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?
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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==
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==Solutions==
{{solution}}
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=== 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==
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https://youtu.be/4Amr36L7otc
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Education, the Study of Everything
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==See also==
 
==See also==
* [[2004 AMC 10A Problems/Problem 16 | Previous problem]]
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{{AMC12 box|year=2004|ab=A|num-b=14|num-a=16}}
* [[2004 AMC 10A Problems/Problem 18 | Next problem]]
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{{AMC10 box|year=2004|ab=A|num-b=16|num-a=18}}
* [[2004 AMC 10A Problems]]
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[[Category:Introductory Algebra Problems]]
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http://www.artofproblemsolving.com/Wiki/index.php?title=2004_AMC_12A_Problems/Problem_16&action=edit&section=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?

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

Solutions

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 $x = 350\Longrightarrow\boxed{\mathrm{(C)}\ 350}$.

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 $2\times100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $\Rightarrow\boxed{\mathrm{(C)}\ 350}$

Video Solution

https://youtu.be/4Amr36L7otc

Education, the Study of Everything


See also

2004 AMC 12A (ProblemsAnswer KeyResources)
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 (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

http://www.artofproblemsolving.com/Wiki/index.php?title=2004_AMC_12A_Problems/Problem_16&action=edit&section=3