secondary

Marshall Cavendish Education Cambridge Lower Secondary Mathematics Student’s Book Year 9
Chapter 2 Fractions, Decimals and Percentages

Ó 2023 Marshall Cavendish Education Pte Ltd

1

Worked Solutions
(The following content has not been through the
Cambridge International endorsement process.)

Practice 2A

1 (a) 34.5 £x < 35.5

(b) 36.15 £x < 36.25

(c) 3.5225 £x < 3.5235

(d) 1.55 £x < 1.65

(e) 35.95 £x < 36.05

(f) 0.035 15 £x < 0.035 25

2 (a) 999.5 £x < 1000.5

(b) 995 £x < 1005

(c) 950 £x < 1050

(d) 500 £x < 1500

TWM.02 Generalising
The range of answers for each case is different
depending on how it has been rounded.

3 (a) $59 500 000 £ amount raised < $60 500 000

(b) 38.5°C £ temperature < 39.5°C

(c) 4 950 000 £ population < 5 050 000

(d) 595 £ number of students enrolled < 605

(e) 650 £ number of passengers < 750

Practice 2B

1 TWM.05 Characterising, TWM.06 Classifying

(a)

!

"# =

$

%, which is in the simplest form.

The denominator 6 is a multiple of 3.
Thus, the decimal representation of

!

"#

is recurring.

(b)

!

&'' is in the simplest form.

The denominator 300 is a multiple of 3.
Thus, the decimal representation of

!

&''

is recurring.

(c)

(

&% =

$

" = 0.25

Thus, the decimal representation of

(

&%

is terminating.

(d)

&$

&% is in the simplest form.
The denominator 36 is a multiple of 3.
Thus, the decimal representation of

&$

&%

is recurring.

(e)

!

#) =

$

" = 0.25

Thus, the decimal representation of

!

#)

is terminating.

(f)

#$

#) =

&

" = 0.75

Thus, the decimal representation of

#$

#)

is terminating.

(g)

&*'

"#' =

*

%, which is in the simplest form.

The denominator 6 is a multiple of 3.
Thus, the decimal representation of

&*'

"#'

is recurring.

(h)

**

%& is in the simplest form.
The denominator 63 is a multiple of 7.
Thus, the decimal representation of

**

%&

is recurring.

2 (a)

&

! = 0. 4̇28571̇

0. 4 2 8 5 7 1 4
7 ) 3.0 0 0 0 0 0 0
2 8
2 0
1 4
6 0
5 6
4 0
3 5
5 0
4 9
1 0
7
3 0

2 8
2

Marshall Cavendish Education Cambridge Lower Secondary Mathematics Student’s Book Year 9
Chapter 2 Fractions, Decimals and Percentages

Ó 2023 Marshall Cavendish Education Pte Ltd

2

(b)

!

% = 1.16̇

(c)

*

$# = 0.416̇

(d)

!

$& = 0. 5̇38461̇

(e)

*

$$ = 0. 4̇5̇

3 TWM.05 Characterising

$

! = 0. 1̇42857̇

#

! = 0. 2̇85714̇

&

! = 0. 4̇28571̇

"

! = 0. 1̇42857̇

*

! = 0. 5̇71428̇

%

! = 0. 8̇57142̇
They have the same set of repeating digits.

Practice 2C

1 (a) 8

$

#´ 1

&

*

=

$!

#´

)

*

=

%)

*

= 13

&

*

Estimate:

8

$

# is more than 8.

1

&

* is close to 1

$

#.

8 ´ 1

$

# = 8 + 4 = 12

8

$

#´ 1

&

* should be more than 12.

Thus, the answer 13

&

* is reasonable.

(b) 3

$

&´ 1

!

$'

=

$'

&´

$!

$'

=

$!

&

= 5

#

&

Estimate:

3

$

& is close to 3.

1

!

$' is close to 2.

3 ´ 2 = 6

Thus, the answer 5

#

& is reasonable.

(c) 7

$

#´ 3

$

*

=

$*

#´

$%

*

= 24

Estimate:

7

$

# is more than 7.

3

$

* is close to 3.

7 ´ 3 = 21

7

$

#´ 3

$

* should be more than 21.

Thus, the answer 24 is reasonable.

