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?
Me, not the dog!
I won't belabor this, because I'm sure someone else has already posted this solution:
The girl rides 5.4 miles and leaves the bike. When the dog gets to the bike he (she? it?) rides it back 0.8 miles toward the start. When the boy gets to the bike he rides it to the end. They all get there at the same time: 2.75 hours
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OK, now that I've read the other posts, I see Dave Domingo got the same answer I did, and SK wants us to show our work. So here is mine (breifly):
Clearly, the problem is symmetric and the boy and girl should each ride the same distance X. Then assuming the "dog rides it back" strategy that I'm using, the time traveled by the girl = X/12+(10-X)/2. The time traveled by the dog will be X/4 + (2X-10)/16 + X/2. Equate and solve for X.
This is certainly minimal for this arrangement of travel, and I'm sure it is the solution sought. But I don't know that I can prove it is the minimal time when all possible combinations are considered.
Edited on February 3, 2004, 7:33 pm
Better late than never...
