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 1== | + | == 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> | ||
− | ==Solution 2== | + | == 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> | 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 14:24, 7 June 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 |
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