Difference between revisions of "2010 AMC 8 Problems/Problem 12"
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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> | ||
| − | ==Solution== | + | ==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 <math>\dfrac{1}{4}</math> of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is <math>\boxed{\textbf{(D)}\ 100}</math>. | 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 <math>\dfrac{1}{4}</math> of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is <math>\boxed{\textbf{(D)}\ 100}</math>. | ||
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| + | ==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 <math>x</math>, so <math>\frac{400-x}{500-x}=\frac{3}{4}</math>. Cross-multiplying gives us <math>1600-4x=1500-3x \implies x=100</math>, so our answer is <math>\boxed{\textbf{(D)}\ 100}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=11|num-a=13}} | {{AMC8 box|year=2010|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:46, 7 November 2016
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. 
