Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (making spacing consistent for problem and solution headings.)
Line 13: Line 13:
  
 
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>
 
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>
 +
 +
== 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 there the percentage of red balls in the urn is 72%.
 +
Therefore, we want
 +
<math>\frac{36}{100-x}=\frac{72}{100}</math>
 +
Solving for <math>x</math> gives that we must remove <math>\boxed{50}</math> blue balls.
  
 
== 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 00:45, 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 there 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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_1&oldid=98255"