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

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<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>. | ||

− | Therefore, Sam was driving <math>\textbf{(D) } 67</math> miles per hour in the third half hour. | + | Therefore, Sam was driving <math>\boxed{\textbf{(D) } 67}</math> miles per hour in the third half hour. |

## Revision as of 15:24, 16 February 2018

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?

## Solution

Let Sam drive at exactly mph in the first half hour, mph in the second half hour, and mph in the third half hour.

Due to , and that min is half an hour, he covered miles in the first mins.

SImilarly, he covered miles in the nd half hour period.

The problem states that Sam drove miles in min, so that means that he must have covered miles in the third half hour period.

, so .

Therefore, Sam was driving miles per hour in the third half hour.