Difference between revisions of "2018 AMC 12A Problems/Problem 1"

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==Problem==
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== 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==
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== 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>
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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>
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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>
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== Solution 3 ==
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There are <math>36</math> red balls out of the total <math>100</math> balls.
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We want to continuously remove blue balls until the percentage of red balls in the urn is 72%.
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Therefore, we want
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<math>\frac{36}{100-x}=\frac{72}{100}</math>
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Solving for <math>x</math> gives that we must remove <math>\boxed{50}</math> blue balls.
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== See Also ==
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{{AMC12 box|year=2018|ab=A|before = First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 01:45, 24 October 2018

Problem

A large urn contains $100$ balls, of which $36 \%$ 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 $72 \%$? (No red balls are to be removed.)

$\textbf{(A)}\ 28 \qquad\textbf{(B)}\  32 \qquad\textbf{(C)}\  36 \qquad\textbf{(D)}\   50 \qquad\textbf{(E)}\ 64$

Solution 1

There are $36$ red balls; for these red balls to comprise $72 \%$ of the urn, there must be only $14$ blue balls. Since there are currently $64$ blue balls, this means we must remove $50 = \boxed{ \textbf{(D)}.}$

Solution 2

There are $36$ red balls and $64$ blue balls. For the percentage of the red balls to double from $36 \%$ to $72 \%$ 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 $50 = \boxed{ \textbf{(D)}}$

Solution 3

There are $36$ red balls out of the total $100$ balls. We want to continuously remove blue balls until the percentage of red balls in the urn is 72%. Therefore, we want $\frac{36}{100-x}=\frac{72}{100}$ Solving for $x$ gives that we must remove $\boxed{50}$ blue balls.

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
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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