Difference between revisions of "2012 AMC 10B Problems/Problem 13"
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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. | 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. | ||
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2012|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:17, 8 February 2014
Solution
Let 
be the speed of the escalator and 
be the speed of Clea. Using 
, the first statement can be translated to the equation 
. The second statement can be translated to 
. Since the same distance is being covered in each scenario, we can set the two equations equal and solve for . We find that 
. The problem asks for the time it takes her to ride down the escalator when she just stands on it. Since 
and , we have 
seconds. Answer choice 
is correct.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
