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 an hour, which | + | Solving for <math>t</math>, we get that Jill gets to the pool in <math>\frac{1}{10}</math> of an hour, which is <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 | + | Jack arrives at the pool in <math>\frac{1}{4}</math> of an hour, which in turn is <math>15</math> minutes. Thus, Jill has to wait <math>15-6=\boxed{\textbf{(D)}~9}</math> |
minutes for Jack to arrive at the pool. | minutes for Jack to arrive at the pool. | ||
Revision as of 20:27, 12 November 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 the swimming pool:
Solving for 
, we get that Jill gets to the pool in 
of an hour, which is 
minutes. Doing the same for Jack, we get that
Jack arrives at the pool in 
of an hour, which in turn is 
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. 
