1 2 3 y 0 1 4 9 ii x 0 1 2 y 0 1 8 b i 0 1 2 3 4 5 6 7 8 9 10 0 2 3 4 5 6 7 8 9 10 y x 1 ii 0 1 2 3 4 5 6 7 8 9 10 0 2 3 4 5 6 7 8 9 10 y x 1 c i y=x2 ii y=x3

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 33 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 2 a i x 2 5 9 11 y 7 28 84 124 ii x 1 3 5 10 y −2 24 122 997 b i y=x2+3 ii y=x3−3 3 a i x −3 1 3 1 2 y 18 2 9 1 2 ii x −5 1 4 1 2 y 100 1 4 1 iii x −4 0 3 y −8 8 125 b Learner’s own discussions. c i y=2x2 ii y=(2x)2 iii y=(x+2)3 d Learner’s discussions. 4 a i x −8 −4 15 y 9 1 400 ii x −2 1 3 1 2 y 12 1 3 3 4 iii x −2 1 4 1 2 y −71 2 33 64 5 8 b i y=(x+5)2 ii y=3x2 iii y = + x3 1 2 5 a x ×4 y 2 b x 1 4 1 3 1 2 1 y 1 4 4 9 1 4 6 a i x −4 −3 3 4 y 16 9 9 16 ii 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 +x and −x, you get x2 . b i x 1 or −1 2 or −2 4 or −4 10 or −10 y 5 20 80 500 ii 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 12 y 8 32 50 288 ii x 7 10 11 13 y 16 49 64 100 b i y=2x2 ii y=(x−3)2 Activity 9.3 Learner’s own answers. 8 a i y=x2 ii x y = ± iii Learner’s own check. b i y=x3 ii x y = 3 iii Learner’s own check. c i y x = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 2 ii x y = ±2 iii Learner’s own check. d i y=x2+3 ii x y = ± 3 − iii Learner’s own check.

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)2 ii x y = + ± 4 iii Learner’s own check. f i y=(2x)3 ii x y = 3 2 iii Learner’s own check. 9 A and iii, B and i, C and v, D and vi, E and iv, F and ii 10 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 x ×8 y 2 x 1 4 1 2 3 or –3 y 1 2 2 72 y=8x2 12 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 1 2 3 because when a negative number is cubed, the answer will be negative. When this is multiplied by 1 2 , the answer will still be negative. Reflection: Learner’s own answers. Check your progress 1 a 3, 4, 11, 116, … b −3, 1, 9, 121, … c 5, 6, 9, 14, … d 40, 38, 34, 28, … 2 a 1 2 3 2 , , 1 , ..., 5 b 8, 11, 16, …, 107 3 a n2 b n2−2 c n 9 4 The number 178 is not a term in this sequence, because when you solve the equation n2+32=178 to fnd the value of n you do not get a whole number. n n n n 2 2 2 32 178 178 32 146 146 12 08 + = = − = = = . ... 5 a i x −2 4 5 9 y 2 8 12 1 2 40 1 2 ii x −15 −8 1 2 1 4 y 49 1 4 81 144 b i y x = 2 2 ii y=(x+8)2 Unit 10 Getting started 1 a $155 b c=20d+35 2 a x −2 −1 0 1 2 3 y −5 −3 −1 1 3 5 b Learner’s own graph; A straight line through (0, −1), (0.5, 0) and (3, 5). c 2 d −1 3 a 40°C b 20°C c At the start Exercise 10.1 1 a $31 b The number of days multiplied by 3 plus 10 for the fxed charge. 2 a 3 days b t=10n +15 3 a 27kg b b=2g − 3 4 a s+f=50 b s+f=52 c s+f=60

