MATHS TEXT BOOK Y5 DLP

SUBTRACTION OF UNITS OF MASS Name Mass
Khairul 34.9 kg
1 C alculate the difference in mass Divesh 41.2 kg
between Khairul and Divesh.
Khairul Divesh

41.2 kg – 34.9 kg = kg

10
3 0 12

4 1 . 2 kg
− 3 4 . 9 kg

6 . 3 kg

41.2 kg – 34.9 kg = 6.3 kg

The difference in mass between Khairul and Divesh is 6.3 kg.

2 How much more is the mass of butter than
the mass of sugar? State the answer in g.
Butter Cake Recipe
Ingredients A 1 kg – 0.1 kg = g
4
• 41 kg of butter
Method 1
• 0.1 kg of sugar
• 5 yolks 1 kg – 0.1 kg = (41 × 1 250 g – (0.1 × 1 000) g
• 230 g of 4
000)
wheat flour
1

= 250 g – 100 g

= g

Method 2

0.25 kg 1 kg – 0.1 kg = 0.25 kg – 0.10 kg
4 1 .00 kg 4
= 0.15 kg
1 −0
Convert 4 kg 10 = (0.15 × 1 000) g

to decimal kg −8 = g
20
first.
− 20
0 Are the answers
for method 1 and
method 2 the same?

The mass of butter is g more than the mass of sugar.

• Carry out activities such as to find the difference in mass between two friends. 193
5.2.3 • Encourage pupils to subtract decimals using vertical form to ensure that the

units are the same and the decimal points are aligned.

3 6 042 g – 1.07 kg – 980 g = kg

Convert unit of g to kg Subtracting consecutively

6 042 g = (6 042 ÷ 1 000) kg 9 18
= 6.042 kg 5 1014 3 8 17

980 g = (980 ÷ 1 000) kg 6 . 0 4 2 kg 4 . 9 7 2 kg
= 0.980 kg – 1 . 0 7 0 kg – 0 . 9 8 0 kg

4 . 9 7 2 kg 3 . 9 9 2 kg

6 042 g – 1.07 kg – 980 g = 3.992 kg

4 10 54 kg – 1 503 g – 3.96 kg = g
Convert unit of kg to g
Subtracting consecutively
10 54 kg = 10 kg + 4 kg ( 45 200
= 5 10 800 g
000) – 1 503 g
(10 × 1 000) g + × 1 g
g
= 10 000 g + 800 g 1 – 3 960 g

= 10 800 g g

3.96 kg = (3.96 × 1 000) g 1 51 kg – 800 g = p kg
= 3 960 g q

Find values of p and q.

1 Calculate. 3
4
a 7.1 kg – 2.54 kg = kg b kg – 0.69 kg = g
c 5 120 g – 1 170 kg – 2.093 kg =
kg

d 10 21 kg – 4 860 g – 3.72 kg = g

2 a Calculate the difference between 38.92 kg and 60 51 kg.

b How much more is 13.051 kg than 7 360 g?

c Subtract 5 52 kg from 9.018 kg. State the answer in g.

194 • Carry out a question and answer session involving subtraction of mass
using words such as calculate the balance, find the difference, and how
5.2.3 much more is a particular mass from another mass.

May I help you,

MULTIPLICATION OF UNITS OF MASS madam?

1 Calculate the mass of 7 packets of hazelnuts. I want 7 packets of
hazelnuts.
7 × 0.25 kg = kg

13

0 . 2 5 kg
× 7 kg

1 . 7 5 kg

7 × 0.25 kg = 1.75 kg
The mass of 7 packets of hazelnuts is 1.75 kg.

2 What is the mass of 15 packets of crisps?

15 × 1 kg = g
2

15 × 1 kg = 15 × (21 × 1 500 g
2
000)

1 2

= 15 × 500 g 15
× 500 g
= 7 500 g 1
1 7 500 g 2
2 kg
15 × kg = 7 500 g

The mass of 15 packets of crisps is 7 500 g.

3 10 × 40 35 kg = g
Method 1
Method 2

40 35 kg = 40 kg + 3 kg 10 × 40 53 kg = 10 × (40 kg + 3 kg)
5 5

= (40 × 1 000) g + (35 × 1 200 g = 10 × (40 kg + 0.6 kg)

000) = 10 × 40.6 kg

1

= 40 000 g + 600 g = 406 kg

= 40 600 g 406 kg = (406 × 1 000) g
= 406 000 g
10 × 40 600 g = 406 000 g
10 × 40 35 kg = 406 000 g

5.2.4 • Encourage pupils to do quick calculation when multiplying any 195
mass by 10, 100 or 1 000.

4 No. Number of packets Brand of cat food Mass per unit

1. 100 Meow Meow 1.2 kg

2. 1 000 Comel 0.85 kg

3. 1 000 Anggun 480 g

Based on the table above, calculate the total mass of:

a M eow Meow. kg b Comel. g
100 × 1.2 kg = 1 000 × 0.85 kg =

100 × 1.2 kg = 120 kg 1 000 × 0.85 kg = 850 kg

100 × 1.2 kg = 120 kg = (850 × 1 000) g

The total mass of Meow Meow = 850 000 g

is 120 kg. 1 000 × 0.85 kg = 850 000 g

The total mass of Comel
is 850 000 g.

Calculate the total mass,
in kg, of Anggun.

1 Calculate. kg b 16 × 4.5 g = kg
a 9 × 1.07 kg = g d 50 × 6.78 kg = g
c 24 × 3 110 kg =
kg b 10 × 4 kg = g
2 Quick calculation. 5

a 10 × 2.06 kg =

c 100 × 1.9 kg = g d 100 × 1 41 kg = g

e 1 000 × 0.003 kg = g f 1 000 × 3 170 kg = kg

196 5.2.4 • Drill multiplication using flash cards. Encourage pupils to calculate
mentally when multiplying any mass by 10, 100 or 1 000.

DIVISION OF UNITS OF MASS The mass of 7
chocolate bars
1 Calculate the mass of one chocolate bar.
0.35 kg ÷ 7 = kg is 0.35 kg.
0 . 0 5 kg
7 0 . 3 5 kg
−0
03
−0
35
− 35
0
0.35 kg ÷ 7 = 0.05 kg
The mass of one chocolate bar is 0.05 kg.

2 12.33 kg ÷ 18 = g 3 16 21 kg ÷ 10 = g

0 0 . 6 8 5 kg 16 21 kg = 16 kg + 1 kg
18 1 2 . 3 3 0 kg 2

−0 = 16 kg + 0.5 kg
12
= 16.5 kg
−0
12 3 16.5 kg ÷ 10 = 1.65 kg
= (1.65 × 1 000) g
−1 0 8 = 1 650 g
1 53
16 21 kg ÷ 10 = 1 650 g
−1 44
90 Try another method by converting
16 21 kg to g first, then divide.
−90
0

0.685 kg = (0.685 × 1 000) g
= 685 g
12.33 kg ÷ 18 = 685 g

• Guide pupils to do a group activity. Pupils will weigh suitable items 197

5.2.5 and divide the number of packets of item according to the same
mass by giving the answers in the required units.

