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2006 AMC 10A Problems/Problem 15

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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 (ProblemsResources)
Preceded by
Problem 14
Followed by
Problem 16
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Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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