AMC8 2010

AMC8 2010 · Q20

AMC8 2010 · Q20. It mainly tests Inclusion–exclusion (basic), Probability (basic).

In a room, $\frac{2}{5}$ of all the people are wearing gloves, and $\frac{3}{4}$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and gloves?

一间屋子里，$\frac{2}{5}$ 的人戴着手套，$\frac{3}{4}$ 的人戴着帽子。屋子里至少有多少人同时戴着帽子和手套？

(A) 3 3

(B) 5 5

(D) 15 15

(E) 20 20

Answer

Correct choice: (A)

正确答案：（A）

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 $x=\boxed{\textbf{(A)}\ 3}$.

设 $x$ 为同时戴帽子和手套的人数。由于戴帽子或手套的人数必须是整数，屋子里的总人数必须是 4 和 5 的公倍数。为了求最小 $x$，取最小公倍数 $lcm(4,5) = 20$。因此假设屋子里有 20 人，每人都至少戴着一件（不戴两件的人会无谓增加总人数）。戴手套的有 $\frac{2}{5} \cdot 20 = 8$ 人，戴帽子的有 $\frac{3}{4} \cdot 20 = 15$ 人。由容斥原理，戴至少一件的总人数为 $8 + 15 - x = 23 - x$。由于每个人都至少戴一件，$23 - x = 20$，因此 $x = \boxed{\textbf{(A)}\ 3}$。

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