Difference between revisions of "2012 AMC 12B Problems/Problem 9"
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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 | + | 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 | ||
| + | ~Extremelysupercooldude (Latex edits) | ||
== See Also == | == See Also == | ||
Revision as of 06:56, 29 June 2023
Contents
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?
Solution 1
She walks at a rate of 
units per second to travel a distance 
. As 
, we find 
and 
, where 
is the speed of the escalator. Setting the two equations equal to each other, 
, which means that 
. Now we divide 
by 
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 
Solution 2
We write two equations using distance = rate * time. Let 
be the rate she is walking, and 
be the speed the escalator moves. WLOG, let the distance of the escalator be 
, as the distance is constant. Thus, our 
equations are 
and 
. Solving for , we get 
. Thus, it will take Clea 
seconds.
~coolmath2017 ~Extremelysupercooldude (Latex edits)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
