The mean of the following data is 42. Find the missing frequencies x and y if the Sum of frequencies is 100.[CBSE 2014]

Class interval	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency	7	10	x	13	y	10	14	9

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 63 + x + y and Σf_ix_i = 2775 + 25x + 45y

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 42)

⇒ 2646 + 42x + 42y = 2775 + 25x + 45y

⇒ 42x – 25x + 42y – 45y = 2775 – 2646

⇒ 17x – 3y = 129 …(i)

As given in the question, frequency(Σf_i) = 100

And as calculated by us, frequency (Σf_i) = 63 + x + y

Equalizing them, we get

63 + x + y = 100

⇒ x + y = 37 …(ii)

We will now solve equations (i) and (ii), multiply eq.(ii) by 3 and then add it to eq.(i), we get

(17x – 3y) + [3(x + y)] = 129 + 111

⇒ 17x – 3y + 3x + 3y = 240

⇒ 20x = 240

⇒ x = 12

Substitute x = 12 in equation (ii),

12 + y = 37

⇒ y = 37 – 12

⇒ y = 25

Thus, x = 12 and y = 25.