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| − | ==Problem==
| + | #redirect [[2010 AMC 12B Problems/Problem 7]] |
| − | 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 1==
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| − | We know that <math>d = vt</math>
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| − | Since we know that she drove both when it was raining and when it was not and that her total distance traveled is <math>16</math> miles.
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| − | We also know that she drove a total of <math>40</math> minutes which is <math>\dfrac{2}{3}</math> of an hour.
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| − | We get the following system of equations, where <math>x</math> is the time traveled when it was not raining and <math>y</math> is the time traveled when it was raining:
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| − | <math>\left\{\begin{array}{ccc} 30x + 20y & = & 16 \\x + y & = & \dfrac{2}{3} \end{array} \right.</math>
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| − | Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:
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| − | <math>-10y = -4 \Leftrightarrow y = \dfrac{2}{5}</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
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| − | So, our answer is <math> \boxed{\textbf{(C)}\ 24} </math>
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| − | ==Solution 2==
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| − | We let <math>t</math> be the time Shelby drives in the rain. This gives us the equation <math> 20t + 30(\frac{2}{3} - t) = 16</math>. Expanding and rearranging gives us <math>10t = 4</math>, or <math>t = 0.4</math> hours. we multiply <math>0.4</math> by <math>60</math>, which gives us <math> \boxed{\textbf{(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}}
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| − | {{MAA Notice}}
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