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

m (Solution 2)
(Video Solution)
Line 23: Line 23:
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
==Video Solution1==
 +
https://youtu.be/HtVPh8AE5dI
 +
 +
~Education, the Study of Everything
 +
  
 
==Video Solution==
 
==Video Solution==

Revision as of 19:38, 6 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 Solution1

https://youtu.be/HtVPh8AE5dI

~Education, the Study of Everything


Video Solution

https://youtu.be/qJF3G7_IDgc

~IceMatrix

Video Solution

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

~The Power of Logic

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png