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? | ||
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<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 18:16, 22 January 2020
Contents
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 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 
of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is 
.
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 
, so 
. Cross-multiplying gives us 
, so our 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. 
