Train Station Meeting Problem, at MathProblems.info

At MathProblems.info is the following problem, entitled “Train station meeting problem”.

Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?

The presented solution is as follows:

This is a hard solution to explain, it will help if you draw a graph.

Suppose another couple (call them couple B) lives with the couple stated in the problem (call them couple A). Normally both men catch the same train home. However today the husband B arrives at the station 20 minutes early and wife B is already there to take him home. Both men will arrive home at the same time, because both arrive 20 minutes earlier than usual.

On their way home at some point couple B will pass wife A picking up husband A and they will drive the rest of the way home side by side, again both arriving 20 minutes early.

Now assume that wife A didn’t see her husband and instead kept going to the train station. She would have arrived exactly 20 minutes after wife B left, because husband B arrived 20 minutes before husband A was expected to arrive, and wife A is always exactly on time. Likewise if wife B had turned around when she saw husband A and went back to the station she would have arrived 20 minutes after she left the station. In other words it would take her 10 minutes to reach husband A and 10 minutes to get back to the station.

So after A walked for 40 minutes wife B would have left the station. 10 minutes later she would have caught up to husband A. So husband A walked a total of 50 minutes.

However, there is a much simpler explanation. The trick is to think of the problem from the wife’s perspective.

Since they arrived home 20 minutes earlier than normal, we know the wife’s roundtrip was 20 minutes shorter than normal. In order for that to be the case, she had to have met her husband 10 minutes earlier than normal (i.e. she saved a 20 minute round trip from where she met him to the station and back). So if she normally meets him at 5:00p, then in this case she met him at 4:50p. Since we know the husband starting walking at 4:00p (i.e. one hour before his wife would normally meet him at the station), he therefore walked for 50 minutes before she picked him up.

Note also that it took the husband 50 minutes to walk a distance that his wife can drive in 10 minutes. So the wife drives 5 times faster than the husband walks. (The ratio of their speeds would make a good corollary question.)

Thanks, I agree that is a better solution. I just updated my site.

I also went back to the "old look" while I was at it.

Thanks for the acknowledgement on the problem's solution at the site - very kind of you. I must admit, it took me awhile before I thought to consider the problem from the wife's perspective.

I originally found the site when it was the "old look", so now I'm waxing nostalgic over its return. :)

It's a great site - I've really enjoyed working on the problems I've attempted to tackle there. Many thanks for that too!

Quote: MathGent

It's a great site - I've really enjoyed working on the problems I've attempted to tackle there. Many thanks for that too!

Thanks for the kind words. You should have been with me in New Zealand last week. I was torturing everybody in my hiking group with math and logic problems.

Quote: Wizard

You should have been with me in New Zealand last week. .

The idle rich and their endless touring.
Yawn..