Difference between revisions of "2010 AMC 8 Problems/Problem 12"
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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>. | ||
| − | ==Solution 2== | + | ==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 <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>. | 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>. | ||
Revision as of 11:01, 28 April 2021
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(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 
of the bag, there must be 400 balls in the bag so 100 red balls must be removed. The answer is 
.
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 
, 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. 
