Difference between revisions of "2010 AMC 10B Problems/Problem 10"

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==Problem==
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Shelby drives her scooter at a speed of <math>30</math> miles per hour if it is not raining, and <math>20</math> miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of <math>16</math> miles in <math>40</math> minutes. How many minutes did she drive in the rain?
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<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30</math>
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==Solution==
 
We know that <math>d = vt</math>
 
We know that <math>d = vt</math>
  
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We know now that the time traveled in rain was <math>\dfrac{2}{5}</math> of an hour, which is <math>\dfrac{2}{5}*60 = 24</math> minutes
 
We know now that the time traveled in rain was <math>\dfrac{2}{5}</math> of an hour, which is <math>\dfrac{2}{5}*60 = 24</math> minutes
  
So, our answer is:
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So, our answer is <math> \boxed{\textbf{(C)}\ 24} </math>
  
<math> \boxed{\mathrm{(C)}= 24} </math>
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==See Also==
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{{AMC10 box|year=2010|ab=B|num-b=9|num-a=11}}

Revision as of 01:54, 26 November 2011

Problem

Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$

Solution

We know that $d = vt$

Since we know that she drove both when it was raining and when it was not and that her total distance traveled is $16$ miles.

We also know that she drove a total of $40$ minutes which is $\dfrac{2}{3}$ of an hour.

We get the following system of equations, where $x$ is the time traveled when it was not raining and $y$ is the time traveled when it was raining:

$\left\{\begin{array}{ccc} 30x + 20y & = & 16 \\x + y & = & \dfrac{2}{3} \end{array} \right.$

Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:

$-10y = -4 \Leftrightarrow y = \dfrac{2}{5}$

We know now that the time traveled in rain was $\dfrac{2}{5}$ of an hour, which is $\dfrac{2}{5}*60 = 24$ minutes

So, our answer is $\boxed{\textbf{(C)}\ 24}$

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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