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

We have got

Σf_i = 35 + f₁ + f₂ and Σf_ix_i = 6150 + 190f₁ + 210f₂

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 188)

⇒ 6580 + 188f₁ + 188f₂ = 6150 + 190f₁ + 210f₂

⇒ 190f₁ – 188f₁ + 210f₂ – 188f₂ = 6580 – 6150

⇒ 2f₁ + 22f₂ = 430 …(i)

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

And as calculated by us, frequency (Σf_i) = 35 + f₁ + f₂

Comparing them, we get

35 + f₁ + f₂ = 100

⇒ f₁ + f₂ = 65 …(ii)

We will now solve equations (i) and (ii), multiply eq.(ii) by 2 and then subtracting it from eq.(i), we get

(2f₁ + 22f₂) – [2(f₁ + f₂)] = 430 – 130

⇒ 2f₁ + 22f₂ – 2f₁ – 2f₂ = 300

⇒ 20 f₂ = 300

⇒ f₂ = 15

Substitute f₂ = 15 in equation (ii),

f₁ + 15 = 65

⇒ f₁ = 65 – 15

⇒ f₁ = 50

Thus, f₁ = 50 and f₂ = 15.