UPSC Prelims 2026 CSAT, question 37

Topic: Arithmetic

Q37

$X$ and $Y$ are two runners who run for the same duration of time on the same circular track. They started running at the same time in the same direction with uniform speeds. When $X$ completed 7 rounds, $Y$ did exactly 5. After completing 5 rounds, $Y$ changed his direction and started running in the opposite direction with speed which is double of his earlier speed. On the other hand, $X$ continued to run with the same speed. They stopped running when $X$ completed exactly 21 rounds. How many times did $X$ and $Y$ meet after they had started and before they finally stopped?

(a) 35Correct answer
(b) 34
(c) 31
(d) 29

Solution.

Answer: (a) 35

In Phase 1 (same direction), while $X$ completes $7$ rounds, $Y$ completes $5$ rounds, so $X$ gains $7 - 5 = 2$ rounds on $Y$ and they meet $2$ times after starting (at $3.5$ rounds and $7$ rounds of $X$ ).
In Phase 2 (opposite directions from the starting line), $X$ 's speed is $7$ units and $Y$ 's doubled speed is $2 \times 5 = 10$ units; while $X$ runs the remaining $21 - 7 = 14$ rounds, $Y$ runs $2 \times 10 = 20$ rounds in the opposite direction.
Since they move in opposite directions in Phase 2, they meet $14 + 20 - 1 = 33$ times strictly before stopping together at the end, giving $2 + 33 = 35$ meetings in total.

Trap: Counting the moment they stop together at $X = 21$ gives 36, which is not an option.