2010 AMC 8 Problems/Problem 20

Problem

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 $x$ be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of 4 and 5. Since we are trying to find the minimum $x$, we must use the least common multiple. $lcm(4,5) = 20$. Thus, we can say that there are $20$ people in the room, all of which are wearing at least a hat or a glove. (Any people wearing neither item would unnecessarily increase the number of people in the room.)

It follows that there are $\frac{2}{5}\cdot 20 = 8$ people wearing gloves and $\frac{3}{4}\cdot 20 = 15$ people wearing hats. Then by applying the Principle of Inclusion and Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is $8+15-x = 23-x$, where $x$ is the number wearing both. Since everyone in the room is wearing at least one item (see above), $23-x = 20$, and so $x=\boxed{\textbf{(A)}\ 3}$.

Video by MathTalks

https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be

Video Solution by WhyMath

https://youtu.be/Ym28CYMKIW8

~savannahsolver

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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