Difference between revisions of "2004 AMC 12A Problems/Problem 15"
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| + | {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #15]] and [[2004 AMC 10A Problems/Problem 17|2004 AMC 10A #17]]}} | ||
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==Problem== | ==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? | + | Brenda and Sally run in opposite directions on a circular track, starting at [[diameter|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? |
<math> \mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 </math> | <math> \mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500 </math> | ||
==Solution== | ==Solution== | ||
| − | Call the length of the race track <math>x</math>. When they meet at the first meeting point, Brenda has run <math>100</math> meters, while Sally has run <math>\frac{x}{2} - 100</math> meters. By the second meeting point, Sally has run <math>150</math> meters, while Brenda has run <math>x - 150</math> meters. Since they run at a constant speed, we can set up a [[proportion]]: <math>\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}</math>. Cross-multiplying, we get that <math>x = 350 \Longrightarrow \mathrm{D}</math>. | + | Call the length of the race track <math>x</math>. When they meet at the first meeting point, Brenda has run <math>100</math> meters, while Sally has run <math>\frac{x}{2} - 100</math> meters. By the second meeting point, Sally has run <math>150</math> meters, while Brenda has run <math>x - 150</math> meters. Since they run at a constant speed, we can set up a [[proportion]]: <math>\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}</math>. Cross-multiplying, we get that <math>x = 350 \Longrightarrow \mathrm{(D)}</math>. |
==See also== | ==See also== | ||
| + | {{AMC12 box|year=2004|ab=A|num-b=14|num-a=16}} | ||
{{AMC10 box|year=2004|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2004|ab=A|num-b=16|num-a=18}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 19:00, 3 December 2007
- 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?
Solution
Call the length of the race track 
. When they meet at the first meeting point, Brenda has run 
meters, while Sally has run 
meters. By the second meeting point, Sally has run 
meters, while Brenda has run 
meters. Since they run at a constant speed, we can set up a proportion: 
. Cross-multiplying, we get that 
.
See also
| 2004 AMC 12A (Problems • Answer Key • 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 | |
| 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 | ||
