2004 AMC 12A Problems/Problem 15

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The following problem is from both the 2004 AMC 12A #15 and 2004 AMC 10A #17, so both problems redirect to this page.

Problem

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 5...$

Solution

Solution 1

Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion: $\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}$ . Cross-multiplying, we get that $x = 350 \Longrightarrow \mathrm{(C)}$ .

Solution 2

The total distance the girls run between the start and the first meeting is one half of the track length.
The total distance they run between the two meetings is the track length.

As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting.

Thus between the meetings Brenda will run $2\cdot 100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $\Rightarrow \mathrm{(C) \ }$

2004 AMC 12A (Problems • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

2004 AMC 10A (Problems • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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