2006 AMC 10A Problem 15

"Passing" on concentric lanes means being on the same radial line, so only angular position matters. Compute each runner's lap time.

Odell: lap length $2\pi \cdot 50 = 100\pi$ m at $250$ m/min, so one lap takes $\dfrac{100\pi}{250} = \dfrac{2\pi}{5}$ min.

Kershaw: lap length $2\pi \cdot 60 = 120\pi$ m at $300$ m/min, so one lap takes $\dfrac{120\pi}{300} = \dfrac{2\pi}{5}$ min.

Identical lap times. Running in opposite directions, the angle between them grows at the rate of two laps per $\tfrac{2\pi}{5}$ minutes, so they line up again every $\tfrac{\pi}{5}$ minutes (half a lap time). They pass at times $\tfrac{\pi}{5}, \tfrac{2\pi}{5}, \dots$, and the number of passes within $30$ minutes is
$$\left\lfloor \frac{30}{\pi/5} \right\rfloor = \left\lfloor \frac{150}{\pi} \right\rfloor = \lfloor 47.7\ldots \rfloor = 47.$$

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

On a circular track, meetings are governed by angular rates, and "opposite directions" means the angular rates add. The problem is designed so that the different speeds and radii cancel to equal lap times; once that is seen, it is a single division and a floor. Always check whether the endpoint ($t = 30$) is itself a meeting; here $48 \cdot \tfrac{\pi}{5} \approx 30.2 > 30$, so it is not.

• Rounding $150/\pi \approx 47.7$ up to $48$, or comparing linear speeds instead of angular speeds (the lanes have different radii, so the lap times, not the speeds, are what match).
• Counting one meeting per lap instead of two, giving $23$ or so, or adding the speeds as $550$ m/min over some single circumference.