Difference between revisions of "2007 AMC 12A Problems/Problem 9"
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==Solution== | ==Solution== | ||
| − | + | 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. | |
| − | + | <math>a = b + 7(a+b)</math> | |
| + | |||
| + | <math>6a = 8b</math> | ||
| + | |||
| + | <math>b/a = 8/6 = 3/4</math> | ||
| + | |||
| + | So the answer is <math>\mathrm{(B)}\ \frac{3}{4}</math> | ||
==See also== | ==See also== | ||
Revision as of 06:39, 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?
Solution
Let the distance from Yan's initial position to the stadium be 
and the distance from Yan's initial position to home be 
. We are trying to find 
, and we have the following identity given by the problem.
So the answer is 
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 | |
