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

Revision as of 11:41, 5 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 $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{\textbf{(E)}\; 64}$

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.

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 ==See Also==
-{{AMC12 box|year=2016|ab=A|num-b=24|num-a=26}}
+{{AMC12 box|year=2016|ab=A|num-b=10|num-a=12}}
 {{MAA Notice}}

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

Revision as of 11:41, 5 February 2016

Problem

Solution

See Also