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2018 AMC 10B Problems/Problem 2: Difference between revisions

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SImilarly, he covered <math>\frac{65}{2}</math> miles in the <math>2</math>nd half hour period.  
SImilarly, he covered <math>\frac{65}{2}</math> miles in the <math>2</math>nd half hour period.  


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


<math>rt = d</math>, so <math>x \cdot \frac{1}{2} = 33 \frac{1}{2}</math>.
<math>rt = d</math>, so <math>x \cdot \frac{1}{2} = 33 \frac{1}{2}</math>.

Revision as of 19:23, 16 February 2018

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

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.

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

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