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

(Created page with "==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 per...")
 
(Problem)
Line 5: Line 5:
 
<math> \textbf{(A)}\ 28 \qquad\textbf{(B)}\  32 \qquad\textbf{(C)}\  36 \qquad\textbf{(D)}\  
 
<math> \textbf{(A)}\ 28 \qquad\textbf{(B)}\  32 \qquad\textbf{(C)}\  36 \qquad\textbf{(D)}\  
 
  50 \qquad\textbf{(E)}\ 64 </math>
 
  50 \qquad\textbf{(E)}\ 64 </math>
 +
 +
==Solution==
 +
 +
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>

Revision as of 13:30, 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

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