CHAPTER 4 RATIOS RATES PROPORTIONS

4.1 RATIOS

Learning standards:
4.1.1 Represent the relation between three

quantities in the form of a : b : c.
4.1.2 Identify and determine the equivalent

ratios in numerical, geometrical or daily
situation contexts.
4.1.3 Express ratios of two and three quantities
in simplest form.



What is ratio??

1) Ratio is used to compare two or more
quantities of the same kind that are
measured in the same unit.

2) For example, the ratio of 5 000 g to 9 kg can
be represented as

5 000 g : 9 kg = 5 kg : 9 kg = 5 : 9

3) The ratio of a to b is written as a : b or a
b

4) Ratios are usually written in lowest terms

Examples…

1) Represent the ratio of 0.02 m to 3 cm to 4.6
cm in the form of a : b : c.

Soln :

0.02 m : 3 cm : 4.6 cm = 2 cm : 3 cm : 4.6 cm

Multiply by 10 = 2 : 3 : 4.6 make sure the
Divide by 2 = 20 : 30 : 46 ratio is in the
= 10 : 15 : 23 whole numbers

and in the
simplest form.

Examples…

2) Represent the ratio of 2 weeks to 16 days to
1 week in the form of a : b : c.

Soln :

2 weeks : 16 days : 1 week = 14 days : 16 days : 7 days

= 14: 16 : 7 make sure the
ratio is in the
same unit and in
the simplest

form.

Examples…

3) During one particular month, the number of
sunny days in Malaysia was 4 days while the
number of rainy days was 12. Write the ratio
of number of sunny days to number of rainy
days.

Soln :

sunny days : rainy days = 4 sunny days : 12 rainy days

= 4 : 12 Simplify by divide 3

= 1:3

What is equivalent ratio??

Observe the equivalent fractions that represents the number of
slices of apple pie above.

1 2 2 2 4

 
2 2 4 2 8

What is equivalent ratio??

TRY!!!

Complete the blanks of the equivalent ratios.



Express Ratio in Simplest Form

To express a ratio in its simplest form, divide the quantities by the
Highest Common Factor (HCF) or multiply the quantities by the
Lowest Common Multiple (LCM).

Examples:

1) Express each of the following ratios in its simplest form.

(a) 800 g : 1.8 kg (b) 32 : 24 : 20 (c) 3 : 7

(d) 0.04 : 0.12 : 0.56 5 10

Soln : Convert to the same unit.
(a) 800 g : 1.8 kg = 800 g : 1 800 g
Divide both parts by 200.
= 800 : 1 800
200 200

=4 : 9

(b) 32 : 24 : 20 = 32 : 24 : 20 Divide the three parts by 4, that is,
444 the HCF of 32, 24 and 20.

= 8: 6 : 5

(c) 3 : 7 = 3 × 10 : 7 × 10 Multiply both parts by 10,
that is, the LCM of 5 and 10.
5 10 5 10

=6:7

(d) 0.04 : 0.12 : 0.56 = 0.04 ×100 : 0.12 ×100 : 0.56 ×100

Multiply the three parts by 100.

= 4 : 12 : 56 Divide the three parts by 4,
= 4 : 12 : 56 that is, the HCF of 4, 12 and 56.

4 44
= 1 : 3 : 14

1. Express each of the following ratios in its simplest form.

(a) 240 g : 1.6 kg (b) 30 : 42 : 48

(c) 2/5 : 8/9 (d) 0.09 : 0.12 : 0.24

4.2 RATES

Aim: To determine the relationship between ratios and rates.

Instruction:
1) State the ratio of two quantities for the measurements involved in each of the

situations given.
2) State the quantities involved and also their units of measurement.

Situation Ratio in the form Quantities Units of
Involved measurement
a
b

A car travels 285 km 285 km Distance and km and hour
in 3 hours. 3 hours time

A plant grows 24 cm
in 4 months.

A baby’s mass
increases by 1.3 kg in
60 days.

Karim’s pulse rate is
75 beats per minute.

What is rates??

1) Rate is a special ratio that compares two
quantities with different units of
measurement.

2) For examples, we can observe from the
Exploration Activity that you did;

(a) 285km (b) 24cm (c) 1.3kg

3h 4months 60days

Conversion of units of rates

1) Rajan is riding his bicycle at a speed of 5 m/s.
Convert 5 m/s to km/h.

