2012 AMC 10A Problems/Problem 16

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?

$\textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000$

Solution

First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let $x$ be the time these runners run in seconds.

$4.8x-4.4x=500 \Rightarrow x=1250$

Because $4.4(1250)=5500$ is a multiple of 5, it turns out they just meet back at the start line.

Now we must find a time that is a multiple of $1250$ and results in the 5.0 m/s runner to end up on the start line. Every $1250$ seconds, that fastest runner goes $5.0(1250)=6250$ meters. In $2(1250)=2500$ seconds, he goes $5.0(2500)=12500$ meters. Therefore the runners run $\boxed{\textbf{(C)}\ 2,500}$ seconds.

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2012 AMC 10A Problems/Problem 16

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