Difference between revisions of "2010 AMC 8 Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | Because they are skating in the same direction, Emily is skating relative to Emerson <math>12-8=4</math> mph. Now we can look at it as if Emerson is | + | Because they are skating in the same direction, Emily is skating relative to Emerson <math>12-8=4</math> mph. Now we can look at it as if Emerson is not moving at all and Emily is riding at <math>4</math> mph. It takes her |
<cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath> | <cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath> | ||
Revision as of 12:07, 8 October 2016
Problem
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction 
mile in front of her. After she passes him, she can see him in her rear mirror until he is mile behind her. Emily rides at a constant rate of 
miles per hour, and Emerson skates at a constant rate of 
miles per hour. For how many minutes can Emily see Emerson?
Solution
Because they are skating in the same direction, Emily is skating relative to Emerson 
mph. Now we can look at it as if Emerson is not moving at all and Emily is riding at 
mph. It takes her
![\[\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}\]](http://latex.artofproblemsolving.com/5/c/3/5c3c6fda6760c790b8771e388c649c6eea3d0835.png)
to skate the mile to reach him, and then the same amount of time to be mile ahead of him. This totals to
![\[2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}\]](http://latex.artofproblemsolving.com/e/1/9/e19d6d55213e448f3e688cf3838fbbdb65a8d60b.png)
Last edited on Nov 11, 2014, 10:23 am.
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
