UPSC Prelims 2020 · CSAT
Question 75: Basic Numeracy
How many pairs of natural numbers are there such that the difference of whose squares is 63?
AnswerUPSC’s answer is (a).
Write a² - b² = (a - b)(a + b) = 63. Both factors must be odd and a - b < a + b. The factor pairs of 63 are 1 × 63, 3 × 21 and 7 × 9. These give (a, b) = (32, 31), (12, 9) and (8, 1). So there are 3 pairs of natural 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) Right: three pairs.
- (b) 63 has no fourth factor pair.
- (c) 63 has no fifth pair either.
- (d) Two pairs miss one valid pair.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2020, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.