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"

m (Fixed MAA box)
(→‎Solution 2)
 
(2 intermediate revisions by one other user not shown)
Line 20: Line 20:
  
 
Our answer is <math>ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}</math>
 
Our answer is <math>ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}</math>
 +
 +
==Solution 2==
 +
An easier way to solve the problem:
 +
Since <math>42</math> students cannot sing, there are <math>100-42=58</math> students who can.
 +
 +
Similarly <math>65</math> students cannot dance, there are <math>100-65=35</math> students who can.
 +
 +
And <math>29</math> students cannot act, there are <math>100-29=71</math> students who can.
 +
 +
Therefore, there are <math>58+35+71=164</math> students in all ignoring the overlaps between <math>2</math> of <math>3</math> talent categories.
 +
There are no students who have all <math>3</math> talents, nor those who have none <math>(0)</math>, so only <math>1</math> or <math>2</math> talents are viable.
 +
 +
Thus, there are <math>164-100=\boxed{\textbf{(E) }64}</math> students who have <math>2</math> of <math>3</math> talents.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=10|num-a=12}}
 
{{AMC12 box|year=2016|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:05, 19 October 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 $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}$

Solution 2

An easier way to solve the problem: Since $42$ students cannot sing, there are $100-42=58$ students who can.

Similarly $65$ students cannot dance, there are $100-65=35$ students who can.

And $29$ students cannot act, there are $100-29=71$ students who can.

Therefore, there are $58+35+71=164$ students in all ignoring the overlaps between $2$ of $3$ talent categories. There are no students who have all $3$ talents, nor those who have none $(0)$, so only $1$ or $2$ talents are viable.

Thus, there are $164-100=\boxed{\textbf{(E) }64}$ students who have $2$ of $3$ talents.

See Also

2016 AMC 12A (Problems • Answer Key • Resources)
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 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=80703"