4 5 1.4 kg ÷ 1 000 = g
1.4 kg ÷ 1 000 = ( × ) g ÷ 1 000
The mass of 100 pieces of = g ÷ 1 000
= g

macaroon is 1 21 kg. Try to solve it.
Calculate the mass, in g, of one

piece of macaroon.
1 21 kg ÷ 100 =
g

1 21 kg ÷ 100 = (32 × 1 500 g ÷ 100

000)

1

=1150000 g 042 1 3

= 15 g Use all the number cards above to
complete the empty boxes below.
1 21 kg ÷ 100 = 15 g
. kg = . kg ÷
The mass of one piece of

macaroon is 15 g.

1 Calculate. kg b 10.2 kg ÷ 6 = kg
a 9.02 kg ÷ 5 =
c 13.51 kg ÷ 14 = g d 7 43 kg ÷ 50 = g
e 21.1 kg ÷ 10 =
g 6.8 kg ÷ 1 000 = g f 36 54 kg ÷ 100 = kg

2 g h 9 170 kg ÷ 1 000 = g

Granulated Based on the diagram on the left,
Sugar complete the empty boxes below and
4.5 kg solve it.

kg ÷ = g

198 5.2.5 • Guide pupils to apply cancellation method or shift the decimal point
when dividing any mass by 10, 100 and 1 000.

CONVERT UNITS OF MILLILITRE AND LITRE

1 a Convert 0.3 to m .

This bottle 0.3 = m
contains
0.3 of juice. This box 0.3 = (0.3 × 1 000) m
conotfajiunisc e 2 1.   = 300 m
0.3 = 300 m

b 21 = m Is 0.5 equal to 1 ?
2
Explain.

21 = (21 × 1 500 m 1.0 1 000 m

000)

1

= 500 m 0.5 500 m

21 = 500 m

2 2 750 m =

in decimal in fraction

2 750 m 2 750 m = 2 000 m + 750 m
= (2 750 ÷ 1 000)
= 2.75 = (2 000 ÷ 1 000) + (1705000)
75 ÷ 25
= 2 + ( 100 ÷ 25 )

= 2 + 34 1 = 1 000 m

=2 43 × 1 000

m

2 750 m = 2.75 or 2 43 ÷ 1 000

• Ask pupils to measure the volume of coloured liquid using 199
5.3.1 measuring apparatus and convert the units into fractions or

decimals.

3 42 9 m = 4 700 m =

42 9 m = 42 + (1 9 ) 700 m = (1700000) Is 170 equal to
000 0.7 ? Discuss.

= 42 + 0.009 =170

= 42.009

42 9 m = 42.009 700 m = 7
10

5 6 025 m = 6 27 53 = m
27 35 = 27 + 53
6 025 m = ( 6 025 ) = (27 × 1 000) m
1 000

= ( 6 000 + 1 25 )
1 000 000

= 6 1 20500÷÷2525 + (53 × 1 200 m

000)

= 1

= m + m

= m

1 Convert to m or vice versa. d e

abc Pomegranate

Juice

0.35 1.2 15 m 200 m 1 400 m

2 State the answers in fractions. c 3 900 m =
a 800 m = b 1 200 m =

3 State the answers in decimals. c 26 40 m =
a 5 m = b 10 080 m =

4 Solve these. m b 5 170 = m
a 6 81 =

200 • Use fractions with denominators 2, 4, 5, 8 and 10 to reinforce
5.3.1 pupils’ understanding involving the conversion of units of volume of
liquid.

ADDITION OF UNITS OF VOLUME OF LIQUID

1

5 21 1.875 125 m

a Calculate the total volume of blue and yellow paints.

1.875 + 5 21 = 0 5.5
2 1 1 .0
1.875 + 5 21 = 1.875 + 121
−0
= 1.875 + 5.5 11

= 7.375 1 −1 0
1.8 7 5 10
+ 5.5 0 0
−1 0
0

1.875 + 5 21 = 7.375 7.3 7 5

The total volume of blue and yellow paints is 7.375 .

b Find the total volume of blue and red paints.

1.875 + 125 m = m

1.875 + 125 m = (1.875 × 1 000) m + 125 m

= 1 875 m + 125 m

= 2 000 m

1.875 + 125 m = 2 000 m

The total volume of blue and red paints is 2 000 m .

The total volume of the three paints is more

than 8 . Is this statement true? Prove it.

5.3.2 • Stress that the unit of volume must be the same before 201
performing addition.

2 9.27 + 13 500 m + 10 52 =

13 500 m = 13 500 m 10 25 = 10 + 25 0 . 4
= 10 + 0.4 5 2 . 0
= (113050000 )
= 13.5 = 10.4 −0
20
11
−2 0
9 .2 7 0
1 3 .5 0
+ 1 0 .4 0
3 3.1 7

9.27 + 13 500 m + 10 52 = 33.17

3 40.08 + 11 190 + 76 m = m

Method 1 11 190 = 11 + 190 Method 2
= 11 + 0.9
= 11.9 76 m = (76 ÷ 1 000)
= (11.9 × 1 000) m =0.076
= 11 900 m
40.0 8 11
40.08 = (40.08 × 1 000) m + 1 1.9 0
= 40 080 m 5 1 .9 8 0
5 1.9 8 + 0 .0 7 6
4 0 0 8 0 m
1 1 9 0 0 m 5 2 .0 5 6
+ 7 6 m
52.056 = (52.056 × 1 000) m
m =m

Are both methods above
correct? Discuss.

Calculate. b 2 34 + 7.265 =
a 3.8 + 9.204 =

c 10.46 + 578 m = m d 190 + 12 304 m =

e 3 51 + 680 m + 0.2 = f 8 645 m + 12 21 + 4.1 =

202
5.3.2 • Identify pupils’ weaknesses and provide more exercises to enhance
their understanding.

SUBTRACTION OF UNITS OF VOLUME OF LIQUID

1 What is the difference in volume of the
liquid in the syringes?
5 m – 3.3 m = m

4 10

5. 0 m
− 3. 3 m

1.. 7 m

5 m – 3.3 m = 1.7 m
The difference in volume of the liquid in the syringes is1.7 m .

2 I have filled up What is the balance of the volume of
2.5 juice the juice in the container?

into a jug. 10 41 – 2.5 =

Full Step 1 Step 2
volume 10 41 = 10 + 41
10 41 = 10 + 0.25 9
0 10 12
= 10.25
1 0 .25
− 2 .50

7 .75

10 41 – 2.5 = 7.75

The balance of the volume of the
juice in the container is 7.75  .

3 12.09 − 780 m = m

12.09 − 780 m = (12.09 × 1 000) m − 780 m 1 10

= 12 090 m − 780 m 1 2 0 9 0 m
− 7 8 0 m
= 11 310 m
1 1 3 1 0 m

12.09 − 780 m = 11 310 m

5.3.3 • In groups, carry out an activity of subtracting two volumes using 203
a game of Dominoes.

