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

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==Solution 2==
 
==Solution 2==
The first third of his distance added to the last one-fifth of his distance equals \frac{8}{15}<math>. Therefore, </math>\frac{7}{15}<math> of his distance is </math>20<math>. Let </math>x<math> be his total distance, and solve for </math>x<math>. Therefore, </math>x<math> is equal to </math>\frac{300}{7}<math>, or </math>E$.
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The first third of his distance added to the last one-fifth of his distance equals <math>\frac{8}{15}</math>. Therefore, <math>\frac{7}{15}</math> of his distance is <math>20</math>. Let <math>x</math> be his total distance, and solve for <math>x</math>. Therefore, <math>x</math> is equal to <math>\frac{300}{7}</math>, or <math>E</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 15:05, 28 December 2015

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 $\frac{8}{15}$. 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 $E$.

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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