2014 AMC 12B Problems/Problem 3: Difference between revisions
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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> | ||
== See also == | |||
{{AMC12 box|year=2014|ab=B|num-b=2|num-a=4}} | {{AMC12 box|year=2014|ab=B|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:22, 22 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)
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
These problems are copyrighted © by the Mathematical Association of America. 
