Difference between revisions of "2010 AMC 10B Problems/Problem 2"
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The total time spent in meetings is <math>45 \text{ min} + 2*45\text{ min} = 2\text{ hours } 15 \text{ min} = 9/4 \text{ hours}</math> | The total time spent in meetings is <math>45 \text{ min} + 2*45\text{ min} = 2\text{ hours } 15 \text{ min} = 9/4 \text{ hours}</math> | ||
| − | Therefore, the percentage is <cmath>\frac{9/4 \text{ hours} }{9 \text{ hours}} = \frac{1}{4} = 25 \% = {(C)} 25}</cmath> | + | Therefore, the percentage is <cmath>\frac{9/4 \text{ hours} }{9 \text{ hours}} = \frac{1}{4} = 25 \% = \boxed{\textbf{(C)} 25}</cmath> |
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2010|ab=B|num-b=1|num-a=3}} | ||
Revision as of 00:49, 26 November 2011
Problem
Makarla attended two meetings during her 
-hour work day. The first meeting took 
minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?
Solution
The percentage of her time spent in meetings is the total amount of time spent in meetings divided by the length of her workday.
The total time spent in meetings is 
Therefore, the percentage is ![\[\frac{9/4 \text{ hours} }{9 \text{ hours}} = \frac{1}{4} = 25 \% = \boxed{\textbf{(C)} 25}\]](http://latex.artofproblemsolving.com/7/3/f/73f3feb942ede81e7f241c1aefb60a12e3e3eaf3.png)
See Also
| 2010 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 | ||
