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

Revision as of 11:01, 1 February 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, $(100 - 70) = 30\%$ must be remaining. 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 Awesome2.1, latex by quacker88

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 12: / Line 12: @@
 Therefore:
-<math>\frac{3}{10} \cdot \frac{2}{3} = \frac{2}{10}= \boxed{\textbf{C) }20\%}</math>
+<math>30\% \cdot \frac{2}{3} = \boxed{\textbf{C) }20\%}</math>
 -Contributed by Awesome2.1, latex by quacker88

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

Revision as of 11:01, 1 February 2020

Problem

Solution 1

See Also