Ls Maths 9 2ed TR Learner Book Answers

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Ls maths 9 2ed tr learner book answers

Masters in business administration (Cambridge Group of Institutions)

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

 Learner’s Book answers Unit 1 Getting started 1  a  144 c 

b  9

8

a  The square root of any any integer between 16 and 25 is a possible answer.

d  4

125

2  a  512

b  128

3  a  15

b  15

7

b  The square root of any integer between 144 and 169 is a possible answer.

3

9

4  a  4 and 3000 and 225

a

14

b  6

b  All of them.

10 a

i 

ii  2

1

 

5  10

b  ( 5 + 1) × ( 5 − 1) = 4, and so on

Exercise 1.1

c  (  N + 1) × ( N − 1) = N − 1

6

1  a  integer 3

b  irrational

c  irrational

d  integer 7

d  Learner’s own answer. 11 a No. It is not a repeating pattern. b

e  irrational  5

1 , 7 , −38 and − 2.25  are rational.

2  a 

12

b 

200 is 200  is the only irrational number.

3  a  integer d  integer

b  surd

c 

surd

e  integer

f  

surd

4  a  irrational because 2  is irrational b  rational because it is equal to 4 = 2 c 

d  rational because it is equal to 3 8 = 2 5  a  Learner’s own answer. For example: 2  and − 2 .

a

i

b

No. It might be a repeating pattern or it might not.

6  a  i   

−

2

1 2 3

true

iii  false

a

3 × 105

b

c 

3.28 × 105

d  3.2871 × 105

a

6.3 × 107

b  4.88 × 108

c 

3.04 × 106

d  5.2 × 1011

a

5400

b  1 410 000

c 

23 370 000 000

d  87 250 000

3.2 × 105

Uranus 2.87 × 109

ii  6

4

iii  10

iv  6

5

a  Russia

d  Learner’s own answer.

6

b  Indonesia

c 

The largest country is approximately 9 times larger than the smallest country.

a

7 × 10−6

b  8.12 × 10−4

c 

6.691 × 10−5

d  2.05 × 10−7

Learner’s own answer.

7  a  7󰂲 = 49 and 8󰂲 = 64

ii

4  Mercury 5.79 × 107 km; Mars 2.279 × 108;

b  They are all positive integers. c 

true

Exercise 1.2

b  Learner’s own answer. For example: 2  and 2

Learner’s own answer.

Reflection:

3

irrational because 4  is irrational

iii  3

b  4󰂳 = 64 and 5󰂳 = 125

1

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

7  a  0.0015 c 

b  0.000 012 34

b 

7  a  36

d  0.000 900 3

0.000 000 079

−25

8  a  30

b  9.11 × 10  kg

8  a 

1

81

b   y

9  a  i 

10  a  65 is not between 1 and 10.

b  i 

9  a  z

b  6.5 × 105 c  4.83 × 107

b 

 

1

1 225

 

b  i 

d 

c  1

d 

1

2

ii 

x = 5

ii  x = 10

4

iii 

4

10  a  i  35   iii  310

11  a  1.5 × 10−2

1

c 

36

   

b  2.73 × 10−3

 

iv  3−2

c 

c 

Learner’s own answers.

5 × 10−8

 

11  a  56

c 

12  a  6−1

b  73

c 

d  4−4

1.75 × 105

13  a  7.6 × 10−6

c 

1.6 × 10−7 7 × 106

iii  4.1 × 10−4

 

ii  3.4 × 107

 

iv  1.37 × 10−3

b  To multiply a number in standa standard rd form by 10, you add 1 to the index.

c 

To multiply a number in standa standard rd form by 1000, you add 3 to the index. To divide a number in standard form by 1000, you subtract 3 from the index.

can write the number without using a lot of zeros zeros.. You can enter them in a calculator.

1 4

1 216

b  e 

 

1 8

c  1

10000

f  

 

1 81 1 32

2  3 , 2  and 4  are equal, 5 , 6 −3

−4

3  a  2−1 d  2−6 4  a  102 d  10−1

−2

−1

0

iii  32

c 

e  20

f   2−3

b  103

c 

100

e  10−3

f  

10−6

4−3

b  x = 6 d  x = 5

x = −2

ii  43 iv  23

b  Learner’s own answers. Learner’s own answers.

15  a  6−3

b  9−1

c 

d  10−5

15−4

16  a  25

b  87

c 

d  122

5−6

17  a  26

b  2−6

c 

36

e  93

f  

9−3

Check your progress 1  a 

rational

b  irrational

c 

rational

d  irrational

26

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

b  8−2

4  a  8.6 × 1010

d  2−6

5  C, D, A, B

6  a  3−4 or 9−2 or 81−1

d  5−6

e  rational

b  2−2

5  a  64−1 c 

11−10

d  3−6

Exercise 1.3 d 

 

c  5−2

14  a  i  22   iii  51 or 5 c 

Reflection: You can compare them easily. You

1  a 

b  52

13  a  x = 4

b  8.02 × 10−5

 

9

v  3−3

b  6.17 × 105

14  a  i 

400

1

d  Learner’s own answers.

12  a  6.1 × 106

c 

1

ii  39 iv  36

ii  3−1

3

9

1 216

6  a 

1 49

 

b  6.45 × 10−6

b 

1

 

81

c 

1 128

b  The three ways in part a.

2

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7  a  53

b  50

8  a  65 c 

c 

5−2

 

b  12−5

4−6

d  152

Unit 2 Getting started 1 

x

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.

 + 7

3

2  a  32 × 34 = 36  5

b 

5  a  x = 1 and y = 14, x = 2 and y = 11, x = 3 and  y = 6

512  = 53 59

b  Learner’s own answer. For example: x = −4 and y = −1, x = −5 and y = −10,  x = −6 and y = −21

c  (7 )  = 7 2

10

3  a  x2 + 2x

b  12 y2 − 21 yw

4  a  4(x + 3)

b  2x(2x + 7)

5  a 

17 5  or 1   12 12

6  a  F  =  = 25 c 

b 

6  or 5

b 

F  a = m

1

c 

1 5

Learner’s own answer. For example: x = −1 and y = 14, x = −2 −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( m + 2 p ) = 4( 2 + 2 × −4 )

= 4( 2 − 8) = 4 × −6 = − 24

a = 6 3

Exercise 2.1 1  a  x − 2 y = 3 − 2 × 5

= −64 + 24 = −40

= 3 − 10 = −7

5

c 

b  x3 + xy = 33 + 3 × 5

= 27 + 15 = 42 c   y2 −

10 x  y

= (5)2  − = 25 −

10 × 3

b  36

c 

16

30

d  64

e  68

f  

−18

5

g  14

h  −25

i 

−7

 j 

= 19 2  a  9

b  4

c 

9

d  8

e  8

f  

30

g  5

h  47

i 

−30

Activity 2.1 Learner’s own answer.

8  Learner’s own counter-exa counter-examples. mples. 3x2 = 3 × 22 = 3 × 4 = 12, and (3x)2 = (3 × 2)2 = 62 = 36, and 12 ≠ 36

3  a  Learner’s own answers. For example: a = 3, b = 10, c = 12, d  = 2

b  For example: When y = 2, (− y)4 = (−2)4 = 16 and − y4 = −24 = −16, and 16 ≠ −16

ii  a = −3, b = −10, c = −12, d  = −2 iii  a = 3, b = 4, c = −36, d  = 3 b  Learner’s own answers. c 

82

a  For example: When x = 2,

−4

i 

5

  p     + ( p)3  =  − 4    + ( −4)3  m     2       = ( −2)5 − 64 = −32 − 64 = −96

7  a  21

5

= 25 − 6

 j 

3

b   p − 3mp = ( −4 ) − 3 × 2 × −4

Learner’s own answers.

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

3

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2

10  5a 2 − 9( b − a ) +

b

5

+ 7ab = 5 × ( −2)2 − 9( −1 − −2) + 2 ( −1)

5

+ 7 × −2 × −1 2

= 5 × 4 − 9 ×1+

−1 = 20 − 9 − 2 + 14

+ 14

b

− 6a3 − (ab )4 +

9 b

2

−a

3

=

−5 × −2 − 6( −2 )3 −1

10

−1

+

( −1)

Reflection: Reflection:   Learner’s own answers.

 

Perimeter = 2 × length + 2 × width = 2 × 8 + 2 × 6 = 28

 

Area = length × width = 8 × 6 = 48

2

9 9

Area = 2x2 + 10x = 2 × 32 + 10 × 3 = 18 + 30 = 48

5  a  i 

− ( −2 )3 9 1+ 8

P = 2x + 10

 

ii  A = 3x + 6

 

iii  When x = 4, P = 18 and A = 18

b  i 

P = 2 y − 4

 

ii  A = 4 y − 24

 

iii  When y = 10, P = 16 and A = 16

c 

i 

 

ii  A = n2 + 4n

P = 4n + 8

 

iii  When n = 6, P = 32 and A = 60 d  i  P = 2 p2 + 8 p

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

a

7x

b

20 − x

c

2x + 9 

d 

x  − 4 6

e

x2 

f  

100 x

g

5(x − 7)

h 

x

i 

x3

 j 

3

l 

(2x)3 − 100 or 8x3 − 100

a

i

2x + 2 y

ii

xy

b

i

6x + 2 y

ii

3xy

c

i

6x + 4 y

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.

 

ii  A = 4 p3

 

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;

 

ii 

both = 12x + 12 9 red + 3 yellow = 18 blue; both = 18x + 18

 

iii  12 red + 4 yellow = 24 blue;

x

k  (3x)2 + 7 or 9x2 + 7

4

Width of rectangle = 2x = 2 × 3 = 6

e  Learner’s own answer.

− 6 × −8 − ( 2 ) 4 +

= 22 + 1 = 23

3

− 9

= −10 + 48 − 16 +

2

 

 

( −2 × −1)4

=

Length of rectangle = x + 5 = 3 + 5 = 8

d  Perimeter = 6x + 10 = 6 × 3 + 10 = 28

= 23 −5a

c 

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

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8  a  3a2 − 7b = 61, 8b – 3a = 31, a2 + 6b = 37,

b  n12

c   p7

d  q5

e  r3

f  

t5

b  133

g  x21

h   y10

i 

z12

c 

 j 

k  5 g 2

l 

−h9

4(a + 3b) = 4 3a2 − 7b + 8b − 3a + a2 + 6b + 4(a + 3b) = 4a2 + 7b − 3a + 4a + 12b = 4a2 + a + 19b

2  a  m14

3  a  Sofia is correct. x2 ÷ x2 = x2−2 = x0 = 1

d  133 e  11 f   Not valid because although the perimeter is positive, positive, three of the side lengths are negative, which is not possible.

9  a  2(3x2 + 4) + 2(5 − x2) or

own answer. b  Learner’s c  x2 ÷ x2 = 1

d  All the answers are 1. Learner’s own explanations. For example:  

When simplified simplified,, all the expressions have an index of 0, and anything to the power of 0 = 1.

 

or Any expression divided by itself, always gives an answer of 1.

3x2 + 4 + 3x2 + 4 + 5 − x2 + 5 − x2

b  2(3x2 + 4) + 2(5 − x2) = 6x2 + 8 + 10 − 2x2 = 4x2 + 18 = 2(2x2 + 9)  

c   

or 3x2 + 4 + 3x2 + 4 + 5 − x2 + 5 − x2 = 4x2 + 18 = 2(2x2 + 9) Arun is correct. Learner’s own explanation.

 

4  a  6x5

b  12 y9

c 

30z7

d  4m7

e  4n13

f  

8 p3

5  a  Learner’s own answer.

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.

 

b  Learner’s own answer. c 

Learner’s own answer.

 

Sasha’s method would be easiest to use to Sasha’s simplify these expressions:

 

4x5 ÷ 6x3 = 3 4 x3

2

or: 2(2(−2)  + 9) = 2(2 × 4 + 9) = 2(8 + 9) = 34 2

and 2(2(2)  + 9) = 2(2 × 4 + 9) = 2(8 + 9) = 34 Perimeter = 4 × 5 = 20 cm Perimeter = 4 × 7 = 28 cm Perimeter = 4 × x  or 4 x

11  a  Volum olumee = x3

b   y × y = y 2

d  w ÷ w = w

8− 6

5

( g )   = g  ×

3 2

f  

=  g 6

5 −1

12 y7 8 y6

= 32 y  and

6 z9

= z6 .

2

36 z 4

5

c 

3t6

d  2u5

e  2v4

f  

5w

2 5 3

b  A 2 y6 5

C k  

d  B 3

1 3

5 12

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.

(h )   = h ×

5 12

= h60

g  5m3 + 3m3 = 8m3

5

,

b  3r4

2+ 4

= w4

= u2 2

4

=  y6

= x9

3

2

8  a  Arun is correct. Learner’s own

1  a  x 4 × x5 = x 4 + 5

e 

3

= 23x

6  a  3q4

c 

Exercise 2.3

c  u ÷ u = u

6x

7  a  D 1 x3 

b  Side length = 3  y

6

5

6z9 ÷ 36z4 = 6

b  Side length = 49 = 7 cm,

8

2

12 y7 ÷ 8 y6 =

10  a  Side length = 25 = 5 cm,

c 

5t7

h  8n2 − n2 = 7n2

b  i   

16x10

 

ii  125 y12

iii  16z28

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

Activity 2.3

4  a   y2 + 6 y + 8

a  Learner’s own spider diagram.

c 

b  There are many possible expression expressions. s.

e   p2 − 11 p + 30

For example:  

2

4

4x  × 9x 36x14 ÷ x2

 

72x20 ÷ 2x8

 

(6x6)2

 

36(x3)4

c 

Learner’s own answers.

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 

1

b  r−2 = 12 r

1 t

d  v−1 = 1

t−5 = 5  

v

10  a  A and iii, B and iv, C and i, D and vii, E and vi, F and v.

(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 c 

B y  y2 − 2 y − 48

that simplifies to give For example:

d  A z2 − 9z + 20

= x2 + 2x + 2x + 4

1 . 6 y7

= x2 + 4x + 4

b  (x − 3)2 = (x − 3)(x − 3)

5 y2 30 y9

= x2 − 3x − 3x + 9

Reflection:   Learner’s own answers. Reflection:

Exercise 2.4

= x2 − 6x + 9

8  a  i   

ii  z2 + 2z + 1

 

iii  m2 + 16m + 64

 

iv  a2 − 4a + 4

2

= x  + 1x + 4x + 4

 y2 + 10 y + 25

 

= x2 + 5x + 4

b  (x − 3)(x + 6)

= x2 + 6x − 3x − 18

 

v   p2 − 8 p + 16

 

= x2 + 3x − 18

 

vi  n2 − 18n + 81

(x + 2)(x − 8)

b  A x2 + 2x − 35

7  a  (x + 2)2 = (x + 2)(x + 2)

b  Learner’s own answer. Any expression

c 

n2 − 30n + 200

b  The plus at the end would change to

9  a  q−3 = 3   q

1  a  (x + 4)(x + 1)

f  

to a minus and the 9 changes to a 1. x2 + 1x − 20

3x  × 12x

   

c 

d  a2 − 7a − 18

5  a  The plus at the end would change

10

8

m2 + m − 12

b  z2 + 14z + 48

= x2 − 8x + 2x − 16

b  (x + a)2 = x2 + 2ax + a2 9  a  (x + 3)(x − 3) = x2 + 3x − 3x − 9 = x2 − 9

 

= x 2 − 6x − 16

d  (x − 4)(x − 1)

= x2 − x − 4x + 4

 

ii  x2 − 25

 

= x2 − 5x + 4

 

iii  x2 – 49

c 

There is no term in x, and the number term is a square number.

2  a  x2 + 10x + 21 c 

x2 + 2x − 15

e  x  − 9x + 14 2

b  x2 + 11x + 10 d  x2 + 4x − 32 f  

2

x  − 14x + 24

b  i 

x2 − 4

d  x2 − 100 e  x2 − a2

3  a  Learner’s own answers and explanations. b  Learner’s own answers and explanations. c 

6

Learner’s own answer.

Activity 2.4 a 

① 33 × 29 = 957, ② 28 × 34 = 952, ③ 957 − 952 = 5

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

b 

① 16 × 12 = 192, ② 11 × 17 = 187,

5  a 

③ 192 − 187 = 5

c 

The answer is always 5.

d 

e 

n

n + 1

n + 5

n + 6

1+ 2 2

c 

5 1  ≠ 1 6 2

 

③ n2 + 6n + 5 − (n2 + 6n) = 2

because the expression is 3 x + y, all divided by 6, not just 3 x divided by 6. x y 3x y 3 x + y + = + = 2

6

6

6

6

n  + 6n + 5 − n  − 6n = 5

e  Learner’s own answer.

 

The answer is always 5.

f  

i 

 

Learner’s own answer.

 

ii  incorrect. Learners should show that

1  a 

2x   5

c 

8   x

e 

2x   5

b 

4x 7

f  

b 

2 5 y

− 251y = 2510y − 251y = 259y

c 

3 y   4

d 

3 y 8

e 

11   9 y

f  

3 y 14

a 2

5a 2 a + a5 = 10 + 10  

=

5 7c

b 

b 4

d 

e 

−

2e 3

= =

21e 16e   24 24 21e − 16e 24 5e 24

= = 4  a  A, D, F c 

f  

x 3

v 

3ab + 40   10 b

iv 

ab + 12 4b

vi 

8ab + 27 18b

5d 6

9 10 f

= 182+ 2 = 202  = 10

b  3 × 3 + 1 = 9 + 1 = 10

7b = 12

ownthe explanation. or example: d  Learner’s He factorises bracket to F give

3 4f

7d  30

10 = 10

2 × bracket, which is then divided by 2. The × 2 and ÷ 2 cancel each other out, leaving just the bracket.

