Difference between revisions of "2012 AMC 10A Problems/Problem 16"
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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? | 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? | ||
<math> \textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000 </math> | <math> \textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000 </math> | ||
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| + | == Solution == | ||
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| + | First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let <math>x</math> be the time these runners run in seconds. | ||
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| + | <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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| + | 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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| + | == See Also == | ||
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| + | {{AMC10 box|year=2012|ab=A|num-b=15|num-a=17}} | ||
Revision as of 00:10, 9 February 2012
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
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 5, 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.
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 | ||
