I always enjoyed math problems that, at first sight, seemed hard to solve but, if you look them from the right perspective, they turned to be really easy. One of my favorites is:

Mr. Greene takes the train home every day after work, but since the train station is a little bit far, his wife go get him at the station by car every time.
On one particular day, Mr. Greene left the work earlier and took a different train home. He didn't tell Mrs. Greene so, when he arrived at the station, one hour earlier than usual, she wasn't there yet. He decided then to proceed home by foot.
Mrs. Greene, unaware of the situation, left her house at the normal time to pick her husband up and, after a while, she met him at the road and took him home where they arrived 20 minutes earlier than usual.

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How long have Mr. Greene been on foot?

  • $\begingroup$ @JaapScherphuis, it's a similar question, yes. However the used values are different so I will leave it. Thanks for point it out! :) $\endgroup$ – Pspl Aug 28 '20 at 12:23

Mr Greene has been on foot for

50 minutes


Given that the round trip for Mrs Greene is 20 minutes quicker, she must meet Mr Greene on the road 10 minutes (=20/2) before the usual train is due to arrive in the station. The usual train arrives 1 hour after Mr Greene's train for today so we just subtract the times to get how long he has been on foot.

  • $\begingroup$ You got it right! :) $\endgroup$ – Pspl Aug 28 '20 at 12:04

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