Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
(→‎Video Solution)
Line 10: Line 10:
 
==Solution 2==
 
==Solution 2==
 
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>.
 
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>.
 +
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/Aaa8O3ko6pQ
 +
 +
~Education, the Study of Everything
 +
 +
  
 
==Video Solution==
 
==Video Solution==

Revision as of 11:05, 2 July 2023

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

~Education, the Study of Everything


Video Solution

https://youtu.be/C4zRNsTVMtY

~savannahsolver

See Also

2014 AMC 10B (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 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_3&oldid=195092"