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Difference between revisions of "2016 AMC 8 Problems/Problem 16"

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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.  
 
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.  
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Someone, please help with the latex. I also don't know why this solution is moved to the bottom of the page. Delete these two lines when the issues have been resolved.
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Revision as of 20:45, 12 November 2019

Annie and Bonnie are running laps around a $400$-meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?

$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$

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 $\boxed{\textbf{(D)}\ 5 }$ laps.

Solution 2

Call x the distance Annie runs. If Annie is 25% faster than Bonnie, then Bonnie will be $\frac{4}{5}$ x. For Annie to meet Bonnie, she must run an extra 400 meters, the length of the track. So x-( $\frac{4}{5}$ )x=400. You get $\frac{4}{5}$ x=400, or x=2000, which is $\boxed{\textbf{(D)}\ 5 }$ laps.

- NoisedHens

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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