Difference between revisions of "2004 AMC 12A Problems/Problem 15"

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==See also==
 
==See also==
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Revision as of 20:19, 28 February 2007

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

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 $x = 350 \Longrightarrow \mathrm{D}$.

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
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