2014 AMC 12B 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

If the first and last legs of his trip account for $\frac{1}{3}$ and $\frac{1}{5}$ of his trip, then the middle leg accounts for $1 - \frac{1}{3} - \frac{1}{5} = \frac{7}{15}$ ths of his trip. This is equal to $20$ miles. Letting the length of the entire trip equal $x$ , we have $\frac{7}{15}x = 20 \implies x=\boxed{\textbf{(E)}\ \frac{300}{7}}$

Video Solution 1 (Quick and Easy)

https://youtu.be/we2FhH_4SIA

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2014 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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2014 AMC 12B Problems/Problem 3

Contents

Problem

Solution

Video Solution 1 (Quick and Easy)

See also