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

Revision as of 09:10, 13 August 2021

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.

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

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

@@ Line 19: / Line 19: @@
 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.
+Solving for <math>x</math> gives that we must remove <math>\boxed{\textbf{(D)}\ 50}</math> blue balls.
 == See Also ==
 {{AMC12 box|year=2018|ab=A|before = First Problem|num-a=2}}
 {{MAA Notice}}

Page

Toolbox

Search