Difference between revisions of "2012 AMC 10A Problems/Problem 16"

Revision as of 19:31, 1 August 2015

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 1

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 500, 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.

Solution 2

Working backwards from the answers starting with the smallest answer, if they had run $1000$ seconds, they would have run $4400, 4800, 5000$ meters, respectively. The first two runners have a difference of $400$ meters, which is not a multiple of $500$ (one lap), so they are not in the same place.

If they had run $1250$ seconds, the runners would have run $5500, 6000, 6250$ meters, respectively. The last two runners have a difference of $250$ meters, which is not a multiple of $500$ .

If they had run $2500$ seconds, the runners would have run $11000, 12000, 12500$ meters, respectively. The distance separating each pair of runners is a multiple of $500$ , so the answer is $\boxed{\textbf{(C)}\ 2,500}$ seconds.

Solution 3

Let $t$ be the time run in seconds, then the difference in meters run between the three runners is $0.2t, 0.4t, 0.6t$ . For them to be at the same location all of them need to be multiples of 500. It is now easy to see that $0.2t=500, 0.4t=1000, 0.6t=1500$ , so $t=\boxed{\textbf{(C)}\ 2,500}$ .

@@ Line 25: / Line 25: @@
 == Solution 3==
-Let <math>t</math> be the time run in seconds, then the difference in meters run between the three runners is <math>0.2t, 0.4t, 0.6t</math>. For them to be at the same location all of them need to be multiples of 500. It is now easy to see that <math>0.2t=500, 0.4t=1000, 0.6t=1500</math>, so t=<math>\boxed{\textbf{(C)}\ 2,500}</math>.
+Let <math>t</math> be the time run in seconds, then the difference in meters run between the three runners is <math>0.2t, 0.4t, 0.6t</math>. For them to be at the same location all of them need to be multiples of 500. It is now easy to see that <math>0.2t=500, 0.4t=1000, 0.6t=1500</math>, so <math>t=\boxed{\textbf{(C)}\ 2,500}</math>.
 == 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

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