2007 AMC 12A Problems/Problem 9

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 $\frac 34\ \mathrm{(B)}$ .

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2007 AMC 12A Problems/Problem 9

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