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100 x
(a) We have, =
60 32
5 x 5
⇒ = ⇒ x = × 32
3 32 3
160
⇒ x = = 53.33 km
3
So, the distance covered in 32 minutes is 53.33 km.

100 250 5 250
(b) Also, = ⇒ =
60 y 3 y
250 × 3
⇒ 5y = 3 × 250 ⇒ y = = 150
5

Therefore, the time taken to cover 250 km is 150 minutes or 2 hours and 30 minutes.
6
Example 8: Suppose 2 kg sugar contains 9 × 10 crystals. How many sugar crystals are there in
5 kg sugar?
Solution: Let the number of sugar crystals in 5 kg of sugar be x.
We put the given information in the form of a table as shown:

Weight of sugar (in kg) 2 5
Number of sugar crystals 9 × 10 6 x

Clearly, more the weight of the sugar, more will be the number of sugar crystals.
So, it is a case of direct proportion.
2 5
Therefore, =
9 × 10 6 x
5 ×× 10 6
9
6
⇒ 2 × x = 5 × 9 × 10 ⇒ x = = 2.25 × 10 7
2
Thus, the required number of sugar crystals in 5 kg of sugar is 2.25 × 10 .
7
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MAths connect

Do you know about the use of scale given at the bottom of a map in your atlas?
A map is a miniature representation of a large region. A map scale is the ratio between the distance on a map
and the corresponding distance on the ground. This scale of a map is based on the concept of direct proportion.
Look at the map of India in your atlas and consider the scale given there. Using the scale, find the actual
distances between major cities.

Practice Time 12A

1. Check whether x and y are in direct variation to each other in each of the following tables.

(a) x 2 4 5 10 25 24 (b) x 12 15 8 10 14 57
y 6 12 15 30 75 72 y 15 30 40 50 56 76

(c) x 6 12 16 10 14 38
y 15 30 40 25 35 95

283 Direct and Indirect Variations