2018 AMC 10B Problems/Problem 2

Revision as of 20:27, 18 September 2021 by MRENTHUSIASM (talk | contribs) (Solution 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 ~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 $60$ 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

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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