Difference between revisions of "2017 AMC 12A Problems/Problem 13"
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==Solution== | ==Solution== | ||
| + | Let total distance be <math>x</math>. Her speed in miles per minute is <math>x/180</math>. Then, the distance that she drove before hitting the snowstorm is \tfrac{x}{3}. Her speed in snowstorm is reduced 20 miles per hour, or \tfrac{1}{3} miles per minute. Knowing it took her 276 minutes in total, we create equation: | ||
| + | <cmath>\frac{\text{Distance before Storm}}{\text{Speed before Storm}} + \frac{\text{Distance in Storm}}{\text{Speed in Storm}} = \text{Total Time} \Longrightarrow \frac{\tfrac{x}{3}}{\tfrac{x}{180}} + \frac{\tfrac{2x}{3}}{\tfrac{x}{180} - \tfrac{1}{3}} = 276</cmath> | ||
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| + | Solving equation, we get <math>x=135</math> <math>\Longrightarrow \boxed{B}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=12|num-a=14}} | {{AMC12 box|year=2017|ab=A|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:47, 8 February 2017
Problem
Driving at a constant speed, Sharon usually takes 
minutes to drive from her house to her mother's house. One day Sharon begins the drive at her usual speed, but after driving 
of the way, she hits a bad snowstorm and reduces her speed by 
miles per hour. This time the trip takes her a total of 
minutes. How many miles is the drive from Sharon's house to her mother's house?
Solution
Let total distance be 
. Her speed in miles per minute is 
. Then, the distance that she drove before hitting the snowstorm is \tfrac{x}{3}. Her speed in snowstorm is reduced 20 miles per hour, or \tfrac{1}{3} miles per minute. Knowing it took her 276 minutes in total, we create equation:
![\[\frac{\text{Distance before Storm}}{\text{Speed before Storm}} + \frac{\text{Distance in Storm}}{\text{Speed in Storm}} = \text{Total Time} \Longrightarrow \frac{\tfrac{x}{3}}{\tfrac{x}{180}} + \frac{\tfrac{2x}{3}}{\tfrac{x}{180} - \tfrac{1}{3}} = 276\]](http://latex.artofproblemsolving.com/5/f/4/5f4fbb98e2aee66ab790657f30cab866e641dd3b.png)
Solving equation, we get 
.
See Also
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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. 