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 35 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 5 a 6×5+6×10=90 b 5f+10t=90 c 8 6 a 12×6+12×4=72+48=120 b 6l+4s=120 c s=2l 7 a The total value is 80 cents. b 4 8 a 3x+2y=50 b 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=100 c 10 d 32 e No. Each pair would have a total of 7 edges and 7 is not a factor of 100. f q=3r−5 Reflection: Two possible ways are x=20 − 2y and 2y=20 − x. Exercise 10.2 1 a x −1 0 1 2 3 y 5 15 25 35 45 b When x=5, then y=10×5+15=65 2 a x −10 0 10 20 30 y −30 −10 10 30 50 b 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 6 y 20 16 8 0 −4 b At (0, 20) and (5, 0) 4 a x 0 10 20 30 40 y 12 8 4 0 −4 b 2×15+5×6=60 5 a x −2 0 2 4 6 y 10 6 10 22 42 b When x=5, then y=52+6=31 6 a x 0 2 6 10 16 y 8 7 5 3 0 b i (16, 0) ii (0, 8) 7 a x 0 1 2 4 6 y 9 7.5 6 3 0 b 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 0 y 0 1 2 3 b Learner’s own graph. A straight line through (15, 0) and (0, 3). 9 a x 0 2 4 6 8 10 y 10 8 6 4 2 0 b 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 5 y 5 4 3 2 1 0 −1 e 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 6 y 12 10 8 6 4 2 0 c 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 12 k on the x-axis.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 36 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 11 a Learner’s own graph. A straight line through (14, 0) and (0, 7). b A straight line through (n, 0) and ( , 0 ) 1 2 n . 12 a x −3 −2 −1 0 1 2 3 y 9 4 1 0 1 4 9 b Learner’s own graph. A parabola with the base at the origin. c x −3 −2 −1 0 1 2 3 y 11 6 3 2 3 6 11 d Learner’s own graph. A parabola with the base at (0, 2). e x −3 −2 −1 0 1 2 3 y 5 0 −3 −4 −3 0 5 f 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 ii 14 a x −5 −4 −3 −2 −1 0 1 2 3 4 5 y 16 7 0 −5 −8 −9 −8 −5 0 7 16 b 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 10.3 1 a gradient 4 and y-intercept −6 b gradient 6 and y-intercept 4 c gradient −6 and y-intercept 4 2 a gradient 0.5 and y-intercept 3 b gradient −1 and y-intercept 8 c gradient 1 4 and y-intercept 0 3 a 3 b 1 c 1 3 4 a − 1 2 b −1 c −4 5 a x 0 2 4 6 8 10 y 5 4 3 2 1 0 b Learner’s own graph. A straight line through (10, 0) and (0, 5). c y = 5 − x 1 2 d gradient − 1 2 and y-intercept 5 e Learner’s own check. 6 a y=15−3x b gradient −3 and y-intercept 15 c x 0 5 2 4 y 15 0 9 3 d Learner’s own graph. A straight line through (0, 15) and (5, 0). e Learner’s own check. 7 a y = 6 − x 3 4 b gradient − 3 4 and y-intercept 6 c x 0 8 4 y 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 − 2x ii y = 9 − x 1 2 iii y=9 − 2x iv y = 3 − x 1 2 b Line Gradient y-intercept 2x+y=18 −2 18 x+2y=18 − 1 2 9 4x+2y=18 −2 9 3x+6y=18 − 1 2 3 c The gradient is − a b and the y-intercept is 18 b .

