Difference between revisions of "2018 AMC 12A Problems/Problem 1"
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− | 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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+ | ==See Also== | ||
+ | {{AMC12 box|year=2018|ab=A|before = First Problem|num-a=2}} | ||
+ | {{MAA Notice}} |
Revision as of 16:21, 8 February 2018
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
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
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by First Problem |
Followed by Problem 2 |
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All AMC 12 Problems and Solutions |
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