Difference between revisions of "2016 AMC 8 Problems/Problem 16"
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==Solutions== | ==Solutions== | ||
===Solution 1=== | ===Solution 1=== | ||
| − | + | 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 <math>\boxed{\textbf{(D)}\ 5 }</math> laps. | |
| − | == | + | == Video Solution by Pi Academy == |
| − | |||
| − | + | https://youtu.be/Sn1QLgCJUAQ?si=UebITjxagXo7KoDI | |
| + | |||
| + | |||
| + | ==Video Solution 2== | ||
| + | https://youtu.be/lRbxzdBZpIY | ||
| + | |||
| + | ~savannahsolver | ||
| − | |||
{{AMC8 box|year=2016|num-b=15|num-a=17}} | {{AMC8 box|year=2016|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:09, 23 November 2024
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
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.
Video Solution by Pi Academy
https://youtu.be/Sn1QLgCJUAQ?si=UebITjxagXo7KoDI
Video Solution 2
~savannahsolver
| 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. 
