2014 AMC 12B Problems/Problem 3: Difference between revisions
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Randy drove the first third of his trip on a gravel road, the next <math> 20 </math> miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip? | Randy drove the first third of his trip on a gravel road, the next <math> 20 </math> miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip? | ||
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ \frac{400}{11}\qquad\textbf{(C)}\ \frac{75}{2}\qquad\textbf{(D) | <math> \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} </math> | ||
==Solution== | ==Solution== | ||
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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> | ||
==Video Solution 1 (Quick and Easy)== | |||
https://youtu.be/we2FhH_4SIA | |||
~Education, the Study of Everything | |||
== 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}} | ||
Latest revision as of 23:39, 14 November 2022
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?
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)
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
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. 
