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Difference between revisions of "2012 AMC 10A Problems/Problem 16"
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<math>4.8x-4.4x=500 \Rightarrow x=1250</math>
 
<math>4.8x-4.4x=500 \Rightarrow x=1250</math>
  
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Because <math>4.4(1250)=5500</math> is a multiple of 5, it turns out they just meet back at the start line.
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Because <math>4.4(1250)=5500</math> 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 <math>1250</math> and results in the 5.0 m/s runner to end up on the start line. Every <math>1250</math> seconds, that fastest runner goes <math>5.0(1250)=6250</math> meters. In <math>2(1250)=2500</math> seconds, he goes <math>5.0(2500)=12500</math> meters. Therefore the runners run <math>\boxed{\textbf{(C)}\ 2,500}</math> seconds.
 
Now we must find a time that is a multiple of <math>1250</math> and results in the 5.0 m/s runner to end up on the start line. Every <math>1250</math> seconds, that fastest runner goes <math>5.0(1250)=6250</math> meters. In <math>2(1250)=2500</math> seconds, he goes <math>5.0(2500)=12500</math> meters. Therefore the runners run <math>\boxed{\textbf{(C)}\ 2,500}</math> seconds.
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==  Solution 2==
 
==  Solution 2==

Revision as of 22:11, 29 June 2013

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.

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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