4 8 25 – 1 360 m – 4.91 =

8 52 – 1 360 m – 4.91

= (8 + 0.4 ) – ( 1 360 ) – 4.91 51 = 0.2 or 200 m
1 000 52 = 0.4 or 400 m

= 8.4 – – 4.91

=

5 20 35 m – 0.948 – 16 81 = m

Convert unit Subtracting consecutively

0.948 = (0.948 × 1 000) m m m
= 948 m
1 9 1 035 18 1087
16 81 = 16 + 81 125
(81 20 35 19 87
= 16 + × 1 000) m − 948 −16 125

19 87 2 962

1
= 16 + 125 m

= 16 125 m

20 35 m – 0.948 – 16 81 = 2 962 m

1 Calculate.

a 10 m – 4.5 m = m b 13 21 – 1.85 =

c 7.025 – 629 m = m d 9 53 – 2 084 m = m

e 12 34 – 960 m – 8.47 =

f 8 110 – 3 640 m – 1.02 = m

2 Solve these.
a 15.24 – 6 81 – 120 m = m
b 6 320 m – 4.5 – 1 21 = m

204 • Carry out a question and answer session involving subtraction of
5.3.3 21 equals
volume of liquid such as “Subtract 0.2 from
how many m ?”.

MULTIPLICATION OF UNITS OF VOLUME OF LIQUID

1 C alculate the total volume of 5 bottles

of cultured milk. Each bottle
5 × 0.08 =
contains 0.08 .

4

0. 0 8

×5

0. 4 0

5 × 0.08 = 0.4
The total volume of 5 bottles of cultured milk is 0.4 .

2 W hat is the total volume of water in the box? 48 x 230 m

48 × 0.23 = m

Method 1 1 Method 2

• Multiply. 1 2
0.2 3
• C onvert 0.23 to m .
× 48 (0.23 × 1 000) m = 230 m

11

1 84 • Multiply. 1
2
+09 20
2 3 0 m
1 1 .0 4
× 48

• Convert 11.04 to m . 1
(11.04 × 1 000) m = 11 040 m
1 840

+ 9200

48 × 0.23 = 11 040 m 1 1 0 4 0 m

The total volume of water in the box is 11 040 m .

3 1 0 × 1 35 =

10 × 1 35 = 2 × 8
5
10

= 16 1 State the answer in m .

10 × 1 35 = 16

5.3.4 • Explain the meaning of cultured milk, which is the dairy products 205
produced when sterilised milk, pasteurised milk, or skimmed milk is
added with specifically cultured bacteria that helps in the digestion
process.

4 BBB
Sanitiser
CLEAN WORLD FACTORY 50 m HAND 0.4 1
HAND 2
SANITISER
SANITIZER
BB
B

a Calculate the volume, in m , of 100 bottles of BB hand sanitiser.

100 × 0.4 = m

Method 1 Method 2

0 .4 40 = (40 × 1 000) m 0.4 = (0.4 × 1 000) m
×10 0 = 40 000 m = 400 m

4 0 .0 100 × 400 m = 40 000 m

100 × 0.4 = 40 000 m

The volume of 100 bottles of BB hand sanitiser is 40 000 m .

b What is the volume, in m , of 1 000 bottles of BBB hand sanitiser?

1 500 × 21 = do cancellation Calculate the
× ) m convert unit volume, in , of 10
000
bottles of
1 =(
B hand sanitiser.
=m

The volume of 1 000 bottles of BBB hand sanitiser
is m .

1 Solve these. b 13 × 5.7 = m c 25 × 0.416 = m
a 16 × 0.39 = m e 100 × 6 170 = m f 1 000 × 7.8 m =
d 48 × 2 41 =
2 Complete these. as 10 × 0.75 as 100 × 43

1 000 × 75 m m



206 • Vary questions such as using a mind map to reinforce
5.3.4 pupils’ understanding.
• Instil moral values such as taking care of hygiene and health.

DIVISION OF UNITS OF VOLUME OF LIQUID

1 W hat is the volume of a bowl of mutton soup
based on the picture given?

2.5 ÷ 4 = I poured 2.5 of mutton soup
equally into 4 bowls.
0.6 2 5
4 2.5 0 0

−0
25

−2 4
10

−8
20

−2 0
0

2.5 ÷ 4 = 0.625
The volume of a bowl of mutton soup is 0.625 .

2 6.93 ÷ 15 = m Method 2

Method 1 6.93 = (6.93 × 1 000) m
0.4 6 2 = 6 930 m

15 6 . 9 3 0 0 4 6 2 m
−0 15 6 9 3 0 m
69
−6 0 −0
93 69
− 90
30 −6 0
−30 93
0
−90
0.462 = (0.462 × 1 000) m 30
= 462 m
6.93 ÷ 15 = 462 m −30
0

5.3.5 • Guide pupils to divide using various strategies to reinforce 207
their knowledge.

3 7 52 ÷ 50 = 0 . 1 4 8
50 7 . 4 0 0
7 52 ÷ 50 = 37 ÷ 50
5 −0
74
=7.4 ÷ 50
−5 0
=0.148 2 40

7 25 ÷ 50 = 0.148 −2 0 0
400

− 400
0

4 What is the volume, in m , of a mini bottle of honey?

2.5 ÷ 100 = m HONEY HONEY HONEY
HONEY HONEY HONEY

2.5 = (2.5 × 1 000) m HONEY HONEY HONEY
=2 500 m
2 501000m = 25 m PARTY SOUVENIRS
2.5 ÷ 100 = 25 m
The volume of a mini bottle of honey is 25 m . contains 100
mini bottles of honey

5 48 ÷ 1 000 = State the
1 40800 = 0.048 answer in m .
48 ÷ 1 000 = 0.048
51 ÷ 100 = ÷ 10
What is the value in ?

1 Calculate.

a 1.2 ÷ 4 = b 9.6 m ÷ 8 = m

c 25 41 ÷ 50 = m d 60 45 ÷ 32 = m
e 19 170 ÷ 100 = f 43.9 ÷ 100 = m

2 a Divide 31 800 m by 1 000. Give the answer in m .

b Calculate 33.36 divided by 60. State the answer in m .

208
5.3.5 • Vary calculation strategies to find the answers.

GROUP ACTIVITY

Tools/Materials product catalogues, manila cards, glue, pens

Task

1 Gather information regarding 2 Cut and paste the information
length, mass, or volume of liquid on the manila card.
from product catalogues (printed
or downloaded from websites) Potato Crisps
or through online shopping sites.

1.5
130 g

3 Construct questions on unit conversion or addition, subtraction,
multiplication, and division operations. Solve them. For example:

Convert 1.5 to: Potato Crisps Calculate the total
a m . 130 g mass, in kg, of 2 cans
of potato crisps.
b in fraction.

a 1.5 = (1.5 × 1 000) m 2 × 130 g = kg
= 1 500 m
130g
1.5 b 1.5 = 1 + 0.5 ×2

= 1 + 5÷5 260g
10 ÷ 5
260 g = (260 ÷ 1 000) kg
=1 21 = 0.26 kg

4 Present the work and discuss. 5 Display the work at the
mathematics corner.

5.1, • Assess pupils’ mastery in terms of knowledge, communication 209
5.2, 5.3 skills, thinking skills, soft skills including attitude and values while

performing the activity.

SOLVE THE PROBLEMS

1 Cliff wants to form a right-angled triangle. 12 cm 1 m
He uses three ropes and 5
the measurements are as shown.

a Calculate the total length, in cm, of the

ropes.

b W hat is the difference, in m, between 0.16 m

the longest rope and the shortest rope?

Understand • The length of the ropes are 12 cm, 0.16 m, and 1 m.
the problem • Calculate the total length of the ropes in cm. 5

• Find the difference between the longest rope and the

shortest rope in m.

