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 "2016 AMC 12A Problems/Problem 11"

(Created page with "==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 thr...")
 
(→‎Solution)
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==
  
 +
Let <math>a</math> be the number of students that can only sing, <math>b</math> can only dance, and <math>c</math> can only act.
 +
 +
Let <math>ab</math> be the number of students that can sing and dance, <math>ac</math> can sing and act, and <math>bc</math> can dance and act.
 +
 +
From the information given in the problem, <math>a + ab + b = 29, b + bc + c = 42,</math> and <math>a + ac + c = 65</math>.
 +
 +
Adding these equations together, we get <math>2(a + b + c) + ab + bc + ac = 136</math>.
 +
 +
Since there are a total of <math>100</math> students, <math>a + b + c + ab + bc + ac = 100</math>.
 +
 +
Subtracting these equations, we get <math>a + b + c = 36</math>.
  
 +
Our answer is <math>ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=24|num-a=26}}
 
{{AMC12 box|year=2016|ab=A|num-b=24|num-a=26}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:06, 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?

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

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 (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_11&oldid=75475"