UPSC Prelims 2026 · CSAT
Question 10: Basic Numeracy
X, Y and Z jump forward 4′, 6′ and 5′, respectively. At 8 AM, they all land on mark 199′. How many times will they all land on the same mark (need not be at the same moment) between mark 195′ and 1000′, if all of them cross mark 1000′ by 9 AM?
AnswerUPSC’s answer is (d).
All three land together only at marks equal to 199 plus a multiple of LCM(4, 6, 5) = 60. The common marks are 199, 259, 319 and so on up to 199 + 13 × 60 = 979; the next, 1039, is beyond 1000. That is 14 marks for k = 0 to 13, including 199 itself, which lies above 195. The 9 AM condition only tells you they keep going past 1000.
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.
This follows UPSC’s provisional key; the final key could change it.
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) Wrong: 11 stops well short of the marks up to 979.
- (b) Wrong: 12 leaves out two common marks.
- (c) Wrong: 13 misses the starting mark 199 or the last mark 979.
- (d) Right: 199, 259, ..., 979 gives 14 common marks.
Only the provisional key (27 May 2026) was found; the final key replaces it when found.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2026, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.