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 37 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 9 a y=4x+8 b straight line, gradient 4, y-intercept 8, passes through (0, 8) and (−2, 0) 10 a i 3 ii 1 2 iii y = +x 1 2 3 b i 3 ii − 1 2 iii y = − x + 1 2 3 Exercise 10.4 1 a 250m b 16s c 12.5m/s d d =12.5t e 625m 2 a 400 b 20 c i 8 ii There are 8 HK dollars to 1 US dollar. d y=8x e 920 HK dollars 3 a 28 dollars b 15 c 1.4 dollars d y=1.4x e 64.12 dollars f 36.5 litres 4 a 1m b Weeks 0 1 2 3 4 5 Height (m) 1 1.2 1.4 1.6 1.8 2 c 0.2m d y=0.2t+1 e 3.2m 5 a 1500m b 750m c 50m/minute d y=1500 − 50x e 350m f 30 minutes 6 a Learner’s own graph. A straight line from (0, 0) through (50, 45). b Dollars 50 20 30 15 Euros 45 18 27 13.5 c i 0.9 ii 1 dollar buys 0.9 euros d y=0.9x e 252 f 170 7 a Learner’s own graph. A straight line from the origin through (25, 40). b 24 dollars c 1.6 d 1.6 dollars e y=1.6x f 152 dollars g 62.5 kg 8 a Learner’s own graph. A straight line from (0, 20) going through (30, 32). b 24°C c 0.4 d y=0.4t+20 e 44°C f 200 seconds g Learner’s own answers. 9 a Learner’s own graph. A straight line from (0, 100) going through (8, 72). b 79 litres c 3.5 litres/hour d y=100 − 3.5h 10 a 4800 b 33 000 11 a The y-intercept is 24. Gradient = = − − 32 24 10 0 0 8. , so the equation of the line is p=0.8t+24. b 36=0.8t+24 so t = = 36 − 24 0 8 15 . ; it takes 15 years. 12 a 8m/s b Marcus. Arun’s speed is 5m/s. 13 The rate for A is 2 cm/minute and the rate for B is 5cm/minute. 14 a 120 10 =12m/s b 280 120 10 16 − = m/s c 400 280 5 24 − = m/s 15 a Decreasing at a rate of 2 litres/hour. b y=18 − 2t c 9 hours 16 Learner’s own answers.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 38 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 Check your progress 1 a 5x+10y=100 b 10y=100 − 5x, then divide both sides by 10. c − 1 2 2 a x 0 1 2 3 4 5 y 15 12 9 6 3 0 b Learner’s own graph. A straight line through (0, 15) and (5, 0). c −3 3 a Learner’s own graph. The usual parabola shape with the bottom at (0, 5). b 5 and −5 4 a 4.5m b 0.3m/year c y=0.3x+3 d 5.7m Unit 11 Getting started 1 a 20:1 b 1: 4 c 1:5 2 a 90 b 108 c 72 3 a 4 7 b 32 4 a Sky blue: 3 4 , Ocean blue: 5 7 b 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 $6.80 Exercise 11.1 1 a Cherries: 2 parts=80g, 1 part=80÷2=40g Sultanas: 5 parts=5×40=200g b Total=80+200=280g 2 a Strawberries: 2 parts=400g, 1 part=400÷2=200g Raspberries: 1 part=200g b Total=400+200=600g 3 a $125 b $200 4 a $18 b $42 5 a Sand: 2 parts=15kg, 1 part=15÷2=7.5kg Cement: 1 part=7.5kg Gravel: 4 parts=4×7.5=30kg b Total=15+7.5+30=52.5kg 6 a 24 and 42 b 120 7 a Learner’s own answers. b Learner’s own answers. c Learner’s own discussions. 8 a 750mL b 1.5L 9 1. Difference in number of parts=4−1=3 2. 3 parts=39g 3. 1 part=39÷3=13g 4. 4 parts=13×4=52 g 5. Total mass=13+52=65g 10 a $70 b 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 frst 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: 3 c Learner’s own discussions. 12 0.18 or 1.28; Check: 0.48: 0.18=8:3 or 1.28: 0.48=8 : 3 13 440g of oats, 220 g of butter and 110g of syrup. Learner’s own method. For example: Butter: 250÷2=125g per part, Oats: 440÷4=110 g per part. Use 110g per part as smallest amount. Syrup: 1×110 g=110 g, Butter: 2×110 g=220g, Oats: 4×110g=440g 14 12g 15 3:4 :5

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 39 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 16 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 children Child: staff ratios Number of children Number of members of staff up to 18 months 3:1 10 10÷3=3.3…=4 18 months up to 3 years 4:1 18 18÷4=4.5=5 3 years up to 5 years 8:1 15 15÷8=1.875=2 5 years up to 7 years 14:1 24 24÷14=1.7…= 2 Total number of members of staff needed=4+5+2+2=13 Reflection: Learner’s own answers. Exercise 11.2 1 a direct proportion b neither c inverse proportion d direct proportion e neither f inverse proportion g neither 2 a $7 b $17.50 c $1.75 d $8.75 3 a 50g b 150g c 1.875L 4 a 4 horses=2 days 1 horse=8 days b 4 horses=2 days 8 horses=1 day 5 a normal speed=36 seconds 1 2 speed=72 seconds b normal speed=36 seconds 3×speed=12 seconds 6 a 20 minutes b 30km/h 7 Number of people 4 12 2 1 6 10 5 Cost per person (€) 300 100 600 1200 200 120 240 8 a–dLearner’s own answers and discussions. 9 2 hours 24 minutes 10 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 96cm c i 0 0 50 100 150 200 250 50 100 150 200 250 300 Height when dropped (cm) Height of bounce (cm) Height of ball before and after bounce ii They are in a straight line. iii Yes iv 225cm 12 a Direct proportion. Learner’s own explanation. For example: The mass:length increase ratio is the same as 5g :3mm for all pairs of values ÷4 ÷2 ×2 ×3 ×4 ×2 ÷2 ÷3

