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Two contestants run a $3-kilometre$ race along a circular course of length $300$ $metres$. If there speeds are in the ratio of$ 4:3$, how often and where would the winner pass the other? (The initial start-off is not counted as passing.) (A) $4 $ times; at the starting point. (B) Twice; at the starting point. (C) Twice; at a distance of $225 $ metres from the starting point. (D) Twice; once at $75 metres$ and again at $225 $ metres from the starting point.

I have solved it in the following way: The race is a total of $10 $ laps ($3km / 300m = 10 $ laps)The faster runner will complete $4 $ laps in the same time the slower runner completes $3 laps$, so they will meet after $4 laps$ and again after $8 $ laps. The faster runner will then complete the race while the slower runner still has $2½$ laps to go. Answer:$ B $ Twice (after 4 laps and 8 laps, at the starting point of each lap)

But i am looking for a more definite and clear method. Any help will be appreciated!

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Your proof is ok, but you should mention that when the contestants meet after 3 and 4 laps this would be the first time they met! This is the only weakness in your proof I can see.

I would suggest a little different approach. Let the speed of faster contestant be $V$. Than the speed of slower one is $3/4*V$. Distance between contestants increases with a speed $1/4 * V$. By the time faster contestant makes $10$ laps the distance would be $2.5$ laps.

But contestants meet whenever the distance between them is a whole number of laps. That meas they would meet when the distance between them is $1$ lap and then $2$ laps. That is 2 times.

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