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

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

Revision as of 00:38, 29 November 2020

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

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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