2010 AMC 8 Problems/Problem 12

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Problem

Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?

$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150$

Solution 1

Since 80 percent of the 500 balls are red, there are 400 red balls. Therefore, there must be 100 blue balls. For the 100 blue balls to be 25% or $\dfrac{1}{4}$ of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is $\boxed{\textbf{(D)}\ 100}$.

Solution 2

We could also set up a proportion. Since we know there are 400 red balls, we let the amount of red balls removed be $x$, so $\frac{400-x}{500-x}=\frac{3}{4}$. Cross-multiplying gives us $1600-4x=1500-3x \implies x=100$, so our answer is $\boxed{\textbf{(D)}\ 100}$.

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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