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 \[\frac{7}{15}x = 20 \implies x=\boxed{\textbf{(E)}\ \frac{300}{7}}\]

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS