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Difference between revisions of "2010 AMC 12B Problems/Problem 1"

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== Solution ==
 
== Solution ==
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The total number of minutes in here <math>9</math>-hour work day is <math>9 \times 60 = 540.</math>
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The total number of minutes in her <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> 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==
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https://youtu.be/8yntxcttyd8
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-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 $9$-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 total number of minutes in her $9$-hour work day is $9 \times 60 = 540.$ The total amount of time spend in meetings in minutes is $45 + 45 \times 2 = 135.$ The answer is then $\frac{135}{540}$ $= \boxed{25\%}$ or $\boxed{(C)}$

Video Solution

https://youtu.be/8yntxcttyd8

-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

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