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

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==Solution==
 
==Solution==
I leave it to you to draw your own diagram.
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Let the distance from Yan's initial position to the stadium be <math>a</math> and the distance from Yan's initial position to home be <math>b</math>. We are trying to find <math>b/a</math>, and we have the following identity given by the problem.
  
* 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>.
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<math>a = b + 7(a+b)</math>
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<math>6a = 8b</math>
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<math>b/a = 8/6 = 3/4</math>
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So the answer is <math>\mathrm{(B)}\ \frac{3}{4}</math>
  
 
==See also==
 
==See also==

Revision as of 07:39, 30 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

Let the distance from Yan's initial position to the stadium be $a$ and the distance from Yan's initial position to home be $b$. We are trying to find $b/a$, and we have the following identity given by the problem.

$a = b + 7(a+b)$

$6a = 8b$

$b/a = 8/6 = 3/4$

So the answer is $\mathrm{(B)}\ \frac{3}{4}$

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