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?
Answer & explanation
Answer: (d) 14
All three touch 199′, and after that their landing marks coincide every LCM(4, 6, 5) = 60 feet. Counting 199, 259, … up to 979 gives 14 common marks in the range.
- LCM of 4, 6 and 5 is 60, so common marks are 199 + 60k.
- 199 − 60 = 139 is below 195, so the first common mark in range is 199′ itself.
- 199 + 60k ≤ 1000 gives k ≤ 13.35, so k = 0, 1, …, 13.
- Number of common marks = 14 (199′, 259′, …, 979′).
Remember · Common landing points of different step lengths repeat every LCM of the steps; count terms of the resulting sequence.
Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·