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

Latest revision as of 20:57, 28 October 2022

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

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

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 $\boxed{\textbf{(D)}\ 50}$ blue balls.

Video Solution 1

https://youtu.be/u26afAqgTKw

~Education, the Study of Everything

2018 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-==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.)
-<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>
-\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>\boxed{\textbf{(D)}\ 50}</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>\boxed{\textbf{(D)}\ 50}</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
+<cmath>\frac{36}{100-x}=\frac{72}{100}.</cmath>
+Solving for <math>x</math> gives that we must remove <math>\boxed{\textbf{(D)}\ 50}</math> blue balls.
+== Video Solution 1 ==
+https://youtu.be/u26afAqgTKw
+~Education, the Study of Everything
+== See Also ==
 {{AMC12 box|year=2018|ab=A|before = First Problem|num-a=2}}
 {{MAA Notice}}

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