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

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~DBlack2021
 
~DBlack2021
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==Solution 3 (One-Line Version)==
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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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~MRENTHUSIASM
  
 
==Video Solution==
 
==Video Solution==

Revision as of 04:12, 17 April 2021

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 $30\%,$ $1 - \frac{1}{3} = \frac{2}{3}$ is left.

Therefore:

$30\% \cdot \frac{2}{3} = \boxed{\textbf{C) }20\%}$

-Contributed by YOur dad, one dude

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 $30\% - 10\% = 20\%$ left $\implies \boxed{\textbf{C}}$.

~DBlack2021

Solution 3 (One-Line Version)

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

See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
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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