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(b) x 8 12 9 24 36 3
y 18 12 16 6 4 48

Solution: We find the product xy for the corresponding values of x and y and compare them.
(a) We have, 4 × 60 = 240, 8 × 30 = 240, 15 × 20 = 300, 10 × 24 = 240, 30 × 12 = 360, 36 × 5 = 180
Since, the products of the values of x and the corresponding values of y are not same
or fixed. So, x and y do not vary inversely.

(b) We have, 8 × 18 = 144, 12 × 12 = 144, 9 × 16 = 144, 24 × 6 = 144, 36 × 4 = 144, 3 × 48 = 144
Clearly, xy = 144

Here, the products of the values of x and the corresponding values of y are same or
fixed, i.e., 144.
So, x and y vary inversely.
Example 10: Given that p and q are in inverse variation from each other in the following table.
Find the constant of variation and hence find the missing corresponding values of p and q.

p 18 8 p 1 36 3 p 2
q 12 27 24 q 1 q 2 54

Solution: Since, p and q are in inverse variation. So first, we find the constant of inverse variation
for p and q.
Here, 18 × 12 = 216 = 8 × 27
So, pq = 216 = k
216
Therefore, p × 24 = 216 ⇒ p = 24 = 9
1
1
216
36 × q = 216 ⇒ q = 36 = 6
1
1
216
3 × q = 216 ⇒ q = 3 = 72
2
2
216
and p × 54 = 216 ⇒ p = 54 = 4
2
2
Example 11: A school has 8 periods in a day each of 45 minutes duration. How long would each
period be, if the school has 9 periods in a day, assuming the number of school hours to be the
same?

Solution: Let the duration of each period be x minutes when there are 9 periods in a day. Thus,
we have the following table:
Number of periods 8 9

Duration of each period (in minutes) 45 x

More the number of periods, lesser will be duration of each period.
So, this is the case of inverse variation.
845
×
Therefore, 8 × 45 = 9 × x (Q x y = x y ) ⇒ 9 = x
1 1
2 2
⇒ x = 40
Hence, the required duration of each period if the school has 9 periods in a day is 40 minutes.

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