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Difference between revisions of "2012 AMC 10B Problems/Problem 13"

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

Revision as of 18:02, 6 January 2013

Solution

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 $\boxed{B}$ is correct.

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