2007 AMC 12A Problems/Problem 9

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(Redirected from 2007 AMC 10A Problems/Problem 13)

The following problem is from both the 2007 AMC 12A #9 and 2007 AMC 10A #13, so both problems redirect to this page.

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:

$\begin{align*}a &= b + \frac{a+b}{7}\\\frac{6a}{7} &= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*}$

Thus $b/a = 6/8 = 3/4$ and the answer is $\mathrm{(B)}\ \frac{3}{4}$

2007 AMC 12A (Problems)
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

2007 AMC 10A (Problems)
Preceded by Problem 12	Followed by Problem 14
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

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2007 AMC 12A Problems/Problem 9

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