# Difference between revisions of "2014 AMC 10B Problems/Problem 3"

## Problem

Randy drove the first third of his trip on a gravel road, the next $20$ miles on pavement, and the remaining one-fifth on a dirt road. In miles how long was Randy's trip? $\textbf {(A) } 30 \qquad \textbf {(B) } \frac{400}{11} \qquad \textbf {(C) } \frac{75}{2} \qquad \textbf {(D) } 40 \qquad \textbf {(E) } \frac{300}{7}$

## Solution 1

Let the total distance be $x$. We have $\dfrac{x}{3} + 20 + \dfrac{x}{5} = x$, or $\dfrac{8x}{15} + 20 = x$. Subtracting $\dfrac{8x}{15}$ from both sides gives us $20 = \dfrac{7x}{15}$. Multiplying by $\dfrac{15}{7}$ gives us $x = \fbox{(E)} \dfrac{300}{7}$.

## Solution 2

The first third of his distance added to the last one-fifth of his distance equals \frac{8}{15} $. Therefore,$\frac{7}{15} $of his distance is$20 $. Let$x $be his total distance, and solve for$x $. Therefore,$x $is equal to$\frac{300}{7} $, or$E\$.

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