Difference between revisions of "2016 AMC 8 Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
Call x the distance Annie runs. If Annie is 25% faster than Bonnie, then Bonnie will be 4/5x. For Annie to meet Bonnie, she must run an extra 400 meters, the length of the track. So x-(4/5)x=400. You get 1/5x=400, or x=2000, which is <math>\boxed{\textbf{(D)}\ 5 }</math> laps. | Call x the distance Annie runs. If Annie is 25% faster than Bonnie, then Bonnie will be 4/5x. For Annie to meet Bonnie, she must run an extra 400 meters, the length of the track. So x-(4/5)x=400. You get 1/5x=400, or x=2000, which is <math>\boxed{\textbf{(D)}\ 5 }</math> laps. | ||
Revision as of 18:49, 12 November 2019
Annie and Bonnie are running laps around a 
-meter oval track. They started together, but Annie has pulled ahead, because she runs 
faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Solution
Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run 
laps.
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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. 
Solution 2
Call x the distance Annie runs. If Annie is 25% faster than Bonnie, then Bonnie will be 4/5x. For Annie to meet Bonnie, she must run an extra 400 meters, the length of the track. So x-(4/5)x=400. You get 1/5x=400, or x=2000, which is laps.
