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?

(In reply to Better late than never... by Brian Wainscott)

You and Dave Domingo have certainly dispelled any initial thought that the trip must take 5 hours. However, it seems wasteful that there should be a backward segment, even though it is a short one employing the fastest means of locomotion. Now the dog alone with no bike riding can do the trip in 2.5 hrs which is .25 hr faster than your solutions. So if the boy and girl can alternate riding the bike and walking at an average rate as great as 4 mph, the trip can be done in 2.5 hrs. While it might at first seem that such an alternating sequence can be constructed, it cannot -- if each walks half the time and rides the other half, the average speed will be only 6/7 of 4 mph and it will take 2.91666 hours for them to do the 10 miles. So maybe the backwards segment is not so bad -- it appears to be the only way that both the boy and girl can use the bike for more than half the distance, and thereby get their average speed above 6/7 x 4 mph. Using the backwards segment, the dog loses a bit of its 4 mph unassisted capability, but the boy and girl gain to the point where all three come out with the same average speed.
Edited on February 4, 2004, 1:01 am

Posted by Richard on 2004-02-03 20:40:21