Difference between revisions of "2010 AMC 8 Problems/Problem 12"
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==Problem== | ==Problem== | ||
| − | Of the <math>500</math> balls in a large bag, <math>80%</math> are red and the rest are blue. How many of the red balls must be removed so that <math>75%</math> of the remaining balls are red? | + | Of the <math>500</math> balls in a large bag, <math>80\%</math> are red and the rest are blue. How many of the red balls must be removed so that <math>75\%</math> of the remaining balls are red? |
<math> \textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150 </math> | <math> \textbf{(A)}\ 25\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 150 </math> | ||
Revision as of 16:54, 24 August 2013
Problem
Of the 
balls in a large bag, 
are red and the rest are blue. How many of the red balls must be removed so that 
of the remaining balls are red?
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 
of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is 
.
See Also
| 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. 