− 35d = 2350d  − 1380d 

−

6×3+2 2

c 

e  When x = 3, 6 × 3 + 1 = 18 + 1 = 19, 19 ≠ 10, so the answer is wrong.  

= 1820− f 15

Learner’s own explanation. For example: The expression shows that 6x + 2 must all be divided by 2.

 

= 203 f 

Arun has only divided the 2 in the numerator by 2, and not the 6x by 2 as well.

f  

Learner’s own answer.

=

b  B, C, E

G; the answe answerr is

 

ii 

= 3b12+ 4b

= −

iii 

7  a 

= 25d30− 18d 

39 35c

 

2a + 9   15

5a + 9b 12

Learner’s own answers.

3b 4b + b3 = 12 + 12

5a + 2a 10

+ 52c = 3255c + 3145c  

a+b 5  

9x − 8 20

Activity 2.5

= 2535+c14

7e 8

iv  incorrect. Learners should show that

b  Learner’s own checks.

7a = 10

c 

 

6  a  i 

4y 3y 7y + 310y = 10 + 10 = 10

3  a 

iii  correct

d  x 4 x

4x − y 10

 

the correct answer is

2 y 5

2  a 

correct

the correct answer is

Exercise 2.5

7

= 32 = 1 12

d  She cannot cancel the 3 with the 6,

① (n + 5)(n + 1) = n2 + 6n + 5,

2

+ 26 = 36 + 26 = 56

b 

② n(n + 6) = n2 + 6n,

 

1 2

18 20 f

−

15 20 f 

8  a  2x + 1 c 

2x − 3

b  x + 2 d  2x − 5

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

9 

6x − 4 2

+ 20x5+ 25 = 2(3x2− 2 ) + 5( 4x5+ 5) =

 

3x − 2 + 4x + 5 = 7x + 3

4  a  Ben’ Ben’ss age is x + 2, Alice’s age is x − 6 b  T  = 3x − 4 d  x =

10  a  2(x + 3) = 2 × x + 2 × 3 = 2x + 6 b  Learner’s own choice and explanation. c 

i 

   

ii  2(x + 2) or 2 x + 4 iii  4(x − 3) or 4 x – 12

 

iv  3(1 − 3x) or 3 − 9x

1  a  S  = 60M 

b  S  =  = 900

95.9 kg (1 d.p.)  y − z 2 2( y + 3h ) 5

 

ii  C x =

F  = 60

ii  F  =  = −78

 

iii  A x = 7k ( y  y − 6)

 

iv  C x = 3ny + m

 

w−y A x = 7

F  m

a = , a = −1.75

v 

b  Learner’s own answer.

3  a 

9  a  t =

3D Shape

N u mb e r of faces

Number Number of of vertices edges

Cube

6

8

12

Cuboid

6

8

12

5

6

9

Triangular prism Triangularbased pyramid Square-based pyramid

V  = E  − F  + 2

 

i 

c 

m−9   7

t = pv − h 

4 5

4 5

6 8

ii  V  = 7

V  = 6

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 five edges and seven vertices. Learner’s own answer.

b  t = 5(k  + m) d  t =

9q + w 5

10  a  A = a2 + bc b  A = 49.5 c 

A = a2 + bc, A − bc = a2,

=

A − bc

d  a = 8 11  a  78.5 cm

b  E  = F  + V  −  − 2, or any equivalent version c 

d  57.3 kg (1 d.p d.p.) .)

B x =

d  M  =  = 22.5

F 

f  

b  49.1 kg (1 d.p d.p.) .)

S    60

b  m = a , m = 12

a = 2

b  60%

7  a  65 kg

8  a  i 

c 

f  

125%

c 

2  a  i 

d  u = 46

u = 27

Exercise 2.6 M  =

b  v = 125

6  a  20% c 

T  = 53

e  x = 22

e  t = 10

Reflection:   Learner’s own answers. Reflection:

c 

T   + +4   3

5  a  v = 87 c 

2(x + 3) or 2 x + 6

c 

c 

b  r =

A π

6.25 cm

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 27.8 °C and 27.8 °C < 30 °C).

14  a  She is not underweight as her BMI is 20.05, which is greater than 18.5.

b  3.7 kg

Check your progress 1  a  39 c 

b  161

12

2  perimeter = 16x + 8, area = 5x(3x + 4) = 15x2 + 20x

8

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

3  a  x5 d  15m9

b  q6

c 

h10

c 

e  2u2

f  

3 p2

d  320 ÷ 100 = 320 ÷ 1 = 320

4  a  x2 + 7x + 10

b  x2 + x − 12

320 ÷ 101 = 320 ÷ 10 = 32

6  a  2.7

b  0.45

d  x2 – 14x + 40

c 

d  0.017

e  x2 − 64

f  

x2 − 12x + 36

e  0.08

2x   3

b 

2 y 15

12 x − y   20

d  3x − 5

c 

5  a  c 

x2 − 3x − 54

6  a  x = 31

b  z =

0.36

f  

0.0248

g  9 h  0.0025 7  a  Learner’s own answer. b  i 

x−y 5

2

, z = 6

c   y = ± x − 5z , y = ±6

6.8 ÷ 10−3 = 6800

 

ii  0.07 ÷ 10−4 = 700

c 

Learner’s own answer.

d  Learner’s own answer. For example: An

Unit 3 Getting started 1  a  8  

e  90

b 

32.5

c 

f  

625

g  700

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

d  0.85

6

h  32

e  Learner’s own answer. 8  a  3.2 ÷ 103 = 3.2 ÷ 1000 = 0.0032

2  B 3  a  15.4

b  640

b  3.2 ÷ 102 = 3.2 ÷ 100 = 0.032

4  a  $345

b  $240

c  3.2 ÷ 10  = 3.2 ÷ 10 = 0.32 d  3.2 ÷ 100 = 3.2 ÷ 1 = 3.2

1

5  63.6 cm2 (3 s.f.)

e  3.2 ÷ 10−1 = 3.2 × 10 = 32 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

3.2 ÷ 10−2 = 3.2 × 100 = 320

g  3.2 ÷ 10−3 = 3.2 × 1000 = 3200 h  3.2 ÷ 10−4 = 3.2 × 10 000 = 32 000 9  a  Yes. Learner’s own explanation.

2  a  3.2 × 103 = 3.2 × 1000 = 3200 b  3.2 × 102 = 3.2 × 100 = 320

b  i 

c 

 

3.2 × 101 = 3.2 × 10 = 32

iii  smaller

10  a  2.5

d  3.2 × 100 = 3.2 × 1 = 3.2

c 

e  3.2 × 10−1 = 3.2 ÷ 10 = 0.32

ii  the same

greater gr

b  47 600 d  8.5

70

−2

f   3.2 × 10−3 = 3.2 ÷ 100 = 0.032 g  3.2 × 10  = 3.2 ÷ 1000 = 0.0032

anyone the secret! 11  Do not tell anyone 12  a  i  40 400 ii  40

h  3.2 × 10−4 = 3.2 ÷ 10 000 = 0.000 32 3  a  Yes. Learner’s own explanation. b  i   

ii  the same

smaller sm

iii  greater

 

iii  4

iv  0.4

 

v  0. 0.04

vi  0.004

b  Smaller c  Smaller

b  7800

c 

240

d  i 

d  85 500

e  65

f  

8000

 

iii  12

iv  120

g  17

h  0.8

i 

0.085

 

v  12 1200

vi  12 000

 j 

k  0.032

l 

1.25

e  Larger

4  a  1300

0.45

5  a  320 ÷ 103 = 320 ÷ 1000 = 0.32

f  

0.12 0.

ii  1.2

Larger

g  Learner’s own answer.

2

b  320 ÷ 10  = 320 ÷ 100 = 3.2

9

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

13  a  0.8 × 10

1

8 ÷ 10

c 

i 

60

ii  30

iii  20

 

iv  15

v  12

vi  10

0

d  i 

ii 

Smaller Sm

Larger

e  Learner’s own answer. 80 × 10 –1

0.08 ÷ 10 –2

=8

8  a  False

b  True

0.008 × 103

800 ÷ 102

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 d ecimal place. Answer = 50.

32 ÷ 102

0.32 × 100

10  a  200

b 

c  3.2 ÷ 101

320 ÷ 103

= 0.32

b  120 d  40

300

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

3.2 × 10 –1

32 × 10 –2

gives an answer of 0.024. For For example: 0.03 × 400 × 0.002

Activity 3.1

c 

Learner’s own answers.

Learner’s own answer.

12  Learner’s own answers and discuss discussions. ions. Reflection:   Learner’s own answers. Reflection:

 

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

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

13  a  123 × 57 = 7011

2  a  0.08 × 0.2

8 × 2 = 16

b  i 

 

0.08 × 0.2 = 0.016

 

8 × 0.2 = 1.6

b  0.4 × 0.007

4 × 7 = 28

 

0.4 × 0.007 = 0.0028

4 × 0.007 = 0.028

B, G, L (1.5); E (15)

b  −50

c 

d  600  

0.81 8 1 × 100 0.0 09 9 × 100

  = 819 = 9

10

vi 

0.070 11

Estimate: 4 × 30 = 120 Accurate:: 119.625 Accurate

0.01

Accurate:: 19 200 Accurate

15  a  0.2 ÷ 0.4 = 0.5 m b  0.45 m

6.4 × 1000 0.00 0 04 × 1000

  = 6400 = 1600 4 b  B

c 

c  C

d  D

0.8 0.

ii 

2.4 2.

iii  4

iv  5. 5.6

v 

7.2 7.

vi  8.8

Larger La

7.011

Accurate: 62 60 × 4 iii  Estimate:  = 24 000

h  −300

b  i 

v 

ii  Estimate: 10 ÷ 0.2 = 50

g  200

 

iv  7.011

   

7  a  i 

70.11

i 

−400

6  a  D

iii 

c 

f  

b 

701.1

b  Learner’s own answer.

e  40 5  a   

−30

ii 

14  a  Learner’s own answer.

3  C, D, I, K (0.015) (0.015);; A, F, H, J (0.15); 4  a  20

701.1

ii  Smaller

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 

240 × 0.95 = $228

200 × 1.2 = $240

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

2  a  Learner’ Learner’ss choice of who they think is correct, with reason.

c 

d  Five years. 10 000 × 0.94 = 6561,

b  Sofia is correct.  

10 000 × 0.95 = 5904.9

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 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.

e  10 000 × 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 

 

ii  265

4  a–e Learner’s own answers. a nswers.

 

iii  274

5  a  i 

195

ii  64.4

d  i 

b  i 

630

ii  108.864

d  Learner’s own answer. 3  a  i 

ii  57.6

57.6

b  = c 

i 

ii  =

=

6  a  1.1235

b  $67.41

7  a  i 

72

ii 

285

ii  48.412

b  i 

8  a  0.7216

52.8

b  $4618.24

9  a  A and iii, B and iv, C and i, E and ii, F and v

10  a  Zara is correct. 1.04 × 1.04 2

is the same as (1.04) , so 5000 × 1.04 × 1.04 = 5000 × (1.04)2 b  5000 × (1.04)3

c 

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 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.

5000 × (1.04)4

d  8. The power on the 1.04 is the number of years.

e  i 

265, 266, 267, 268, 269, 270, 271, 272, 273, 274

 

3  a  i 

b  D and 0.81

5  a–c Learner’s own answers and discussions. 6  a  3.5 ⩽ x < 4.5

12

5000 × (1.04)

b  11.5 ⩽ x < 12.5

 

ii  5000 × (1.04)

c 

 

iii  5000 × (1.04)

d  669.5 ⩽ x < 670.5

f  

15 years

20 n

11  a  i 

355.5 ⩽ x < 356.5

7  a  15 ⩽ x < 25

10 000 × 0.9

 

ii  10 000 × 0.9

 

iii  10 000 × 0.93

c 

4745 ⩽ x < 4755

b  335 ⩽ x < 345 d  6295 ⩽ x < 6305

2

b  The population after 5 years years..

11

The population after 10 years.

8  a  250 ⩽ x < 350 c 

4650 ⩽ x < 4750

b  1850 ⩽ x < 1950 d  7950 ⩽ x < 8050

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

9  Learner’s own answers and discuss discussions. ions. a  i 

ii  5

0.5 0.

iii  50

b  The lower and upper bounds of a

Exercise 4.1 1  a 

rounded number number will always be +/− half of the degree of accuracy.

10  a  i 

b  1555 cm  x < 1565 cm ii  172.5 cm

171.5 cm

c 

b  171.5 cm ⩽ x < 172.5 cm

2 y 3 2 y 3

12  A, i and e; B, i and f; C, ii and b; D, iii and a;

8

 x =

= 11 + 5

1  a  74 500

 y =

b  12

e  0.0728

f  

g  37

h  18

= 24

5

b  3.6 d  600

e  300

f  

c   y = 4

d   y = 8

e  a = −6

f  

g  7.5

h  0.11

c 

3 2 3

a = −1 z = 4

h

Learner’s own answers.

ii  20 000 × (1.08)

 

iii  20 000 × (1.08)3

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.

b  The value of the painting after 5 years years.. The value of the painting after 20 years years..

d  6 years years.. 20 000 × (1.08)5 = 29 386.561 54, 20 000 × (1.08)6 = 31 737.486 46

e  20 000 × (1.08)n 7150 m2

 

 

When he brought the −3x to the left-hand side he forgot to make it +3x.

 

When he brought +8 to the right-hand

c 

side he forgot to make it −8. 2 x + 16 = 18 − 3x

ii  7250 m2

5 x

b  7150 m2 ⩽ x < 7250 7250 m2

1  a  x = 5

b  x = 9

c   y = 25 2  a  5 3  a  2x > 10 c   y + 5 ⩾ 13

 

d   y = 25 b  7

+ 16 = 18 5 x = 2  x =

Unit 4 Getting started

12

5

a  Substitute x = 26 back into the original

20 000 × 1.08 2

5  a  i 

 y =

9

example:

 

c 

15

4  Learner’s own answers and explanations. For

3  $265.20 4  a  i 

 

g 3  a, b x = 15

9

5

b  x = −3

  x = 2 −1.6 −0.0028

 y =

 y = 1

2  a  x = −11

d  59

0.046

2

3

9 y = 15

= 16

48

  −6 = −10

d  6 y + 3 y = 22 − 7

2 y = 48

Check your progress

2  a  c 

− 10 x = 9 −10 x = 9 − 15 −10 x = − 6

2 y = 16 × 3

E, ii and c; F, iii and d

c 

15

 x = − 2

⩽

11  a  i 

b 

−16

 x =

ii  1565 cm

1555 cm

= −30 + 14 8 x = − 16

8 x

c 

5, 6, 7

b  4x < 36 d   y − 5 ⩽ −11

2 5

= 0.4

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.

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

6  a 

42

=7   42 = 7c 42 =c 7 c=6

b 

c

c 

21

= 15 12 = 15d  d 

12 15

i, ii and iii Learner’s answers and discussions. a  10x − 8 = 5x + 12, x = 4

= d 

b  12(x − 5) = 4( x + 1), x = 8

12

4

15

5

d  =   =

 

Activity 4.1

12

c 

  =7

e+ 2

5x − 4 = 2x + 20, x = 8

 

5

d e 

9

7

14  a 

= e+2

c 

3−2 = e 3

7  a  a = 27 b  b = 7   c  c = 3

54

 

i 

 

ii  The two shorter sides of a rectangle

d  d  = 11

explanations. 3

1

ii  x = 6   5

x =14   4

1

iii  x = − 5 ii  A − 6

A + 10

 

iii  There are x sweets in bag A. There are five fewer sweets in bag B than bag A. The sweets in bag B are shared between 180 people. Each person gets 15 sweets. How many sweets are in bag A?

A = 22

10  a  2(x + 3) + 7x − 5 + 5(7 − x) = 48 OR 4x + 36 = 48

b  i 

b  x = 3 c 

A quadrilateral has sides of length x cm, 2(x + 1) cm, cm, 3(x + 2) cm, cm, and 4(x + 3) cm. The perimeter is 80 cm. Work out the value of x. have side lengths of 6(3a − 4) and 3(4a − 3). Work Work out the value of a.

b  A + 10 = 2(A − 6) c 

=   x270 −4

54 °, 54 °, 72 °

8  a, b, c and e Learner’s own answers and

9  a  i 

= =  126 , x = 7 2 x

15  a  Learner’s own problem. For example:

e =1

d  i 

, x = 8

b  x = 9

3= e+2

 

75

 x + 7

21 = 7 ( e + 2) 21

 

12 cm, 16 cm, 20 cm

x = 6

 

ii  a = 2.5

 

iii  x = 17

11  a  9a = 4a + 20 b  a = 4

Exercise 4.2

c 

1 

Triangle sides 12 cm, rectangle sides 7 cm and 11 cm

1  Work out x.

12  a  B and D b  A  xx =

 1 15

; B x = 15; C x = 8640;

D x = 15; E  xx =

 

13  a  c 

 x =

 1 15

 y 85

 y

=5

b 

3

=6

2  Work out y.  y =

−3 = 5×6−3 = 30 − 3 = 27

 

3  Check values are correct.  y =

 

4  Write the answers: x = 6 and y = 27

152

  =8

 y + 2

= 5 → y = 855 = 17 and

18

 

There are 15 sectors in the pie chart. 85

− 3 = 2x + 15 5 x − 2 x = 15 + 3 3 x = 18 5 x

 5x

+ 15 = 2 × 6 + 15 = 12 + 15 = 27

152

  = 8 → 152 =  y + 2 → 19 = y + 2 → y = 17 8

 y + 2

d  Learner’s own answer.

