A boy, a girl and a dog go for a 10 mile walk. The boy and girl can walk at 2 mph and the dog can trot at 4 mph. They also have a bicycle which only one of them (including the dog!) can use at a time. When riding, the boy and girl can travel at 12 mph while the dog can pedal at 16 mph. What is the shortest time in which all three can complete the trip?

First note that there's no apparent way to benefit from letting either the boy or girl ride the bike longer than the other. Any solution which gets the boy there faster, must involve him using the bike (forward) more; similarly for the girl. Thus the bike must go backwards more for it to remain within the 10-mile route. Thus the dog won't make it there in time. So the solution assumes they ride the bike for the same amount of time.

Also note that there's no apparent way to benefit from letting any of the three arrive at the finish ahead of the others. If they do, they can probably take time out to help the others. So the solution assumes they all finish at the same time.

The boy starts off on the bike, and travels 5.4 miles. At this point, he drops the bike and completes the rest of the trip on foot. The dog eventually reaches the bike, and takes it **backward** .8 miles (so the girl gets to it sooner) and then returns to trotting. Finally, the girl makes it to the bike and rides it to the end. The answer is 2.75 hours.

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