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?
- (a) 11
- (b) 12
- (c) 13
- (d) 14 ✓ UPSC's answer
Why the answer is (d)
• X lands on 199 + 4k, Y on 199 + 6k and Z on 199 + 5k. A mark is shared by all three when the distance from 199 is a multiple of LCM(4, 6, 5) = 60.
• The common marks are 199, 259, 319, …, 199 + 60k. They need not land at the same moment, so 199 itself counts, and it lies between 195 and 1000.
• The largest such mark below 1000 is 199 + 60 × 13 = 979; the next, 1039, is beyond 1000.
• k = 0 to 13 gives 14 common marks: (d).
Why the other options are wrong
- (a) 11
- 11 misses several common marks between 199 and 979.
- (b) 12
- 12 misses common marks; there are 14 from 199 to 979.
- (c) 13
- 13 leaves out the starting mark 199, which also lies in the range.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2026, held on 24 May 2026. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.