1.1 6 6
6 )7.0 0 0
6
1 0
6
4 0
3 6
4 0
3 6
4

0.4 1 6 6
12 )5.0 0 0 0
4 8
2 0
1 2
8 0
7 2
8 0
7 2
8

0. 5 3 8 4 6 1 5
13 ) 7.0 0 0 0 0 0 0
6 5
5 0
3 9
1 1 0
1 0 4
6 0
5 2
8 0
7 8
2 0
1 3
7 0

6 5
5

0.4 5 4 5
11 )5.0 0 0 0
4 4
6 0
5 5
5 0
4 4
6 0
5 5
5

4

8

3

Marshall Cavendish Education Cambridge Lower Secondary Mathematics Student’s Book Year 9
Chapter 2 Fractions, Decimals and Percentages

Ó 2023 Marshall Cavendish Education Pte Ltd

3

(d) 3

$

&÷ 1

$

(

=

$'

&÷

$'

(

=

$'

&´

(

$'

= 3

Estimate:

3

$

& is close to 3.

1

$

( is close to 1.

3 ÷ 1 = 3

Thus, the answer 3 is reasonable.

(e) 5

$

"÷ 1

&

"

=

#$

"÷

!

"

=

#$

"´

"

!

= 3

Estimate:

5

$

" is close to 5.

1

&

" is close to 2.

5 ÷ 2 = 2.5

Thus, the answer 3 is reasonable.

(f) 7

$

"÷ 2

$

*

=

#(

"÷

$$

*

=

#(

"´

*

$$

=

$"*

""

= 3

$&

""

2 (a) 3

$

"÷

&

"

=

$&

"÷

&

"

=

$&

"´

"

&

=

$&

&

= 4

$

&

Estimate:

&

" is less than 1.

So, 3

$

"÷

&

" should be greater than 3

$

".

Thus, the answer 4

$

& is reasonable.

(b) 4

$

*÷

$!

#'

=

#$

*÷

$!

#'

=

#$

*´

#'

$!

=

)"

$!

= 4

$%

$!

Estimate:

$!

#' is less than 1.

So, 4

$

*÷

$!

#' should be greater than 4

$

*.

4

$%

$! is greater than 4

$

*.

Thus, the answer 4

$%

$! is reasonable.

(c) 7

$

&÷

##

&'

=

##

&÷

##

&'

=

##

&´

&'

##

= 10

Estimate:

##

&' is less than 1.

So, 7

$

&÷

##

&' should be greater than 7

$

&.

Thus, the answer 10 seems quite reasonable.

(d) 6

$

&´

&

&)

=

$(

&´

&

&)

=

$

#

Estimate:

6

$

& is close to 6.

&

&) is close to

&

&%, or

$

$#.

6 ´

$

$# =

$

#

Thus, the answer

$

# is reasonable.

(e) 4

#

!´

$"

"*

=

&'

!´

$"

"*

=

"

&

= 1

$

&

Estimate:

4

#

! is close to 4.

$"

"* is close to

$"

"#, or

$

&.

4 ´

$

& = 1

$

&

Thus, the answer 1

$

& is reasonable.

3

3

4

10

2

2

2

3

Marshall Cavendish Education Cambridge Lower Secondary Mathematics Student’s Book Year 9
Chapter 2 Fractions, Decimals and Percentages

Ó 2023 Marshall Cavendish Education Pte Ltd

4

(f) 3

"

(´

$)

&$

=

&$

(´

$)

&$

= 2

Estimate:

3

"

( is close to 3

$

#.

$)

&$ is close to

$)

&%, or

$

#.

3

$

#´

$

# is half of 3

$

#, which is

!

" or 1

&

".

Thus, the answer 2 is reasonable.

