Difference between revisions of "2010 AMC 12B Problems/Problem 1"
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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> or <math>\boxed{(C)}</math> | The answer is then <math>\frac{135}{540}</math> <math>= \boxed{25\%}</math> or <math>\boxed{(C)}</math> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/8yntxcttyd8 | ||
| + | |||
| + | -Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 15:59, 1 August 2022
- The following problem is from both the 2010 AMC 12B #1 and 2010 AMC 10B #2, so both problems redirect to this page.
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 her -hour work day is 
The total amount of time spend in meetings in minutes is 
The answer is then 
or 
Video Solution
-Education, the Study of Everything
Video Solution
https://youtu.be/uAc9VHtRRPg?t=82
~IceMatrix
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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
