Difference between revisions of "2017 AMC 10B Problems/Problem 11"

m (Solution)
(Solution)
Line 3: Line 3:
  
 
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math>
 
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%</math>
==Solution==
+
12%
<math>60\% \cdot 20\% = 12\%</math> of the people that claim that they like dancing say they dislike it, and <math>40\% \cdot 90\% = 36\%</math> of the people that claim that they dislike dancing actually dislike it. Therefore, the answer is <math>\frac{12\%}{12\%+36\%} = \boxed{\textbf{(D) } 25\%}</math>.
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:37, 6 January 2021

Problem

At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?

$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%$ 12%

Video Solution

https://youtu.be/93lThricxLE

~savannahsolver

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 10 Problems and Solutions

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