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 40 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 b Mass (g) 20 10 15 30 25 20 25 30 35 40 45 50 Length of increase (mm) Length of increase of string when different masses are added c Use your graph to work out i 27mm ii 33 g −34g (accurate answer is 33 1 3 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 progress 1 a 750g b 1050g or 1.05kg 2 a 24, 30 and 42 b 114 3 Sugar=50 g, Butter=100 g, Flour=400g 4 a $6 b $18 c $4.50 5 a 12 days b 3 days c 6 people Unit 12 Getting started 1 0.85 2 a H 1 H 2 H 3 H 4 H 5 H 6 T 1 T 2 T 3 T 4 T 5 T 6 b i 1 12 ii 3 12 1 4 = 3 a 13 50 = 0 2. 6 b 1 5 or 0.2 4 a 4 5 b 3 25 c 5 8 d 3 32 Exercise 12.1 1 25% 2 a 1 6 b 3 6 1 2 = c 4 6 2 3 = 3 a 4 10 2 5 = b 3 10 c 7 10 d 3 10 4 a 0.3 b 0.45 c 0.7 d 0.25 5 a 0.084 b 0.916 6 a 0.85 b 0.7 c 0.05 7 a 0.4 b 0.52 c 0.6 d 0.48 8 a 1 8 b 2 8 1 4 = c 2 8 1 4 = d 5 8 e 3 8 9 a 0.45 b 0.7 10 P(A)= 4 7 ; P(B)= 2 7 ; P(C)=1 7 11 a 0.2 b 0.95 c 0.4 12 a 0.1 b 0.09 c 0.19 d 0.81 13 a 1 12 b 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 12.2 1 P(S) is always 1 2 whether the frst spin is a head or a tail. 2 If A happens, the number is 2, 3 or 4 and then P(1 or 2)=1 3; if A does not happen, the number is 1, 5 or 6 and then P(1 or 2)=1 3; as these are the same, the events are independent. 3 No. If the frst two spins are tails then the probability that all three are=P(Y)=1 2. If the frst 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 1 2 . The coin has no memory of the previous throws! 5 Fog will decrease the probability that the fight will leave on time because the fight could be cancelled. 6 a If R happens, the number is 1, 2, 3, 4, 5 or 6 and P(even)= 3 6 1 2 = . If R does not happen, the number is 1, 2, 3 or 4 and P(even)= 2 4 1 2 = . The probabilities are the same and so the events are independent.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 41 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 b If B happens, the number is 1, 2, 3 or 4 and P(2)=1 4. If B does not happen, the number is 1, 2, 3, 4, 5 or 6 and P(2)=1 6. The probabilities are not the same, so the events are not independent. 7 a They are independent. If the frst ball is replaced then the situation is exactly the same both times. b They are not independent. If the frst ball is black, the probability that the second ball is black is smaller than if the frst ball is white. 8 a Learner’s own explanation. For example: Arun and Sofa are not friends and do not travel together and there are no external factors such as weather or traffc. b Learner’s own explanation. For example: Arun and Sofa 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 2 3 . 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 1 2 . These probabilities are different, so the events are not independent. Exercise 12.3 1 a 1 4 b 1 4 c 1 4 2 a 1 36 b 1 12 c 1 12 3 a 1 6 b 1 9 c 25 36 4 a 0.09 b 0.49 c 0.21 d 0.21 5 a 0.48 b 0.32 c 0.12 d 0.08 6 a i 0.015 ii 0.085 iii 0.135 iv 0.765 b Learner’s own explanation. For example: They are mutually exclusive and one of them must happen. 7 a First 5 not 5 Second 5 not 5 5 not 5 Outcome 5, 5 5, not 5 not 5, 5 not 5, not 5 1 9 1 9 8 9 8 9 1 9 8 9 × = 1 81 1 9 1 9 × = 8 81 1 9 8 9 × = 8 81 8 9 1 9 × = 64 81 8 9 8 9 b i 1 81 ii 64 81 iii 8 81 iv 8 81 c Not getting a 5 either time.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 42 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 8 a First red not red Second red not red red not red Outcome red, red red, not red 0.3 × 0.4 = 0.12 0.3 × 0.6 = 0.18 not red, red not red, not red 0.7 × 0.4 = 0.28 0.7 × 0.6 = 0.42 0.3 0.7 0.6 0.4 0.6 0.4 b i 0.18 ii 0.28 iii 0.12 iv 0.42 c Learner’s own explanation. For example: They are mutually exclusive and one of them must happen. 9 a Blackbird Yes No Robin Yes No Yes No Outcome Yes, Yes Yes, No No, Yes No, No 0.9 0.1 0.8 0.2 0.8 0.2 0.9 × 0.8 = 0.72 0.9 × 0.2 = 0.18 0.1 × 0.8 = 0.08 0.1 × 0.2 = 0.02 b i 0.72 ii 0.02 c 0.98 10 a First Blue Yellow Second Blue Yellow Blue Yellow Outcome Blue, Blue Blue, Yellow Yellow, Blue Yellow, Yellow × = = 2 12 2 3 1 4 1 6 × = = 6 12 2 3 3 4 1 2 × = 1 12 1 3 1 4 × = = 3 12 1 3 3 4 1 4 1 4 3 4 2 3 1 3 1 4 3 4 b i 1 6 ii 1 4 iii 1 2 iv 3 4 v 5 6

