2023 CSAT, all questions

UPSC Prelims 2023 · CSAT

Question 47: Basic Numeracy

What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated ?

a7998
b8028
c8878
d9238

AnswerUPSC’s answer is (a).

Numbers below 2000 start with 1, so the other three places use 2, 3 and 4 in any order: 3! = 6 numbers. The thousands place contributes 6 x 1000 = 6000. Each of 2, 3, 4 appears twice in each of the hundreds, tens and units places, so each place contributes 2 x (2 + 3 + 4) = 18, giving 18 x 111 = 1998. The sum is 6000 + 1998 = 7998.

Why each option is right or wrong
  • (a) Right: 6000 + 1998 = 7998.
  • (b) Wrong: the sum is 6000 + 1998 = 7998.
  • (c) Wrong: the sum is 6000 + 1998 = 7998.
  • (d) Wrong: the sum is 6000 + 1998 = 7998.

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

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Source: UPSC, Civil Services (Preliminary) Examination 2023, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.