UPSC Prelims 2021 · CSAT
Question 59: Basic Numeracy
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 ?
AnswerUPSC’s answer is (c).
Add the units digits: P + P + P + P = 4P must end in 2, so 4P = 12 or 32, giving P = 3 or 8. For P = 3: 33 + 43 + 33 + 33 = 142, so R = 1 and Q = 4, all different from P = 3. For P = 8: 38 + 48 + 88 + 88 = 262, so R = 2 and Q = 6, different from 8. Both work. The mean is (142 + 262) ÷ 2 = 202. Remember: in digit puzzles start from the units column and test each digit that satisfies it.
Why each option is right or wrong
- (a) 102 is below both sums; the mean of 142 and 262 is 202.
- (b) 120 is below both sums, 142 and 262.
- (c) Right: (142 + 262) ÷ 2 = 202.
- (d) 220 lies between the two sums but is not their mean, 202.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2021, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.