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

## 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 1

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 $\boxed{\textbf{(C)}\ 24}$

## Solution 2

We let $t$ be the time Shelby drives in the rain. This gives us the equation $20t + 30(\frac{2}{3} - t) = 16$. Expanding and rearranging gives us $10t = 4$, or $t = 0.4$ hours. we multiply $0.4$ by $60$, which gives us $\boxed{\textbf{(C)}\ 24}$.