Plan the strategy 12 cm 1 m 0.16 m
Solve 5

total length

a 12 cm + 1 m + 0.16 m = cm b 1 m – 12 cm = m
5 5
= 20 cm – 12 cm
12 cm + 1 m + 0.16 m = 8 cm
5
20
= 12 cm + (51 × cm + (0.16 × 100) cm = (8 ÷ 100) m
100) = 0.08 m

1
= 12 cm + 20 cm + 16 cm

= 48 cm

Check b 0.08 m + 12 cm

a 4 8 cm 3 2 cm = 0.08 m + (12 ÷ 100) m
– 1 6 cm – 2 0 cm
= 0.08 m + 0.12 m
3 2 cm 1 2 cm
= 0.2 m

12 cm + 1 m + 0.16 m = 48 cm 1 m – 12 cm = 0.08 m
5 5

The total length of the ropes The difference between the longest
is 48 cm. rope and the shortest rope is 0.08 m.

210 5.4.1 • Guide pupils to find keywords in the questions and check
the answers using a calculator.

Ashley’s 8 41 km school
2 home Jeera’s

home

Based on the diagram above, the distance of Ashley’s home to school is

3 times the distance of Jeera’s home to school. What is the distance, in m,

from Jeera’s home to school?

Solution The distance of 8 41 km Draw the
Ashley’s home diagram. Write

The distance of the number
Jeera’s home ? sentence.

8 41 km ÷ 3 = m

Convert 8 41 km 8 41 km = ( 343 × 1 250 m
to m.
000)

1 2 750m

1 3 8 250m Check
1 −6
22 21
250m −2 1
× 33 15 2 750m
− 15 ×3
1 750 00
−0 8 250m
+7 500 0
8 41 km ÷ 3 = 2 750 m
8 250m

The distance from Jeera’s home to school is 2 750 m.

3 In conjunction with the recycling campaign, Underline the
5 Meteor pupils collected 4.67 kg of cans. The mass important information.
of paper collected is 12 times the mass of cans.
Calculate the mass, in g, of paper collected.

Solution 12 × 4.67 kg = g 1 1

Convert 4.67 kg = (4.67 × 1 000) g 4 670g
× 12
4.67 kg to g. = 4 670 g
11
The mass of paper collected is g.
9 340

+46 700

56 040g

• Show various problem-solving strategies such as constructing 211

5.4.1 a table or simulation.
• Discuss the method to check the answer for example 3.

4 The table on the right shows the Product Volume
volume of some products. Puan Ani date milk 200 m
bought two products. She bought
10 bottles of each product. The total pomegranate 0.25
volume of the products bought juice 21
is 7 . What are the two possible
products bought by Puan Ani? olive oil

hand gel 450 m

Solution Calculate the total Identify two
volume of 10 bottles. volume with the
Convert all volume in decimal .
10 × 0.2 = 2.0 total of 7 .
200 m = (200 ÷ 1 000) 2.0 + 5.0 = 7
= 0.2 10 × 0.25 = 2.5
21 = 0.5 2.5 + 4.5 = 7
450 m = (450 ÷ 1 000) 10 × 0.5 = 5.0
= 0.45 10 × 0.45 = 4.5

The two possible products bought by Puan Ani are date milk and
olive oil or pomegranate juice and hand gel.

Puan Ani wants to fill up one bottle of pomegranate juice equally into
2 glasses. Which glass should she choose? Explain.

P Q

120 m 130 m

212 5.4.1 • Guide pupils to calculate mentally and estimate efficiently to identify
the total of a certain quantity.

Solve the following problems.

a P uan Anita bought a piece of white cloth and batik cloth. The length

of the white cloth is 20 34 m. The length of the batik cloth is 1.5 m more
than the length of the white cloth. Calculate the length, in cm, of the

batik cloth bought by Puan Anita.

b T he diagram below shows the location of some places on a straight
road.

Azam’s Ajay’s recreational swimming
home home park complex

36 21 km
Ajay’s home is located 19.7 km away from Azam’s home and
9 km 80 m from the recreational park.

i Calculate the distance, in km, from the recreational park to the
swimming complex.

ii A zam goes to Ajay’s home twice a week for a group study.
Calculate the total distance of a round-trip, in m, taken by Azam.

c 2.1 kg

The diagram above shows some rubber balls of the same type and
size on a balanced weighing scale. Calculate the mass, in g, for one
rubber ball.

d T he table on the right shows the volume of three Water Volume
drinking water containers prepared by Lay Ting. container
i Calculate the total volume, in , of drinking
P 3.18

water in the three containers. Q 750 m

ii The drinking water in container R was filled R 3 21
equally into 30 glasses. Each glass contains

150 m of drinking water. Is it a true statement?

Prove it.

5.4.1 • Construct more questions as in (a) to (d) and ask pupils to solve the 213
questions in groups.

1 Convert the measurements to the required units.
a 0 .8 cm = mm b 3 41 m = cm c 9 002 m = km

d 17 m 3 cm = m e 126 mm = cm f 45 km 9 m = km

2 Calculate.

a 10.9 cm + 8.21 cm = cm b 6 35 m + 0.78 m + 94.1 cm = cm
d 13 81 km – 6.03 km – 40 m = km
c 9 110 m – 65 cm = cm f 100 × 0.46 km = m
e 8 × 1 54 m = m
h 76 m 9 cm ÷ 1 000 = cm
g 9.1 km ÷ 26 = m

3 Solve these.
a Add 24.6 cm and 37 cm. State the answer in mm.
b Is 18 × 7 21 cm equal to 5.4 m ÷ 4? Show the calculation.

4 Complete the unit conversion.

Measurement In decimal In fraction
a 4 200 g kg kg

b g 3.9 kg kg
c 600 m
d m 8.7

5 Calculate.

a 1 kg + 0.15 kg = kg b 47.07 kg + 13 080 g + 9 34 kg = g
2 d 12 190 kg – 6.98 kg – 1 120 g = g
c 7 41 kg – 0.26 kg = kg f 2 87 kg ÷ 100 = g

e 18 × 4.5 g = kg

6 Find the answers. m b 6 41 + 1 140 m + 3.9 =
a 0.375 + 900 m = d 40.308 – 17 21 – 580 m =
f 8 92 m ÷ 1 000 = m
c 15.01 – 860 m = m
e 27 × 3 51 = m

214 5.1, • Encourage pupils to use various calculation methods to find
5.2, 5.3 the answers.

7 Solve these.

a How much should be subtracted from 3 206 g to become 1.507 kg?

b F ind the quotient of 16 34 and 25. State the answer in .

8 Solve the following problems.

a Aishah used 2 41 m of red ribbon and 85 cm of blue ribbon to make
a handicraft.

i Calculate the length, in m, of the blue ribbon.

ii Calculate the total length, in m, of the two ribbons.

b Based on the diagram on the right, 1 25 km school
the distance from the school to the

bookshop is 4 times the distance Adam’s
home
from Adam’s home to the school. 5.09 km
Adam cycles from his home to the

bookshop passing by the school.

