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?

Ideally, the bicycle would be in use all the time and the three hikers would arrive at the destination at the same time.

Let x be the total number of hours for the trip.
Let b be the hours that the boy is riding the bike.
Let g be the hours that the girl is riding the bike.
Let d be the hours that the dog is riding the bike.

If the bicycle is being ridden the whole time then
x=b+g+d

They each travel 10 miles:
10=2x+10b, where 10 is the incremental speed added by the bike.
10=2x+10g
10=4x+12d

These translate to
12b+2g+2d=10
2b+12g+2d=10
4b+4g+16d=10

The last two give d=(10g-5)/6, and then the last one gives
104g = 70, with b=g.

The total time, x=b+g+d, would be 85/52 of an hour, the the amounts of time each rides the bike would be 35/52 hour each for the boy and the girl and 15/52 hour for the dog.

But there's no way the bike could be ridden all the while, as the bike rider would get ahead of the non-riders and would have to leave the bike in place until a walker catches up.

But no solution can be lower than 98.08 minutes.

Posted by Charlie on 2004-02-03 13:40:35