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 43 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 11 a Learner’s own diagram. For example: The best way to do this is with a tree diagram. First Yes No Second Outcome Yes No Yes No Yes, Yes Yes, No No, Yes No, No 0.4 0.6 0.9 0.1 0.9 0.1 0.4 × 0.9 = 0.36 0.4 × 0.1 = 0.04 0.6 × 0.9 = 0.54 0.6 × 0.1 = 0.06 b Miss the frst time, and get a basket the second time. c 0.94 Exercise 12.4 1 a 3 25 or 0.12 b 7 25 or 0.28 c 1 5 or 0.2 2 a Red 0.39; white 0.27; blue 0.34 b 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 90 100 Total frequency 2 4 5 8 9 10 11 16 17 18 Relative frequency 0.2 0.2 0.167 0.2 0.18 0.167 0.157 0.2 0.189 0.18 b 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 80 100 Frequency of heads 8 19 30 38 44 Relative frequency 0.4 0.475 0.5 0.475 0.44 b 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 80 100 120 140 160 180 200 Frequency 10 14 27 36 42 50 55 62 70 79 Relative frequency 0.5 0.35 0.45 0.45 0.42 0.417 0.393 0.388 0.389 0.395

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 44 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 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. For example: 8 black and 12 white. 7 a Digits 20 40 60 80 100 Frequency of 0 2 5 7 7 8 Relative frequency 0.1 0.125 0.117 0.088 0.08 b Learner’s own graph. Check that the relative frequency values from the table in part a have been plotted correctly. c Digits 20 40 60 80 100 Frequency of 0 2 6 8 9 15 Relative frequency 0.1 0.15 0.133 0.113 0.15 d Learner’s own graph. Check that the relative frequency values from the table in part c have been plotted correctly. e Digits 100 200 300 400 500 Frequency of 0 11 27 40 52 60 Relative frequency 0.11 0.135 0.133 0.13 0.12 f 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. Sofa has the closest fnal value. You might expect her fnal value to be close because she has the largest sample size. 8 Learner’s own answers and experiments. Check your progress 1 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)=1 3 . If X does not happen then the number is 1, 3 or 5 and again P(Y)=1 3 . b If X happens the numbers are 2, 4 or 6 then P(Z)=1 3 . If X does not happen the numbers are 1, 3 or 5 then P(Z)= 2 3 . Different probabilities so they are not independent. 3 a 0.36 b 0.16 4 a 0.2 b 0.22 c The probability is 0.2. The relative frequencies are the same or similar. Unit 13 Getting started 1 a b c N B A 220° d N B A 305° 2 a 16km b 30 cm 3 a (8, 8) b (5, 8) 4 a − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 5 6 b 5 6 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ N B A155° N B A 60°

