2018 AMC 10B Problems/Problem 2

The following problem is from both the 2018 AMC 12B #2 and 2018 AMC 10B #2, so both problems redirect to this page.

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

Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.

Recall that a half hour is equal to $30$ minutes. Therefore, Sam drove $60\cdot0.5=30$ miles during the first half hour, $65\cdot0.5=32.5$ miles during the second half hour, and $x\cdot0.5$ miles during the last half hour. We have $\begin{align*} 30+32.5+x\cdot0.5&=96 \\ x\cdot0.5&=33.5 \\ x&=\boxed{\textbf{(D) } 67}. \end{align*}$ ~Haha0201

Solution 2

Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.

Note that Sam's average speed during the entire trip was $\frac{96}{3/2}=64$ mph. Since Sam drove at $60$ mph, $65$ mph, and $x$ mph for the same duration ( $30$ minutes each), his average speed during the entire trip was the average of $60$ mph, $65$ mph, and $x$ mph. We have $\begin{align*} \frac{60+65+x}{3}&=64 \\ 60+65+x&=192 \\ x&=\boxed{\textbf{(D) } 67}. \end{align*}$ ~coolmath_2018 ~MRENTHUSIASM

Video Solution

https://youtu.be/77dDIzKprzA

~savannahsolver

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/VN88m4xUHM0

~Education, the Study of Everything

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.

It will be very appreciated if you don't grief any answer worked hard on by our fellow Art of Problem Solvers!

Page

Toolbox

Search