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

Home > Just Math
Race Resolution (Posted on 2013-10-08) Difficulty: 3 of 5
Kane challenges Tom to a race between two red lights and back. They both start from the same red light but Kane starts when Tom is already halfway across. Kane catches up with Tom at a point 50 meters from the other red light.

Kane continues to the end and turns back and meets Tom only 20 meters from the same red light. He continues running to the starting point and starts running back again to the other end while Tom reaches the other end and starts his return trip.

Assuming they were running at constant speeds, how far from the first red light do Kane and Tom meet for the third time?

No Solution Yet Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 1 of 2

K goes 70 meters in the time it takes T to go 30 meters between their first and second meetings, so the ratio of their times is 7/3.

Let K take 3 arbitrary units of time to cross between the lights and T take 7 units of time.

K reaches start 9.5 units of time after T first set out, while T reached the other light 7 units of time after his start, so it was 2.5 units of time later when K was starting over toward the second light, so T advanced toward the start 2.5/7 or 5/14 of the way across the gap by this time. At that instant T was still 9/14 of the way from the first light to the second. Let t be the time after this instant that they met for the third time to close this 9/14 distance gap:

1/7 * t + 1/3 * t = 9/14

t = 189/140 time units

during which K traveled (189/140)/3 = 63/140 of the distance between the lights, this being the sought distance for the solution, so we need only the distance between the lights.

The time of their first meeting was let's say t1 units of time after K set out and T was already halfway across. Since they met at that time,

1/2 + 1/7 * t1 = 1/3 * t1

t1 = 21/8

d1 = 21/8 * 1/3 = 7/8

That means the remaining 50 meters was 1/8 the distance between the lights, which was therefore 400 meters.

400 * 63/140 = 180 meters, the answer.

  Posted by Charlie on 2013-10-08 18:09:07
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information