 2x

2  x = 5, y = 9

13

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3  x = 4, y = 13

11  a

4  x = 7, y = −5 5  a 

1 Add the two equations.

 y = 3x + 1

0 3 6 1 10 19

  x  y

 y x

 

 =  + 9

 

b 

 2x +  y = 50

0 9

 

3 x + 0 y = 54

 

3 6 12 15

3x = 54,  x =

 y = 3x + 1

 y = x + 9

14

10

6

0

1

2

3

 

4

5

6

x + 4 × 9 = 41   18 2

=9

3

(4, 13)

solution of simultaneous equations is the point of intersection of the straight-line graphs.

1 Subtract the two equations.

x = 5 and y = 9

2

Substitute y = 4 into first equation

3x + 2 y = 38

3x + 2 × 4 = 38

− 3 x −  y = 26

x = 2, y = 6

3x = 38 − 8

0 x + 3 y = 12 3 y = 12,  y =

Learner’s own answers and explanations.

3x = 30,  x = 12 3

=4

3

4

ii  x = 6, y = 2

9  a  x = 5, y = 2 x = 7, y = 4

3

= 10

Check in second equation

x = 10 and y = 4

12  a  Learner’s own answer.

b  Learner’s own answers. ii  x = 10, y = 8

30

3 × 10 − 4 = 26

x = 2, y = 7

x = 9, y = 4

Check in second equation

c

b  x = 2, y = 6

8  a  i 

=5

4

ii  x = 2, y = 6

7  a  i 

x = 41 − 36

5 + 2 × 9 = 23

e  Learner’s own answer. For example: The

 

i 

b  x = 16, y = 3

 

ii  Subtract to elimina eliminate te the xs.

d  x = 3, y = 6

 

iii  Add to eliminate the ys.

 

iv  Subtract to elimina eliminate te the ys.

b  i   

x = 2, y = 3

ii  x = 4, y = 8

10  Sofia is correct, x = −3 −3 and y = 6. Zara got the signs round the wrong way.

14

Substitute y = 9 into first equation

x

equations; x = 4 and y = 13

c 

2

0 x + 2 y = 18 2 y = 18,  y =

d  The coordinate coordinatess give the solution of the

 

x = 18 and y = 14

 

2

6  a  i 

4

−   x + 2 y = 23

4

 

Check in second equation

3

  x + 4 y = 41

8

0

= 14

b 1 Subtract the two equations.

12

c 

3

= 18

y = 50 − 36

18 − 14 = 4

16

 

54

 y

20 18

c 

2 × 18 + y = 50

+   x −  y = 4

x  y

Substitute x = 18 into first equation

2

You can add or subtract. If you you add, you eliminate the ys, if you subtract you eliminate the xs.

b  Learner’s own answer.

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c 

Learner’s own answer. For example: Subtract to eliminate one of the letters when the coefficients coefficients of that letter are the same number and both positive or both negative. negativ e. Add to eliminate one of the letters when the coefficients coefficients of that letter are the same number and one positive and one negative. x = 9, y = 6

ii  x = −3, y = 2

iii  x = 8, y = 3

iv  x = 9, y = 5

d  i   

5  a 

2

3

4

 –5

 –4

 –3

 –2

 –1

0

 –4

 –3

 –2

 –1

0

Exercise 4.3 1  a  x ⩽ 2

b  x > −2 d  x < −20

x ⩾ 10

e  −2 ⩽ x < 2 2  a 

f  

1

2

–3

3

4

–1

0

–2

–1

 

3x + 6 ⩽ 2x − 5

 

3x − 2x ⩽ −5 − 6

1

d  –5

0

5

6

x = −12

 

3(−12 + 2) ⩽ 2 × −12 − 5

 

−30 ⩽ −29

 

True

2

3

4

3(−11 + 2) ⩽ 2 × −11 − 5

 

−27 ⩽ −27

 

True

8

9

iii  x = −10

 

3(−10 + 2) ⩽ 2 × −10 − 5

 

−24 ⩽ −25

 

False

 

7

ii  x = −11

 

e 8  a 

⩽

For    −11 values that xare true the andsubstitutions when x > −11give it gives a false value.

 –3

 –2

 –1

0

1

3  a  7 c 

15

3

4

b  −4

−2, −1, 0 or 1

4  a  x > 2 c 

2

x < −3

b  x ⩽ 4 d  x ⩾ −3

5

+ 3) − 5 y < 18 − y 8 y + 12 − 5 y < 18 − y 8 y − 5 y + y < 18 − 12 4 y < 6  y < 1.5

(

4 2 y

f   –4

5

x ⩽ −11

b  i 

 

0

c 

 –20  –15  –10

3(x + 2) ⩽ 2x − 5

 

−10 < x ⩽ 15

b   –  4

 

 

Reflection:   Learner’s own answers. Reflection:

1

1

incorrectly.

d  x = 7, y = 1

4 × 2 + 2 × 2 = 8 + 4 = 12

0

0

7  a  He has multiplied out the bracket

b  3 × 2 + 2 = 6 + 2 = 8 and

 –1

 –1

b, c Learner’s own answers.

14  a  x = 2, y = 2

 –2

4

d 

b  x = 5, y = −2

x = 2, y = 4

 –5

3

6  a  x < 3

13  a  x = 9, y = 4

0

2

c 

All answers should be x = 6, y = 18

c 

1

b 

Activity 4.2

c 

0

6

b  i 

 y = 1

 

4(2 × 1 + 3) − 5 × 1 < 18 − 1

 

15 < 17 True

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ii   y = 1.5

 

4(2 × 1.5 + 3) − 5 × 1.5 < 18 − 1.5

 

16.5 < 16.5

 

False

 

iii   y = 2

When x = 5, 3 × 5 − 7 < 4 × 5 − 11

8 −27

10  a  5n + 5 ⩽ 30 c 

c 

c 

b  n ⩽ 5

3 < n < 9 2

 

5, 12 and 13

5

3

4

5

6

7

8

9

10

11

7

8

d  −3 < m < 6

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 final inequality with the x on the left and so he has to change the < to >. > . To make the x positive, Natalia divides both sides by −1, but this has the effect of changing the < to >.

b  Learner’s own answers.

 –4  –3  –2  –1

1  a  x = −4 d   y = 9

2(x − 8) ⩾ 4x − 26

3  x = 19, y = 7

 

2x − 16 ⩾ 4x − 26

4  a  a < 2

 

2x − 4x ⩾ −26 + 16 10 ⩾ 2x

 

5 ⩾ x

 

x ⩽ 5

3

4

5

6

b  a = −2.5

c 

x = 2.4

e  m = 16

f  

n = 10

2  x = 5, y = 19

 

 

2

Learner’s own checks for each solution.

Learner’s own answer. For example:

− 2x ⩾ −10

1

Check your progress

c 

 

0

c   

b  b ⩾ 5 d  d  ⩾ −5

c > −1

Learner’s own checks for each solutio solution. n.

5  a  −1 < x ⩽ 2  

12  a  x > −4 or −4 < x

 –2

 –1

0

1

2

3

4

–2

–1

0

1

b  −4 < n < 1

b  x ⩾ 5 or 5 ⩽ x c 

 –5

x > 6 or 6 < x

d  x ⩽ −  −13 13 or −13 ⩾ x

 –3

e  x < 4 or 4 > x

Unit 5 Getting started

f  

1  140 °

x ⩾ −  −2 2 or −2 ⩽ x

13  a  3x − 7 < 4x − 11

b  For example:

− 7 < 4 x − 11 −7 + 11 < 4 x − 3x 4 <  x  x > 4 3 x

 

16

 –  4

2

2  62 ° 3  a  a and d  OR  OR b and e OR c and f  b  c and d  c 

a and c OR d  and  and f 

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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 ha ve a sum of 180 °.

8  a  Six triangles; 6 × 180 ° = 1080 ° b  Eight triangles; 8 × 180 ° = 1440 ° 9  a

5  a  Learner’s own diagram.

Polygon

b  Each angle should be 37.5 °. c 

triangle quadrilateral pentagon hexagon octagon decagon

Learner’s own check.

Exercise 5.1 1  60 °, 25 °, 95 ° 2  a  x = 36, y = 50 c 

b  122 °

A + B  + C  + D = 116 ° + 72 ° + 122 ° + 50 ° = 360 °

3  a = 40 °, b = 30 °, c = 70 °, d  =  = 120 °

b  A = 60 °, B  =  = 120 °, C  =  = 135 °, D = 45 °  = 40 °, B  = D = 100 °, A = 120 ° 6  C  = 7  a  54 ° (angle of isosceles isosceles triangle AOB )

Sum of interior angles

3 4 5 6 8 10

180 ° 360 ° 540 ° 720 ° 1080 ° 1440 °

b  The sum of the angles = (n − 2) × 180 ° c 

4  75 5  a  Trapezium. One pair of parallel sides.

N u mb e r of sides

7 × 180 ° = 1260 °; correct because there are seven triangles.

10  a  100 °

b  135 °

11  144 ° 12  a, b There are two ways:

b  36 ° (angle BOC  is  is 108 ° and triangle OBC   is isosceles)

c 

90 ° = 54 ° + 36 °

8  x = 65 ° (angles on a straight line);

  The second way could be drawn in a reflected form.

 y = 45 ° = 115 ° (corresponding angles) − 70 ° (alternate angles)

c 

9  105 ° Reflection: Learner’s own answer 10  a  45 ° + 51 ° = 96 ° b  A + B  + C  + D = 96 ° + 65 ° + 127 1 27 ° + 72 ° = 360 °

if it is reflected, but it is still the same arrangement.

Exercise 5.2

13  a  Learner’ Learner’ss own diagram of a regular arrangement of triangles triangles..

1  110 °

b  Learner’ Learner’ss own diagram of a regular

2  40 °

arrangement of hexagons hexagons..

3  136 °

c 

4  a  103 °

b  128 °

5  a  88 °

b  128 °

6  a, b Learner’  Learner’ss own diagram of a hexagon split 4 × 180 ° = 720 °

7  a  109 °

17

d  120 °

Because 108 ° is not a factor of 360 °.

d  Learner’s tessellati tessellations ons based on the two drawings in Question 12.

e  Learner’s own diagram: two octagons (135 ° angle) and one square (90 °) at every point.

into four triangles.

c 

There is no other way way.. Either the two squares are adjacent or they t hey have have one triangle between them on one side and two triangles between them on the other side. This way will look different

f  

Learner’s own answer.

b  100

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Reflection:   In this case Reflection: case,, subtract the 360 ° at the

Reflection:   Yes they do. Check with some values Reflection:

centre. 5 × 180 − 360 = 540 gives the same answer.

for n. It is easier to see if you write ( n − 2) × 180 ÷ n  as (180n − 360) ÷ n

Exercise 5.3 1  a – c Learner’s own diagram and explanation. The explanation is the same as for a pentagon. Walking round theand hexagon youisturn each angle in turn t he total the 360 through °.

2  a = 99 °; b = 112 °; c = 125 ° 3  a  Yes, vertically opposit oppositee 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 °

5  a  360 °

c 

72 °

b  360 ÷ 8 = 45 °

6  a

Re gu l ar po l y go n

Sides

Equilateral triangle Square Regular pentagon Regular hexagon Regular octagon Regular decagon

Exterior angle

3 4 5 6 8 10

120 ° 90 ° 72 ° 60 ° 45 ° 36 °

b  The exterior angle = 360 ÷ n degrees c 

i 

b  i 

Exercise 5.5 1  a  10 cm

b  13 cm

c 

17 cm

2  a  4.3 cm

b  12.1 cm

c 

14.2 cm

3  a  12 cm

b  4.8 m

c 

75 mm

4  a  6.6 cm 5  a  2

b  5.0 cm b  3

c  c 

13.5 m 4 =2

d  Learner’ Learner’ss own diagram. A continua continuation tion of the spiral pattern.

e  The 3rd hypotenuse is 2, the 8th hypotenuse is 3 and the t he 15th hypotenuse is 4.

150 °

ii  160 °

iii  170 °

12

ii  18

iii  36

c 

20

d  24

b  Learner’s own diagram. 12  (360 − 60) ÷ 2 = 150 ° is the interior angle 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 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.

+ 702   = 80 cm to the nearest cm.

1052

3.502

+ 582  = 120 cm to the nearest cm.

− 0.912  = 3.38 m to the nearest cm.

8  a  Learner’s drawing. b  5.12 + 6.82 = 8.52, so it is a right-angled triangle.

c  b  12

392

6  a 

7 

b  140 °

11  a  360 − 2 × 135 = 90

18

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.

b 

9  15 sides 10  a  8

The answers to all the questions in this exerc exercise ise are diagrams.. Each question asks the learner to check diagrams their accuracy either by measuring themselves or by asking a partner to measure.

ii  18 °

30 °

7  a  9 8  a  i 

Exercise 5.4

5.12 + 6.82 = 72.25 = 8.52. The triangle satisfies Pythagoras’ theorem, theorem, and so is right-angled.

9  Either 152 + 20 2 = 25 cm or 202

− 152 = 13.2 cm to 1 d.p.

10  a  90 + 40 = 130 m b  130 − (902 + 402 ) = 31.5  m to 1 d.p. 11  a  Square perimeter perimeter = 4 × 25 = 100 mm, rectangle perimeter = (2 × 20) + (2 × 30) = 40 + 60 = 100 mm

b  Diagonal of square = 35.4 mm; diagonal of rectangle = 36.1 mm

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c 

3  A number is assigned to each person. 50

Learner’s diagram and value.

d  The values so far support Sofia’s conjecture and any further values should too. The square has the minimum diagonal for a given perimeter. All the examples here are for a perimeter of 100 mm, but it is true for any given perimeter.

12  There are two possible answers. Either the two shorter sides are 1 and 4 OR the hypotenuse is 9 and one of the other sides is 8. 2

numbers between 1 and 632 are generate generated. d. Any number that is a repeat is ignored.

Exercise 6.1 These are suggested answers but there are many other possibilities. possibilities. It is not possible to give a complete list of answe answers. rs.

1  Learner’s own answers. a  For example: Can boys estimate more

2

13  a  7.5  + 5.5  = 86.5 and so length of

accurately than girls? Can learners accurately estimate acute angles more accurately than obtuse angles? Can learners accurately estimate how long one minute is?

diagonal = 86.5 .

b  x2 + 5.52 = x2 + 30.25 and so length of diagonal = x 2

c  d =

x2

+

14  a  i 

72

+ 72 =

ii 

98

=

 

b 

x2

+ 30.25 .

y2

+ x2 =

b  For example: Girls can estimate the length

49 × 2 2x2

of a short line more accurately than boys. boys. Older learners can estimate an obtuse angle more accurately than younger learners.. Learners tend to underestima learners underestimate te one minute of time time..

98

=

=

72

×2 =7

2

x 2

Check your progress

c 

1  a = 65. The reason could use corresponding angles and the exterior angle of a triangle.

2  116 ° (x = 106)

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.

3  10 sides

d  Learner’s own answer and explanation explanation..

4  a  Learner’s own diagram.

e  Learner’s own answer.

b  Each side should be 8.5 cm. 5  35 m or 35.3 m or 35.36 m are possible answers. answers. 6  x = 10 10 and y = 24

f  

Learner’s own generalis generalisation, ation, 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

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 answe answers. rs.

1  Learner’s own answers. a  For example: length or width.

school start earlier in the day? b  For example: Learners want longer lessons.. Learners want fewer lessons in a lessons day. Learners would prefer to start school one hour later.

c 

b  For example: number of doors or passenger seats.

c 

For example: colour or manufacturer.

2  Learner’s own answer. For example: Using random numbers of position on the register. It could include a specific number from each year group.

19

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 explanation.. e  Learner’s own answer. f  

Learner’s own generalis generalisation, ation, depending on their data.

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3  Learner’s own answers.  

 

7  a  i 

a, b For example: Questions and predictions

ii  Wha Whatt do you think is the cause of

could be about lengths of words, lengths of sentences,, lengths of articles or vocabulary sentences vocabulary used.

 

c, d, e Learner’s own answers. This will

 

ii  Wha Whatt is a fair price for entry to this

c 

i 

 

ii  The question is too personal. A better

depend on the t he predictions. predictions.

f  

Learner’s own generalis generalisation, ation, depending on their data.

Reflection:   Learner’s own suggestions about Reflection:

global warming?

b  i 

making predictions and choosing a sample to test them.

Exercise 6.2

d  i 

2  a  To encourage people to buy Supremo Shampoo.

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 c 

b  26

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%

exhibition? People will not want to admit they are overweight.

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?

b  For example: Sample choice, asking a question suggesting a particular answe answerr,

People are likely to say ‘yes’.

question would be, for example, ‘Do you weigh less than …’ and give a particular value.

1  17 girls and 13 boys

8  People are more likely to reply if they have 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. day. These are the numbers required:

Under 30 30 or more

Me n 15 45

Women 15 45

Ask the first question about age. When the required number 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:

b  The people who reply might all have a

The scale does not start at zero, which makes the proportional differences between men and women look greater than they really are.

similar opinion and not be representative.

6  a  The questioner is suggesting the answer they want, i.e. ‘yes’.

b  For example: Do not let the person know which drink is the new recipe. Ask ‘Which drink do you prefer?’. Arrange for half the people to have the original drink first firs t and for half of the people to have the original drink second.

20

If you you ask people to agree with you, they might do so just to avoid conflict.

c 

Learner’s own diagram. The vertical axis should start at 0, and they should use a uniform scale.

