2016 AMC 8 Problems/Problem 16

Revision as of 15:40, 28 November 2021 by Lorax123 (talk | contribs) (Solution 1)

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

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

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.

- idiot1234

See Also

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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