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 standing still and Emily is skating <math>4</math> mph. It takes her | |
| − | + | <cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath> | |
| + | |||
| + | to skate the <math>1/2</math> mile to reach him, and then the same amount of time to be <math>1/2</math> mile ahead of him. This totals to | ||
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| + | <cmath>2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}</cmath> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=7|num-a=9}} | {{AMC8 box|year=2010|num-b=7|num-a=9}} | ||
Revision as of 18:09, 25 December 2012
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 standing still and Emily is skating 
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)
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 | ||
