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 $\boxed{\textbf{(D)}\ 100}$ .

Solution 3

Before any balls are taken out of the bag, there are 400 red balls. Let the number of red balls we take out of the bag be $x$ . Since we want the number of red ball to be 75% of the remaining balls, we can set up the equation: $0.75(500–x)=400–x$ 75% of the balls are equal to the remaining red balls. Solving we get $x=100$ , so our answer is $\boxed{\textbf{(D)}\ 100}$ .

Video by MathTalks

https://www.youtube.com/watch?v=6hRHZxSieKc

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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