Difference between revisions of "2016 AMC 8 Problems/Problem 16"
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Call <math>x</math> the distance Annie runs. If Annie is <math>25\%</math> faster than Bonnie, then Bonnie will run a distance of <math>\frac{4}{5}x</math>. For Annie to meet Bonnie, she must run an extra <math>400</math> meters, the length of the track. So <math>x-\left(\frac{4}{5}\right)x=400 \implies x=2000</math>, which is <math>\boxed{\textbf{(D)}\ 5 }</math> laps. | Call <math>x</math> the distance Annie runs. If Annie is <math>25\%</math> faster than Bonnie, then Bonnie will run a distance of <math>\frac{4}{5}x</math>. For Annie to meet Bonnie, she must run an extra <math>400</math> meters, the length of the track. So <math>x-\left(\frac{4}{5}\right)x=400 \implies x=2000</math>, which is <math>\boxed{\textbf{(D)}\ 5 }</math> laps. | ||
| − | - | + | - idiot1234 |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2016|num-b=15|num-a=17}} | {{AMC8 box|year=2016|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:39, 28 November 2021
Problem
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?
Solutions
Solution 1
For Annie and Bonnie to meet again, Annie needs to run another lap to overtake Bonnie. Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. That means Annie will have run 
laps.
Solution 2
Call 
the distance Annie runs. If Annie is faster than Bonnie, then Bonnie will run a distance of 
. For Annie to meet Bonnie, she must run an extra meters, the length of the track. So 
, which is laps.
- idiot1234
See Also
| 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. 
