UPSC Prelims 2022 · CSAT
Question 18: Basic Numeracy
How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5 ?
AnswerUPSC’s answer is (b).
A number divisible by 5 ends in 0 or 5; every digit is odd, so the last digit is 5 (zero is even and cannot appear). The other two digits come from 1, 3, 7 and 9 without repetition: 4 choices for the first place and 3 for the middle. So 4 × 3 = 12 numbers.
How you could have got this
No shortcut here: in a “how many” question, ruling out options doesn’t help. Every statement has to be judged.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) 8 undercounts; the first two places give 4 × 3 ways.
- (b) Correct: last digit 5, then 4 × 3 = 12.
- (c) 16 would allow repeated digits (4 × 4).
- (d) 24 is 4 × 3 × 2, choosing all three digits from 1, 3, 7, 9 and ignoring that the last must be 5.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2022, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.