Difference between revisions of "2015 AMC 8 Problems/Problem 3"
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<math>1=10t</math> | <math>1=10t</math> | ||
− | Solving for <math>t</math>, we get that Jill gets to the pool in <math>\frac{1}{10}</math> of | + | Solving for <math>t</math>, we get that Jill gets to the pool in <math>\frac{1}{10}</math> of an hour, which translates to <math>6</math> minutes. Doing the same for Jack, we get that |
Jack arrives at the pool in <math>\frac{1}{4}</math> of an hour, which in turn translates to <math>15</math> minutes. Thus, Jill has to wait <math>15-6=\boxed{\textbf{(D)}~9}</math> | Jack arrives at the pool in <math>\frac{1}{4}</math> of an hour, which in turn translates to <math>15</math> minutes. Thus, Jill has to wait <math>15-6=\boxed{\textbf{(D)}~9}</math> |
Revision as of 14:52, 7 July 2016
Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of miles per hour. Jack walks to the pool at a constant speed of miles per hour. How many minutes before Jack does Jill arrive?
Solution
Using , we can set up an equation for when Jill arrives at swimming:
Solving for , we get that Jill gets to the pool in of an hour, which translates to minutes. Doing the same for Jack, we get that
Jack arrives at the pool in of an hour, which in turn translates to minutes. Thus, Jill has to wait
minutes for Jack to arrive at the pool.
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.