2010 AMC 8 Problems/Problem 20

Problem 20

In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?

$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

Solution

Let $n$ be the total number of people in the room and $x$ be the number wearing both a hat and a glove. It is possible that there are people in the room who are wearing neither a hat or a glove. However, as we are trying to minimize x, we also want to have as few people in the room as possible and still satisfy the requirements of the problem. Thus, it is safe to assume that everyone in the room is either wearing a hat or a glove or both. Then by applying the Principle of Inclusion Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is $\frac{2}{5}n + \frac{3}{4}n - x = \frac{23}{20}n - x$ . Since $n$ is also the number of people in the room that are wearing at least a hat or a glove or both, it follows that $\frac{23}{20}n - x = n$ . Solving for x, we get $x = \frac{3}{20}n$ . Since $x$ must be a whole number and we want the smallest such $x$ , it follows that $n=20$ is the least number of people that must be in the room. Substituting $n=20$ into the inequality, we get $x = 3$ . Thus, $\boxed{\textbf{(A)}\ 3}$ is the correct answer.

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2010 AMC 8 Problems/Problem 20

Problem 20

Solution

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