Difference between revisions of "2018 AMC 10B Problems/Problem 2"

Revision as of 17:31, 18 September 2021

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 ~MRENTHUSIASM

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 the speeds $60$ 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 (Solution)

~MRENTHUSIASM (Revision)

Video Solution

https://youtu.be/77dDIzKprzA

~savannahsolver

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.

@@ Line 33: / Line 33: @@
 ~coolmath_2018 (Solution)
-~MRENTHUSIASM (Minor Edits)
+~MRENTHUSIASM (Revision)
 == Video Solution ==

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