A few weeks after his first confusing journey home from the train station, Smith again finishes work ahead of schedule and takes an early train home. This time he arrives at his suburban station half an hour early. Again, rather than wait for the chauffeur, he starts walking home. And as before, he meets his chauffeur on the road, who picks him up promptly and takes him home. How many minutes early do they reach the house this time?
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In the first puzzle we learned that Smith takes 55 minutes to walk a distance that the car can cover in 5 minutes. That means the car travels 11 times as fast as Smith. Today, if Smith has been walking for t minutes when he meets the chauffeur on the road, then he has spared the chauffeur an additional journey of t/11 minutes from that meeting point to the station. Since the driver would have arrived at the station at the normal meeting time after that t/11 final portion, 30 minutes after Smith had started walking, t + t/11 = 30. That means t = 27.5, and the total time saved, 2t/11, is five minutes.
(From Edward J. Barbeau et al., Five Hundred Mathematical Challenges, 1995.)
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