Difference between revisions of "2010 AMC 10B Problems/Problem 2"

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== Problem ==
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Makarla attended two meetings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?
 
Makarla attended two meetings during her <math>9</math>-hour work day. The first meeting took <math>45</math> minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?
  
<cmath> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 </cmath>
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<math>\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35</math>
 
 
  
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== 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 percentage of her time spent in meetings is the total amount of time spent in meetings divided by the length of her workday.
  

Revision as of 20:11, 22 January 2011

Problem

Makarla attended two meetings during her $9$-hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35$

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 $45 \text{ min}  + 2*45\text{ min} = 2\text{ hours } 15 \text{ min} = 9/4 \text{ hours}$

Therefore, the percentage is \[\frac{9/4 \text{ hours} }{9 \text{ hours}} = \frac{1}{4} = 25 \% = \textbf{(C)}\]