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

Latest revision as of 19:39, 26 May 2020

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-We know that <math>d = vt</math>
+#redirect [[2010 AMC 12B Problems/Problem 7]]
-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.
-We also know that she drove a total of <math>40</math> minutes which is <math>\dfrac{2}{3}</math> of an hour.
-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:
-<math>\left\{\begin{array}{ccc} 30x + 20y & = & 16 \\x + y & = & \dfrac{2}{3} \end{array} \right.</math>
-Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:
-<math>-10y = -4 \Leftrightarrow y = \dfrac{2}{5}</math>
-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:
-<math> \boxed{\mathrm{(C)}= 24} </math>