3 (a) –2

$

"´ (–

"

()

= 2

$

"´

"

(

=

(

"´

"

(

= 1

(b) 6

$

#´ (–

$$

#%)

= – 6

$

#´

$$

#%

= –

$&

#´

$$

#%

= –

$$

"

= –2

&

"

(c) –2

$

*´ 1

#

$$

= –

$$

*´

$&

$$

= –

$&

*

= –2

&

*

(d) 2

$

*÷ (–

&

$')

= – 2

$

*÷

&

$'

= –

$$

*÷

&

$'

= –

$$

*´

$'

&

= –

##

&

= –7

$

&

(e) –4

$

&÷ (–

#%

(')

= 4

$

&÷

#%

('

=

$&

&÷

#%

('

=

$&

&´

('

#%

=

&'

#

= 15

(f) –5

$

&÷

)

(

= –

$%

&÷

)

(

= –

$%

&´

(

)

= –6

4 (a) (–2.28) ´ (–3.1)

= 2.28 ´ 3.1

= 7.068

(b) (–1.5) ´ (–8)

= 1.5 ´ 8

= 12

(c) –3.5 ´ 6.2

= –(3.5 ´ 6.2)

= –21.7

(d) 6.2 ´ (–4.1)

= –(6.2 ´ 4.1)

= –25.42

(e) –3.14 ÷ (–2.1)

= 3.14 ÷ 2.1

= 1.50 (to 2 d. p.)

(f) –6.283 ÷ (–1.1)

= 6.283 ÷ 1.1

= 5.71 (to 2 d. p.)

5 60 ÷ 1

$

#

= 60 ÷

&

#

= 60 ´

#

&

=

$#'

&

= 40
40 plates are needed.

6 TWM.05 Characterising, TWM.07 Critiquing,
TWM.08 Improving
As the divisor

&

" is less than 1, the result of 3

)

(÷

&

"

should be greater than 3

)

(.

3

)

(÷

&

"

=

&*

(÷

&

"

=

&*

(´

"

&

=

$"'

#!

= 5

*

#!

The correct answer is 5

*

#!.

2

2

2

30

2

3

2

Marshall Cavendish Education Cambridge Lower Secondary Mathematics Student’s Book Year 9
Chapter 2 Fractions, Decimals and Percentages

Ó 2023 Marshall Cavendish Education Pte Ltd

5

Practice 2D

1 (a)

=

=

=

=

(b)

=

=

=

=

(c)

=

=

=

=

=

=

(d)

=

=

=

(e)

=

=

=

=

= 5

$

"('

(f) –

!

(´ (–

$)

#$) + (2

$

$# –

$

$')

=

!

(´

$)

#$+ (2

*

%' –

%

%')

=

#

&+ (1

%*

%' –

%

%')

=

#

&+ 1

*(

%'

=

"'

%'+

$$(

%'

=

$*(

%'

= 2

&(

%'

= 2

$&

#'

2 (a) 6 ´ (0.2 +

!

$') ÷ 18 + 1

= 6 ´ (0.2 + 0.7) ÷ 18 + 1

= 6 ´ 0.9 ÷ 18 + 1

= 5.4 ÷ 18 + 1

= 0.3 + 1
= 1.3

(b) (7.5 +

$

#) ´ 2 – 3.6 ÷ 0.1

= (7.5 + 0.5) ´ 2 – 3.6 ÷ 0.1

= 8 ´ 2 – 36

= 16 – 36
= –20

(c) (0.125 +

&

)) ´ ( –22 – 1.5)

= (0.125 + 0.375) ´ ( –4 – 1.5)

= (0.125 + 0.375) ´ ( –4 – 1.5)

= 0.5 ´ (–5.5)

= –2.75

(d) 2.6 – 3 ´ (

$

# + 1.5) ÷ 0.01

= 2.6 – 3 ´ (0.5 + 1.5) ÷ 0.01

= 2.6 – 3 ´ 2 ÷ 0.01

= 2.6 – 6 ÷ 0.01

= 2.6 – 600
= –597.4

132

245
-´

13

210
-

53

1010
-

2
10
1

5

2121
1
3294
´+´

2125

3294
´+´

15

318

+

65

1818
+

11
18

1422
2

1

2933

æö
-´÷ç÷
èø

5425

2933

æö
-´÷ç÷
èø

583

2275
-´

58

245
-

22516

9090
-

209
90
29
2 90

4114
1

55105

æ

ö
÷+-ç

÷
è

ø

45118

511010

æ

ö
´+-ç

÷
è

ø

3
4
10
+

3
410

5227
1
71452
´+´

5277

71452
´+´

549

4910
+

502401

490490

+

2451
490

2

2

9

7

2

3

Marshall Cavendish Education Cambridge Lower Secondary Mathematics Student’s Book Year 9
Chapter 2 Fractions, Decimals and Percentages

Ó 2023 Marshall Cavendish Education Pte Ltd

6

3 TWM.01 Specialising, TWM.08 Improving
By trial and error, the expression with the largest
value is 3 ´ 6

#

& + 2 ÷

$

$'' –

!

