Difference between revisions of "2007 AMC 12A Problems/Problem 9"

 
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==Problem==
 
==Problem==
Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
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Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the [[ratio]] of Yan's distance from his home to his distance from the stadium?
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<math>\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78</math>
  
 
==Solution==
 
==Solution==
I leave it to yo to draw your own diagram.
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I leave it to you to draw your own diagram.
  
* Label home H, Yan Y, and the stadium S. If he rides 7 times as fast as he walks, when he goes home and gets his bike, he must be 6/7 to the stadium. The place he started from, therefore, is (6/7)/2=3/7 away from the stadium. Therefore, the ratio of the distance from home to the stadium must be 3/4.
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* Label home H, Yan Y, and the stadium S. If he rides 7 times as fast as he walks, when he goes home and gets his bike, he must be <math>\displaystyle \frac 67</math> to the stadium. The place he started from, therefore, is <math>\frac{6/7}{2} = \frac 37</math> away from the stadium. Therefore, the ratio of the distance from home to the stadium must be <math>\frac 34\ \mathrm{(B)}</math>.
  
 
==See also==
 
==See also==
* [[2007 AMC 12A Problems/Problem 8 | Previous problem]]
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{{AMC12 box|year=2007|ab=A|num-b=8|num-a=10}}
* [[2007 AMC 12A Problems/Problem 10 | Next problem]]
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* [[2007 AMC 12A Problems]]
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[[Category:Introductory Algebra Problems]]

Revision as of 14:57, 9 September 2007

Problem

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?

$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$

Solution

I leave it to you to draw your own diagram.

  • Label home H, Yan Y, and the stadium S. If he rides 7 times as fast as he walks, when he goes home and gets his bike, he must be $\displaystyle \frac 67$ to the stadium. The place he started from, therefore, is $\frac{6/7}{2} = \frac 37$ away from the stadium. Therefore, the ratio of the distance from home to the stadium must be $\frac 34\ \mathrm{(B)}$.

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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