2020 AMC 12A Problems/Problem 1

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

-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 $30\% - 10\% = \boxed{\textbf{(C)}\ 20\%}$ 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

2020 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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

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2020 AMC 12A Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

Solution 3 (One Sentence)

Video Solution

Video Solution

See Also