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Check your progress

2  a  diameter = 16 cm

1  a  Which cake do you think tastes best? Which cake looks most attractive? Do you dislike any 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: Includin Including g random numbers or using registers and a particular number from each year.

b  diameter = 9 cm

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  37.70 cm

b  21.99 m

c 

diameter = 2.6 m

2  4.8 cm or 48 mm 3  a  34 cm2

b  44 m2

4  Group 1: A, D, G, H; Group 2: B, F; Group 3: C, E

5  a  320 000 c 

b  560 000 000 d  4.5

6.82

Exercise 7.1 = π r 2 = 3.14 × 22 = 3.14 × 4 = 12.6 cm 2 (1 d.p.) 2  A = π r  = 3.14 × 92 14 × 81 = 3.14 = 254.3 cm 2 (1 d.p.) 2  A = π r  = 3.14 × 4.22 = 3.1144 × 17.6644 = 55.4 m 2 (1 d.p.)  A

b  radius = 9 cm

c 

3  a  153.938 cm2 b  i 

1  a  radius = 2 cm

radius = 4.2 m

= d  ÷ 2 = 16 ÷ 2 = 8 cm 2  A = π r  = 3.14 142 × 82 = 3.11442 × 64 = 201 201.09 cm2 (2 d. d.p. p.)) r = d  ÷ 2 = 9÷2 = 4.5 cm 2  A = π r  = 3.14 142 × 4. 52 1 42 × 20.25 25 = 3.14 = 63.63 c  cm m 2 (2 d.p.) d.p.) r = d  ÷ 2 = 2. 6 ÷ 2 = 1.3 m 2  A = π r  = 3.14 142 × 1.32 = 3.11442 × 1.6699 = 5.31 m 2 (2 d p.)  p . r

153.86 cm2 

 

iii  154 cm2

c 

i 

 

iii  0.04%

0.05%

ii  153.958 cm2 ii  0.01%

d   π  =  = 3.142 e  Learner’s own answers and explanations. For example: It is best to use the  π  button  button for the most accurate answer, but if you have to use an approximation, then  π  = 3.142 is the best to choose as it gives an approximate answer closest to the accurate answer.

4  a  113 cm2 c 

415 cm2

b  56.7 m2 d  18.1 m2

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 multipl multiplying ying the radius by 2, rather than squaring the radius.

21

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b 

3.14

× 1.7 = 3.14 × 2.89 = 9.074 0746 Area = 9.07  m (3 s.f.)

15  a  Learner’s own answers and explanations.

2

b  i 

2

6  Learner’s own answers. 2

 

 A

    d        = π      or  A =

 

iii  45 π  cm

c 

i 

 

8  a 

ii 

ii  C  =  = 35.2 cm A = 804.2 mm

Estimate: A ≈

Accurate:  A =

c 

r = 7.35 cm;

 

Estimate: A ≈ Accurate:  A =

1

1

× 3 × 6 = × 3 × 36 = 54 cm 2

2

2

Estimate: A ≈ Accurate:  A =

1 2

× π × 6.2 = 60.38 cm 2

1

1

2

2

2

1 2

1

85 = 346.40 40 m × π × 14.85 2

1

49 = 73.5 cm × 3 × 7 = × 3 × 49 2

2

2

× π × 7.35 = 84.86 cm 2

2

;

2

 

Total area = 20 + 22 = 42 cm2

b 

Area A

 

Area B = l  × w = 12 × 3 = 36

 

Total area = 36 + 36 = 72 cm2

c 

Area A = l  × w = 5 × 12 = 60

 

Area B =

1

1

× 3 × 10 = × 3 × 100 = 150 m 2

2

2

1 2

64 = 145. 97 97 m × π × 9.64 2

2

A = 245.4 m2

ii 

P = 64.3 m

A = 831.0 mm2

ii 

P = 118.3 mm

2

2

2

1 π

2

r 2

1

= × π  × 6 = 56.55 2

2

;

 

Area circle = πr2 = π × 32 = 28.27

 

Shaded area = 28.27 − 6 = 22.27 cm2

2  a  i 

3 cm

ii  68 cm2

b  i 

7 cm, 8 cm

ii  98 cm2

c 

7 cm

ii  138 cm2

 

i 

7 × 4 + 0.5 × 7 × 5 = 45.5 cm2

ii  48.1 cm2

b  i  3 × 3 + 0.5 × 3 × 1.52 = 12.375 m2   ii  10 m2 c 

i 

 

ii  50.5 cm2

d  i 

b  Learner’s own answers and explanations. ii  2.4 m

iii  9.0 mm

12  a, b A and v, B and i, C and vi, D and iii, E and iv, F and ii

1

Total area = 60 + 56.55 = 116.55 cm2

3  a  i 

11  a  Learner’s own answers and explanations. 3.3 cm

1

 = × b × h = × 12 × 6 = 36

d  Area rectangle = l  × w = 4 × 1.5 = 6

Area of semicircle = 10.618 cm2, Area of quarter-circle = 9.0792 cm2 and 10.618 > 9.0792.

14  84 m2

2

1

+ d  = × π × 24 + 24 = 12π + 2  4 m

Area B = l  × w = 11 × 2 = 22

 

2

10  Marcus is correct.

13  16.44 m

πd

 

2

1

Learner’s own answers.

i 

1

=

2

Activity 7.1

c 

 P

2

1  a  Area A = l  × w = 5 × 4 = 20

;

2

d  r = 9.64 m;

 

2

2

Exercise 7.2

b  Estimate: A ≈ × 3 × 152 = × 3 × 225 = 337.5 m 2 ;

b  i 

2

1

2

Reflection:   Learner’s own answers. Reflection:

ii  C  =  = 100.5 mm

9  a  i 

1

= × π × 12 = × π × 144

2

π r 

2

Accurate:  A =

 

1

=

iv  400 π  cm2

= 72π  m

 π d 

4  2   A = 98.5 cm2

b  i 

 A

  2

7  a  i   

ii  144 π  mm2

25 π  mm

 

0.5 × 4 × 10 + 0.5 × 3 × 52 = 57.5 cm2 0.5 × 3 × 302 + 0.5 × 3 × 15 2 = 1687.5 mm2. The following could be accepted as an alternative: 0.5 × 3 × 302 + 0.5 × 3 × 152 = 1687.5

ii  1539.4 mm2

4  a  Learner’s own answer. b  Learner’s own answers and explanations. c 

Learner’s own discuss discussions. ions.

5  a  34 cm2 b  34.365 cm2 c  187.56 mm2

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6  Sofia is correct, the two shaded areas are

 

approximately the same size.  

Area of 1st shape = 86.31 cm2, Area of 2nd shape = 87.96 cm2

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.

Activity 7.2 Learner’s own answers. 7  a  i  18( π   π  −  − 2) cm2  

 π  − iii  72( π   − 2) cm2

ii 

 π  − 50( π   − 2) cm2

answer is always a number times the bracket π − 2. The number outside the bracket is always half of the square of the radius. 1 2

1 kilolitre = 1000 litres which is the same as 1 kL = 1 × 103 L.

 

You can also say that there are one thousand litres in a kilolitre or 1 litre is one thousandth of a kilolitre.

 π  − iv  4.5( π   − 2) cm2

b  Learner’s own answer. For example: The

c 

 

r   r 2  (π  −   − 2)

b  A gigametre is a very large measure of length. It is represented by the letters Gm.  

1 gigametre = 1 000 000 000 metres which is the same as 1 Gm = 1 × 109 metres.

 

You can also say that there are one billion metres in a gigametre or that 1 metre is one billionth of a gigametre. gigametre.

discussions. ions. d  Learner’s own discuss

8  Learner’s own answers and explanations. For example: The shaded areas are the same as they are both ‘Area of square of side length 10 cm cm − Area of circle of radius 5 cm’. The areas of both are 21.46 cm2.

9  a  When radius = 4, Area of circle = π  × 42 = 16 π .  

 

When radius = 4, side length of square = 4 × 2 = 8 cm. Area of square = 8 × 8 = 64.

3  a  8 micrometres, 8 millimetres, 8 centimetres, 8 metres, 8 kilometres, 8 gigametres

b  8 μm, 8 mm, 8 cm, 8 m, 8 km, 8 Gm 4  a  Learner’s own answers and explanations. For example:  

Marcus is correct. 1 tonne = 1000 kg. Also 1 kg = 1000 g and 1 Mg = 1 000 000 g = 1000 kg = 1 t.

 

Arun is incorrect. 1 litre = 1000 mL and 1 litre = 100 100 cL, so 1000 mL = 100 cL →10 mL = 1 cL, not 100 mL = 1 cL

Shaded area = 64 − 16 π  =  = 16(4 − π ) cm2.

b  i 

25(4 − π ) cm2

ii  9(4 − π ) cm2

b  Learner’s own discuss discussions. ions.

iv  100(4 − π ) cm2

c 

 

iii  36(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

d  Learner’s own discuss discussions. ions. 5  a  2.5 Mm to m → 1 Mm = 1 000 000 m, m, so 2.5 Mm = 2.5 × 1 000 000 = 2 500 000 m

b  0.75 GL to L →1 GL = 1 000 000 000 L, so 0.75 GL = 0.75 × 1 000 000 000 = 750 000 000 L

c 

1  a  A milligram is a very small measure of mass.. It is represented by the letters mg. mass  

1 milligram = 0.001 grams which is the same as 1 mg = 1 × 10−3 g.

 

You can also say that there are one thousand milligrams in a gram or 1 milligram is one thousandth of a gram.

b  A nanometre is a very small measure of length. It is represented by the letters nm.  

23

Learner’s own answers and explanations.

13.2 hg to g → 1 hg = 100 g, so 13.2 hg = 13.2 × 100 = 1320 g

6  a  364 cL to L → 100 cL = 1 L, L, so 364 cL = 364 ÷ 100 = 3.64 L

b  12 000 mg to g → 1000 mg = 1 g, so 12 000 mg = 12 000 ÷ 1000 = 12 g

c 

620 000 μm to m → 1 000 000 μm = 1 m, so 620 000 μm = 620 000 ÷ 1 000 000 = 0.62 m

1 which is nanometre t he same as=10.000 the nm =000 1 ×001 10−9metres  m.

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7 

From Ea From Eart rth h to: to: Mars Jupiter Saturn Uranus

Dista Dis tanc nce e in in … 78.34 Gm 628.7 Gm 1.28 Tm 2.724 Tm

Unit 8 Getting started

Neptune

4.35 Tm

3  a  68

8  A and v, B and iv, C and i, D and iii, E and ii 9  a  Learner’s own answers and explanations.

1  a  b  2  a 

4  a  5  a 

For example:  

Sofia is correct. 300 000 000 × 60 × 60 × 24 × 365.25 = 9.467 28 × 10 15, which rounds to 9.47×1015.

b  299 792 458 × 60 × 60 × 24 × 365.25

1  a   

6

5

.  = 0.83 recurring 1

b 

3

6

1

c 

2

6

 5 12

b  10

1 3 1

b 

2

1 4

7 20

 7

b 

1

c 

10

15

d 

1 2

= 0  .25 which is a terminating decimal

2

1

= 2 × = 2 × 0.25 = 0.5 which is a

4

9 460 000 000 000 000

 

4

= 5.676 × 1016 Learners own discuss discussions. ions.

3

1

= 3 × = 3 × 0.25 = 0.75 which is a

4

4

terminating decimal

= 56 760 000 000 000 000

1

b   

= 0  .2 which is a terminating decimal 2 1 = 2 × = 2 × 0.2 = 0.4 which is a

5

5

10  a  D, B, C, A

5

terminating decimal

b  2 147 483 648 bytes c 

5

= 0  .625 terminating

terminating decimal

d  6 × 9 460 000 000 000 000

e 

8

Exercise 8.1

= 9.460 730 473 × 10 15

c 

5

 

10 880 photos

d  1864 films 11  Learner’s own answers and explanations. For For example: Magnar is incorrect. The fastest is model B because 10 ns is quicker than 40 ns and 60 ns. ns.

Reflection:   Learner’s own answers. Reflection:

2  a 

1  a  39.27 cm

b  21.36 m

2  a  123 cm2

b  36.3 m2

c   

   

3  a 

3  49.1 cm2

1

= 4 × = 4 × 0.2 = 0.8 which is a

5

5

terminating decimal . 1  = 0.1 9

b  Recurring decimal.

 

Check your progress

4

All recurring decimals. .   i  2  = 0.2 9 . 4   iii   = 0.4

.  = 0.3 9 . 5 iv   = 0.5

ii 

9

 = 0.6. 9 . 8 vii   = 0.8 9 . 3 1 =  = 0.3 and

v 

9

6

3

 

6 9

=

2 3

3

9

vi 

7 9

 = 0.7.

.  = 0.6

b  Learner’s own discuss discussions. ions. Their answers . 9 are not different because  = 0.9 = 1.

4  170 cm2

9

5  a  5 nanograms, 5 micrograms, 5 milligrams, 5 grams, 5 kilograms, 5 tonnes

b  5 ng, 5 μg, 5 mg, 5 g, 5 kg, 5 t

24

4  a 

1 8

= 0  .125

b  Terminating decimal.

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c 

They are all termin terminating ating decimals.

 

i 

 

iii 

 

v 

2

= 0  .25 8

 

ii 

= 0  .5

 

iv 

= 0  .75 8

 

vi 

4 8 6

7  a  Always true: 7 is odd and a prime number number,, so all fractions with a denominator of 1 7 cannot be simplified.  is a recurring 7 decimal, so all fractions with a denominator of 7 are recurring.

3

= 0  .375 8 5 8

= 0  .625

7

= 0  .875 8

b  Sometimes true: For example: 1 , the 6

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 , the

own answers. The following d  Learner’s three fractions can be simplified. s implified. ci v 

6 8

2 8

1

4

1

4

8

2

= = 0  .25, iii  = = 0  .5 and

4

3

denominator is a multiple of 2, and the fraction is a terminating decimal.

= = 0  .75 4

3

1

6

2

5  a  No, =

= 0  .5 which is not a recurring

c 

b  Yes. Learner’s own explanations. For example: 6 is even, so it can be halved. So 3 = 1 . However, 7 is odd and so it 6

2

cannot be halved, so there is not an equivalent equivale nt fraction such that ? 1 . 7

2

which is a power of 2 is a terminating 1 1 1 decimal. = 0.5, 2 = 0.25, 3 = 0.125, 

For example: If the denominator is even, then there will be a fraction such

2

1 2

 1

that =  which will not be a recurring 2 ? 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 1 exceptions excep tions such as:  is recurring, but 15 3 terminating. minating. = 1 = 0  .2 which is ter

15

= 0.0625, 2 = 0.03125, etc. Each decimal 5

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.

explanations. For example: The

For example: E is not written in its

denominators mul numerators areare all multiples 1. tiples of 3. The

simplest form,1but when it is, it is equivalent equivale nt to  which is recurring. So it 14 doesn’t change the answer to part a.

own explanations. For example: The fractions that can be cancelled down still have ha ve a denominator with a multiple of 3, and once cancelled are not even.

c 

Learner’s own explanations. For example: B is now

3 1 1 , D is now = , E is now = 12 4 6 2 3

3

= 15 . These are all terminating decimals. 15 d  No. Learner’s own discuss discussions. ions.

9  a  b  c 

25

2

8  a  Learner’s own answers and explanations.

b  They are still recurring decimals. Learner’s

c 

4

2

1

can be divided by 2 to get the next decimal in the sequence, so they will all be terminating.

5

6  a  Recurring decimals. Learner’s own

, the

d  Never true: A fraction with a denominator

= answers. d  Learner’s own investigations and ?

1 20

denominator is a multiple of 10, and the fraction is a terminating ter minating decimal. decimal. However, it is not always true because they can also be recurring decimals 1 e.g. , the denominator is a multiple of 30 10, and the fraction is a recurring decimal.

decimal.

c 

Sometimes true: For example:

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. 20

1  recurring = 60 3 36 60

= 35  terminating

45

3  terminating = 60 4

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55

11  recurring = 60 12

d 

b 

 2  recurring 1 = 15 60  8

e 

1

f  

3

21

7 3  terminating = 60 20

Multiplication:

5

 

Rewrite 10:

10 9

b  terminating

 

Subtraction:

c  recurring

d  terminating

c 

5

 

Brackets:

 

Division:

 

Addition:

Dave

 

21 168

  35 168

=

=

1

Bim

8 5 24

Enid

 

28 168

 

40 168

= =

1 6 5 21

Caz Fin

  32 168  

42 168

= =

12

7 5 9 − = 12 12 12

÷  43 +   23     

2

4

 

For example:

 

Abi and Fin – the fractions they work are terminating decimals.

 

Bim, Caz, Dave and Enid – the fractio fractions ns they work are recurring decimals.

 

OR

 

Abi, Bim and Fin – the fractions they work are unit fractions. Caz, Da Dave ve 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.

2  a 

2

c 

2

3

+ 94 = 609 + 94 = 649 = 7 19

 5 16

b 

3

d 

3

1 4 3 4

d  Learner’s own discuss discussions. ions.  3

4  a  i 

9 − (2 + 4) = 9 − 6 = 3

ii 

3

b  i 

8 + (2 − 1) = 8 + 1 = 9

ii 

9

c 

5 + 2 × 16 = 5 + 32 = 37

ii 

2

− × = 16 − = 15   ii 

15

i 

d  i 

16

1

1

1

3

2

2

4

4

40  5 24 4 9 11 24

5  a  Learner’s own answers. b  Learner’s own answers. For example: It

25

−   5 19 +  8157     or 25 − 5 19 − 8 157

explanation.. 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 which is not possible in a triangle triangle..

 3 −  1  +  5 2    3 2

Addition:

20

 1 b  7 12 c  Learner’s own answers and explanations.

Exercise 8.2 Brackets:

÷ 34 = 5 × 43 = 20 3

7 + 3 − (6 − 3) = 10 − 3 = 7

6  a 

Reflection:   Learner’s own answers. Reflection:

 

5

might be easier to work with the whole numbers and fractions separately and not convert into improper fractions.

Learner’s own answers.

5

= 23 ×  23 = 23 ×× 32 = 94

3  a  Learner’s own answers. For example:

Activity 8.1

1  a 

 2     3    

1

the friends into two groups.

