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## Question

Doris and Forrest start out on a bike trip, but on the way Doris' bicycle breaks down. They agree to share the remaining bicycle. They begin simultaneously, Forrest on foot, Doris riding the good bike. After a certain distance, Doris dismounts, leaves the bicycle behind and continues on foot. When Forrest reaches the waiting bicycle, he mounts it and rides until he catches up with Doris. He then dismounts, and the process repeats until they reach their goal.

How far from their destination should the the bicycle be left behind the final time to ensure they reach their destination at precisely the same moment? The distance from the breakdown to the destination is 60 miles, and they both walk at a constant rate of 5 mph and ride at a constant rate of 15 mph.

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1) Doris ride for 15miles (say from A), and leaves the bicycle (say at B), starts on foot from there.and walks on foot for 10 miles,

By this time (3 hrs), Forrest reaches bicycle (at B) and ride
15 miles for 1hr (say up to C). Doris also walks further 5miles and both meet at C.
2) Above is repeated from this point, Forrest leaves bicycle, starts on foot; Doris rides for 15 miles, leaves the bicycle (say up to D), which is finally 15 miles away from destination, and starts on foot. Forrest reaches bicycle rides and reaches the destination, simultaneously Doris reaches the destination on foot

So Finally the bicycle is left 15 miles behind the destination.

Edited by bhramarraj
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The bike should be left at half the remaining distance. Since their walking and riding speeds are the same they ride the bike for equal distance and the walk equal distance each iteration.

The total distance and their absolute speeds don't matter as long as their speeds are the same. If Doris leaves the bike x miles away from the starting point they will meet again 2x miles from the starting point and the process repeats with the new starting point.

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Take this answer a bit farther.

Is there a dropping off place that will not result in their arriving simultaneously?

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