2010 AMC 8 第 20 题

Question

一个房间里，

\frac{2}{5}

的人戴手套，

\frac{3}{4}

的人戴帽子。房间里至少有多少人同时戴帽子和手套？

In a room,

\frac{2}{5}

of 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?

\ 3

\ 5

\ 8

\ 15

\ 20

答案：A

知识点：容斥原理最小公倍数

难度评级：1610

小提示：

要最小化重叠，如果可能，让每个人都至少属于两个群体之一

Minimize overlap by making everyone counted by at least one of the two groups if possible.

大提示：

总人数必须使两个分数对应的人数都是整数

The total number of people must make both fractions whole numbers.

解答：

因为房间中

\dfrac 25

的人戴手套，所以总人数必须是

5

的倍数。因为

\dfrac 34

的人戴帽子，所以总人数必须是

4

的倍数。因此总人数必须是

20

的倍数。

另外，由容斥原理可以写出下面的关系：同时戴两者的比例 = 戴手套的比例 + 戴帽子的比例 − 至少戴其中一种的比例。

所求比例等于

\dfrac{2}{5} + \dfrac 34 -

至少戴一种物品的比例。要使同时戴两者的人数最少，就让至少戴一种物品的比例尽可能大，最多为

1

。所以同时戴两者的比例至少为

\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}

。

取最小的正总人数

20

，在这

20

人中，同时戴两者的人数为

3

。

所以正确答案是 A。

Since our room has

\dfrac 25

of the people wearing gloves, the number of people must be a multiple of

5.

Since our room has

\dfrac 34

of the people wearing hats, the number of people must be a multiple of

4.

Therefore, the people in the room must be a multiple of

20.

Now, we can also use the following formula by the principle of inclusion exclusion: Fraction of people wearing both = Fraction of people wearing gloves + Fraction of people wearing hats - Fraction of people wearing either.

This makes our desired fraction equal to

\dfrac{2}{5} + \dfrac 34 -

Fraction of people who wear either. If we wish to minimize the number who wear both, we maximize the fraction of people who wear either, up to

1.

Therefore, the fraction of people that wear both is

\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}.

Since our number is a (positive) multiple of

20,

we have the number of people wearing both as

3

if we choose to have just

20

people.

Therefore, A is the correct answer.

2010 AMC 8 第 20 题

答案：A

难度评级：1610

小提示：

大提示：

解答：

其他年份的第 20 题