Then, he returned home by the shortest route.

i Calculate the distance, in m, from the school Bookshop
to the bookshop.
bookshop
ii Calculate the total distance, in km, taken by Adam.

c Encik Sulaiman donated two boxes that contained 96 bottles of
mineral water altogether to the flood victims. The total volume
of water is 24 . What is the volume, in , for one bottle of
mineral water? State the answer in fraction.

d The table shows the mass of three types of Cake Mass
cake that is sold in a shop. 1 21 kg

i What is the mass, in kg, for one Chocolate 50 g more than
carrot cake
strawberry cake? 1.2 kg
ii The shopkeeper sold 6 kg of cake in one Strawberry

afternoon. State the number and types of Carrot
cakes that might have been sold by him.

e I used 125 mm of I used 0.5 cm less than
wire to form a regular the length of wire you
pentagon. used to form a square.

State the length of sides, in cm, for:

Jacky i pentagon. ii square. Farah

5.53..52,.35, .4•• AInddgrsoimuppsl,ecqaurreystoiount squtoizizmepsrtoovseoplvreob“TlerymI-st oAlgvianign”sqkiullse.stions. 215

Tools/Materials coins, markers , game cards, papers, pens

Participants 2 players and 1 referee

S & A GAME SOLVE AND ARRANGE

0.84 cm + 37 mm + 1 cm 7 m – 34 cm – 5.6 cm 70 × 3 21 km = m
2 8 1 000 × 17.8 =

= mm = cm

2 51 kg ÷ 8 = marker station

g

Multiply 3 kg by 15. Give Divide 9.018 km by 9.
4 Give the answer in m.

the answer in g.

A piece of robe needs The mass of 1 teabag is Aida drinks 14.7 of
3 34 m of cloth. What water in one week. In
is the length, in cm, 0.002 kg. The mass of one day, she drinks

for 6 pieces of robes? 100 teabags is g. m of water.

How to play

1 Put all the questions cards face down.
2 Toss the coin to decide the first player.
3 The first player opens a question card and answer the question.
4 Put a marker on any circles in the marker station if the answer is correct.

Only one marker is to be placed for each correct answer.
5 Take turns. Repeat steps 3 and 4 until all the question cards are answered.
6 The first player to place all the markers in these forms

, , or , wins.

216 5.1, 5.2,
5.3, 5.4 • Modify questions on the game cards to suit pupils’ ability.

6 SPACE Let’s identify the
characteristics of
REGULAR POLYGONS regular polygons.

1

Wow! This mural is
beautiful.

There is a Yes, there are regular
pentagon! polygons and

irregular polygons.

a straight side b

teacher friends

diagonals all straight sides
are of equal length

closed all angles

corner interior angle shape are of equal
characteristics size
of regular
Characteristics of a
regular pentagon the number polygons

of angles the number of

• 5 straight sides are of equal length equals the symmetrical
• 5 interior angles
• 5 corners number of axes equals
• 5 diagonals
• 5 symmetrical axes corners the number of

reference straight sides

books websites

• Carry out polygon-shaped folding activities for pupils using pieces of 217
paper or card to explore the characteristics of regular polygons.

6.1.1 • Explain to pupils that corners are synonymous with vertices. Review the
characteristics of regular polygons. Explain the meaning of interior angle.

2 Name this regular polygon. Explain the corner A diagonal is a
characteristics of this regular polygon. diagonal straight line drawn
from one corner to
corner the opposite corner.

diagonal

corner corner

1 Name each polygon based on its characteristics.

a •• 6walshl tisrcathriagaihgrtehstcisdloiedssee s dof b • 3 straight sides c
of equal length • 7 corners
•• 1oa4flledsqitaruagaioglnhleatnlssgidthes
• 3 equal interior

equal length axes angles
• 6 symmetrical • 0 diagonal

2 How many diagonals a b
are there in each
of these regular
polygons?

3 Complete the table below.

Regular Number Number Number of Number Number
polygon of straight of symmetrical of angles of

equilateral sides corners axes diagonals
triangle 3
square 4 3
pentagon 6 5
hexagon 4
heptagon 8 7 5
octagon 69

7 14
8 20

218 • Create a mobile project (dangling ornament) of regular polygons with their
6.1.1 characteristics and display them at the mathematics corner.
• Guide pupils to identify the number of diagonals for regular polygons.

MEASURING INTERIOR ANGLES

1 The size of all interior angles in a regular pentagon are equal.
What is the value of angle a ?

A protractor is used to Ways to measure the value of angle a:
measure the value of
1 Place the centre of 2 Make sure line
angle a.
the protractor at AE overlaps the

corner A. baseline of the

B C protractor.

3 Read the inner

scale from 0°

(line AE) to the

inner scale B D side (line AB).

outer scale 4 The value of a
is 108o or
108o 110o − 2o = 108o.
a
A The value of angle a is one

E hundred and eight degrees.

baseline centre of protractor 108o

2 State the value of angle p.

angle p corner P
P
Make sure the centre
p of the protractor

overlaps the corner
of the angle that is
to be measured, p.

The value of angle p is .

Can you state the types of angles for a and p? Why?

• Introduce the characteristics of protractor to pupils such as the 219

6.2.1 inner scale, outer scale, baseline, and centre of protractor.
• Demonstrate how to measure angles using a protractor in detail.

3a bS

b projection
a line

p

R

The value of angle a is .

The value of angle b is . Outer scale reading in S: 0°

Outer scale reading in R: 120°

The value of angle p = −

Name the polygon which has a right =
angle 90° and acute angle.

Copy the regular polygon diagrams below.
Measure and write down the value of the interior angles.
a b Qc

q

p
w

pw

220 • Guide pupils to solve the ‟Try These” question while ensuring that they
6.2.1 measure the value of the angles using the correct technique.
• Carry out a project of measuring angles of a polygon mask to reinforce
pupils’ understanding.

PERIMETERS OF COMPOSITE SHAPES

1 Shanti, let’s decorate
the outline of the
Indung, we bookmark with this
have made green-coloured
this bookmark paper.
out of these
two regular
polygons.

How do we determine the length of the green-coloured paper
that is needed?

The length of the 9 cm 9 cm
green-coloured
paper is the length 9 cm
of the outer sides or
the perimeter of the
composite shape.

9 cm

9 cm

1 Measure all the 9 cm 9 cm
outer sides of
the composite Perimeter = (9 + 9 + 9 + 9 + 9 + 9 + 9) cm
shape. = 63 cm

2 Total up the OR
length of all the Perimeter = 7 × 9 cm
outer sides of the = 63 cm
composite shape. The length of the green-coloured paper
needed is 63 cm.
9 cm

Is the perimeter of the composite shape of this
regular polygon 63 cm as well? Discuss.

• Provide each group with a few regular polygon shapes. Each group must 221
6.3.1 form a composite shape from two polygons and determine the perimeter

of the composite shape.

2 4m
4 m 4 m end 10 m

4m 4m start Determine one
4m 4m 6m 10 m side as a starting
point to calculate.

Method 1 10 m

Perimeter = (10 + 10 + 6 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 10) m
=64 m

Method 2 Perimeter = Perimeter of the outer + Perimeter of the outer

sides of the octagon sides of the square

= (7 × 4 m) + (10 m + 10 m + 10 m + 6 m)

=m + m

=m

3 Determine the perimeter of the shaded region.

start Perimeter of the shaded region Total up the
= (6 + 6 + 5 + 6 + 5 + 6 + 6) cm surrounding sides of
6 cm = cm the shaded region.

