All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
A Boy, a Girl, a Dog, and a Bicycle (Posted on 2004-02-03) Difficulty: 3 of 5
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?

See The Solution Submitted by DJ    
Rating: 4.4000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution Comment 21 of 21 |
(In reply to answer by K Sengupta)

Since all the three contestants must finish the trip, it follows that if one of the three contestants gets to the finish line ahead of the other, he would still have to go back and help the other two finish the trip. Thus, the three contestants must act in a concerted manner to ensure that they finish at the same time.

Since the respective walking spped and the cycling speed of the boy and the girl are equal, it follows that the respective distances traversed by the boy and the girl in walking and cycling must be equal. let the distance traversed by the boy and the girl in cycling be x miles so that the respective distances traversed in walking is (10-x) miles. This is seemingly a contradiction since the boy and the girl start at the same time and the cycle can only be used by one rider at a time.

Since the dog's respective cycling speed and walking speed is greater than the other two, it follows that the dog must lend its assistance in the matter. Accordingly, precisely one of the boy and the girl must proceed initially on foot, while the other must ride the bicycle. Without loss of generality, let us suppose that the boy initially rode the bicycle for x miles, and finished the remaining (10-x) miles on foot. The girl starts by walking (10-x) miles on foot and then ride the remaining x miles on the bicycle. This is only possible, if the dog, who is faster than either of them, starts at the same time as the boy and the girl, trots to the bicycle left by the boy at the x mile location, brings it back to the (10-x) mile mark by riding atop it for x - (10-x) = 2x - 10 miles for the girl to pick it up. The dog then trots the remaining x miles.

Accordingly, the time taken by each of the boy and the girl to reach the finish line is (x/12 + (10-x)/2) hours, while the total time taken by the dog to reach the finish line is (x/4 + (2x-10)/16 + x/4) hours.

Thus, by the given conditions, we must have:

x/12 + (10-x)/2) = x/4 + (2x-10)/16 + x/4 ........(*)

or, 4(60-5x) = 3(10x-10)

or,x = 5.4

Substituting this value back in (*), we observe that each of the lhs and the rhs is equal to 2.75

Consequently, the shortest time in which all three can complete the trip is 2.75 hours or, 2 hours 45 minutes. 


  Posted by K Sengupta on 2008-06-06 01:46:51
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information