2004 AMC 12A Problems/Problem 15

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?

$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500$

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

Sally runs twice as much the second time they met as she ran the first time they met. Therefore, she ran 75 before they first met. 100+75=175, and that's 1/2 of the length. $175*2=350 \Rightarrow \mathrm{(C) \ }$

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 12 Problems and Solutions

2004 AMC 10A (Problems • Answer Key • 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
All AMC 10 Problems and Solutions

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2004 AMC 12A Problems/Problem 15

Contents

Problem

Solution

Solution 1

Solution 2

See also