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 45 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 5 a x y 4 2 0 2 3 4 6 7 8 10 y = 4 1 5 9 3 1 5 6 b x y 4 2 0 2 3 4 6 7 8 10 x = 6 1 5 9 3 1 5 6 6 Rotation, 90° clockwise, centre (−1, 0). 7 Exercise 13.1 1 Distance on scale drawing=800÷100=8 cm N 8cm 50° 2 a N 5 cm J R 7cm 140° 230° b Learner’s own measurement. Answer in range 85m–88m. 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 N Ship 8cm A B 42° 152° 5 a Teshi’s sketch is incorrect. He has drawn Yue south of Jun instead of Jun south of Yue.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 46 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 b N (8km) 4 cm (6km) 3cm Yue Jun 4.1cm = 8.2km 70° 137° 6 a N N 4 cm (80km) 5cm (100km) 30° 120° b Learner’s own measurement. Answer in range 125km–130km (Accurate answer is 128km 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 N N 6 cm (12km) 45° 275° 8cm(16km) b Learner’s own measurement. Answer in range 12km–13km (Accurate answer is 12.4km 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 N P Q N Shop Farm house café b ii Learner’s own measurements. In the range 14km–14.5km and 275 °–280 °. c ii Learner’s own measurements. In the range 6.5km–7km, 140 °–145 °. d ii Learner’s own measurements. In the ranges: Distance from P=11.5km– 12km, Distance from Q=1.2km– 1.6km. 9 a P L N Ship 7.5cm (75km) N b Learner’s own measurements. In the range 46km–47km. c Learner’s own measurements. In the range 53km–54km.

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 47 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 10 a i 120 °±2 ° ii 247°±2° iii 351 °±2 ° b 20km c $1120 Reflection: Learner’s own answers. Exercise 13.2 1 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 i 3 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 fnding 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 1 4 of the distance OE and so DE lies 3 4 of the distance OE. This means the ratio OD:DE is 1 4 3 4 : : = 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 : 4 e Learner’s own discussions. 9 Difference in x-coordinates=9−3=6, 2 3 × 6 4 = Difference in y-coordinates=13−4=9, 2 3 × 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, bLearner’s own diagram. Check that the points and diagonals are drawn accurately. c E 3 3 1 2 1 2 , ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ d 2 5 2 5 2 2 7 2 7 2 1 2 1 2 3 3 ⎛ + + ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ = = ⎟ , , , e Difference in x-coordinates of AC=5−1=4 5 8 1 2 × 4 2 = Difference in y-coordinates of AC=5−1=4 5 8 1 2 × 4 2 = E =A(1, 1)+ 2 2 1 2 1 2 , ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 2 1 2 1 2 1 2 + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ , = 3 3 1 2 1 2 , ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 12 J (13, 13). Learner’s own working. For example: Difference in x-coordinates is 17−5=12 Difference 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 L x-coordinates 5 7 9 11 13 15 17 y-coordinates 1 4 7 10 13 16 19

CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE 48 Cambridge Lower Secondary Mathematics 9 – Byrd, Byrd & Pearce © Cambridge University Press 2021 Exercise 13.3 1 a and iii, b and i, c and ii 2 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 fnishing position. iii Learner’s own discussions. iv Learner’s own transformations. For example: Refection in line y=−2, then refection in line x=3. v Learner’s own checks. 5 a Refection in the y-axis. b Refection in the x-axis. c Refection in the line y=1. d Refection in the line x=1. 6 a 2 −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ b 5 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ c − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 6 4 d 5 −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ e ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ 4 4 f 1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 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 refection in the line y=1 OR translation 0 −3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟. ii Translation 2 −4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ OR rotation 180 °, centre (2.5, 3) iii Refection in the line x=4.5 OR rotation 180 °, centre (4.5, 1) OR translation 2 0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟. 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 refect 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 − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 5 . iii For example: Rotation 90 ° anticlockwise, centre (3, 0) followed by a translation ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ 4 2 .