Difference between revisions of "2018 AMC 10B Problems/Problem 2"
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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?
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.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
