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

We have got

Σf_i = 32 + f₁ + f₂ and Σf_ix_i = 1940 + 30f₁ + 70f₂

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 57.6)

⇒ 1843.2 + 57.6f₁ + 57.6f₂ = 1940 + 30f₁ + 70f₂

⇒ 57.6f₁ – 30f₁ + 57.6f₂ – 70f₂ = 1940 – 1843.2

⇒ 27.6f₁ – 12.4f₂ = 96.8

⇒ 69f₁ – 31f₂ = 242 …(i)

As given in the question, frequency(Σf_i) = 50

And as calculated by us, frequency (Σf_i) = 32 + f₁ + f₂

Comparing them, we get

32 + f₁ + f₂ = 50

⇒ f₁ + f₂ = 18 …(ii)

We will now solve equations (i) and (ii), multiply eq.(ii) by 31 and then adding to eq.(i), we get

(69f₁ – 31f₂) + [31(f₁ + f₂)] = 242 + 558

⇒ 69f₁ – 31f₂ + 31f₁ + 31f₂ = 800

⇒ 100f₁ = 800

⇒ f₁ = 8

Substitute f₁ = 8 in equation (ii),

8 + f₂ = 18

⇒ f₂ = 18 – 8

⇒ f₂ = 10

Thus, f₁ = 8 and f₂ = 10.