2004 AMC 10A Problem 17

Because they run toward each other, the distances they cover add up to the gap closed. From the start to the first meeting the gap is half the track, $\tfrac{L}{2}$. From the first meeting to the second, they must together cover a full lap, $L$.

The second stretch requires twice the combined distance of the first, so at constant speeds it takes twice as long, and each girl runs twice as far as she did in the first stretch. Brenda ran $100$ meters before the first meeting, so she runs $200$ meters between the meetings. Sally runs $150$ meters in that same stretch, and together they cover one lap:
$$L = 200 + 150 = 350.$$

The answer is $\boxed{\textbf{(C)}\ 350}$.

For runners moving in opposite directions, "they meet" means the sum of their distances equals the distance between them, so meetings happen at combined distances of $\tfrac{L}{2}, \tfrac{L}{2} + L, \dots$. Constant speeds make each runner's distance proportional to elapsed time, so comparing two time intervals compares all the distances at once, with no need to introduce actual speeds.

Set up the proportion directly: in the first interval Brenda runs $100$ and Sally runs $\tfrac{L}{2} - 100$; in the second, Sally runs $150$ and Brenda runs $L - 150$. Equal speed ratios give $\frac{100}{\frac{L}{2} - 100} = \frac{L - 150}{150}$, so $15000 = \left(\tfrac{L}{2} - 100\right)(L - 150)$. Expanding, $\tfrac{L^2}{2} - 175L = 0$, giving $L = 350$.

• Assuming the runners cover only half a lap between consecutive meetings, which gives a track of 250 meters.
• Measuring Sally's $150$ meters from the start rather than from the first meeting point.