Difference between revisions of "2007 AMC 12A Problems/Problem 9"
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<math>\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78</math> | <math>\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78</math> | ||
− | ==Solution== | + | ==Solution 1== |
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: | 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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Thus <math>b/a = 6/8 = 3/4</math> and the answer is <math>\mathrm{(B)}\ \frac{3}{4}</math> | Thus <math>b/a = 6/8 = 3/4</math> and the answer is <math>\mathrm{(B)}\ \frac{3}{4}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Another way of solving this problem is by setting the distance between Yan's home and the stadium, thus filling in one variable. | ||
+ | Let us set the distance between the two places to be <math>42z</math>, where <math>z</math> is a random measurement (cause life, why not?). The distance to going to his home then riding his bike, which is <math>7</math> times faster, is equal to him just walking to the stadium. So the equation would be: | ||
+ | Let <math>x=</math> the distance from Yan's position to his home. | ||
+ | Let <math>42z=</math> the distance from Yan's home to the stadium. | ||
+ | <math>42-x=x+42/7</math> | ||
+ | <math>42-x=x+6</math> | ||
+ | <math>36=2x</math> | ||
+ | <math>x=18</math> | ||
+ | But we're still not done with the question. We know that Yan is <math>18z</math> from his home, and is <math>42z-18z</math> or <math>24z</math> from the stadium. | ||
+ | <math>18z/24z</math>, the <math>z</math>'s cancel out, and we are left with <math>3/4</math>. | ||
+ | Thus, the answer is <math>\mathrm{(B)}\ \frac{3}{4}</math> | ||
+ | ~ProGameXD | ||
==See also== | ==See also== |
Revision as of 12:19, 23 July 2017
- The following problem is from both the 2007 AMC 12A #9 and 2007 AMC 10A #13, so both problems redirect to this page.
Contents
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 1
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:
Thus and the answer is
Solution 2
Another way of solving this problem is by setting the distance between Yan's home and the stadium, thus filling in one variable. Let us set the distance between the two places to be , where is a random measurement (cause life, why not?). The distance to going to his home then riding his bike, which is times faster, is equal to him just walking to the stadium. So the equation would be: Let the distance from Yan's position to his home. Let the distance from Yan's home to the stadium. But we're still not done with the question. We know that Yan is from his home, and is or from the stadium. , the 's cancel out, and we are left with . Thus, the answer is ~ProGameXD
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 |
2007 AMC 10A (Problems • Answer Key • Resources) | ||
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.