 

= 9 12 12

4 21

a, b Learner’s own decisions on how to sort

 

6

2

form are: Abi

× 107 = 65××107 = 35 = 127 60

 

10  a  recurring

11  The fractions fractions written in their simplest

26

−  65 × 107

10

3

1 6 5 1 − = − = 5 2 10 10 10

5

2 3

20 23 + 101 = 5 30 + 303 = 5 30

c 

11

19 45

. Learner’s own answer and

explanation. For example: Yes, the third side is less than the total of the other two sides.

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

7 

1

kg   8  Division: 6 ÷ 3

2  a 

4

5 30 6× = = 5 4 4

 

Multiplication: 3

 

Addition:

9  a 

1

7

 m

30 4

13

×5=

4

4

×5 =

b 

65

b 

8

8

1

 cm

c  c 

2

8

1

3

 m

d 

2

11

2

b 

not give the same answer. This can be shown using a multiplication box.

1 1 2

 

1

1

1

1 2

1 2 1

2

c  5  a  1

2

4

1 4

c 

. Marcus’s

method gets this answer.  

Arun’s method only gets the 1 and the , it 4

doesn’t get the other two s. 2

General rule: change the mixed number to an improper fraction. Square the numerator, square the denominator. Change the answer back to a mixed number. 3

3

b 

4

29

5 9

6  a 

 5 18

c 

18

3

1  a  b  c  d 

4 5 7 4 5 3 8

× 12 = 1

5

× 28 = 1

7

4

× 45 =

1

× 72 =

× 12

4

5

3 1

3

5

20

3

3

9

=6

7

63

2

2

× 45 = × 9 =

10

2 1

= 17

3

2 3

= 31

1 2

24

=

2

×

16

11 24 3

 

= 2×

11

22

=

3

3

=7

1 3

highest common factor

2

b  140

1

d 

2

10 21

b 

8

d 

39 6

1

1 2

5 16 2 3

8

b 

8

22

3

f  

35

1

3 3   × 2 5

a 

1

b 

2

c 

1

e  3

3

1 4 1

3

E st i ma t e

d 

 m2

3

4

 

1 4

× 3 23  

1

= 3× 3 = 9

f  

1

× 3 16  

1

5 1   × 3 22

3

8

3

3

4

3

3 × 4   4 5 4 7

× 2 165 

×4=6 2

 

× 3 12 = 7

 

× 3 12 = 3 12

2

1

4

5

2 5

8

1 4

 

3

 9 16

× 1 = 3 12

 

4

× 4 12 = 18

 

17

2

4

Accurate

1

2

5

Exercise 8.3 3

11

×

16

8

2

18

52

more than 180 km.

1 1     1   1 1 1 12  a   2    + 2 × 5  or 2 ×   2 +  5     3 3 2   3   3 2

b 

4

1

7  Lewis is correct, he travels 183  km which is

b  Learner’s own discuss discussions. ions.

11  a 

e 

1

1

 

13

4  a  84

1

2

2

2

× 8 = × 4= 7

× 45 =

2

= 13

d  Learner’s own discuss discussions. ions.

×1 = 1+ + + = 2

2

c 

1 4

1

7

 

 

6

27

She cancelled using a common factor of 4, but she should have cancelled using the highest common factor of 8.

2

1 2

3

3

3  a  Learner’s own discuss discussions. ions. For example:

multiplies 1 by 1 and 1  by 1 , which does

1

5

×8 =

10

For example: They get different answers. Marcus is correct. His method does 1 1 multiply 1  by 1 . Arun’s method

×

5

9

× 39 = × 13 =

9

3

6

2

4

× 39 =

9

10  a  Learner’s own answer and explanation explanation..

2

4

 

× 36 = × 9 =

8

2

4

+ 654 = 954 = 23 34

2

3

× 36 =

8

4

1

3

1 2

× 2 = 10

1 2 1 4

 

10

4 7

4

× 28 = 5 × 4 = 20 9

× 45 = 4 × 9 = 36 9

× 72 = 3 × 9 = 27

9  a  Learner’s own working. For example:  

8

 

5 9

× 12 = 4 and 4 < 8, 4 12 × 23  = 3 and 3 < ×

3 10

1

1 2

,

5

1 =  and 6 < 9 6

8

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

b  i 

When you multiply any number by an improper fraction, the answer will always alwa ys be greater than the original number.

Esti Es tima matte

5 

ii  When you multiply any number by a

 

mixed number, the answer will always be greater than the original number.

c 

a 

1

b 

2

c 

4

1

c 

2  1

1

12

3

11  a  A 17 , B 1 , C 9 , D 1 , E 1  5 , F 2 33

1

d 

2

e 

5

f  

4

2 3

16

3

9

1 2 4 5

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.

1

g 

 

A, B, C, D and F have a denominator which is a multiple of 3; E does not have a denominator which which is a multiple of 3, etc.

3

h 

5

3

Exercise 8.4

21

c 

14

d 

8

2

 

÷ 74 =

4

16

÷ 53 =

7

÷ 92 =

7

4

÷ 11 =

×

7 4

2

4

÷7= 9

3  a 

21 ×

14

×

8

×

= 4 × 7 = 28

1

5

= 7 × 5 = 35

31 9

= 7 × 9 = 63

21

11 41

= 2 × 11 = 22

2

8

×

9

= 29 ×× 17 = 149 = 1 59

7 41

35 ÷ 25 = 79 × 52 = 97 ×× 25 = 18 = 117 9 18 6

c 

7

3

÷ 14 =

2

1

6 7

×

14 3

2

24 ÷ 15 = 56 × 15 = d  6 24 20

1

1

1

4  C 10 , D  21, B  9 , A 

=

1

5

9

 

2÷2=1

3 1

 

4

1

20

÷ 5 =  4

 

99

5

1

124  

3÷3=1

4

 7

32 39

÷2 ÷2

1

÷

÷2

3

6÷3=2

4 2

 

5

3

10

1

÷ 3 = 1 23

 

1

 

1

4 5

 

1÷1=1

11

10

2

1

3 8

 

÷ 2 =  12

2 7

1

× 24 6 × 15

5

1

 9 31

×2   1×1

2

=4 4

3

= 43 = 113

1

1

2

1

2

2

3

4

 and

> 1 12 , 85 ÷ 16 = 3 34 and 3 34 > 58 4

1

When you divide any number by an improper fraction, the answer will always alwa ys be smaller than the original number.

ii  When you divide any number by a

Learner’s own discuss discussions. ions.

7  a  bigger, 9 1

3

b  smaller, 4 c 

smaller, 2

 2 21

8  Learner’s own answer and explanation. For example: His conjecture 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 1 ÷ 3 1 = 10 2

9  a 

7

b 

÷3

6

mixed number, the answer will always be smaller than the original number number..

c 

2  A and iii, B and i, C and iv, D and v, E and ii 8

2

5

÷ = 6 and 6 > 3, 1 ÷ = 2

Reflection:   Learner’s own answers. Reflection:

b 

5

6

4

b  i 

OR

16

÷5

 

2÷2=1

6  a  Learner’s own working. For example:

 

1  a 

÷1

4

4

3

1

bigger, 1

1

8

b  bigger, 8 1

4

÷1

2

Learner’s own discuss discussions. ions.

10  a  smaller, 2

1

Accu cura rate te

c  e 

14 15

1

1 7

11 27

b 

4

2

6 7

d 

1

f  

1

1 9 1

11

13

28

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

 

10  a  Learner’s answer and explanation. For example: π  ≈  ≈ 3, and diameter = circumference ÷  π . 15 cm is slightly more than the circumference, 15 ÷ 3 = 5, so the diameter will be just under 5 cm.

b 

14

11  a  12  13   

92

1

2

1

÷ 22 = 99 ×

7

7

2

b  Learner’s own answers and explanations.

= 9 = 41

22

b 

3 1

7

7

1

1

2

2

c 

2

1

Dae’s method – advantage: can work on one step at a time and could easily do this t his method mentally, disadvantage: method is longer (which learners might not like).

1

c 

Learner’s own answers and explanations.

 

i 

 

ii  For example: Akeno’s method

For example: Dae’s method because when 14 is multiplied by 2.5 it gives a whole number.

3

 km/h

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.

2

5 1    +   + 1 6 ÷ 5 2  is greater 3   2

2

34  and − 54    ÷ 15 = 34 = 27 2 36

2

   + 1 5 ÷ 5 1   6 3   2

1

2

28 = 79 = 36

6  a

Exercise 8.5

0.28 × 5

 0.28 = ⇒  0.

2

1  a 2

   − 1 ⇒  1 +  5 1    = (6)   = 36 36 ⇒ + 5 .5   2 2         2 2

36

,5

100

1.3 × ( 4

3

3

1  1   3 −     = (3)   = 27 − 0 .2    + 23 ⇒            5 5   5 ⇒ 27 + 23 = 50 1

2

3      ⇒    3 3  + 7   = −    = 4 ⇒  3 10 + 0.7   10 10         ⇒ 36 − 4 = 32

2  a  48

2 7

1

= 30  ⇒ 7

1

7

10

−

71

3

=3

6

−

4 = 60 ⇒ 13 10 1

×

b  49

9

b 

c 

10

60

= 7788

6

2

c 

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.

= 36

12

b  Learner’s own answers and explanations. 2

b  35 8 1

×

56

7

= 35 × 7 = 245

c 

Learner’s own discuss discussions. ions.

d 

0.8

 

1

4

2

5

  ×7 = 4 2

; example strategy: 2

  × 7 12 =   45      × 152

0.8

2

c  3

11

4

4

× 18 ⇒ 2 × 18 =

4  a  126

× 18 ⇒

b  108

11 2

4

9

× 18 = c 

99 2

= 49

8

1

=

2

5

16 25

= 85 ×× 13

105

=

5  a  Learner Learner’s ’s own answers. For example: Akeno’s method – advantage: shorter, disadvantage: involves involves changing decimals to improper fractions and cancelling before multiplying (which learners might not like).

4

For example: Fraction, because  and  are 9 3 both recurring decimals so it is easier to write them as fractions fractions..

10

× 3 12 × 56 ⇒ 54 × 72 = 358 ⇒

41

9  a  Learner’s own answers and explanations.

1

2.75

30

8  2 m2

× 2.5 × 40 ⇒ 32 × 52 = 154 ⇒ 154 × 40 = 15 × 10 = 150

1.25

×

3

13 , 4 4)  ⇒ 1.3 = 10

7  a  1

3  a  1.5

10

,4

3

c  6

7

1

× 25 = 7

100

c  3

3

28

= 25 ⇒ 4

7

− 1 = 35

b 

2

0.7 = b  0.7 × 4 2 ⇒  0.7

2

1

28

2

5

= 4 45

 

10 

24

1 3

m  

×

15

3

21

29

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

11  a  Learner’s own answers and explanations.

2  a  5, 7, 9, …

For example: Write the decimal as a fraction, square root the fraction then t hen complete the calculation using fractions fractions..

b  add 2 c  Pattern 4

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 fractions..

c 

Learner’s own discuss discussions. ions.

d 

4.25

 ×

1

7

= 5 23 ; example strategy:

9

  ×

4.25

1

7 9

1

16

= 44 × =

17 1

=

4

9 4

×

 

1

3

 

3

2

×2

=

1

2

d 

v

=

 

4 1

=

4

4

=   

 K

3

term

5

7

9 11

2 × position number

2

4

6

2 × position nu number + 3

5

7

9 11

2 1 2 4

1

1

2

2

= 89, but v =113 ≠ 89

m

m

x

6

11

18

 y

6

81

1 12 15

=

× 18 25

36

=

25

6

2

25 2

−15 2

 y

 

× 36 = 18 25

Activity 8.5

1 2

 x

b  i 

 y

25 2

1 8 11

2

1

= 5 = 1 5 and

= 12 mv = 12 × 25 ×   65      =

2

1 −2 1

x 2

= +3

2

35 52  

1 2

ii   y = 5(x − 1)

2

Learner’s own answers.

Exercise 9.1

Reflection:   Learner’s own answers. Reflection:

1  a  linear

b  linear

Check your progress

c  non-linear

d  non-linear

e  linear

f  

1  a  recurring c  recurring

g  linear

h  non-linear

1

2  a 

5

3  a 

12

b  terminating d  terminating b 

4

2

 7 12

1

b 

2

4  a  48

b 

5  a  80

b  50

c 

4

c 

1

b  i 

add

2 5

subtract 0.3

 

29 30

4 15 3 4

3

i   

3

ii  5 , 6 5

ii  7.3, 7

non-linear

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, … c 

Unit 9 Getting started 1  a  i 

8

b  24 − 5n

ii

2 K 

2 K 

4

Position-to-term rule is: term = 2 × position number + 3

4  a  3n + 5

2

 

2

5  a  i

2

v

1

b  0.5, 4.5, 8.5, …, 36.5

12  a  K  = 2

c 

Position number

3  a  12 , 13, 1133 , …,  17

17

= 5 23

b  v =

d 

1

2

3

3

4 , 3 , 3, .. ...

e  1.25, 3.25, 7.25, …

b  2, 5, 11, … d  40, 18, 7, … 1

f   1, 2 , 7, … 2

30

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

3  A and iii, B and i, C and iv, D and ii

9  Zara is correct. Learner’s own explanation.

4  a  4, 5, 14, …

 

For example: The first term is 3 and when you cube 3 you get 27. Then:

 

 If

 

 If

 

 If

c 

b  2, 7, 52, … d  0, 9, 144, …

5, 9, 49, …

5  a  3, 3, 3, … All the terms of the sequence are the same.

b  Learner’s own two sequences.  

For example: First term is 5, term-to-term rule is square and subtract 20.

 

first term is 16, term-to-term rule is subtract 12 and square square..

c 

Learner’s own answers.

 

For example: It is not possible if the numbers are positive 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 fractio However fractions, ns, it is possible 1

2

1

d  Learner’s own discuss discussions. ions. 2

4

6

1

3

explanation. For example: She needs to reverse the term-to-term rule to find the previous terms in the sequence s equence,, 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. 3rd term = (11 − 6) × 2 = 10, 10, 2nd term = (10 − 6) × 2 = 8

5

3

1

1

3

1

4

2

4

4

2

b  90, 84 , 79 , 74 , 69, 63 , 58

−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 fifth term, which which is 126 382 570 (fourth term = 11 242 which is less than one million)

8  a  3, 4, 6, 9, … b  6, 8, 12, 18, … c 

 

12  4th term = (11.5 − 6) × 2 = 11,

6  a  2, 3 7 , 4 7 , 5 7 , 7 7  , 8 7 , 9 7

c 

2, 4, 244, …

11  Tania’ Tania’ss method is incorrect. Learner’s own

9

It is also possible if you add negative numbers – e.g. first term is 2, term-to-term rule is ‘square and add −2’, or first term is 9, term-to-term rule is ‘add −6 and square’.

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 negativee number – e.g if you subtract 25, the negativ sequence will be 3, 2, −17, −4938, …

c 

‘square and add 4 ’, or first term is 9 , 2 term-to-term rule is ‘add  and square’.  

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. e.g. if you subtract 23, the sequence will be 3, 4, 41, 68 898, …

10  a  4, 8, 216, … b  −6, −8, −8, −64, …

1

 – e.g. e.g. first term is , term-to-term rule is

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.

20, 19, 16, 11, …

13  3 Reflection:   Learner’s own answers. Reflection:

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

Activity 9.1

c 

1st term =  

Learner’s own questions and discussions.

 

d  100, 90, 75, 55, …

1

2nd term =

3

3

2

3 4

1

= 1 

4th term = 3 = 1 3

d  1st term = 1  = 1

2nd term = 23 = 8

 

3rd term =

3 3

3rd term = 33 = 27

4th term = 43 = 64

31

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

2  a  7, 11, 15, …, 43 1

1

2

2

b  −3, −1, 1, …, 15

c  3 , 4, 4 , …, 8

1

d 

,

5

2 5

,

3 5

n

8  a  i 

, …, 2

7

15

c 

× 20 + 33 = 49

 

+1

 

c 

 

16,

+5

+2

23,

+7

+2

ii  2,

 

+2

+2

+2

7,

13,

20,

+6

+1

iii  3,

5,

 

+2 +6

 

+4

+7 +1

11,

28,

 

+1

21,

35,

+4

37, …

when n = 1, 4 when n = 2, 4

12  a  c 

53, …

+ 18

9

In each sequence the second differences are all the same.

e  i 

quadratic qu

ii  linear

 

iii  ne neither

iv  linear

 

v 

vi  quadratic

f  

Learner’s own discuss discussions. ions.

2

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 positive.. You cannot have a first term of −1 as this is a negative number not a positive number, so it

2 1 2 1 2

1

− × 1 = 4, 2

1

1

2

2

− ×2= 3

4   1 2

,

1

− × 3 = 3, etc. 2

− n

b  20.2 − 0.2n d  −3.5 − 1.5n

n

Exercise 9.3 1  a  i

 y

0 0

1 1

2 4

x

0

1

2

 y

0

1

8

x

ii  

b  i 

3 9

0

1

2

3

4

5

6 7

8

9 10

0

1

2

3

4

5 6

7

8

9 10

0

1

2

3

4

5

6 7

8

9 10

0

1

2

3

4

5 6

8

9 10

x

 y

b  n2 + 10 d  n2 − 9

1

1

−1 −

 

ii  x

7  Learne Learner’s r’s own explanati explanation. on. For example: When

cannot be in the sequence.

1

+4

 

n2 − 1

No, when n = 16, 163 = 4096, when n = 17, 173 = 4913 and 4896 lies between 4096 and 4913, so cannot be in the sequence.

when n = 3, 4

+9

+1

6  a  n2 + 3

 

For example:

+2

+8

neither ne

OR

11  Marcus is correct. Learner’s own explanations.

+ 10 + 14 +4

 

+9

+2

+5

 

c 

ii  No, 3 4896 = 16.98 …, so not a whole 9 8 …,

32, …

5, 7, 11, 17, 25, 35, … + 2 + 4 + 6 + 8 + 10

 

5

number.

