Swimming pool problem: Time required to empty the swimming pool

Question

In a swimming pool, 6 swimmers have to swim such that 3 swimmers start from end A at intervals of 1 minute and the remaining 3 start from end B at intervals of 2 minutes where A and B are opposite ends of the pool & length of the pool is 120 meters. The speed of each swimmer is 20m/min. Whenever two swimmers meet, each of them reverse their direction & start swimming in the opposite direction without any time delay. Each swimmer stops when he reaches one of the ends. If the first swimmer from each start simultaneously, after how many minutes from that time will there be no swimmer in the pool?

I have tried to map the problem, but it becomes confusing after some time.

Any ideas?

The answer given is 10 min. Is it right?

Answer 1

Imagine the swimmers are indistinguishable. Then when two swimmers meet, you wouldn't be able to tell whether they both reversed direction, or if they kept going. So suppose they just keep following their original trajectories. In that case, the swimming ends when the last person to enter the pool finishes. That swimmer begins at t=4 and finishes 6 minutes later.

Answer 2

I don't see any good way to do this problem except slowly and carefully write it step by step. Drawing it is a great strategy. If you draw every time two new swimmers meet you should be able to work through the problem. I'll try explaining it out as best I can.

I'll call the swimmers A1, A2, A3 and so on. Imagine looking at the pool from a top view, with side A on the bottom and side B on the top. Side A will be 0 meters, side B will be 120 meters. So a swimmer at 100 meters would be 20 meters from side B.

First, both sides send a swimmer at 0 minutes. Everybody swims at the same speed, so these two will meet in the middle at 60 meters. As they swim towards the middle, the other swimmers start coming out.

3 mins: Swimmers A1, B1 are at 60 meters. Swimmer A2 is at 40 meters, swimmer A3 at 20 meters. Swimmber B2 is at 100 meters, and swimmer B3 won't get in the pool until minute 4. A1 and B1 reverse directions.

3mis 30 seconds: A1 and A2 are 20 meters apart, and if they head towards each other they hit after only 30 seconds. So we have A1, A2 at 50 meters, A3 at 30 meters, B1 at 70 meters, and B2 at 90 meters.

4mins : A1 at 60 meters heading up, A2 and A3 at 40 meters, B1 and B2 at 80 meters, and B3 at 120 meters is just entering the pool. A3 is going to reverse direction and start heading towards the edge of the pool. We only want to know when the last person leaves the pool, and A3 clearly won't be the last person left, so lets ignore him for the rest of the problem.

4mins 30sec: A1 and B1 at 70 meters, A2 at 50 meters, B2 at 90 meters, and B3 at 110 meters.

5mins: A1, A2 at 60 meters. A2 now reverses direction and heads towards the wall, so we can forget about him. B1, B2 at 80 meters. B3 at 100 meters.

5mins 30sec: A1, B1 70 meters. B2, B3 at 90 meters. Both A1 and B3 now both are heading towards the wall with no oncoming traffic, so we can ignore them for the rest of the problem.

6min: B1, B2 at 80 meters. Now both B1 and B2 are heading for the wall, B2 is heading up and is only 40 meters away from the wall, while B1 heads down and is a full 80 meters away. B1 will be the last out of the pool, so ignore B2.

10min: B1 swims from 80 meters to 0 meters and comes to rest at side A.

Answer 3

The A member of the third couple will be the last one to get out from the water. He will stay 2 minutes out of water, then 2 mins in the water when his B couterpart starts swimming. There is a distance of 80 meters between them. They have to swim 40 meters, i.e. 2 minutes to meet. Then the A guy will have to swim back 80 meters to get out. This will take 4 minutes.

Summary

2 minutes: A is waiting.

2 minutes: B jumps in.

2 minutes: they meet.

4 mins: A gets back.

Total is: 2+2+2+4=10.