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?
There are five different time intervals:
t1:boy has bike
t2:boy leaves bike for dog
t3:dog has bike and rides back toward the start
t4:dog leaves bike for girl
t5:girl has bike
For the best result everyone (and everything) finishes at the same time. The four following equations can be formulated:
boy: 12*t1 + 2*t2 + 2*t3 + 2*t4 + 2*t5 = 10
dog: 4*t1 + 4*t2 - 16*t3 + 4*t4 + 4*t5 = 10
girl: 2*t1 + 2*t2 + 2*t3 + 2*t4 + 12*t5 = 10
bike: 12*t1 - 16*t3 + 12*t5 = 10
After the boy leaves the bike, the bike is (12-4)*t1=8*t1 miles away from the dog. The dog reaches it in t2 hours after the boy drops the bike, therefore 8*t1=4*t2.
When the situation is played in reverse, 8*t5=4*t4 by a similar argument, and t1=t5 and t2=t4
This system of eight equations has exactly one solution:
t1 = 9/20 = 27 min
t2 = 9/10 = 54 min
t3 = 1/20 = 3 min
t4 = 9/10 = 54 min
t5 = 9/20 = 27 min
t1+t2+t3+t4+t5 = 55/20 = 165 min = 2 hr 45 min
The boy takes the bike for 27 min, leaves it at the 5.4 mile mark and finishes the trip walking.
The dog trots for 81 min, finds the bike at the 5.4 mile mark and pedals back toward the start for 3 min, leaves the bike at the 4.6 mile mark and trots the rest of the way.
The girl walks for 138 min, finds the bike at the 4.6 mile mark and finishes the trip on the bike.
Edited on February 4, 2004, 2:43 pm