UPSC Prelims 2019 · CSAT

Question 60: Basic Numeracy

Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is

AnswerUPSC’s answer is (d).

Units: 6 + 7 = 13, write 3 and carry 1. For the hundreds, 1 + 5 plus any carry must give 7, so the tens column must carry 1, which means 3 + B + 1 is 10 or more, so B is at least 6 and A = B - 6. For 7A3 to divide by 3, 7 + A + 3 = 10 + A must be a multiple of 3, so A = 2 and B = 8. Check: 587 + 136 = 723 = 3 × 241.

Why each option is right or wrong

(a) Wrong: B = 2 gives no carry into the hundreds.
(b) Wrong: B = 5 gives no carry into the hundreds.
(c) Wrong: B = 7 gives 713, which is not divisible by 3.
(d) Right: B = 8 gives 723, divisible by 3.

Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.

Practise this paper timedPrevious question Next question

Source: UPSC, Civil Services (Preliminary) Examination 2019, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.