Difference between revisions of "2004 AMC 12A Problems/Problem 15"
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<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== | + | ==Solutions== |
| − | 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{(C)}</math>. | + | === Solution 1 === |
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| + | 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\boxed{\mathrm{(C)}\ 350}</math>. | ||
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| + | == Solution by Alcumus == | ||
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| + | 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 <math>2\times100=200</math> meters. Therefore the length of the track is <math>150 + 200 = 350</math> meters <math>\Rightarrow\boxed{\mathrm{(C)}\ 350}</math> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/4Amr36L7otc | ||
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| + | Education, the Study of Everything | ||
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==See also== | ==See also== | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | http://www.artofproblemsolving.com/Wiki/index.php?title=2004_AMC_12A_Problems/Problem_16&action=edit§ion=3 | ||
Latest revision as of 21:27, 15 August 2023
- 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?
Solutions
Solution 1
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 
.
Solution by Alcumus
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 
meters. Therefore the length of the track is 
meters 
Video Solution
Education, the Study of Everything
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 | ||
