2010 AMC 8 Problems/Problem 12

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(logical solution)

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(algebra solution)

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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