5 cm 6 cm

5 cm

4 4m Discuss the error in
5m calculation of this perimeter.
1m
Perimeter of the shaded region
2m (4 + 5 + 2 + 1 + 1) m = 13 m

222 6.3.1 • Carry out exploration activities of two composite shapes using regular
polygons including triangles and quadrilaterals to find perimeter.

MY DESIGN

Tools/Materials MS Word, A4 papers, printer, rulers, pens
Participants 4 pupils in a group

Task

1 Launch MS Word. EXAMPLES OF
COMPOSITE SHAPES
2 Click Insert. Click Shapes. Select two regular polygon
shapes up to six sides including right-angled triangles,
isosceles triangles, and rectangles.

3 Make a few composite shapes from any two regular
polygons using your own creativity by adjusting the
size, sequence, and colour for each shape.

4 Print and measure the perimeter of each shape that
you have made.

5 Present your completed work through Gallery Walk.

1 Calculate the perimeter for each of the composite shapes below.
3 cm 3 cm
a 3 cm b 4m

3 cm

3 cm 12 m 5m

3 cm 3 cm

3 cm
3 cm

2 Find the perimeter of the shaded region.

ab

7 m 10 cm
6 cm
5 cm

5 m 3.5 cm 4.5 cm

• Carry out activity in groups to answer the ‟Try These” questions. 223
6.3.1 • Each group presents their work and other groups can suggest other ways

when appropriate.

AREA OF COMPOSITE SHAPES

1 Calculate the area of the composite shape coloured by Chin.

4 units 4 units Write down the
AB measurement of its

length, breadth,
base, and height.

2 units
4 units

Area of the square, A:
4 units × 4 units = 16 units2

Area of the right-angled triangle, B:

Area of a square 1 × 1 units × 4 units = 4 units2
= length × breadth 2
2

1

Area of a triangle Area of the composite shape = 16 units2 + 4 units2
= 20 units2
= 1 × base × height
2

The area of the composite shape coloured by Chin is 20 units2.

2 2m What is the area of the painted wall?

3m The area of the painted wall
A = surface A + surface B
3m B = (3 m × 2 m) + (3 m × 5 m)
5m = 6 m2 + 15 m2
= 21 m2

The area of the painted wall is 21 m2.

224 • Discuss the role of the inner sides of the composite shape to calculate

6.3.2 the area of the composite shape.
• Carry out group activities to calculate areas based on two given

composite shapes. Each group presents their work.

10 cm3 Area of the composite shape
60 cm 10 cm
40 cm = area + area
= (21 × 60 cm × 40 cm) + (21 ×
100 cm = cm2 + cm2 cm × 100 cm)
= cm2

4 Dad planted grass on the shaded region as shown. Find the area of
the grassy compound.
6m
Area of the grassy compound
= area of the rectangle – area of the square

4.8 m = (6 m × 4.8 m) – (2 m × 2 m)
2 m = m2 – m2

= m2
2m

1 Calculate the area of each of the composite shapes below.

a 7m b

5m

4m 10 cm

7m 6 cm 6 cm

1 m 2 cm 6 cm
18 cm
2 Calculate the area of the shaded region.

ab

80 m 100 m

3 cm

50 m 50 m 5 cm

• Relate the activity of calculating the area of the composite 225

6.3.2 shapes to daily life situations such as the area of group table’s
surface, the area of playground, and the area of tiled floor.

VOLUME OF COMPOSITE SHAPES 1 unit
1 unit
1
We will learn how to 1 unit
determine the volume
of composite shapes. The volume of
the small cube
is 1 cubic unit.

What is the volume of the composite shape?

Method 1 Count the number of cubes
of 1 cubic unit.

6 cubes 54 cubes Volume of the
1 cubic unit 1 cubic unit composite shape
54 units3 = 6 units3 + 54 units3
6 units3 = 60 units3

Method 2 Volume of the composite shape
1 unit 2 units
3 units = Volume + Volume
= (2 units × 1 unit × 3 units)
3 units
+ (9 units × 2 units × 3 units)
9 units = 6 units3 + 54 units3
2 units = 60 units3

The volume of the composite shape is 60 units3.

Build a combination of cube and cuboid from 59 cubic units. For each of
the composition of shapes, state the volume of the cuboid and the cube.

226 • Emphasise that the volume of each of the composite shape is the

6.4.1 same even though they are in different arrangements.
• Provide cubes and cuboids of various sizes for pupils to explore

various compositions of two shapes to find its volume.

2 Volume of the cuboid = 10 cm × 8 cm × cm

17 cm =cm3
8 cm
Volume of the cube = cm × cm × cm

=cm3

8 cm Volume of the composite shape
8 cm
8 cm = cm3+ cm3

= cm3 Complete the
calculation.
10 cm

3 Calculate the volume of the remaining block after the middle part
is removed.

The volume of the remaining block 6m

= (4 m × 5 m × 6 m) – (2 m × 4 m × 3 m) 3m 5m
= 120 m3 – 24 m3

= 96 m3 4m

The volume of the remaining block is 96 m3. 2m 4m

Calculate the volume of the following blocks. 8 cm

ab
2m

4 cm 4 cm 8 cm

10 m

3m

3 m 12 m 3m 8 cm

• Guide pupils to calculate the unknown length of a side in 227

6.4.1 ‟Try These” question.
• In groups, conduct a Think-Pair-Share session on the volume of

two composite shapes.

SOLVE THE PROBLEMS 1m 8m1m

1 Zariq designed an ornamental fish pond by
composing two shapes. He will build a fence
around the pond. What is the length of the fence?

20 m

Understand the problem 10 m

• T he composition of a rectangle and regular octagon.
• T he pond measurements as in the plan.
• F ind the length of the fence around the pond.

Plan the strategy the length

• Mark the length of the outer sides of the pond. of the outer
• Add the length of all outer sides to find the
8m sides of the
pond
length of the fence.

Solve 20 m

The length of the fence 10 m

= length of the outer sides + length of the outer sides of

of the rectangular pond the regular octagon pond

= (1 m + 20 m + 10 m + 20 m + 1 m) + (7 × 8 m)

= 52 m + 56 m

= 108 m

Check Perimeter of the pond
= (2 × 20 m) + (2 × 10 m)
Perimeter of the pond = 40 m + 20 m
=8×8m = 60 m
= 64 m

The perimeter of the outer sides of the pond = 64 m + 60 m − (2 × 8 m)

= 124 m − 16 m

= 108 m

The length of the fence is 108 m.

228 • Ask pupils to individually create an object such as a picture frame and
6.5.1 a greeting card using a composition of two regular polygons. Ask them
to measure the perimeter of the object.

14 cm

2 Jafri sketches the parts of a 16 cm 14 cm
wooden block to be cut. The 18 cm
sketch would be in the shape of
a cuboid. Calculate the volume of 23 cm
the wooden block after being cut.

Understand the problem 14 cm

• I dentify the length, breadth, and 14 cm
height of the original wooden block 18 cm
and the shape that has been cut.
23 cm
• C uboid-shaped cuts.
• F ind the volume of the wooden 16 cm

block after being cut.

Plan the strategy

• The volume of the original wooden block = 23 cm × 16 cm × 18 cm.