(

= 3 ´

#'

& + 2 ´ 100 –

!

(

= 20 + 200 –

!

(

= 220 –

!

(

= 219

#

(

4 TWM.07 Critiquing
1

"

! is less than 2.

2

$

* is less than 3.

Since (2 + 3) ´ 3 = 15, the value of (1

"

! + 2

$

*) ´ 3

should be less than 15. So, Roland's answer is
incorrect.

Practice 2E

1 (a) 100% ® $30 (original price)

1% ® $

&'

$'' = $0.30

70% ® $0.30 ´ 70 = $21

The new price is $21.

(b) 100% ® $60 (original price)

1% ® $

%'

$'' = $0.60

80% ® $0.60 ´ 80 = $48

The new price is $48.

2 Amount after 10% service charge
= 110% of $12
=

$$'

$''´ $12

= $13.20

Amount after 5% tax

= 105% of $13.20

=

$'*

$''´ $13.20

= $13.86

Jim needs to pay $13.86 for the meal.

3 TWM.07 Critiquing, TWM.08 Improving
No. I disagree with Marlene. All the products are
sold at 90% of the original prices. Since members
get additional 15% off from 90% of the original
price, it is incorrect to simply add the two
percentages.

Non-members pay 100% – 10% = 90% of the
original price. Members pay 100% – 15% = 85% of
the discounted price, or 85% of 90% of the original
price.

85% of 90% = 0.85 ´ 0.9

= 0.765

= 76.5%

The total percentage discount for members is

100% – 76.5% = 23.5%.

The correct statement should be:
The total discount for members is 23.5%.

4 Selling price of drink on weekend
= 93% of the original price

Price after 10% service charge
= 110% of the discounted selling price

= 110% of 93% of the original price

= 1.1 ´ 0.93 of the original price

= 1.023 of the orginal price

Percentage change from the original price

= (1.023 – 1) ´ 100%

= 0.023 ´ 100%
= 2.3% (increase)

5 (a) TWM.04 Convincing

The price of the car increased by 10% in
2021. The price in 2021 was 110% of the
original.

In 2022, the price increased by another 5%
based on the price in 2021. In other words,
the price increased by 5% from 110% of the
original.

Price in 2022
= 10% of original price + 5% of 110% of
original price

As 110% of original price > 100% of original
price, the price in 2022 should be more than
15% of the original.

(b) Price of car in 2021

= 110% of original price
= 1.1 ´ original price

Price of car in 2022

= 105% of price in 2021

= 1.05 ´ price in 2021

= 1.05 ´ 1.1 ´ original price

= 1.155 ´ original price

Net percentage increase in price

= (1.155 – 1) ´ 100%

= 0.155 ´ 100%

= 15.5%

Marshall Cavendish Education Cambridge Lower Secondary Mathematics Student’s Book Year 9
Chapter 2 Fractions, Decimals and Percentages

Ó 2023 Marshall Cavendish Education Pte Ltd

7

6 Amount owed at the end of the first year

= 105% of $5000

= 1.05 ´ $5000

= $5250

Amount owed at the end of the second year

=105% of amount owed at the end of the first year
= 1.05 ´ $5250

= $5512.50

He will need to pay $5512.50 at the end of the

second year for the car loan.

Chapter 2 Revision

1 (a) The range of values of the length l is

46.15 £l < 46.25.

The range of values for the breadth b is

21.15 £b < 21.25.

(b) Smallest possible area

= 46.15 ´ 21.15

= 976.0725 cm2

2 (a)

$

%' is in the simplest form.
The denominator 60 is a multiple of 3.
Thus, the decimal representation of

$

%'

is recurring.

(b)

$*

%' =

$

" = 0.25

Thus, the decimal representation of

$*

%'

is terminating.

