A neat TSD trick from Deepak Mehta (CAT 100 percentiler)

Ram and Shyam start at same end of a swimming competition.The length of pool is 50m.The race is for completing 1000m. If Ram beats Shyam and meets him 17 times during the race what could be the speed of Shyam if speed of Ram is 5m/s?

Every time Ram and Shyam meet, they (combined) cover a distance of 100m (2x the length of the pool).

To illustrate this, consider the first time they meet. Ram will be on the second leg (back from the other end), and Shyam will be on his first (since he is slower).

You can see that Ram’s and Shyam’s combined trajectory is 2x the length of the pool; i.e. 100m.

Now consider the second time they meet. Depending on Shyam’s speed, he could still be in his 1st lap (in which case Ram would be on his 4th), or he could be in his 2nd (with Ram being in his 3rd).

In either case, we have 4 laps in total; i.e. a distance of 200m.

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So, if they meet 17 times, total distance covered is 17∗100=1700m.

Out of which Ram covers 1000m (since he wins the race). So Shyam covers 1700–1000=700m.

Since Ram’s speed is 5m/s, Shyam’s is: 5 ∗ 700/1000=3.5m/s.