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

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{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #9]] and [[2007 AMC 10A Problems/Problem 13|2007 AMC 10A #13]]}}
 
==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?
 
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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==Solution==
 
==Solution==
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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.
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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:
  
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<math>a = b + \frac{a+b}{7}</math>
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<cmath>\begin{align*}a &= b + \frac{a+b}{7}\\
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\frac{6a}{7} &= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*}</cmath>
  
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<math>\frac{6a}{7} = \frac{8b}{7} \Longrightarrow 6a = 8b</math>
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Thus <math>b/a = 6/8 = 3/4</math> and the answer is <math>\mathrm{(B)}\ \frac{3}{4}</math>
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<math>b/a = 6/8 = 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 18:00, 5 January 2008

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 $\mathrm{(B)}\ \frac{3}{4}$

See also

2007 AMC 12A (Problems • Answer Key • Resources)
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
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