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?

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

Solutions

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 $\boxed{\textbf{(D) } 67}$ . ~coolmath_2018

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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