Difference between revisions of "2015 AMC 8 Problems/Problem 3"
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minutes for Jack to arrive at the pool. | minutes for Jack to arrive at the pool. | ||
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+ | ==Solution 2== | ||
+ | T=D/s | ||
+ | Jill: (1/10)x60 because in minutes, is equal to 6 min | ||
+ | Jack:(1/4)x60 is 15 minutes. | ||
+ | 15-6 is 9, so our answer is <math>15-6=\boxed{\textbf{(D)}~9}</math> - TheNerdWhoIsNerdy. | ||
==Video Solution (HOW TO THINK CRITICALLY!!!)== | ==Video Solution (HOW TO THINK CRITICALLY!!!)== |
Latest revision as of 19:24, 15 September 2024
Contents
[hide]Problem
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.
Solution 2
T=D/s Jill: (1/10)x60 because in minutes, is equal to 6 min Jack:(1/4)x60 is 15 minutes. 15-6 is 9, so our answer is - TheNerdWhoIsNerdy.
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
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.