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 ?
- (a) 102
- (b) 120
- (c) 202 ✓ UPSC's answer
- (d) 220
Why the answer is (c)
• 3P + 4P + PP + PP = (30 + P) + (40 + P) + 11P + 11P = 70 + 24P.
• The sum RQ2 ends in 2, so 24P must end in 2: P = 3 (24 × 3 = 72) or P = 8 (24 × 8 = 192).
• P = 3 gives 142 (R = 1, Q = 4) and P = 8 gives 262 (R = 2, Q = 6); in both, P, Q and R are different digits.
• Mean of the possible sums = (142 + 262) ÷ 2 = 202 — option (c).
Why the other options are wrong
- (a) 102
- 102 is not the mean of the possible sums 142 and 262.
- (b) 120
- 120 is not the mean of the possible sums 142 and 262.
- (d) 220
- 220 is not the mean of the possible sums 142 and 262; the mean is 202.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2021, held on 10 October 2021. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.