Difference between revisions of "2018 AMC 12A Problems/Problem 1"
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− | ==Problem== | + | == Problem == |
A large urn contains <math>100</math> balls, of which <math>36 \%</math> 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 <math>72 \%</math>? (No red balls are to be removed.) | A large urn contains <math>100</math> balls, of which <math>36 \%</math> 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 <math>72 \%</math>? (No red balls are to be removed.) | ||
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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> | ||
− | ==See Also== | + | == Solution 2 == |
+ | |||
+ | 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> | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | There are <math>36</math> red balls out of the total <math>100</math> balls. | ||
+ | We want to continuously remove blue balls until the percentage of red balls in the urn is 72%. | ||
+ | Therefore, we want | ||
+ | <math>\frac{36}{100-x}=\frac{72}{100}</math> | ||
+ | Solving for <math>x</math> gives that we must remove <math>\boxed{50}</math> blue balls. | ||
+ | |||
+ | == 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}} |
Latest revision as of 01:45, 24 October 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 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
Solution 3
There are red balls out of the total balls. We want to continuously remove blue balls until the percentage of red balls in the urn is 72%. Therefore, we want Solving for gives that we must remove blue balls.
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 |
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