Difference between revisions of "2010 AMC 12B Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | The total number of minutes in here <math>9</math>-hour work day is <math>9 \times 60 = 540</math> | + | The total number of minutes in here <math>9</math>-hour work day is <math>9 \times 60 = 540.</math> |
| − | The total amount of time spend in meetings in minutes is <math>45 + 45 \times 2 = 135</math> | + | The total amount of time spend in meetings in minutes is <math>45 + 45 \times 2 = 135.</math> |
| − | The answer is then <math>\frac{135}{540} = | + | The answer is then <math>\frac{135}{540} = \boxed{25%}</math> <math>\LongRightArrow</math> <math>(C)</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=1|num-a=2|ab=B}} | {{AMC12 box|year=2010|num-b=1|num-a=2|ab=B}} | ||
Revision as of 16:54, 6 February 2013
Problem 1
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 total number of minutes in here -hour work day is 
The total amount of time spend in meetings in minutes is 
The answer is then $\frac{135}{540} = \boxed{25%}$ (Error compiling LaTeX. Unknown error_msg) $\LongRightArrow$ (Error compiling LaTeX. Unknown error_msg) 
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 2 |
| 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 | |