The second differenc differences es are all the same (+2).

 

 

+3

+2

d  i 

 

11,

= 79 , C 129 = 34

the 13th term.

5  a  7, 8, 11, 16, 23, 32, … 8,

6

10  a, b, c Learner’s own answers. d  i  Yes, when n = 13, 13, 132 − 76 = 93, so s o 93 is

Learner’s own answer.

b  7,

18

9

A: 8th term = 82 − 14 = 50, 5

14

5

4

4  a  Learner’s own answer and reason. b  Card A has the greater value. B: 20th term =

8

n

b  C 3 , B 7 , A 4

3  A and iv, B and iii, C and i, D and ii

4

iii 

 

b  Learner’s own discuss discussions. ions. 9  a  A 12 = 4 , B

e  0, 3, 8, …, 99

n

ii 

 

 y

c 

i 

 y = x2

 

7

ii   y = x3

32

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

2  a  i

2 7

x  y

 

ii

 

5 9 11 28 84 124

1 3 5 10 −2 24 122 997

x  y

18

ii x

 y

1 3

1 2

2 9

1 2

−5

1 4

1 2

100

1 4

1

   

−4 −8

x  y

1 2

1

 y

1 4

4 9

1

4

x

−4 −3 −3

3

4

 y

16

9

16

0 8

x = −4 and 4 have the same y-value. x = −3 and 3 have the same y-value.

iii  Learner’s own discussi discussions. ons. For

 

example: Yes, when you square +x  and −x, you get x2.

b  i

3 125

1 or −1 2 or −2 4 or −4 10 or −10 5

 y

b  Learner’s own discuss discussions. ions.

 

i 

 

ii   y = (2x)2

 

iii   y = (x + 2)3

 

 

example: You could say that either all the x-values are positive or that all the x-values are negative.

−8 −4 15 9 1 400 −2

x

12

 y

 

7  a  i

1 3

1 2

1 3

3 4

 

ii

x

−2

1 4

1 2

 y

  −7 21

33 64

5 8

 

 

ii   y = 3x2 iii 

 y

 

  1

=x +

4 32

7 10 16 49

 y = 2x2

5 12 12 50 288 11 64

13 100

 

ii   y = (x − 3)2

 

ii  x = ± y

Learner’s own answers.

 y = (x + 5)2

3

2 8

Activity 9.3 8  a  i 

 

x  y

b  i 

iii

b  i 

x  y

ii

500

iii  Learner’s own discussi discussions. ons. For

d  Learner’s discuss discussions. ions.

 y

80

ii  Learner’s own answer. For example:

 y = 2x2

x

20

There are two possibilities for x for each y-value.

c 

4  a  i

9

ii  Learner’s own answer. For example:

 

x

iii

1 3

ii   y = x  − 3

−3

 y

1 4

3

 y = x  + 3 x

x

6  a  i

2

b  i  3  a  i

b

iii  Learner’s own check.

b  i   

2

 y = x2

 y = x3

 

ii  x = 3 y

iii  Learner’s own check. 2

5  a 

  x

 

2

×4

 

y

c   

 

  x    i   y =  2    ii    iii  Learner’s own check.

d  i   

 y = x2 + 3

iii  Learner’s own check.

x = ±2 y

ii  x = ± y − 3

33

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

 

 

e  i 

 y = (x − 4)2

ii  x = ± y + 4

 

 

iii  Learner’s own check.

f  

i 

 

iii  Learner’s own check.

 y = (2x)3

ii 

 

 x

=

  3

4  The number 178 is is not a term term in this sequence sequence,, because when you solve the equation n2 + 32 = 178 to find the value of n you do not get a whole number.

 y

2 2

+ 32 = 178 n = 178 − 32 2

9  A and iii, B and i, C and v, D and vi, 

n2

= 146 n = 146 = 1 2. 08 . . .

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.

5  a  i x

−2

4

5

 y

2

8

12

11   

x

2

×8

 

y

  x

1 4

 y

1 2

 

ii

−15 − 8

x

1 2

3 or –3

2

72

2 1 4

49

 y

1

9 1

40

2

1 2

1

4

81

144

  2

 x

 

b  i

2

 y = 8x

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 answer.. This is then multiplied by two to still give a positive answer.

Reflection:   Learner’s own answers. Reflection:

1  a  $155 2  a

 y

2

2 2

3  a  n

, . .... , 5

0 −1

1 1

2 3

3 5

b  Learner’s own graph; A straig straight ht line through (0, −1), (0.5, 0) and (3, 5).

c 

2

d  −1 3  a  40 °C

10 for the fixed charge.

d  40, 38, 34, 28, … 3

−2 −1 −5 −3 −3

b  The number of days multiplied by 3 plus

5, 6, 9, 14, …

, 1,

x

b  c = 20d  + 35

1  a  $31

b  −3, 1, 9, 121, …

1

ii   y = (x + 8)

Exercise 10.1

1  a  3, 4, 11, 116, …

2  a 

 

b  20 °C c  At the start

Check your progress

c 

=

2

2

Unit 10 Getting started

 1

He is incorrect for the function  y = x 3  because when a negative number 2 is cubed, the answer will be negative. When this is multiplied by 1 , the answer will still be 2 negative.

 y

b  8, 11, 16, …, 107 b  n2 − 2

c 

n 9

2  a  3 days

b  t = 10n +15

3  a  27 kg

b  b = 2 g  −  − 3

4  a  s + f  = 50 b  s + f  = 52 c 

s + f  = 60

34

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

6  a

5  a  6 × 5 + 6 × 10 = 90 b  5 f  + 10t = 90 c 

0 8

x  y

8

b  i 

6  a  12 × 6 + 12 × 4 = 72 + 48 = 120

7  a

0 9

 y

c  s = 2l  7  a  The total value is 80 cents.

b  i  c 

b  4 8  a  3x + 2 y = 50

8  a

b  If 3x = 21 then 2 y = 19 and that is

c 

b  3r + 4q = 100 d  32

10

9  a

c 

and 2 y = 20 − x.  

Exercise 10.2

x

−1 0 5 15

 y

−10 −30

d

1 25

2 35

3 45

3  a

 y

f  

0 −10

10 10

20 30

30 50

x  y

0 12

x  y

−2 10

10 1

5 2

0 3

0 10

2 8

4 6

6 4

8 2

10 0

Learner’s own graph. A straig straight ht line through 7 on each axis. x

−1 5

0 4

1 3

2 2

3 1

4 0

5 −1 −1

A straight line through c on each axis. others through the origin.

10  a  Learner’s own graph. A straig straight ht line through 12 on each axis.

1 16

3 8

5 0

6 −4

10 8

20 4

30 0

40 −4

 

b

x

c 

0 6

0 12

1 10 10

2 8

3 6

4 4

5 2

6 0

Learner’s own graph. A straig straight ht line through 12 on the y-axis and 6 on the x-axis.

d  Learner’s own graph. A straig straight ht line through 12 on the y-axis and 4 on the x-axis.

e  Learner’s own graph. A straig straight ht line

b  2 × 15 + 5 × 6 = 60 5  a

15 0

g  Learner’s own graph. A line parallel to the

b  At (0, 20) and (5, 0) 4  a

x

 y

0 20

ii  (0, 9)

through 4 on each axis.

When x = 23, 23, then y = 2 × 23 23 − 10 = 36, so (23, 36) is on the graph. x

6 0

e  Learner’s own graph. A straig straight ht line

b  At (0, −10) c 

x

 y

b  When x = 5, 5, then y = 10 × 5 + 15 = 65 2  a

4 3

through 10 on each axis.

Reflection:   Two possible ways are x = 20 − 2 y  Reflection:

 y

2 6

b  Learner’s own graph. A straig straight ht line

q = 3r − 5

x

1 7.5

Learner’s own graph. A straig straight ht line through (6, 0) and (0, 9).

 y

edges and 7 is not a factor of 100.

1  a

ii  (0, 8)

through (15, 0) and (0, 3).

e  No. Each pair would ha have ve a total of 7 f  

16 0

b  Learner’s own graph. A straig straight ht line

9  a–d Learner’s own answers. 10  a  32

10 3

(6, 0)

 y

impossible if y is a whole number.

6 5

(16, 0) x

b  6l  + 4s = 120

2 7

2 10

4 22

6 42

2

5, then y = 5  + 6 = 31 b  When x = 5,

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.

35

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

5

11  a  Learner’s own graph. A straig straight ht line

a

 y

 1

b  A straight line through (n, 0) and (0, n). x  y

−3 −2 −1 9 4 1

0 0

1 1

2 4

c 

x  y

−3 −2 − 2 −1 −1 11 6 3

0 2

1 3

2 6

3 11

x  y

f  

−3 −2 −1 0 1 5 0 −3 −4 −3

2 0

  1

 

2

x

c

0

5

2

4

 y   15

0

9

3

x

 

d  Learner’s own graph. A straig straight ht line through (0, 15) and (5, 0).

Learner’s own graph. A parabola with the base at (0, −4).

e  Learner’s own check. 7  a 

c). symmetry andnot theexpected lowest point at (0,the (Learners are to know word parabola.)

 

  3

 y

= 6−

 

−5 −4 − −3 3 −2 −2 − −1 1 0 1 2 3 16 7 0 −5 −8 −9 −8 −5 0

4 5 7 16

through (0, 6) and (8, 0).

e  Learner’s own check. 8  a  i 

 y = 18 − 2x

i 

 

iii  (20, 391)

 

ii 

 

iv  (−3, 0) or (3, 0)

 

iii   y = 9 − 2x

 

v  (6, 27) or (−6, 27)

 

iv 

Exercise 10.3

 

b

1  a  gradient 4 and y-intercept −6 b  gradient 6 and y-intercept 4 c 

gradient −6 and y-intercept 4

= 9−

2

x

  1

 y

= 3−

2

x

Li n e 2x + y = 18

Gradient −2

 

4x + 2 y = 18 3x + 6 y = 18

b  gradient −1 and y-intercept 8

 

  y-intercept

18

− 21

9

−2

9

− 21

3

1

gradient  and y-intercept 0 4

3  a  3 4  a  −

  1

 y

x + 2 y = 18

2  a  gradient 0.5 and y-intercept 3 c 

3

Learner’s own (correct) values in the last Learner’s column.

c 

(−10, 91)

ii  (8, 55)

0

d  Learner’s own graph. A straig straight ht line

b  Learner’s own graph. A parabola with the bottom at (0, −9).

x

6

 y

14 a

4

b  gradient − 3  and y-intercept 6 4 c x 0 8 4

13  A and iii, B and iv, C and i, D and ii

 y

= 5−

 y

b  gradient −3 and y-intercept 15

3 5

g  A curve with the y-axis as a line of

x

10 1 0 0

6  a   y = 15 − 3x

base at (0, 2).

e

8 1

e  Learner’s own check.

d  Learner’s own graph. A parabola with the  

6 2

d  gradient − 12  and y-intercept 5

base at the origin.

c

4 3

through (10, 0) and (0, 5).

3 9

b  Learner’s own graph. A parabola with the  

2 4

b  Learner’s own graph. A straig straight ht line

2

12  a

0 5

x

through (14, 0) and (0, 7).

1 2

b  1

c 

b  −1

c 

1 3

−4

c 

a

18

b

b

The gradient is −  and the y-intercept is

.

36

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

9  a   y = 4x + 8

7  a  Learner’s own graph. A straig straight ht line from the origin through (25, 40).

b  straight line line,, gradient 4, y-intercept 8, passes through (0, 8) and (−2, 0) 1

10  a  i 

3

ii 

iii 

 y

b  i 

3

1 ii  −   iii 

 y

2

 

2

=

1 2

x

+3

1

=−

+

Exercise 10.4

d  1.6 dollars

e   y = 1.6x

f  

g  62.5 kg

own graph. (30, A straig straight 8  a  Learner’s (0, 20) going through 32).ht line from

12.5 m/s

b  16 s

c 

d  d =12.5t

d   y = 0.4t + 20

2  a  400

f  

b  20

200 seconds

g  Learner’s own answers.

c 

i 

 

ii  There are 8 HK dollars to 1 US

8

9  a  Learner’s own graph. A straig straight ht line from (0, 100) going through (8, 72).

dollar.

d   y = 8x

b  79 litres

e  920 HK dollars

c 

3  a  28 dollars c  1.4 dollars e  64.12 dollars

b

c 

d   y = 100 − 3.5 h 10  a  4800

f  

11  a  The y-intercept is 24.

36.5 litres

Weeks Height (m)

0 1 2 3 4 5 1 1.2 1.4 1.6 1.8 2

0.2 m

b  750 m d   y = 1500 − 50x

50 m/minute

e  350 m

f  

30 minutes

6  a  Learner’s own graph. A straig straight ht line from

13  The rate for A is 2 cm/minute and the rate for B is 5 cm/minute. cm/minute. 120

= 12 m/s 280 280 − 120 120 = 16 m/s b  10

14  a 

(0, 0) through (50, 45).

50 45

20 18

30 27

c 

i 

 

ii  1 dollar buys 0.9 euros

0.9

d   y = 0.9x e  252 f  

170

32

b  Marcus. Arun’ Arun’ss speed is 5 m/s.

5  a  1500 m

Dollars Euros

=

12  a  8 m/s

e  3.2 m

b

b  33 000

− 24 = 0.8, so the equation of 10 − 0 the line is p = 0.8t + 24. 36 − 24 b  36 = 0.8t + 24 so = = 15; it takes 15 0.8 years. Gradient

d   y = 0.2t + 1

c 

3.5 litres/hour

b  15 d   y = 1.4x

4  a  1 m

 

0.4

e  44 °C

e  625 m

 

152 dollars

1.6

b  24 °C

1  a  250 m c 

c 

3

x 2

b  24 dollars

15 13.5

c 

10

400

− 28 0 5

= 24 m/s

15  a  Decreasing at a rate of 2 litres/hour litres/hour.. b   y = 18 − 2 t c 

9 hours

16  Learner’s own answers.

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

Check your progress

3  a  $125

b  $200

1  a  5x + 10 y = 100

4  a  $18

b  $42

b  10 y = 100 − 5x, then divide both sides

5  a  Sand: 2 parts = 15 kg, 1 part = 15 ÷ 2 = 7.5 kg

by 10.

c 

−1

2  a

x  y

0 15

1 12

2 9

3 6

4 3

5 0

b  Learner’s own graph. A straig straight ht line through (0, 15) and (5, 0).

c 

b  5 and −5 c   y = 0.3x + 3

6  a  24 and 42

7  a  Learner’s own answers. c 

8  a  750 mL

b  1.5 L

2. 3 parts = 39 g

d  5.7 m

 

3. 1 part = 39 ÷ 3 = 13 g

 

4. 4 parts = 13 × 4 = 52 g

 

5. Total Total mass = 13 + 52 = 65 g

b  1 : 4

c 

1:5

2  a  90

b  108

c 

72

4

b  32

7 3

5

4

7

4  a  Sky blue: , Ocean blue:

10  a  $70 b  Moira gets $21 and Non gets $49. 11  a  There are two possible solutio solutions. ns. The numbers are either 6 and 9 or 4 and 6.

b  i 

b  Sky blue is lighter. Learner’s own method.

 

For example:

Learner’s own answers.

ii  There are two possible solutio solutions. ns. Either the first number is 6 or the second number is 6.

Sky blue 1 : 3 = 2 : 6 = 2 parts blue blue and 6 parts white

 

iii  6 : 9 → dividing both numbers by 3 gives g ives 2 : 3

Ocean blue 2 : 5 = 2 parts blue and 5 parts white

 

There is more white in sky blue, so this shade is lighter lighter..

c 

5  $6.80

4 : 6 → dividing both numbers by 2 gives g ives 2 : 3 Learner’s own discuss discussions. ions.

12  0.18 or 1.28; Check: 0.48 : 0.18 = 8 : 3 or 1.28 : 0.48 = 8 : 3

Exercise 11.1

13  440 g of oats oats,, 220 g of butter and 110 g of

1  a  Cherries: 2 parts = 80 g,

syrup. Learner’s own method. For example: Butter: 250 ÷ 2 = 125 g per part, Oats: 440 ÷ 4 = 110 g per part.

1 part = 80 ÷ 2 = 40 g Sultanas: 5 parts = 5 × 40 = 200 g

b  Total = 80 + 200 = 280 g 2  a  Straw Strawberries: berries: 2 parts = 400 g, 1 part = 400 ÷ 2 = 200 200 g  

Learner’s own discuss discussions. ions.

 

1  a  20 : 1

 

b  120

b  0.3 m/year

Unit 11 Getting started

 

Gravel: 4 parts = 4 × 7.5 = 30 kg

9  1. Difference Difference in number number of parts = 4 − 1 = 3

4  a  4.5 m

 

 

b  Learner’s own answers.

shape with the bottom at (0, 5).

 

Cement: 1 part = 7.5 kg

b  Total = 15 + 7.5 + 30 = 52.5 kg

−3

3  a  Learner’s own graph. The usual parabola

3  a 

 

2

Raspberries: Raspberrie s: 1 part = 200 g

b  Total = 400 + 200 = 600 g

 

Use 110 g per part as smallest amount.

 

Syrup: 1 × 110 g = 110 g, Butter: 2 × 110 g = 220 g, Oats: 4 × 110 g = 440 g

14  12 g 15  3 : 4 : 5

38

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

16  Learner’s own working. For For example:  

7

When working out the number of members of staff the number must be rounded up to make sure there are enough members of staff staff..

Age of children

Child : staff ratios

up to 18 months

Number Number of of members of children staff

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

 Learner’s own answers and discussions.

