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==
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https://youtu.be/uAc9VHtRRPg?t=82
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~IceMatrix
  
 
== 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 02:59, 26 September 2020

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 here $9$-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/uAc9VHtRRPg?t=82

~IceMatrix

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
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

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