2006 AMC 10A Problem 15

On two lanes around the same center, the runners "pass" when they are on the same line from the center. So what matters is how fast each one goes around, not the meters per minute.

1. Odell's lap. One lap is $2\pi \times 50 = 100\pi$ meters. At $250$ meters per minute it takes
$$\frac{100\pi}{250} = \frac{2\pi}{5} \text{ minutes}.$$
2. Kershaw's lap. One lap is $2\pi \times 60 = 120\pi$ meters. At $300$ meters per minute it takes
$$\frac{120\pi}{300} = \frac{2\pi}{5} \text{ minutes}.$$
3. Same lap time. They go around equally fast, but in opposite directions. So each one covers half a lap by the time they first meet, in $\frac{\pi}{5}$ minutes. After that they meet again every $\frac{\pi}{5}$ minutes.
4. Count the meetings. The number of meetings in $30$ minutes is how many whole times $\frac{\pi}{5}$ fits into $30$:
$$30 \div \frac{\pi}{5} = \frac{150}{\pi} \approx 47.7.$$
So there are $47$ meetings; the $48$th would come at $48 \times \frac{\pi}{5} \approx 30.2$ minutes, after they stop.

The answer is $\boxed{\textbf{(D)}\ 47}$.

On a circular track, compare runners by how long each lap takes. When they go in opposite directions, they meet each time their combined laps add up to one more full lap. This problem was built so the different speeds and lane sizes give equal lap times; after spotting that, it is one division. Always check whether the last meeting falls before or after the end time.