A railway track runs parallel to a road until a bend
brings the road to a level crossing. A cyclist rides along to work along the
road every day at a constant speed of 12 miles per hour.
He normally meets a train that travels in the same direction
at the crossing.
One day he was late by 25 minutes and met the train 6 miles
ahead of the level crossing. Can you figure out the speed of the train?
Answer:
Let’s
assume that the man and the train normally meet at the crossing at 8 A.M. then
the usual time of the cyclist at the bend is 8 A.M. and he is 6 miles behind at
7.30 A.M. But when the cyclist is late, he arrives at the bend at 8.25 A.M. and
therefore he is six miles behind at 7.55 A.M. Since the train takes 5 minutes
to travel the six mile run, the speed of the train is 72 m.p.h.
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