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2016 AMC 12A Problems/Problem 11

Revision as of 20:16, 5 February 2016 by FractalMathHistory (talk | contribs) (→‎Solution)

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 sing, there are $100-65=35$ students who can. And $29$ students cannot sing, 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 $164-100=\boxed{\text{bf(E) }64}$ 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

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