Difference between revisions of "2010 AMC 12B Problems/Problem 1"
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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}</math> <math>= \boxed{25\%}</math> <math>\ | + | The answer is then <math>\frac{135}{540}</math> <math>= \boxed{25\%}</math> or <math>\boxed{(C)}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2010|before=First Question|num-a=2|ab=B}} | {{AMC12 box|year=2010|before=First Question|num-a=2|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:58, 13 July 2015
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 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 
or 
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Question |
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