Soln :

5m/s 1 km = 1000 meter
1 hour = 60 X 60 seconds

 5m
1s

 5m  1s

 5 km  1 hour
1000 60  60

 5 km  60  60 hour
1000 1

 18 km/h

Conversion of units of rates

1) The density of a type of metal is 2 700 kg per m3.

State the density of this metal in g per cm3.

Soln :

2700 kg/m 3 1 kg= 1000 gm
1 m3 = 100 x 100 x 100 cm3
 2700kg
1m 3

 2700 1000
100 100 100

 2700000
1000000

 2.7g/cm 3

 2.7g per cm 3



4.3 PROPORTIONS

Learning standards:
4.3.1 Determine the relationship between ratios and proportions.
4.3.2 Determine an unknown value in a proportion.

What is proportions?

1) Proportion is a relationship that states that the two ratios or two
rates are equal.

2) Proportion can be expressed in the form of fraction.
3) Example:

If 10 beans have a mass of 17 g, then 30 beans have a mass
of 51 g.

Soln :

10 beans  30 beans
17 grams 51grams

How do you determine the unknown value
in a proportion?

Electricity costs 43.6 sen for 2 kilowatt-hour (kWh). How
much does 30 kWh cost?

Soln : Unitary method Proportion method

The cost of electricity for 2 kWh Let the cost of electricity for 30 kWh
= 43.6 sen
be x sen.

Then, X 15

The cost of electricity for 1 kWh 43.6 sen  x sen
= 43.6 sen 2 kWh 30 kWh

2
= 21.8 sen

X 15

The cost of electricity for 30 kWh x  43.6 15
= 30 × 21.8
= 654 sen x  654sen

Alternative Method





4.4 RATIOS, RATES & PROPORTIONS

Learning standards:
4.4.1 Determine the ratio of three quantities, given two or more

ratios of two quantities.
4.4.2 Determine the ratio or the related value given

(i) the ratio of two quantities and the value of one quantity.
(ii) the ratio of three quantities and the value of one

quantity.
4.4.3 Determine the value related to a rate.
4.4.4 Solve problems involving ratios, rates and proportions,

including making estimations.

4.4 RATIOS, RATES & PROPORTIONS

Determine the ratio of three quantities, given two or
more ratios of two quantities.

Examples:

1) If p : q = 2 : 9 and q : r = 9 : 7, find the ratio of p : q : r.

Soln :

p:q=2:9 q:r=9:7

make sure the values of q are same

Therefore, the ratio of p : q : r is 2 : 9 : 7.

2) At the reading corner of Class 1D, the ratio of the
number of storybooks to the number of reference
books is 2 : 5. The ratio of the number of reference
books to the number of magazines is 3 : 2. Find the
ratio of the number of storybooks to the number of
reference books to the number of magazines.

Soln :

storybooks : reference books reference books : magazines

2 x3 : 5 x3 3 x5 : 2 x5

= 6 = 15 = 15 = 10

the value of reference books not same,
so find the LCM to make it same

Therefore, the ratio of storybooks : reference books : magazine

is 6 : 15 : 10.

4.4 RATIOS, RATES & PROPORTIONS

Determine the ratio or the related value.

Examples:

1) The ratio of the price of a baju kebaya to the price of a baju kurung is 7 : 4.
If the price of the baju kebaya is RM84, find the price of the baju kurung.

Soln :

a) Unitary method

Let the price of the baju kurung is x,

baju kebaya : baju kurung

= 7 :4

= RM 84 : RM x

7 parts = RM 84 RM x = 4 x RM 12

1 part = RM 84 x = RM 48

7

= RM 12 Thus, the price of baju kurung is RM 48

4.4 RATIOS, RATES & PROPORTIONS

Soln :
b) Cross multiplication method
Let the price of the baju kurung is x,

baju kebaya : baju kurung
= 7 :4
= RM 84 : RM x

baju kebaya : baju kurung
7 4

RM 84 RMx
RMx  7  RM 84  4

RMx  RM 336
7

x  RM 48

Thus, the price of baju kurung is RM 48.

4.4 RATIOS, RATES & PROPORTIONS

Determine the value related to a rate.
Examples:
1) Mr Tan jogs at a steady rate on a treadmill and his heart beats

420 times in 4minutes. Find the number of times his heart will
beat if he jogs on the treadmill at the same rate for 12 minutes.

Soln :
Heart rate = 420 times

4 minutes
Let the number of heartbeats be x times in 12 minutes.

*we can solve this problem by using unitary method, proportion
method or cross multiplication method.

