UPSC Prelims 2022 · CSAT
Question 10: Basic Numeracy
The digits 1 to 9 are arranged in three rows in such a way that each row contains three digits, and the number formed in the second row is twice the number formed in the first row; and the number formed in the third row is thrice the number formed in the first row. Repetition of digits is not allowed. If only three of the four digits 2, 3, 7 and 9 are allowed to use in the first row, how many such combinations are possible to be arranged in the three rows?
AnswerUPSC’s answer is (c).
Each row is a three-digit number, the second twice and the third thrice the first, and together they use the digits 1 to 9 once. Testing gives only four first rows: 192, 219, 273 and 327 (for example 273, 546, 819). The first row must be made of three of 2, 3, 7 and 9 only. 192 and 219 each contain a 1 and just two of those digits, so they fail. 273 and 327 use 2, 3, 7, which gives 2 arrangements.
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) 4 counts all valid arrangements, including two that put the digit 1 in the first row.
- (b) 3 has no support; only 273 and 327 qualify.
- (c) Correct: 273 (546, 819) and 327 (654, 981).
- (d) 1 counts only one of the two first rows.
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.