2010 AMC 8 Problem 20

Question

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

Answer: A

Concepts:inclusion-exclusion least common multiple

Difficulty rating: 1610

Small hint:

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

Big hint:

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

Solution:

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.