2006 AMC 10A Problems/Problem 15

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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draw((-5,0){up}..{right}(0,5),red);
draw((5,0){down}..{left}(0,-5),red);
draw((-5,0){down}..{right}(0,-5),red);
draw((6,0){up}..{left}(0,6),blue);
draw((-6,0){up}..{right}(0,6),blue);
draw((6,0){down}..{left}(0,-6),blue);
draw((-6,0){down}..{right}(0,-6),blue);
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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 $\Longrightarrow \mathrm{(D)}$ .

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2006 AMC 10A Problems/Problem 15

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