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 "2018 AMC 10B Problems/Problem 2"
Line 5: Line 5:
  
  
−
==Solution==
+
==Solution 1 ==
  
 
Let Sam drive at exactly <math>60</math> mph in the first half hour, <math>65</math> mph in the second half hour, and <math>x</math> mph in the third half hour.
 
Let Sam drive at exactly <math>60</math> mph in the first half hour, <math>65</math> mph in the second half hour, and <math>x</math> mph in the third half hour.
Line 18: Line 18:
  
 
Therefore, Sam was driving <math>\boxed{\textbf{(D) } 67}</math> miles per hour in the third half hour.
 
Therefore, Sam was driving <math>\boxed{\textbf{(D) } 67}</math> miles per hour in the third half hour.
 +
 +
==Solution 2 (Faster)==
 +
 +
The average speed for the total trip is <cmath>\text{avg. speed} = \frac{96}{\frac{3}{2}} = 64.</cmath> Therefore the average speed for the total trip is the average of the average speeds of the three intrevals. So we have <math>64 = \frac{60 + 65 + x}{3}</math> and solving for <math>x = 67</math>. So the answer is <math>\boxed{\textbf{(D) } 67}</math>.
 +
  
 
==Video Solution==
 
==Video Solution==

Revision as of 22:24, 26 November 2020

Problem

Sam drove 96 miles in 90 minutes. His average speed during the first 30 minutes was 60 mph (miles per hour), and his average speed during the second 30 minutes was 65 mph. What was his average speed, in mph, during the last 30 minutes?

$\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68$


Solution 1

Let Sam drive at exactly $60$ mph in the first half hour, $65$ mph in the second half hour, and $x$ mph in the third half hour.

Due to $rt = d$, and that $30$ min is half an hour, he covered $60 \cdot \frac{1}{2} = 30$ miles in the first $30$ mins.

SImilarly, he covered $\frac{65}{2}$ miles in the $2$nd half hour period.

The problem states that Sam drove $96$ miles in $90$ min, so that means that he must have covered $96 - \left(30 + \frac{65}{2}\right) = 33 \frac{1}{2}$ miles in the third half hour period.

$rt = d$, so $x \cdot \frac{1}{2} = 33 \frac{1}{2}$.

Therefore, Sam was driving $\boxed{\textbf{(D) } 67}$ miles per hour in the third half hour.

Solution 2 (Faster)

The average speed for the total trip is \[\text{avg. speed} = \frac{96}{\frac{3}{2}} = 64.\] Therefore the average speed for the total trip is the average of the average speeds of the three intrevals. So we have $64 = \frac{60 + 65 + x}{3}$ and solving for $x = 67$. So the answer is $\boxed{\textbf{(D) } 67}$.


Video Solution

https://youtu.be/77dDIzKprzA

~savannahsolver

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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
2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 12 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=2018_AMC_10B_Problems/Problem_2&oldid=138579"