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 discuss discussions. ions.

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.

Reflection:   Learner’s own answers. Reflection:

b  96 cm c  i

Exercise 11.2

   ) Height of ball before and after bounce   m   c250    (   e   c200   n   u150   o    b    f 100   o    t 50    h   g 0    i   e 0 50 100 150 200 250 300    H

1  a  direct proportion b  neither inverse proportio proportion n

d  direct proportion e  neither f  

 

8 a–d 9  2 hours 24 minutes

Activity 11.2

Total number of members of staff needed = 4 + 5 + 2 + 2 = 13

c 

Number 4 12 12 2 1 6 10 10 5 of people Cost per 300 100 600 1200 200 120 240 person (€)

inverse proportio proportion n

Height when dropped (cm)

g  neither

 

ii  They are in a straight line line..

2  a  $7

b  $17.50

 

iii  Yes

c 

d  $8.75

 

iv  225 cm

$1.75

3  a  50 g 4  a   

b   

÷4 ×2

5  a   

b   

b  150 g

c 

1.875 L

explanation. For example: The mass : length increase ratio is the same as 5 g : 3 mm for all pairs of values

4 horses = 2 days 1 horse = 8 days

×4

4 horses = 2 days 8 horses = 1 day

÷2

normal speed = 36 seconds ÷2

1 2

×3

 speed = 72 seconds

×2

normal speed = 36 seconds 3 × speed = 12 seconds

6  a  20 minutes

b  30 km/h

12  a  Direct proportio proportion. n. Learner’s own

÷3

39

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

b     )   m30   m    (   e25   s   a   e   r   c20   n    i    f   o 15    h    t   g   n   e10    L

5  a  0.084 Length of increase of string when different masses are added

6  a  0.85 c 

d  0.48

e  30

35

40

45

c  b  0.52

c  25

b  0.7

7  a  0.4 8  a 

20

b  0.916

50

0.6 1

b 

8 2 8

1

=4

d 

=1

8 5

4

8

3 8

9  a  0.45

Mass (g)

2

b  0.7

4

1

c 

Use your graph to work out

10  P(A) = 7 ; P(B) = 2 ; P(C) =

 

i 

11  a  0.2

b  0.95

c 

0.4

 

ii  33 g −34 g (accurate answe answerr is 33  g)

12  a  0.1

b  0.09

c 

0.19

7

27 mm 1

3

13  a 

d  True. Learner’s own explanation explanation.. For example: Because one set of values is a multiple of the other, so the gradient of the line is constant.

Exercise 12.2

b  1050 g or 1.05 kg

1

1  P(S) is always 2  whether the first spin is a head or a tail.

2  If A happens, happens, the number is 2, 3 or 4 and and

4  a  $6

b  $18

c 

$4.50

5  a  12 days

b  3 days

c 

6 people

1

then P(1 or 2) = 3 ; if A does not happen, the

Unit 12 Getting started

1

number is 1, 5 or 6 and then P(1 or 2) = 3 ; as these are the same, the events are independent.

3  No. If the first two spins are tails then the 1

1  0.85

probability that all three are = P(Y) = 2 . If

2  a

the first two spins are not both tails then Y is impossible and P(Y) = 0.

H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6 1

b  i  13 50

= 0  .26

12 1

b  3

b 

5

3

ii 

12

4

4  a 

5

1

=4

the probability is always . The coin has no 2

memory of the previous throws!

 or 0.2

5

c 

25

4  They are independent. The coin is fair and so

1

d 

8

3 32

Exercise 12.1

5  Fo Fog g will decrease the probability that the flight will leave on time because the flight could be cancelled.

6  a  If R happens happens,, the number is 1, 2, 3, 4, 5 3

1  25%

3  a 

1 12

any multiples of these these..

3  Sugar = 50 g, Butter = 100 g, Flour = 400 g

2  a 

d  0.81

smallest possible numbers are black 3, white 8, yellow 1. Or learners could have

2  a  24, 30 and 42 b  114

3  a 

7

b  Learner’s own answers. For example:  The

Check your progress 1  a  750 g

0.05

1

b 

6 4

2

= 5   b  10

4  a  0.3

3 10

 

b  0.45

3

= 6

1

c 

2

c  c 

7 10

 

0.7

d 

4

2 = 6 3 3 10

d  0.25

1

or 6 and P(even) = = . If R does not 6 2 happen, the number number is 1, 2, 3 or 4 and 2 1 P(even) = = . The probabilities are the 4 2 same and so the events are independent.

40

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

1

b  If B happens, happens, the number is 1, 2, 3 or 4 and P(2) = 4 . If B does not happen, the number is 1, 2, 3, 4, 1

5 or 6 and P(2) = 6 . The probabilities are not the same, so the events are not independent.

7  a  They are independent. If the t he first ball ball is replaced replaced then then the situation situation is exactly exactly the same both times times.. b  They are not independent. independent. If the first ball is black, the proba probability bility that the second ball is black is smaller than if the first ball is white. For example: Arun and Sofia are not friends and do not travel together 8  a  Learner’s own explanation. For and there are no external factors such as weathe weatherr or traffic.

b  Learner’s own explanation. For For example: Arun and Sofia are brother and sister and travel travel to school together.

9  If X happens happens then one of the cards must be A, C or D. D. Of these, 2 out of 3 are in the word word CODE, so 2

the probability of Y is . If X does not happen the card must be B or E. Then 1 out of 2 is in the word 3 1 CODE, so the proba probability bility is . These probabilities are differe different, nt, so the events are not independent. 2

Exercise 12.3 1  a  2  a 

1

b 

4

1

b 

36

1

c 

4

1

c 

12

1 4

1 12

1

3  a 

b 

6

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

ii  0.085

iii  0.135

6  a  i 

0.015

iv  0.765

b  Learner’s own explanation. For For example: example: They are mutually exclusive exclusive and one of them must happen. First

7  a 

1 9

Second

Outcome

1 9

5

5, 5

1 9

×

1 9

=

1 81

8

not 5

5, not 5

1 9

×

8 9

=

8 81

1 9

5

not 5, 5

8 9

×

1 9

=

8 81

8 9

not 5

not 5, not 5

8 9

×

8 9

=

64 81

5

9

8 9

not 5

b  i  c 

1 81

 

Not getting a 5 either time time..

ii 

64 81

 

iii 

8 81

 

iv 

8 81

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8  a 

First 0.6

Second

Outcome

red

red, red

0.3 × 0.6 = 0.18

not red

red, not red

0.3 × 0.4 = 0.12

red

not red, red

0.7 × 0.6 = 0.42

not red

not red, not red

0.7 × 0.4 = 0.28

red 0.3

0.4

0.7

0.6 not red 0.4

b  i  c 

ii  0. 0.28

0.18 0.

iii  0. 0.12

iv  0.42

Learner’s own explanation. For For example: example: They are mutually exclusive exclusive and one of them must happen.

9  a 

Blackbird

Robin

Outcome

Yes

Yes, Yes

0.9 × 0.8 = 0.72

No

Yes, No

0.9 × 0.2 = 0.18

Yes

No, Yes

0.1 × 0.8 = 0.08

No

No, No

0.1 × 0.2 = 0.02

0.8 Yes 0.9

0.2

0.1

0.8 No 0.2

b  i  c 

ii  0.02

0.72 0.

0.98

10  a 

First

Second

Outcome

1 4

Blue

Blue, Blue

2 3

×

1 4

=

2 12

=

1 6

3 4

Yellow

Blue, Yellow

2 3

×

3 4

=

6 12

=

1 2

1 4

Blue

Yellow, Blue

1 3

×

1 4

=

 1 12

3 4

Yellow

Yellow, Yellow

1 3

×

3 4

=

3 12

=

1 4

Blue 2 3

1 3

Yellow

b  i 

1 6

 

ii 

1 4

 

iii 

1 2

 

iv 

3 4

 

v 

5 6

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11  a  Learner’s own diagram. For For example: The best way to do this is with a tree diagram. First 0.9

Second

Outcome

Yes

Yes, Yes

0.4 × 0.9 = 0.36

No

Yes, No

0.4 × 0.1 = 0.04

Yes

No, Yes

0.6 × 0.9 = 0.54

No

No, No

0.6 × 0.1 = 0.06

Yes 0.4 0.1

0.9

0.6 No

0.1

 

b  Miss the first time time,, and get a basket the second second time. time. c 

0.94

Exercise 12.4  

 

3

1

7

 or 0.12

25

1 a b  2  a  Red 0.39; white 0.27; blue 0.34

25

 or 0.28

c 

5

 or 0.2

b  The probabili probability ty of each colour is 0.333. Blue is closest to this, this, white white is furthest from this. 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  4  a

Line through 0.167 on vertical axis.

Flips

20

40

60

80

100

Frequency of heads Relative frequency

8 0.4

19 0.475

30 0.5

38 0.475

44 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 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 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 probability is 0.583 and the estimate could be close to this.

d  Learner’s own discuss discussions. ions. 6  a Draws 20

Frequency Relative frequency

10 0.5

40

60

80

100

120

140

160

180

200

14 27 36 42 50 55 62 70 79 0.35 0.45 0.45 0.42 0.417 0.393 0.388 0.389 0.395

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

b  Learner’s own graph. Check that the

2  a  If X happens then the number is 2, 4 or 6 1

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.

and P(Y) = . If X does not happen then 3

3

b  If X happens the numbers are 2, 4 or 6 1

then P(Z) = . If X does not happen 3 the numbers are 1, 3 or 5 then

7  a

Digits 20 40 60 Frequency 2 5 7 of 0 Relative 0.1 0.125 0.117 frequency

80

100

7

8

0.088

0.08

P(Z) = 2 . Different probabilities so they 3 are not independent.

3  a  0.36

b  0.16

4  a  0.2

b  0.22

c 

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 0.2. The relative frequencies are the same or similar.

Unit 13 Getting started 1  a 

Digits Frequency of 0

20

40

60

80

100

2

6

8

9

15

Relative frequency

0.1

1

the number is 1, 3 or 5 and again P(Y) = .

N

b 

N

B A155°

60°

0.15 0.133 0.113 0.15

d  Learner’s own graph. Check that the

A

relative frequency values from the table in part c have been plotted correctly. B

e 

Digits Frequency of 0 Relative frequency f  

100

200

300

400

500

11

27

40

52

60

0.11 0.135 0.133 0.13

N

c 

N

d  B

0.12

A A

220°

305°

Learner’s own graph. Check that the relative frequency values from the table in part e have been plotted correctly.

g  The probability is 0.1. The probabilities vary around this value. Sofia has the closest final value. You might expect her final value to be close because she has the largest sample size.

8  Learner’s own answers and experiments.

Check your 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.

B

 

2  a  16 km

b  30 cm

3  a  (8, 8)

b  (5, 8)

 −5 

4  a      −6 

b 

    6  5

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

5  a 

2  a 

 y

N

6 5  y = 4

4 3 2

140°

1 230° 0

b 

1

2

3

4

5

6

7

8

9 10 x

5 cm

7 cm

 y

6 J

5 4

R

3

b  Learner’s own measurement measurement.. Answer in

2

range 85 m–88 m.

1

3  Yes they could meet. Learner’s own answers 0

1

2

3

4

5

6

7

8

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’tt know if the yacht and the speedboat will don’ meet because you don’t know their speeds, but if they do meet it will be at this point.

9 10 x

x = 6

6  Rotation, 90 ° clockwise, centre (−1, 0). 7  4 

N

A

152°

Exercise 13.1 1  Distance on scale drawing drawing = 800 ÷ 100 = 8 cm 8 cm

N

Ship 8 cm

42°

50°

 

B

5  a  Teshi’ Teshi’ss sketch is incorrect incorrect.. He has drawn Yue south of Jun instead of Jun south of Yue.

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

b  Learner’s own measurement measurement.. Answer in

N

b 

range 12 km–13 km (Accurate answer is 12.4 km to 3 s.f.)

c 

137°

Yue

Learner’s own measurement measurement.. Answer in range 140 °–145 ° (Accurate answer is 143 ° to 3 s.f.)

4.1 cm = 8.2 km

Activity 13.1 Learner’s own question and discussions.

(8 km) 4 cm

8  a, b i, c i, d i N

70°

(6 km) 3 cm

Jun

Farm house

6  a  N

N

Q

café N

P 5 cm (100 km)

120° 30°

Shop

4 cm (80 km)

b  ii  Learner’s own measurement measurements. s. In the range 14 km–14.5 km and 275 °–280 °.

b  Learner’s own measurement measurement.. Answer in

c 

range 125 km–130 km (Accurate answer is 128 km to 3 s.f s.f.) .)

c 

range 6.5 km–7 km, 140 °–145 °.

d  ii  Learner’s own measurement measurements. s. In the

Learner’s own measurement measurement.. Answer in range 246 °–252 ° (Accurate answer is 249 ° to 3 s.f.)

d  Learner’s own discuss discussions. ions.

ii  Learner’s own measurement measurements. s. In the

ranges: Distance from P = 11.5 km–  km–  12 km, Distance from Q = 1.2 km–  km–  1.6 km.

9  a N

7  a 

Ship N

N

8 cm (16 km) 275° N

45°

6 cm (12 km)

 

P

7.5 cm (75 km)

L

b  Learner’s own measurement measurements. s. In the range 46 km–47 km.

c 

Learner’s own measurement measurements. s. In the range 53 km–54 km.

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10  a  i   

ii  247 ° ± 2 °

120 ° ± 2 °

iii  351 ° ± 2 °

7  R (12, 15) 8  a  Learner’s own explanation. For example:

b  20 km

c 

The point (1, 2) is not on the line. It is two units to the left of where the line starts at point A.

$1120

Reflection:   Learner’s own answers. Reflection:

b  Learner’s own explanation. For example:

Exercise 13.2 1  Learner’ Learner’ss own diagram. Check that all of the

She needs to add (1, 2) on to the coordinates of A (3, 2).

points are plotted and labelled correctly.

c 

a  A (0, 2) and B (3, 2)

The coordinates coordinates of C are (3 + 1, 2 + 2) = (4, 4). Learner’s own check.

b  (1, 2)

d  i 

c 

e  Learner’s own discuss discussions. ions.

(2, 2)

ii  1 : 4

1:5

9  Difference in x-coordinates = 9 − 3 = 6, 2 × 6 = 4

d  C (4, 0) and D (4, 8)

3

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

 

H = F(3, 4) + (4, 6) = (3 + 4, 4 + 6) = (7, 10)

10  a  L (10, 11) 11  a, b Learner’s own diagram. Check that the

Chesa’s method will work as she takes into account the position of S. When S movess she will add her distances on to the move coordinates of S. Tefo’s method will not work as he is just finding the fraction of the coordinates of T. When S moves this will give the incorrect answer.

c 

points and diagonals are drawn accurately.

c 

2

5

AC = 5 − 1 = 4  

b  A (12, 9)

8

1 

1  

d  B (8, 12)  

J (2 × 10, 3 × 10) = (20, 30)

× 4 = 2 12

Difference in y-coordinates of 5

1 × 4 = 2 8 2

AC = 5 − 1 = 4

b  C (6, 9)

  1     2  3 1  ,  3 1  =    2 2   

E = A (1, 1) +  2 ,  2

  =  1  + 2 1 ,  1 + 2 1         2    2 2  

1

12  J (13, 13). Learner’s own working. working. For

d  P (2 × 16, 3 × 16) = (32, 48)

example:

e  (2 × 20, 3 × 20) = (40, 60) f  

 1     2  

 2 + 5   5 + 2      7   7     

5  a  B (4, 6) c 

1

E 3 ,3

d    2 , 2    =  2 ,  2   =  3 2 , 3 2    e  Difference in x-coordinates of

Learner’s own discuss discussions. ions.

C (2, 3)

3

b  Learner’s own check using a diagram.

e  Learner’s own answer. For example:

4  a  B (4, 3)

×9= 6

Difference in y-coordina -coordinates tes = 13 − 4 = 9,

3  a – d Learner’s own answers.

f  

2

 

Difference in x-coordinates is 17 − 5 = 12

Coordinates of the point label Coordinates labelled led 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

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.

1

example: OD lies  of the distance OE 4 3 and so DE lies  of the distance OE. This 4

1

means the ratio OD : DE is not 1 : 4.

4

:

3 4

= 1 : 3 and

Points: x-coordinates  y-coordinates

F 5 1

G H I J K L 7 9 11 13 15 17 4 7 10 13 16 19

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Exercise 13.3 1  a and iii, b and i, c and ii 2  a  Learner’s own diagram. The image should have ha ve vertices (2, 0), (4, 0), (4, 1) and (3, 1).

b  Learner’s own diagram. The image should c 

have vertices (3, 0), (4, 0), (4, 1) and (3, 2). have Congruent. Congruen t. 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 ha ve vertices (2, 1), (4, 3) and (1, 3).

b  Learner’s own diagram. The image should have ha ve vertices (−2, 0), (−5, 0) and (−3, ( −3, 2).

c 

Learner’s own diagram. The image should have ha ve vertices (−2, 1), (−2, 4) and (0, ( 0, 3).

d  Learner’s own diagram. The image should

5  a  Reflection in the  yy-axis. b  Reflection in the x-axis. c 

Reflection in the line y = 1.

d  Reflection in the line x = 1.   2 

6  a       −1 

5   −6  d   −4  −1      1   −4   e       f    3      4   7  a  90 ° clockwise, centre (3, 3)

b  90 ° anticlock anticlockwise, wise, centre (3, 0) c 

180 °, centre (3, 0)

d  90 ° clockwise, centre (−1, 0) e  90 ° anticlock anticlockwise, wise, centre (−1, −1) 8  a  i 

Learner’s own diagram. The image should have vertices vertices (−1, −3), (1, −3), (0, −2) and (0, −3).