• The volume of the cuboid to be cut = 14 cm × 16 cm × 14 cm.
• The volume of the wooden = The volume of the – The volume of the

block after being cut original wooden block cuboid to be cut

Solve

The volume of the original The volume of the cuboid to be cut

wooden block 2 1
1 56
2 3 cm 1 4 cm 2 2 4 cm2
× 1 6 cm 3 6 8 cm2 × 1 6 cm × 1 4 cm
× 1 8 cm
138 11 1 11
+ 230
2 9 44 84 896
3 6 8 cm2 + 3 6 80 + 140 +224 0

6 6 2 4 cm3 2 2 4 cm2 3 1 3 6 cm3

The volume of the 11 Check
wooden block
after being cut 5 1 14 11

6 6 2 4 cm3 3 4 8 8 cm3
− 3 1 3 6 cm3 + 3 1 3 6 cm3

3 4 8 8 cm3 6 6 2 4 cm3

The volume of the wooden block after being cut is 3 488 cm3.

6.5.1 • Guide pupils to find the volume of a single shape before calculating the 229
volume of the composite shape.

2m

3 Dad wants to build a brick footpath 4 m
footpath in the house compound (B)
as shown in the diagram. He has
allocated RM4 000 for this. The grass grass
cost of building 1 m2 brick footpath
is RM100. Is there enough money? footpath (A) 4m

Solution 6m

Identify the length Find the Calculate Conclude
and the breadth of total area the total cost whether the
footpaths A and B. of footpaths for the areas allocation of
Calculate the area of A and B. of footpaths RM4 000 is
footpaths A and B.
A and B. enough.

Area of footpath A = 6 m × 4 m
= 24 m2

Area of footpath B = 2 m × 4 m
= 8 m2

Total area of footpaths A and B = Area of footpath A + Area of footpath B
= 24 m2 + 8 m2
= 32 m2

Total cost of the footpath = Total area of footpaths A and B
× The cost of 1 m2 brick footpath
= 32 × RM100
= RM3 200

Conclusion
Yes. No. The allocation of RM4 000 is enough not enough .

230 • Get each group of pupils to give their opinions on the method of calculation.
6.5.1 • Instil entrepreneurship and the value of making a budget before
making decisions.

10 m

Solve the problems below.

a Uncle Naim dug a ditch dpiatcrhit
around his farm as shown
in the diagram. Calculate 10 m 6.7 m
the length of the ditch. 2.5 m
3.5 m
4.8 m

10.2 m

b A card with a rectangular  50 cm 50 cm
surface has been cut as shown
in the diagram. 40 cm
i Find the perimeter of the
card that has not been cut.
ii What is the area of the card
that has not been cut?

30 cm 30 cm

c T he shape of Seng Huat’s 15 m 15 m
house compound is 15 m 15 m
shown in the diagram. He 25 m
planted grass in the whole
compound. What is the area
of the grassy compound?

d In a design project, a cuboid was cut 5 cm
10 cm 5 cm
and removed from a cube.

i Calculate the volume of the 20 cm
cuboid removed.

ii Is the volume of the remaining 20 cm

cube 7 750 cm3? Prove it. 20 cm

• Get pupils to complete the ‟Try These” questions in groups. 231
6.5.1 • Next, carry out activities on a station basis. Each group will show the

steps of their calculations.

1 Copy the following pictures of regular polygons. Write down
the characteristics and names of the regular polygons in the
empty boxes.

ab

2 Measure the interior angles of these regular polygons. Write down

the value of the angles.

ab

r   r  

3 Calculate the perimeters of these composite shapes.

ab

5 cm

1 cm 6m 3m
7 cm 2m
7 cm

232 6.1.1, 6.2.1,
6.3.1

4 Determine the area of these composite shapes.

a 2m b

80 cm

40 cm

6m

2m 1m 60 cm

5 Calculate the volume of the blocks below. 20 mm

a 14 cm b

14 cm

18 mm 35 mm

18 mm

14 cm 8 cm 16 mm
6 cm 36 mm
18 mm

10 cm

6 Solve these problems. Dad cuts off the middle part of a
cuboid-shaped block. The middle part is
a 6 cm 6 cm cube-shaped as shown in the diagram.

6 cm 12 cm i What is the volume of the cube?
6 cm 6 cm
18 cm ii Calculate the volume of the original
cuboid.

iii Is the volume of the remaining
block 936 cm3? Prove it.

b Pak Ali’s cultivated paddy field is 15 m

shaded as shown in the diagram.

i Calculate the area of Pak Ali’s 12 m 13 m

paddy field.

ii What is the length of the irrigation

canal surrounding Pak Ali’s paddy irrigation 5m
field? canal

6.3.2, 6.4.1, 233
6.5.1

SPACE BUNTING

Tools/Materials 5 task cards, 5 manila cards, pens, a 1 m roll of
ribbon, a paper puncher, polygons, composite

shapes of cubes and cuboids

Participants 5 groups of pupils Example of
Steps completed work

1 Each group is given one task card
to complete. For example:

Card 1 Card 2

Choose a regular Choose a regular
polygon and paste polygon and paste
it on a manila card. it on a manila card.
Write down the Label all the values of
characteristics and the interior angles.
the name of the
regular polygon.

Card 3 Card 4

Combine and Combine two regular
paste two regular polygons, or one regular
polygons. Calculate polygon and a triangle.
the perimeter of the Calculate the area of the
composite shape. composite shape.

Card 5

Choose a composite shape of cube
and cuboid. Calculate the volume of the
composite shape.

2 C arry out a Goldfish Bowl session.
Each group will present their work
while members of other groups will
ask questions to obtain the required
information.

3 C ollect the work of all groups to be
turned as a bunting.

4 D isplay the bunting at the
mathematics corner.

6.1.1, 6.2.1,
234 6.3.1, 6.3.2,

6.4.1

7 COORDINATES, RATIO,
AND PROPORTION

DISTANCE BETWEEN TWO COORDINATES

Horizontal distance and vertical distance from the origin

The Cartesian plane shows several areas in a zoo.

Ticket counter is atythe origin.

The distance of (0, 5)
each grid is 1 unit. 5

The distance vertical axis 4 (5, 3)
along the (1, 4) (5, 2)
y-axis is
vertical 3
distance. 2

ticket 1
counter
O 1 unit 1 23 (4, 0) x
horizontal axis 45 6

The distance along the x-axis is horizontal distance.

a The horizontal distance of the parrot The horizontal distance
from the ticket counter is 4 units.
from the origin is x unit.
b T he vertical distance from the ticket
counter to the jetty is 5 units. ( x, y )

c T he horizontal distance from The vertical
the origin to the lion is 5 units. distance from the
The vertical distance is . origin is y unit.

• Start the lesson by recalling the coordinates of a point in the first 235

7.1.1 quadrant of the Cartesian plane.
• Discuss the horizontal distance and vertical distance of other places

in the Cartesian plane above.

Horizontal distance and vertical distance between two coordinates
The Cartesian plane shows the interior of Rashwin’s home.

a W hat are the horizontal distance and vertical distance
from the kitchen to room 2?
y

6 room 2 (6, 5)
toilet (4, 6)
Start moving from
5 the kitchen and
room 1 (2, 5) stop at room 2.

