Difference between revisions of "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 $50 = \boxed{ \textbf{(D)}.}$

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 $50 = \boxed{ \textbf{(D)}}$

See Also

 2018 AMC 12A (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 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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