Difference between revisions of "2016 AMC 12A Problems/Problem 11"

(Solution)
(Problem)
Line 4: Line 4:
  
 
<math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64</math>  
 
<math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64</math>  
 
==Problem==
 
 
Each of the <math>100</math> students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are <math>42</math> students who cannot sing, <math>65</math> students who cannot dance, and <math>29</math> students who cannot act. How many students have two of these talents?
 
 
<math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64</math>
 
  
 
==Solution==
 
==Solution==

Revision as of 16:07, 4 February 2016

Problem

Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64$

Solution

Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act.

Let $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act.

From the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$.

Adding these equations together, we get $2(a + b + c) + ab + bc + ac = 136$.

Since there are a total of $100$ students, $a + b + c + ab + bc + ac = 100$.

Subtracting these equations, we get $a + b + c = 36$.

Our answer is $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}$

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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