4 5 units

3
living room (3, 3)

2

dini1ng area 2 34 5 6x
(0, 1) kitchen (4, 0)
O1

2 units

To go to room 2 Place Distance from origin
from the kitchen,
move 2 units to the kitchen Horizontal distance Vertical distance
right and 5 units up. room 2
4 units 0 unit

6 units 5 units

the difference 6 units – 4 units = 2 units
of distances

the difference 5 units – 0 unit = 5 units
of distances

The distance from the kitchen to room 2 is
2 units horizontally and 5 units vertically.

Rashwin wants to go to the toilet from the
living room. Describe the distance he has to travel.

236 7.1.1 • Emphasise the importance of the starting and ending points to
determine the exact horizontal distance and vertical distance.

b Calculate the horizontal distance and the vertical distance from room 1

to the dining area. Horizontal distance

2 units – 0 unit = 2 units From room 1,
Room 1 coordinate (2, 5) Dining area coordinate (0, 1) move 2 units to
the left and
units downwards

Vertical distance to the dining area.

units – 1 unit = units

The distance from room 1 to the dining area is
2 units horizontally and units vertically.

State the horizontal distance and vertical distance between:
a room 1 and room 2.

b the living room and dining area.

The horizontal distance and vertical distance from point L (a, b) to point
M (1, 8) is 2 units horizontally and 3 units vertically. What are the values
of a and b?

y B K Based on the Cartesian plane on the
16 Q left, state the horizontal distance and
CT vertical distance:
5 P 456 a of point A from the origin.
4 b of point B from the origin.
c of point C from the origin.
A d from point P to point T.
3 e from point Q to point P.
2 x f from point P to point K.
1

O 123

2 Calculate the horizontal distance and vertical distance from:

a point L (3, 6) to point J (4, 9). b point M (1, 8) to point N (5, 7).

237
7.1.1

RATIO BETWEEN TWO QUANTITIES

1 The picture shows the uniformed units’ camping activity.

The Malaysian Red
Crescent Society (MRCS)
has 5 members. How
many scouts are here?

Ratio is the comparison of
measurements with other
measurements or values

with other values.

We have 3 members.

a What is the ratio of the number of scouts to the number of
MRCS members?

scouts MRCS

the ratio of three to five
3:5

The ratio of the number of scouts to the number of MRCS
members is 3 : 5.

b State the ratio of the number of MRCS members to the total number of
MRCS members and scouts.

5:8

The ratio of the number of MRCS members to the total number of
MRCS members and scouts is 5 : 8.

c T he ratio of the total number of scouts and MRCS members to the
number of scouts is : .

State the ratio of the number of boys to the
number of girls in the picture above.

238 7.2.1 • Emphasise how to write and pronounce the correct ratio.
(i), (ii), (iii) For example, 5 : 8 pronounced as the ratio of five to eight.
the
• Prepare appropriate questions to state ratios and ensure that

answers cannot be simplified.

2 The following is part of the recipe to make dodol by Puan Maslina.

10 kg of glutinous flour We use 11 kg of
11241k  kkgkgggooooffffribgccroeroacfwnolounnuluarsttuemgdailksrugar granulated sugar

and 10 kg of
glutinous flour.

a What is the ratio of the mass of granulated sugar to the mass of
glutinous flour? the mass of granulated sugar the mass of glutinous flour

1 kg 1 kg

the ratio of eleven to ten
11 : 10

The ratio of the mass of granulated sugar to the mass of glutinous
flour is 11 : 10.
b T he ratio of the mass of glutinous flour to the total mass of flour
is 10 : .
c T he ratio of the total mass of sugar to the mass of brown sugar
is : .

3

Alif Raudah Whose answer
is correct?

Why?

7.2.1 • Emphasise that units of measurement must be the same to state 239
(i), (ii), (iii) the ratio of two quantities.

• Get other recipes from websites for activities to state ratios.

The mass of each watermelon is
the same. State the ratio of:
a the mass of a watermelon to

the mass of a pineapple.
b the mass of a pineapple to the

total mass of the fruits.

4 kg 3 kg

1 Based on the picture, state the ratio of:
a the number of roosters to the
number of hens.
b the number of roosters to the total
number of chickens.
c the total number of chickens to the
number of hens.

2 The table shows the length of fabric bought by Raysha’s mother.

Fabric colour Yellow Green Blue
Length 2m 300 cm 4m

State the ratio of:
a the length of yellow fabric to the length of green fabric.
b the length of blue fabric to the total length of all fabrics.
c the total length of all fabrics to the length of the yellow fabric.

3 T he picture on the right shows the volume of liquids 15 m 11 m 9 m 10 m
in four test tubes. State the ratio of:

a the volume of liquid in test tube A

to the volume of liquid in test tube B.

b the volume of liquid in test tube B to the total

volume of liquids. A B CD

240 7.2.1 • Guide pupils to carry out exercises and if necessary, repeat the
(i), (ii), (iii) learning activities using manipulative materials for them to master
the skills.

PROPORTION TO FIND A VALUE

1 I got the recipe for
a batik cake, mum.

• 600 g Marie biscuits

• 200 g butter

• 1 cup of cooking chocolate

• 1 cup of condensed milk

• 1 cup of cocoa powder
2

• 1 cup of boiled water
2

• 2 eggs

• Vanilla essence

a I want to make two batik cakes.
What is the mass of butter for
2 cups of cooking chocolate?

1 cup of cooking chocolate 200 g butter

2 cups of cooking chocolate 2 × 200 g = 400 g

The mass of butter for 2 cups of cooking chocolate
is 400 g.

b W hat is the mass of Marie biscuits needed when 300 g of butter
is used?

200 g butter for 600 g of Marie biscuits.

Divide by 2 to find the mass of Marie biscuits Total up the mass of 200 g
for 100 g butter. Marie biscuits and
100 g butter.
100 g 300 g

200 g butter for 600 g Marie biscuits.
2 2
11

300 g butter The mass of Marie
200 g + 100 g biscuits needed

600 g + 300 g = 900 g

The mass of Marie biscuits needed when
300 g of butter is used is 900 g.

• Emphasise the use of arrows in proportional representations. 241

7.3.1 • Construct other appropriate questions from the recipe above to
enhance pupils’ understanding.

2

1 kg GREEN BARLEY 1 kg
RM8.50 BEANS RM7.30

a Calculate the price for 2 kg 400 g of green beans.
2 kg 400 g = 2 kg + 400 g

Step 1 Step 2

1 kg RM8.50 1 000 g RM8.50

2 kg 2 × RM8.50 100 g RM8.50 ÷ 10 = RM0.85

= RM17.00 400 g 4 × RM0.85 = RM3.40

Step 3 400 g 2 kg 400 g
2 kg

RM17.00 + RM3.40 = RM20.40
The price for 2 kg 400 g of green beans is RM20.40.
b Is RM30 enough to buy 4 kg of barley?

Method 1 RM7.30 + RM7.30 + RM7.30 + RM7.30 = RM29.20

1 kg + 1 kg + 1 kg + 1 kg = 4 kg

Method 2 RM7.30 1 kg
2 × RM7.30 = RM14.60 2 × 1 kg = 2 kg
2 × 2 kg = 4 kg
2 × RM14.60 = RM29.20

Yes, RM30 is enough to buy 4 kg of barley.

Find the price for 3 kg 500 g of barley.

242 7.3.1 • Vary questions involving daily life situations such as the relationship
between the volume of petrol and its price.