2010 AMC 8 Problem 20

Attempt Problem 20 of the 2010 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 8 solutions, or check the answer key.

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20.

In a room, 2/52/5 of the people are wearing gloves, and 3/43/4 of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and gloves?

 3 \ 3

 5 \ 5

 8 \ 8

 15 \ 15

 20 \ 20

Answer: A
Concepts:inclusion-exclusionleast common multiple
Difficulty rating: 1610
Solution:

Since our room has 25\dfrac 25 of the people wearing gloves, the number of people must be a multiple of 5.5. Since our room has 34\dfrac 34 of the people wearing hats, the number of people must be a multiple of 4.4. Therefore, the people in the room must be a multiple of 20.20.

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 25+34 \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 1.1. Therefore, the fraction of people that wear both is 25+341=320.\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}.

Since our number is a (positive) multiple of 20,20, we have the number of people wearing both as 33 if we choose to have just 2020 people.

Therefore, A is the correct answer.

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