Difference between revisions of "2015 AMC 8 Problems/Problem 3"
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Jill arrives in <math>\dfrac{1}{10}</math> of an hour, which is <math>6</math> minutes. Jack arrives in <math>\dfrac{1}{4}</math> of an hour which is <math>15</math> minutes. Thus, the time difference is <math>\boxed{\textbf{(D)}~9}</math> minutes. | Jill arrives in <math>\dfrac{1}{10}</math> of an hour, which is <math>6</math> minutes. Jack arrives in <math>\dfrac{1}{4}</math> of an hour which is <math>15</math> minutes. Thus, the time difference is <math>\boxed{\textbf{(D)}~9}</math> minutes. | ||
| − | == | + | ==Solution 2== |
Using <math>d=rt</math>, we can set up an equation for when Jill arrives at swimming: | Using <math>d=rt</math>, we can set up an equation for when Jill arrives at swimming: | ||
Revision as of 09:55, 26 November 2015
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
Jill arrives in 
of an hour, which is 
minutes. Jack arrives in 
of an hour which is 
minutes. Thus, the time difference is 
minutes.
Solution 2
Using 
, we can set up an equation for when Jill arrives at swimming:
Solving for t, we get that Jill gets to the pool in 
of on 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: answer 
.
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. 