4.4 RATIOS, RATES & PROPORTIONS

a) Unitary method b) Cross multiplication method
4 minutes  420 beats
1 minute  420 beats Let x be the number of beats in 12
4 minutes;
 105 beats
420 beats  x beats
Thus,12 minutes  105 beats 12 4 minutes 12 minutes
 1260 beats
420 12  4  x
Mr Tan’s heart will beat 1 260 420 12  x

times in 12 minutes. 4
x  1260 beats

Thus, Mr Tan’s heart will beat 1 260
times in 12 minutes.

4.4 RATIOS, RATES & PROPORTIONS

c) Proportion method

x3

420 beats  x beats
4 minutesx 3 12 minutes

x  4 20  3
x  1260 beats

Mr Tan’s heart will beat 1 260 times in

12 minutes.

4.5 Relationship between Ratios, Rates and Proportions, with
Percentages, Fractions and Decimals

Learning standards:
4.5.1 Determine the relationship between percentages and

ratios.
4.5.2 Determine the percentage of a quantity by applying the

concept of proportions.
4.5.3 Solve problems involving relationship between ratios, rates and

proportions with percentages, fractions and decimals.

The relationship between percentages and ratios

4.5 Relationship between Ratios, Rates and Proportions, with
Percentages, Fractions and Decimals

The relationship between percentages and ratios.
Examples:

1) In a class, the ratio of the number of girls to the number of boys is 3 : 2.
Find the percentage of the girls in the class.

Soln :

The ratio of the number of girls to the total number of students = 3 : 5 = 3

Change to a 5

3  3  20 fraction with a Alternative
5 5  20 denominator of 100 Method:
 60
3 100%
100 5
 60%

Thus, the percentage of the girls in the class is 60%.  60%

4.5 Relationship between Ratios, Rates and Proportions, with
Percentages, Fractions and Decimals

The relationship between percentages and ratios.

Examples:

2) It is known that 35% of the hard disk of a computer has been filled with data.
Find the ratio of the capacity that has been filled with data to the capacity
that has not been filled with data.

Soln :
Hard disk with data = 35% = 35

100

Hard disk without data = 65% (100% - 35%) = 65

100

Hard disk with data : Hard disk without data = 35 : 65

100 100

= 35 : 65
= 7 : 13
Thus, the ratio of the capacity that has been filled with data to the capacity
that has not been filled with data is 7 : 13

4.5 Relationship between Ratios, Rates and Proportions, with
Percentages, Fractions and Decimals

Determine the percentage of a quantity by applying the concept of proportions.

Examples:

At a sale carnival, Encik Rosli chooses a shirt from a rack which displays ‘45%

price reduction’. The original price of the shirt is RM85. 00. When Encik Rosli

scans the price tag of the shirt, the scanner shows that the price is RM57. 80. By

applying the concept of proportions, determine whether this percentage

discount corresponds to the percentage reduction displayed. Give a reason for

your answer.

Soln :

Let p be the percentage discount of the original price.

p  selling price
100 original price

p  57.80 The selling price is 68% of is original price.
100 85.00 Percentage discount = 100% - 68% = 32%
85  p  57.80 100 Thus, the percentage obtained by En. Rosli is less

p  5780 than percentage discount reduction displayed.
85

p  68

4.5 Relationship between Ratios, Rates and Proportions, with
Percentages, Fractions and Decimals

Solving problems involving ratios, rates and proportions.

Examples:

1) There are 40 passengers on a bus. At the next bus stop, 8 passengers get off and 18
passengers get on the bus.
(a) By applying the concept of proportions, determine the percentage of the passengers
who get off the bus compared to the number of passengers originally on the bus.
(b) What is the ratio of the passengers who get on the bus at the bus stop compared to
the new total number of passengers on the bus?

Soln :
(a) Let x be the percentage of the passengers get off

passengers get off  8
total passengers 40

X2.5 Thus, the percentage of the passengers get off is 20%.

8 x
40 100

X2.5

x  8  2.5
x  20

4.5 Relationship between Ratios, Rates and Proportions, with
Percentages, Fractions and Decimals

(b) What is the ratio of the passengers who get on the bus at the bus stop
compared to the new total number of passengers on the bus?

Soln :
(b) Number of the passengers get on the bus = 18

New total number of the passengers on the bus = 40 -8 + 18 =50

Ratio of the passengers who get on the bus to the new total passengers on the bus
is
= 18 : 50
= 9 : 25