Consider the following addition problem:
3P + 4P + PP + PP = RQ2; where P, Q and R are different digits.
What is the arithmetic mean of all such possible sums?
Answer & explanation
Answer: (c) 202
In place-value form the four numbers add to 70 + 24P, and a units digit of 2 allows only P = 3 or P = 8. The sums 142 and 262 both keep P, Q and R different, and their mean is 202.
- 3P + 4P + PP + PP = (30 + P) + (40 + P) + 11P + 11P = 70 + 24P.
- The sum ends in 2, so 24P ends in 2, i.e. 4P ends in 2: P = 3 or P = 8.
- P = 3: 70 + 72 = 142, so R = 1, Q = 4 — the digits 3, 4, 1 are different.
- P = 8: 70 + 192 = 262, so R = 2, Q = 6 — the digits 8, 6, 2 are different.
- Mean of the possible sums = (142 + 262) ÷ 2 = 202.
Remember · Write letter-numbers in place-value form (3P = 30 + P, PP = 11P), then shortlist with the units digit.
Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·