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

(Solution)
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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>
 
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

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 $50 = \boxed{ \textbf{(D)}}$

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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