2017 CSAT, all questions
UPSC Prelims 2017 · CSAT
Question 73: Basic Numeracy
A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder?
a9
b27
c36
d45
AnswerUPSC’s answer is (d).
Let the number be 10a + b with a > b; reversed it is 10b + a. Their difference is 9(a - b). When the quotient is 1, the remainder is 9(a - b), and it must be less than the divisor 10b + a. Trying a = 9: b = 4 gives 94 / 49 = 1 remainder 45; b = 5 gives remainder 36; b = 3 gives 93 / 39 = 2 remainder 15. Wider gaps fail the divisor test, so the largest remainder is 45.
Why each option is right or wrong
- (a) 9 is smaller than the 45 that 94 / 49 gives.
- (b) 27 occurs (96 / 69) but is smaller than 45.
- (c) 36 occurs (95 / 59) but is smaller than 45.
- (d) Correct: 94 divided by 49 leaves remainder 45.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2017, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.