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Difference between revisions of "2020 AMC 12A Problems/Problem 1"

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==Solution 1==
 
==Solution 1==
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If Carlos took <math>70\%</math> of the pie, there must be <math>(100 - 70)\% = 30\%</math> left. After Maria takes <math>\frac{1}{3}</math> of the remaining <math>30\%, \ 1 - \frac{1}{3} = \frac{2}{3}</math> of the remaining <math>30\%</math> is left.
  
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If Carlos took <math>70\%</math> of the pie, there must be <math>(100 - 70) = 30\%</math> left. After Maria takes <math>\frac{1}{3}</math> of the remaining <math>30\%, </math>
+
Therefore, the answer is <math>30\% \cdot \frac{2}{3} = \boxed{\textbf{(C)}\ 20\%}.</math>
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<math>1 - \frac{1}{3} = \frac{2}{3}</math> is left.
 
  
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Therefore:
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~Awesome2.1 (Solution)
  
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<math>30\% \cdot \frac{2}{3} = \boxed{\textbf{C) }20\%}</math>
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~quacker88 (<math>\LaTeX</math> Adjustments)
−
 
 
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-Contributed by YOur dad, one dude
 
  
 
==Solution 2==
 
==Solution 2==
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Like solution 1, it is clear that there is <math>30\%</math> of the pie remaining. Since Maria takes <math>\frac{1}{3}</math> of the remainder, she takes <math>\frac{1}{3} \cdot 30\% = 10\%,</math> meaning that there is <math>30\% - 10\% = \boxed{\textbf{(C)}\ 20\%}</math> left.
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Like solution 1, it is clear that there is <math>30\%</math> of the pie remaining. Since Maria takes <math>\frac{1}{3}</math> of the remainder, she takes <math>\frac{1}{3} \cdot 30\% = 10\%</math> meaning that there is <math>30\% - 10\% = 20\%</math> left <math>\implies \boxed{\textbf{C}}</math>.
 
  
 
~DBlack2021
 
~DBlack2021
  
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==Solution 3 (One-Line Version)==
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==Solution 3 (One Sentence)==
 
We have <cmath>\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{\textbf{(C)}\ 20\%}</cmath> of the whole pie left.
 
We have <cmath>\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{\textbf{(C)}\ 20\%}</cmath> of the whole pie left.
  
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~IceMatrix
 
~IceMatrix
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 +
==Video Solution==
 +
https://www.youtube.com/watch?v=1fkJ2Mm55Ls
 +
 +
~The Power of Logic
 +
 +
==Video Solution==
 +
https://youtu.be/HtVPh8AE5dI
 +
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

Latest revision as of 03:17, 7 October 2022

Problem

Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?

$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$

Solution 1

If Carlos took $70\%$ of the pie, there must be $(100 - 70)\% = 30\%$ left. After Maria takes $\frac{1}{3}$ of the remaining $30\%, \ 1 - \frac{1}{3} = \frac{2}{3}$ of the remaining $30\%$ is left.

Therefore, the answer is $30\% \cdot \frac{2}{3} = \boxed{\textbf{(C)}\ 20\%}.$

~Awesome2.1 (Solution)

~quacker88 ($\LaTeX$ Adjustments)

Solution 2

Like solution 1, it is clear that there is $30\%$ of the pie remaining. Since Maria takes $\frac{1}{3}$ of the remainder, she takes $\frac{1}{3} \cdot 30\% = 10\%,$ meaning that there is $30\% - 10\% = \boxed{\textbf{(C)}\ 20\%}$ left.

~DBlack2021

Solution 3 (One Sentence)

We have \[\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{\textbf{(C)}\ 20\%}\] of the whole pie left.

~MRENTHUSIASM

Video Solution

https://youtu.be/qJF3G7_IDgc

~IceMatrix

Video Solution

https://www.youtube.com/watch?v=1fkJ2Mm55Ls

~The Power of Logic

Video Solution

https://youtu.be/HtVPh8AE5dI

~Education, the Study of Everything

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
First Problem 		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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