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

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==Solution==
 
==Solution==
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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>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2014|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:16, 20 February 2014

Problem 3

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

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 $x = \fbox{E) \dfrac{300}{7}}$ (Error compiling LaTeX. ! Missing $ inserted.).

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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