Difference between revisions of "2014 AMC 12B Problems/Problem 3"
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<cmath>\frac{7}{15}x = 20 \implies x=\boxed{\textbf{(E)}\ \frac{300}{7}}</cmath> | <cmath>\frac{7}{15}x = 20 \implies x=\boxed{\textbf{(E)}\ \frac{300}{7}}</cmath> | ||
| − | Solution by kevin38017 | + | (Solution by kevin38017) |
Revision as of 16:25, 20 February 2014
Problem
Randy drove the first third of his trip on a gravel road, the next 
miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ \frac{400}{11}\qquad\textbf{(C)}\ \frac{75}{2}\qquad\textbf{(D)}}\ 40\qquad\textbf{(E)}\ \frac{300}{7}$ (Error compiling LaTeX. Unknown error_msg)
Solution
If the first and last legs of his trip account for 
and 
of his trip, then the middle leg accounts for 
ths of his trip. This is equal to miles. Letting the length of the entire trip equal 
, we have
![\[\frac{7}{15}x = 20 \implies x=\boxed{\textbf{(E)}\ \frac{300}{7}}\]](http://latex.artofproblemsolving.com/9/f/9/9f9d4790f1d1b91761a14e562aba2427d5346705.png)
(Solution by kevin38017)
