P and Q walk along a circular track. They start at 5:00 am. 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.?

1. A. 12
2. B. 13
3. C. 14
4. D. 15

Answer: B

Explanation

When two individuals move in opposite directions on a circular track, their relative speed is the sum of their individual speeds. Here, relative speed = 5 rounds/hour + 3 rounds/hour = 8 rounds/hour. This means they will meet 8 times in one hour.
Time taken for one meeting = 60 minutes / 8 meetings = 7.5 minutes.

They start at 5:00 a.m. The meetings occur at:
5:00 (start, not a crossing), 5:07:30, 5:15:00, 5:22:30, 5:30:00, …, 6:52:30, 7:00:00.

We need to count the crossings *between* 5:20 a.m. and 7:00 a.m. This means strictly after 5:20 a.m. and strictly before 7:00 a.m.

Meetings before or at 5:20 a.m. (excluding the start at 5:00 a.m.):
– 5:07:30
– 5:15:00
There are 2 such meetings.

Total meetings from 5:00 a.m. to 7:00 a.m. (120 minutes) = 120 / 7.5 = 16 meetings. This includes the start at 5:00 a.m. and the end at 7:00 a.m. if they meet exactly then.

If we count meetings from the first crossing (5:07:30) up to the last crossing before 7:00 a.m. (6:52:30):
Total meetings from 5:07:30 to 7:00:00 (112.5 minutes) = 112.5 / 7.5 = 15 meetings.
From these 15 meetings, we exclude the 2 meetings that occurred before 5:20 a.m. (5:07:30 and 5:15:00).
So, 15 – 2 = 13 meetings.

Alternatively, list the meetings: 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. There are 13 such meetings. This question tests the application of time, speed, and distance concepts, particularly relative speed in circular motion, which is a standard topic in UPSC CSAT.