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A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7.30 a.m. They shall first cross each other at what time?

  1. 7.42 a.m.
  2. 7.48 a.m.
  3. 8.10 a.m.
  4. 8.30 a.m.

In my book, it is solved using relative speed concept as follows:

Relative speed of A and B = 6-1 = 5 rounds per hour.

Time taken to complete one round at this speed = 1/5 hr = 12 min.

But I don't understand why we have done so and can this question be solved without using relative speed concept? If yes, then please help me to solve it. Otherwise, explain how to solve this question by using the book method of relative speed.

Thank you.

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It can be solved without, but it is overcomplication, in my opinion. Let $t$ be in hours. We want to know when B has run 1 round more than A, so $6t=1+t$, therefore $5t=1$, so t = 0.2 h = 12 min. We have $6t$ because the number of rounds B has run is 6 times as more as the time.

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Correct answer is 1st option.

Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two Relative speed of A and B = (6-1) = 5 rounds per hour. Time taken to complete one round at this speed = 1/5 hr = 12 min

time =7:30 am+ 12 min = 7:42 am

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