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

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==Solutions==
 
==Solutions==
 
===Solution 1===
 
===Solution 1===
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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 <math>\boxed{\textbf{(D)}\ 5 }</math> laps (harder daddy).
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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.  
  
 
===Solution 2===
 
===Solution 2===
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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. I like them creamy and thick that's what she said. So <math>x-\left(\frac{4}{5}\right)x=400 \implies x=2000</math>, which is <math>\boxed{\textbf{(D)}\ 5 }</math> laps.
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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.
  
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- idiot1234
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- NoisedHens
  
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==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 11:15, 30 November 2021

Problem

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?

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

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

Solution 2

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

- NoisedHens

2016 AMC 8 (Problems • Answer Key • Resources)
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Problem 15 		Followed by
Problem 17
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