2018 AMC 12A Problems/Problem 1

Problem

A large urn contains $100$ balls, of which $36 \%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72 \%$ ? (No red balls are to be removed.)

$\textbf{(A)}\ 28 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 64$

Solution 1

There are $36$ red balls; for these red balls to comprise $72 \%$ of the urn, there must be only $14$ blue balls. Since there are currently $64$ blue balls, this means we must remove $\boxed{\textbf{(D)}\ 50}$ .

Solution 2

There are $36$ red balls and $64$ blue balls. For the percentage of the red balls to double from $36 \%$ to $72 \%$ of the urn, half of the total number of balls must be removed. Therefore, the number of blue balls that need to be removed is $\boxed{\textbf{(D)}\ 50}$ .

Solution 3

There are $36$ red balls out of the total $100$ balls. We want to continuously remove blue balls until the percentage of red balls in the urn is 72%. Therefore, we want $\frac{36}{100-x}=\frac{72}{100}.$ Solving for $x$ gives that we must remove $\boxed{\textbf{(D)}\ 50}$ blue balls.

Video Solution 1

https://youtu.be/u26afAqgTKw

~Education, the Study of Everything

2018 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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