P and Q walk along a circular track. They start at 5:00 a.m. from the same point in opposite directions. P walks at an average speed of 5 rounds per hour and Q walks at an average speed of 3 rounds per hour. How many times will they cross each other between 5:20 a.m. and 7:00 a.m. ?

(a) 12
(b) 13 ✓ UPSC's answer
(c) 14
(d) 15

Why the answer is (b)

• Since P and Q walk in opposite directions, their relative speed is the sum of their individual speeds: 5 + 3 = 8 rounds per hour, meaning they close one full round of the track together every 1/8 hour, i.e., every 7.5 minutes.

• Starting together at 5:00 a.m., their successive meetings occur at 5:07:30, 5:15:00, 5:22:30, 5:30:00, 5:37:30, 5:45:00, 5:52:30, 6:00:00, 6:07:30, 6:15:00, 6:22:30, 6:30:00, 6:37:30, 6:45:00, 6:52:30, and 7:00:00.

• The question asks for crossings strictly between 5:20 a.m. and 7:00 a.m., so the meeting at 7:00:00 is excluded and any meeting at or before 5:20:00 is also excluded.

• The meetings that fall in this open interval are: 5:22:30, 5:30:00, 5:37:30, 5:45:00, 5:52:30, 6:00:00, 6:07:30, 6:15:00, 6:22:30, 6:30:00, 6:37:30, 6:45:00, and 6:52:30.

• Counting these gives exactly 13 crossings, which corresponds to option (b).

Why the other options are wrong

(a) 12: 12 would result from mistakenly excluding one of the valid meetings in the interval, for example by treating the 6:00:00 crossing as outside the window or by miscounting the 7.5-minute spacing.
(c) 14: 14 would result from including the 7:00:00 meeting, but the question specifies 'between 5:20 a.m. and 7:00 a.m.', which excludes the endpoint 7:00.
(d) 15: 15 would result from including both the 7:00:00 meeting and an extra meeting, or from using a relative speed of 9 rounds/hour (adding speeds incorrectly) which would shift all meeting times and produce an incorrect count.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.