2014 AMC 10B 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 1

Let the total distance be $x$ . We have $\dfrac{x}{3} + 20 + \dfrac{x}{5} = x$ , or $\dfrac{8x}{15} + 20 = x$ . Subtracting $\dfrac{8x}{15}$ from both sides gives us $20 = \dfrac{7x}{15}$ . Multiplying by $\dfrac{15}{7}$ gives us $x = \boxed{{\textbf{(E) }\dfrac{300}{7}}}$

Solution 2

The first third of his distance added to the last one-fifth of his distance equals $\frac{8}{15}$ . Therefore, $\frac{7}{15}$ of his distance is $20$ . Let $x$ be his total distance, and solve for $x$ . Therefore, $x$ is equal to $\frac{300}{7}$ , or $E$ .

Video Solution (CREATIVE THINKING)

https://youtu.be/Aaa8O3ko6pQ

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Video Solution

https://youtu.be/C4zRNsTVMtY

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2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 10 Problems and Solutions

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

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