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Difference between revisions of "2006 AMC 10A Problems/Problem 15"

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Odell and Kershaw run for 30 minutes on a [[circle|circular]] track. Odell runs [[clockwise]] at 250 m/min and uses the inner lane with a [[radius]] of 50 meters. Kershaw runs [[counterclockwise]] at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial [[line]] as Odell. How many times after the start do they pass each other?  
 
Odell and Kershaw run for 30 minutes on a [[circle|circular]] track. Odell runs [[clockwise]] at 250 m/min and uses the inner lane with a [[radius]] of 50 meters. Kershaw runs [[counterclockwise]] at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial [[line]] as Odell. How many times after the start do they pass each other?  
  
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<center><asy>
 
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/begin{figure}
 
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unitsize(1cm);
 
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draw((5,0){up}..{left}(0,5),red);
 
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draw((-5,0){up}..{right}(0,5),red);
 
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draw((5,0){down}..{left}(0,-5),red);
 
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draw((-5,0){down}..{right}(0,-5),red);
 
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draw((6,0){up}..{left}(0,6),blue);
 
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draw((-6,0){up}..{right}(0,6),blue);
 
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draw((6,0){down}..{left}(0,-6),blue);
 
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draw((-6,0){down}..{right}(0,-6),blue);
 
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\end{figure}
 
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</center></asy>
 
  
 
<math>\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad</math>
 
<math>\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad</math>

Revision as of 13:01, 27 July 2010

Problem

Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?


$\mathrm{(A) \ } 29\qquad\mathrm{(B) \ } 42\qquad\mathrm{(C) \ } 45\qquad\mathrm{(D) \ } 47\qquad\mathrm{(E) \ } 50\qquad$

Solution

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Since $d = rt$, we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$ minutes, while Kershaw also runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$ minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are $\frac{30}{\frac{2\pi}{5}} \approx 23.8$ laps run by both, or $\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 = 47$ meeting points $\Longrightarrow \mathrm{(D)}$.

See Also

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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