 

b  i 

 

Learner’s own diagram. The image should have vertices vertices (3, −3), (5, −2), (5, −1) and (4, −1).

ii  Learner’s own diagram. The image should have vertices vertices (−1, −3), (1, −2), (1, −1) and (0, −1).

c 

i 

Learner’s own diagram. The image should have vertices vertices (−2, 2), (−2, 4), (−3, 3), (−4, 3) and (−4, 2).

 

ii 

Learner’s own diagram. The image Learner’s should have vertices (−2, −4), (−2, −6), (−4, −6), (−4, −5) and (−3, −5).

d  i 

The positions of the shapes are different, even even though the elements of the transformations are the same same..

 

 

iii  Learner’s own discuss discussions. ions.

 

iv  Learner’s own transfor transformations. mations. For example: Reflection in line y = −2, then reflection in line x = 3.

v  Learner’s own checks.

Rotation 180 °, centre (−2, 1) OR reflection in the line y = 1 OR 0  translation  . −3 

  2 

ii  Translation    OR rotatio rotation n 180 °,  −4  centre (2.5, 3)

 

iii  Reflection in the line x = 4.5 4.5 OR

rotation 180 °, centre (4.5, 1) OR 2  translation 0   . b  Learner’s own discuss discussions. ions. 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 shape.. The shapes can all also be rotated 180 °. For the two pairs where the translation is either horizontal only or vertical only, it is also possible to reflect the shapes in a vertical or horizontal mirror line.

c   

ii  Yes. Learner’s own explanation. For example: A different order often results in a different finishing position.

 

 

ii  Learner’s own diagram. The image should have vertices vertices (−3, 5), (−5, 5), (−4, 4) and (−4, 5).

    1  5

c 

have ha ve vertices (2, −1), (4, −3) and (2, ( 2, −4).

4  a  i 

b 

 

i 

For example: Rotation 180 °, centre 1  (3, 5) followed by a translation . −4    ii  For example: Rotation 90 ° anticlockwise, centre (1, 4) followed  −2  by a translation   .  −5  iii  For example: Rotation 90 ° anticlockwise, centre (3, 0) followed  −4  by a translation   2   .

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d  Learner’s own discussi discussions. ons. For example: Yes, for all of them there is more than one combined transformation. Each object can be rotated about any point to get it in the same orientation as the image, and then you can use a translation to move it into the correct position. In part i, you could also

Exercise 13.4 1 

use a reflection in any vertical or horizontal line and then you can use a translation to move it into the correct position.

9  a  They are both correct. When you start with triangle G and follow Sofia’s instructions,, the final image is triangle instructions H. When you start with triangle G and follow Zara’s instructions, the final image is triangle H.

2  a 

b  For example: Reflection in the line x = 3 −6  then translation −2   .

 

 −8  For example: Translatio ranslation n  −2    then reflection in the line x = −4.

c 

There are an infinite number of combined transformations. Learner’s own explanation. For example: G can be reflected in any line x = ‘a number’ then th en translated to H.

10  a  i 

 

b 

Learner’s own diagram. Shape B with vertices (6, 4), (8, 5), (8, 2) and (6, 2). Shape C with vertices (2, 5), (4, 6), (4, 8) and (2, 8).

ii  Reflection in the line y = 5.

b  i 

 

Scale factor 2

Learner’s own diagram. Shape D with vertices (5, 8), (8, 8), (8, 10) and (6, 10). Shape E with vertices (2, 5), (5, 5), (5, 7) and (3, 7).

Scale factor 3

c 

ii  Rotation 90 ° anticlock anticlockwise, wise, centre (2, 5).

Activity 13.3 Learner’s own answers and discussions.

11  a  A to B c 

B to D

b  A to C d  C to E Scale factor 4

Reflection: a  It is the same shape and size. b    

•  •  • 

corresponding lengths are equal corresponding angles are equal the object and the image are congruent

3  a  Learner’s own explanation. For example: She hasn’t enlarged the shape correctly from the centre of enlargeme enlargement. nt. She has incorrectly used the centre as one of the vertices of the triangle.

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7  Perimeter = 60 cm, Area = 150 cm2

b 

8  Shape G is an enlargeme enlargement nt of shape F, scale factor 3 and centre of enlargement at (1, 2).

9  a  Enlargeme Enlargement nt scale factor 2, centre (−5, 2). b  Enlargement scale factor 4, centre (−6, −2). 10  Enlargement scale factor 3, centre (4, −5). 11  Learner’s own own answers answers and justificati justification. on. For For example: Arun is incorrect. When one shape is an enlargement of another, and the centre of enlargement is inside the shapes, you can use ray lines to find the centre of enlargement.

Activity 13.4 Learner’s own enlargements and discussions.

12  Enlargement scale factor 3, centre (6, 5).

4  a  Learner’s own diagram. Check that the

13  Enlargement scale factor 2, centre (4, 4).

shape has been enlarged correctly. Vertices of the image should be at (1, 7), (5, 7), (5, 3) and (1, 3).

b  Learner’s own diagram. Check that the

Check your progress 1  a  N

shape has been enlarged correctly. Vertices of the image should be at (2, 6), (8, 6), (8, 0) and (2, 0).

c 

5  a  i   

9 cm (90 km)

b  Answe Answerr in range 148 km–152 km (accurate answer 150 km).

Scale Squares factor of

Ratio of

Ratio Ratio of of periareas enlargement lengths meters 2 1:2 1 : 2 1 : 4 = 1 : 22

A:C

3

1:3

1:3

1 : 9 = 1 : 32

A:D

4

1:4

1:4

1 : 16 = 1 : 42

c 

Answer in the range 264 °–270 ° Answer (accurate answers 267 ° to 3 s.f.)

2  a  (5, 3)

4  a  i 

 

of triangle C should be at (3, 3), (4, 2) and (4, 4).

b  i   

6  Perimeter of R = 14 cm → Perimeter of T = 14 × 3 = 42 cm  

Area of R = 10 cm2 

  Are Area a of

→

2

T = 10 × 3  = 90 cm

2

Learner’s own diagram. The vertices of triangle B should be at (3, 3), (5, 3) and (4, 4).

ii  Learner’s own diagram. The vertices

e  Yes. Yes. Learner’s own discuss discussions. ions.

b  (6, 10)

3  L (4, 10)

ratio of lengths = ratio of perimeters. perimeters.

d  ratio of lengths squared = ratio of areas. f  

12 cm (120 km)

50°

ii  Areas: A = 4 cm2, B = 16 cm cm2,

b

c 

N

Perimeters: A = 8 cm, B = 16 cm, C = 24 cm and D = 32 cm C = 36 cm2 and D = 64 cm2

A:B

140°

Learner’s own diagram. Check that the shape has been enlarged correctly. Vertices of the image should be at (1, 9), (9, 9), (9, 1) and (1, 1).

Rotation of 180 °, centre (3, 4).

ii  Rotation 90 ° anticlock anticlockwise, wise, centre (2, 3).

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5 

4  Learner’s own answers. For example: a  Yes. The cross-sect cross-section ion is a circle. b  Area of circle × height c 

V  = πr2h

d  Learner’s own discuss discussions. ions. 5  Learner’s own explanation. For example: The

radius and height are in different units. She needs to change the 5 mm to cm cm or change the 2 cm to mm before she works out the volume. Volume = 1570 15 70 mm3 (3 s.f.) or 1.57 cm 3 (3 s.f.)

6  a  942.5 cm3 b  353.4 cm3

6  Scale factor 3, centre of enlargement enlargement at (10, 4).

c 

7  Perimeter = 54 cm and area = 180 cm2.

17 592.9 mm3

Activity 14.1

Unit 14 Getting started

Learner’s own cylinders, answers and discussions.

1  25.13 cm

7 2

2

2  a  27 mm 3  a  120 cm3

2

b  21 cm c  78.5 m b  158 cm2 a

4  a  480 cm3 b  Learner’s own diagram. Any correct net. c 

528 cm2

5  a  1

Radius of circle

b  2

c 

6

d  0

2.5 m b 6 cm c 2.52 m d 4.56 mm mm 8  a  5.5 cm

Exercise 14.1 b  130 cm3

a b c

Length of  Volume prism of prism 10 cm 120 cm3 8.5 cm 204 cm3 6.2 m 114.7 m3

3  a  Learner’s own explanation. For example:   Yusaf hasn’t used the correct crosssection. Instead of using the trapezium as the cross-section, he has used the side rectangle (which is not the cross-section of the prism). 1

b  Area of trapezium = (8 + 14 14) × 4 = 44  cm2 2

 

c 

2.1 cm

 

Volume of cylinder: V  = πr2h = π  × 62 ×18 = 2035.75 cm3 (2 d.p.)

 

Volume of cube: V  = 83 = 512 cm3

 

Volume of water: 1.5 litres litres = 1500 mL = 1500 cm3

 

Volume of cube + 1.5 litres litres = 512 + 1500 3 = 2012 cm

 

The total volume of the cube and water is less than the volume of the cylinder, cylinder, so the water will not come over the top of the cylinder. cylinder. 2012 cm3 < 2035.75 cm3

134.4 cm3

Area of cross-section 12 cm2 24 cm2 18.5 m2

b  4.2 cm

example:

1  a  120 cm

2

19.63 m2 113.10 cm2 20 m2 65.25 mm mm2

9  Learner’s own methods and answers. For

3

c 

Area of circle

Height  Volume of of cylinder cylinder 4.2 m 82.47 m3 4.48 cm 507 cm3 2.5 m 50 m3 16 mm 1044 mm3

Volume of prism = 44 × 20 = 880 cm3

Reflection:   Learner’s own explanations. Reflection:

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Exercise 14.2

e  Add 1 to the number in front of the r, then double it. This gives you the number in front of the  πr2. So, 19 + 1 = 20, 20 × 2 = 40, so SA = 40 πr2.

1  Answer using rounded intermediate values: Area of circle =

 

r 2

π 

otal area = 2

×

6  Learner’s own methods and answers. For example:

a  i   

c 

408 cm2

b  SA = 1188 mm2

2

r 

π 

ii  Pythagora Pythagoras’ s’ theorem

7  a  SA = 660 cm2

Answer using accurate intermediate values: Area of circle =

The hypote hypotenuse nuse of the triangular cross-section.

b  Learner’s own discuss discussions. ions.

78.54 + 377.04

= 534 534 ccm m 2 (3 ss.f .f.) .)

 

Learner’s own discuss discussions. ions.

5  226 cm2 (3 s.f.)

= 377.0 377.04 4 cm2 (2 d.p. d.p.))

 

 

f  

= π × 52 .p.) = 78.54 cm 2 (2 d.p.) Circumference of circle = π d  = π × 10 = 31.42 cm (2 d.p.) Area of rectangle = 31.42 × 12

c 

= π × 5 5398 . . .  cm = 78.5398 Circumference of circle = π d  = π × 10 = 31.4159 4159. . .  cm Area of rectangle = 31.4159. .. × 12

SA = 23.3 m2

2

 

Activity 14.2

2

a, b Learner’s own shapes. For example:

A cuboid with length 10 cm, width 10 cm and height 8 cm (V = 800 cm3, SA = 520 cm2); A triangular prism of length 33 cm with a right-angled cross-section with base length 6 cm, height 8 cm and hypotenuse 10 1 0 cm (V = 792 cm3, 2 SA = 840 cm ); A cylinder with height 16 cm cm and cross-section radius 4 cm (V = 804 cm3, SA = 503 cm2).

= 376.9911... cm 2

  otal area = 2

×

78.5398. 78.5398. .. + 376.9 376.9911. 911. ..

= 534 534 cm2 (3 s.f. s.f.)

 

2  a  SA = 477.5 cm2

c 

b  SA = 322.0 cm2 c 

d  Learner’s own discuss discussions. ions.

SA = 4272.6 mm2

8  754 cm2

3  The pyramid has a greater surface area than 2

9  15 labels is the maximum using Method 1

2

the cylinder. 132 cm  > 125.66 cm .

below.

 

 1     + 6 × 6 = 11332 cm Pyramid: SA = 4 ×  × 6 × 8  +  2  

 

Cylinder: SA = π  × 22 × 2 + π  ×  × 4 × 8 = 125.66 cm2

2

 

Method 1:

 

120 ÷ 23.6 = 5 whole lengths

 

35 ÷ 10 = 3 whole lengths

 

Number of labels = 5 × 3 = 15

 

Method 2:

2 πr(r + h)

 

120 ÷ 10 = 12 whole lengths

SA = 2 πr(r + h) = 2 πr(r + 2r) = 2 πr × 3r = 6 πr2

 

35 ÷ 23.6 = 1 whole length

 

Number of labels = 12 × 1 = 12

4  Learner’s own methods and answers. For example:

a  SA = πr2 + πr2 + 2 πrh b  SA = πr2 + πr2 + 2 πrh = 2 πr2 + 2 πrh = c 

d  i   

Learner’s own answers and explanations.

SA = 8 πr2

iii  SA = 12 πr2

 

ii  SA = 10 πr2

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Exercise 14.3 1  a, b and c Learner’s own drawings. Check that the planes of symmetry are drawn correctly. correctly. Shapes a and b hav  havee vertical planes of symmetry. Shape c has a horizontal plane of symmetry.

2  a, b Learner’s own drawings. Check that the planes of symmetry are drawn correctly. Shape a has one vertical and one horizontal plane of symmetry. Shape b has two vertical and one horizontal plane of symmetry symmetry..

3  a, b Learner’s own drawings. Check that the plane of symmetry is drawn correctly. The plane of symmetry should be vertical.

c 

6  a

2D regular polygon Triangle Square Pentagon Hexagon Octagon

Number of planes of symmetry Triangular 4 Square 5 Pentagonal 6 Hexagonal 7 Octagonal 9

Number of lines of 3D prism symmetry 3 4 5 6 8

b  Learner’s own answers and explanations. For example:  

The plane of symmetry is a vertical plane of symmetry symmetry..

4  a, b Learner’s Learner’s own lines of symmetry symmetry.. Any of these:

c 

Number of planes of symmetry = numbe numberr of lines of symmetry + 1. This happens because the planes of symmetry can be drawn, the length of the prism, in the same place as the lines of symmetry on the cross-section of the prism. There is then the extra plane of symmetry that divides the prism halfway along its length. i  11 ii  13

d  Learner’s own discuss discussions. ions. 7  a, b Learner’s own diagram. Check that the plane of symmetry passes through the circular ends of the cylinder, cylinder, dividing the circular cross-section into two identical semi-circles.

c  c 

A cube has a total of nine planes of symmetry.

d  Learner’s own answers and explanations.

d  Learner’s own justificati justification. on. All nine diagrams shown in the answer to part  b.

e  Learner’s own discussions. 5  a  There are two vertical and one horizontal planes of symmetry symmetry..

b 

Learner’s own diagram. Check that the plane of symmetry passes halfway along the height, splitting the cylinder into two identical cylinders. For example:

 

It has an infinite number of planes of symmetry. A circle has an infinite number of lines of symmetry, so this is the same in 3D for the cylinder. When the cylinder is placed upright there is alwa always ys one horizontal plane of symmetry symmetry,, but an infinite number of vertical ones. ones.

Reflection:   Learner’s own answers. Reflection:

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Check your progress

3  a

Mean Median Mode Range

1  120 cm3 2  14 m2

History

12.9

13

16

7

3  452 cm3

Chemistry

14

16

18

15

4  The square-based pyramid has the greater surface area.

 

b  The Chemistry group has better marks on average, because the mean, median and mode are all greater than for the History group.

Pyramid: SA = 340 cm2, Cylinder: SA = 320.44 cm2, 340 > 320.44

5  a  The shape has two vertical, one horizontal

c 

and two diagonal planes of symmetry symmetry..

b  Learner’s own diagrams showing the five planes of symmetry correctly as described in the answer to part a.

The History group has more consistent marks because the range is lower.

Exercise 15.1 1  a

Height, h ( cm cm) 140 ⩽ h < 150 150 ⩽ h < 160 160 ⩽ h < 170 170 ⩽ h < 180

Unit 15 Getting started 1  a

Age, a (y  (ye e ar s ) 10 < a ⩽ 15

Frequency 3

15 < a ⩽ 20

6

20 < a ⩽ 25

7

25 < a ⩽ 30

4

b  Learner’s own diagram. Frequency diagram showing the data in part a. Make sure the axes are labelled correctly and that a sensible scale is used. Make sure the bars are the correct width and height.

c 

Class 9P test results

polygon with points (145, 7), (155, 13), polygon (165, 6) and (175, 2) joined with straight lines. Make sure that the axes are labelled correctly and that a sensible scale is used.

2  a

Mass, m (  (kkg) 40 ⩽ m < 50 50 ⩽ m < 60 60 ⩽ m < 70

1

2 4 6 7 8 9

2

2 3 4 4 6 7 8

3

0 1 6 8 9 9 9

c 

4

0 0

d 

1 5

d  14

Midpoint 45 55 65

polygon with points (45, 4), (55, 12) and polygon (65, 8) joined with straight lines. Make

3 8

b  32%

Frequency 4 12 8

b  Learner’s own diagram. Frequency

0

9

Key: 0 3 mea means ns 03 ma mark rks s

c 

b  Learner’s own diagram. Frequency

11

2  a 

Frequency Midpoint 7 145 13 155 6 165 2 175

sure that athe axes are labelled correctly and that sensible scale is used. 24 2 3

e  Arun is incorrect incorrect.. Learner’s own explanation. For example: You do not know how heavy the heaviest student is. You only know that their mass is in the interval 60 kg ⩽ m < 70 kg.

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CAMBRIDGE LOWER SECONDARY MATHEMATICS 9: TEACHER’S RESOURCE

3  a  Learner’s own frequency table. For

c 

example:

Age, a (y  (yea ears rs)) Fr Freq eque uenc ncyy

 

10 ⩽ a