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(b) We have 6 = 1 12 = 3 15 = 1 20 = 2 14 = 2 24 = 3
,
,
,
,
,
18 3 20 5 30 2 50 5 21 3 80 10
x x
Here, 1 ≠ 2 . Thus, the ratio of the corresponding values of x and y is vary every
y 1 y 2
time. So, here x and y do not vary directly.
Example 2: Given that x and y are in direct variation from each other in the following table. Find
the constant of variation and hence find the missing values of x and y.
x 8 18 x 1 10 32 x 2
y 12 27 21 y 1 y 2 81
x
Solution: We first find the constant of variation, i.e., k = .
y
8 2 18
We have, = =
12 3 27
2
So, k =
3
x 2 2
Now, 1 = ⇒ x = × 21 = 14
1
21 3 3
10 2 3
= ⇒ y = × 10 = 15
1
y 1 3 2
32 2 3
= ⇒ y = × 32 = 48
2
y 2 3 2
x 2 2
and 2 = ⇒ x = × 81 = 54
2
81 3 3
Example 3: If the cost of 8 notebooks is `280, find the cost of 15 such notebooks.
Solution: Let the cost of 15 notebooks be `x.

Number of notebooks 8 15
Cost of notebooks (in `) 280 x
More number of notebooks, more is the cost. This is the case of direct variation.
8 15
So, = ⇒ 8 × x = 15 × 280
280 x
15 × 280
⇒ x = = 525
8
Thus, the cost of 15 notebooks is `525.
3
Example 4: If 4 cardboard boxes occupy 2.5 m space, then how many such boxes can be kept in
3
56.25 m space?
Solution: Let x number of such boxes can be kept in the given space.

Number of boxes 4 x
3
Space occupied (in m ) 2.5 56.25
As the space required to keep the boxes increases, the number of boxes also increases.

So, this is a case of direct proportion.
4 x 4 × 5625
.
So, = ⇒ x × 2.5 = 4 × 56.25 ⇒ x = = 90
.
25 56 25 25
.
.
3
Thus, 90 such boxes can be kept in 56.25 m space.
281 Direct and Indirect Variations