Q. Consider the following addition problem : 3P+4P+PP+PP=PQ2; where P, Q and R are different digits.

Explanation : 

3P + 4P + PP + PP = RQ2

Breaking it down:
(30 + P) + (40 + P) + (10P + P) + (10P + P) = 100R + 10Q + 2

Simplifying:
24P + 70 = 100R + 10Q + 2

Further Breaking Down:
20P + 70 + 4P = 100R + 10Q + 2

Finding Value of P:
– Unit digit is 2
– Must come from 4 × P
– Therefore, P must be 3 or 8

Case 1 (P = 3):
24P + 70 = (24 × 3) + 70
= 72 + 70
= 142

Case 2 (P = 8):
24P + 70 = (24 × 8) + 70
= 192 + 70
= 262

Arithmetic Mean:
(142 + 262) ÷ 2 = 202

Answer: 202