2012 AMC 10A Problems/Problem 16
Contents
Problem
Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
Solution 1
First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let 
be the time these runners run in seconds.
Because 
is a multiple of 500, it turns out they just meet back at the start line.
Now we must find a time that is a multiple of 
and results in the 5.0 m/s runner to end up on the start line. Every seconds, that fastest runner goes 
meters. In 
seconds, he goes 
meters. Therefore the runners run 
seconds.
Solution 2
Working backwards from the answers starting with the smallest answer, if they had run 
seconds, they would have run 
meters, respectively. The first two runners have a difference of 
meters, which is not a multiple of 
(one lap), so they are not in the same place.
If they had run seconds, the runners would have run 
meters, respectively. The last two runners have a difference of 
meters, which is not a multiple of .
If they had run 
seconds, the runners would have run 
meters, respectively. The distance separating each pair of runners is a multiple of , so the answer is seconds.
See Also
| 2012 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
