Difference between revisions of "2010 AMC 8 Problems/Problem 8"
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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 | 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 | ||
− | <cmath>2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}</cmath> | + | <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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:25, 13 September 2018
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 both moving 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[on his skateboard] and Emily is riding at mph. It takes her
to skate the mile to reach him, and then the same amount of time to be mile ahead of him. This totals to
.
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 |
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