Difference between revisions of "2014 AMC 10B Problems/Problem 3"

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<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>
 
<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==
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==Solution 1==
 
Let the total distance be <math>x</math>. We have <math>\dfrac{x}{3} + 20 + \dfrac{x}{5} = x</math>, or <math>\dfrac{8x}{15} + 20 = x</math>. Subtracting <math>\dfrac{8x}{15}</math> from both sides gives us <math>20 = \dfrac{7x}{15}</math>. Multiplying by <math>\dfrac{15}{7}</math> gives us <math>x = \fbox{(E)} \dfrac{300}{7}</math>.
 
Let the total distance be <math>x</math>. We have <math>\dfrac{x}{3} + 20 + \dfrac{x}{5} = x</math>, or <math>\dfrac{8x}{15} + 20 = x</math>. Subtracting <math>\dfrac{8x}{15}</math> from both sides gives us <math>20 = \dfrac{7x}{15}</math>. Multiplying by <math>\dfrac{15}{7}</math> gives us <math>x = \fbox{(E)} \dfrac{300}{7}</math>.
  

Revision as of 20:35, 11 March 2014

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 $x = \fbox{(E)} \dfrac{300}{7}$.

Solution 2

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

See Also

2014 AMC 10B (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 10 Problems and Solutions

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