(c)

&

&( =

$

$&

= 1 ÷ 13
= 0.0̇76923̇

Thus, the decimal representation of

&

&(

is recurring.

(d)

!

"( =

$

!, which is in the simplest form.

The denominator 7 is a multiple of 7.
Thus, the decimal representation of

!

"(

is recurring.

3 (a) 2

$

*´ 3

$

&

=

$$

*´

$'

&

=

##

&

= 7

$

&

(b) 2

$

*÷ 1

"

*

=

$$

*÷

(

*

=

$$

*´

*

(

=

$$

(

= 1

#

(

(c) 1

$

$'´

*

##

=

$$

$'´

*

##

=

$

"

(d) (–1

$

*) ´ (–

*

()

= 1

$

*´

*

(

=

%

*´

*

(

=

#

&

(e) –2

$

"÷ (–

&

#)

= 2

$

"÷

&

#

=

(

"÷

&

#

=

(

"´

#

&

=

&

#

= 1

$

#

4 (a) 2 ´ (1

$

# –

$

&) – 3

= 2 ´ (

&

# –

$

&) – 3

= 2 ´ (

(

% –

#

%) – 3

= 2 ´

!

% – 3

=

!

& – 3

=

!

& –

(

&

= –

#

&

0.0 7 6 9 2 3 0 7
13 )1.0 0 0 0 0 0 0 0
9 1
9 0
7 8
1 2 0
1 1 7
3 0
2 6
4 0
3 9
1 0
0
1 0 0
9 1
9

2

2

2

2

3

3

2

Marshall Cavendish Education Cambridge Lower Secondary Mathematics Student’s Book Year 9
Chapter 2 Fractions, Decimals and Percentages

Ó 2023 Marshall Cavendish Education Pte Ltd

8

(b) 4 + (2

$

& –

#

&) ´ 3

= 4 + (

!

& –

#

&) ´ 3

= 4 +

*

&´ 3

= 4 + 5

= 9

(c) 4 ÷ (2

$

& – 1

*

%) – 3 ´ (–

$

#)

= 4 ÷ (

!

& –

$$

%) – 3 ´ (–

$

#)

= 4 ÷ (

$"

% –

$$

%) – 3 ´ (–

$

#)

= 4 ÷

&

% – 3 ´ (–

$

#)

= 4 ÷

&

% + 3 ´

$

#

= 4 ÷

$

# +

&

#

= 4 ´ 2 +

&

#

= 8 +

&

#

= 8 + 1

$

#

= 9

$

#

(d)

"

* –

#

&´ (1

$

& –

*

%)

=

"

* –

#

&´ (

"

& –

*

%)

=

"

* –

#

&´ (

)

% –

*

%)

=

"

* –

#

&´

&

%

=

"

* –

$

&

=

$#

$* –

*

$*

=

!

$*

(e) 1.2 ´ 5 – (

&

" + 1.5) ´ 82

= 1.2 ´ 5 – (0.75 + 1.5) ´ 64

= 1.2 ´ 5 – 2.25 ´ 64

= 6 – 144

= –138

5 (a) TWM.04 Convincing

The additional 5% discount is applied on the
discounted price which is 90% of the original.
5% of 90% of the original is less than 5% of
100% of the original. Hence, the total
discount is less than 15%.

(b) Price of petrol after 10% discount

= 90% of original price

Price of petrol on weekdays

= 95% of discounted price

= 95% of 90% of original price

= 0.95 ´ 0.9 of original price

= 0.855 of original price

Total percentage discount

= (1 – 0.855) ´ 100%
= 0.145 ´100%

= 14.5%

6 Price of watch after 15% delivery charge

= 115% of $100

= 1.15 ´ $100

= $115

Price of watch after 7% government tax

= 107% of $115

= 1.07 ´ $115

= $123.05

Kelly needs to pay $123.05 for the watch.

7 Amount after 1 year

= 107% of $5000

= 1.07 ´ $5000

= $5350

Amount after 2 years

= 107% of $5350

= 1.07 ´ $5350

= $5724.50

Amount after 3 years

= 107% of $5724.50

= 1.07 ´ $5724.50

= $6125.215
= $6125.22 (to the nearest cent)

I will get $6125.22 after 3 years.

3