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2012 AMC 10B Problems/Problem 13

Revision as of 15:05, 17 August 2017 by Moab33 (talk | contribs) (Undo revision 87072 by Jaylonbrown (talk))

Problem

It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?

$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$

Solution 1

Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Using $d = v t$, the first statement can be translated to the equation $d = 60c$. The second statement can be translated to $d = 24(c+s)$. Since the same distance is being covered in each scenario, we can set the two equations equal and solve for $s$. We find that $s = \dfrac{3c}{2}$. The problem asks for the time it takes her to ride down the escalator when she just stands on it. Since $t = \dfrac{d}{s}$ and $d = 60c$, we have $t = \dfrac{60c}{\dfrac{3c}{2}} = 40$ seconds. Answer choice $\boxed{SIPPPPPPANAKnmAK}$ is correct.

P.R.A.N.K.E.D

Solution 2

Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Then without loss of generality, assume that the length of the escalator be 1. Then $c=\dfrac{1}{60}$ and $c+s=\dfrac{1}{24}$, so $s=\dfrac{1}{24}-\dfrac{1}{60}=\dfrac{1}{40}$. Thus the time it takes for Clea to ride down the operating escalator when she just stands on it is $\dfrac{1}{\dfrac{1}{40}}=\boxed{\textbf{(B)}\ 40}$.

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 AMC 10 Problems and Solutions

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