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Difference between revisions of "2012 AMC 12B Problems/Problem 9"

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==Problem==
 
==Problem==
  
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It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How seconds would it take Clea to ride the escalator down when she is not walking?
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It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?
  
 
<math>\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52 </math>
 
<math>\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52 </math>
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==Solution 2==
 
==Solution 2==
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We write two equations using distance=rate * time. Let r be the rate she is walking, and e be the speed the escalator moves. WLOG, let the distance of the escalator be 120, as the distance is constant. Thus, our 2 equations are 120=60r and 120=24(r+e). Solving for e, we get e=3. Thus, it will take Clea 120/3=40 seconds.
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We write two equations using distance = rate * time. Let <math>r</math> be the rate she is walking, and <math>e</math> be the speed the escalator moves. WLOG, let the distance of the escalator be <math>120</math>, as the distance is constant. Thus, our <math>2</math> equations are <math>120 = 60r</math> and <math>120 = 24(r+e)</math>. Solving for <math>e</math>, we get <math>e = 3</math>. Thus, it will take Clea <math>\dfrac{120}{3} = \boxed{\textbf{(B)}\ 40}</math> seconds.
  
 
~coolmath2017
 
~coolmath2017
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~Extremelysupercooldude (Latex edits)
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==Solution 3==
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Clea covers <math>\dfrac{1}{60}</math> of the escalator every second. Say the escalator covers <math>\dfrac{1}{r}</math> of the escalator every second. Since Clea and the escalator cover the entire escalator in <math>24</math> seconds, we can use distance <math>=</math> rate <math>\cdot</math> time to get <math>24\left(\dfrac{1}{60} + \dfrac{1}{r}\right) = 1</math>. Solving gives us <math>r = 40</math>, so if Clea were to just stand on the escalator, it would take her <math>\boxed{\textbf{(B)}\ 40}</math> seconds to get down. ~Extremelysupercooldude
  
 
== See Also ==
 
== See Also ==

Latest revision as of 07:14, 29 June 2023

Problem

It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?

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

Solution 1

She walks at a rate of $x$ units per second to travel a distance $y$. As $vt=d$, we find $60x=y$ and $24*(x+k)=y$, where $k$ is the speed of the escalator. Setting the two equations equal to each other, $60x=24x+24k$, which means that $k=1.5x$. Now we divide $60$ by $1.5$ because you add the speed of the escalator but remove the walking, leaving the final answer that it takes to ride the escalator alone as $\boxed{\textbf{(B)}\ 40}$

Solution 2

We write two equations using distance = rate * time. Let $r$ be the rate she is walking, and $e$ be the speed the escalator moves. WLOG, let the distance of the escalator be $120$, as the distance is constant. Thus, our $2$ equations are $120 = 60r$ and $120 = 24(r+e)$. Solving for $e$, we get $e = 3$. Thus, it will take Clea $\dfrac{120}{3} = \boxed{\textbf{(B)}\ 40}$ seconds.

~coolmath2017 ~Extremelysupercooldude (Latex edits)

Solution 3

Clea covers $\dfrac{1}{60}$ of the escalator every second. Say the escalator covers $\dfrac{1}{r}$ of the escalator every second. Since Clea and the escalator cover the entire escalator in $24$ seconds, we can use distance $=$ rate $\cdot$ time to get $24\left(\dfrac{1}{60} + \dfrac{1}{r}\right) = 1$. Solving gives us $r = 40$, so if Clea were to just stand on the escalator, it would take her $\boxed{\textbf{(B)}\ 40}$ seconds to get down. ~Extremelysupercooldude

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 12 Problems and Solutions

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