Difference between revisions of "2018 AMC 12A Problems/Problem 1"
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50 \qquad\textbf{(E)}\ 64 </math> | 50 \qquad\textbf{(E)}\ 64 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
There are <math>36</math> red balls; for these red balls to comprise <math>72 \%</math> of the urn, there must be only <math>14</math> blue balls. Since there are currently <math>64</math> blue balls, this means we must remove <math>50 = \boxed{ \textbf{(D)}.}</math> | There are <math>36</math> red balls; for these red balls to comprise <math>72 \%</math> of the urn, there must be only <math>14</math> blue balls. Since there are currently <math>64</math> blue balls, this means we must remove <math>50 = \boxed{ \textbf{(D)}.}</math> | ||
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| + | ==Solution 2== | ||
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| + | There are <math>36</math> red balls and <math>64</math> blue balls. For the percentage of the red balls to double from <math>36 \%</math> to <math>72 \%</math> of the urn, half of the total number of balls must be removed. Therefore, the number of blue balls that need to be removed is <math>50 = \boxed{ \textbf{(D)}}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|before = First Problem|num-a=2}} | {{AMC12 box|year=2018|ab=A|before = First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:45, 8 February 2018
Contents
Problem
A large urn contains 
balls, of which 
are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be 
? (No red balls are to be removed.)
Solution 1
There are 
red balls; for these red balls to comprise of the urn, there must be only 
blue balls. Since there are currently 
blue balls, this means we must remove 
Solution 2
There are red balls and blue balls. For the percentage of the red balls to double from to of the urn, half of the total number of balls must be removed. Therefore, the number of blue balls that need to be removed is 
See Also
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
