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Difference between revisions of "2010 AMC 8 Problems/Problem 20"

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==Problem 20==
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==Problem==
 
In a room, <math>2/5</math> of the people are wearing gloves, and <math>3/4</math> of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?  
 
In a room, <math>2/5</math> of the people are wearing gloves, and <math>3/4</math> of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?  
  
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==Solution==
 
==Solution==
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Let <math>n</math> be the total number of people in the room and <math>x</math> be the number wearing both a hat and a glove. It is possible that there are people in the room who are wearing neither a hat or a glove. However, as we are trying to minimize x, we also want to have as few people in the room as possible and still satisfy the requirements of the problem. Thus, it is safe to assume that everyone in the room is either wearing a hat or a glove or both. Then by applying the Principle of Inclusion Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is  <math>\frac{2}{5}n + \frac{3}{4}n - x = \frac{23}{20}n - x</math>. Since <math>n</math> is also the number of people in the room that are wearing at least a hat or a glove or both, it follows that <math>\frac{23}{20}n - x = n</math>. Solving for x, we get <math>x = \frac{3}{20}n</math>. Since <math>x</math> must be a whole number and we want the smallest such <math>x</math>, it follows that <math>n=20</math> is the least number of people that must be in the room. Substituting <math>n=20</math> into the inequality, we get <math>x = 3</math>. Thus, <math>\boxed{\textbf{(A)}\ 3}</math> is the correct answer.
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Let <math>x</math> 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 <math>x</math>, we must use the ''least'' common multiple. <math>lcm(4,5) = 20</math>. Thus, we can say that there are <math>20</math> 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.)
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It follows that there are <math>\frac{2}{5}\cdot 20 = 8</math> people wearing gloves and <math>\frac{3}{4}\cdot 20 = 15</math> 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  <math>8+15-x = 23-x</math>, where <math>x</math> is the number wearing both. Since everyone in the room is wearing at least one item (see above), <math>23-x = 20</math>, and so <math>x=\boxed{\textbf{(A)}\ 3}</math>.
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==Video by MathTalks==
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https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
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==Video Solution by WhyMath==
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https://youtu.be/Ym28CYMKIW8
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC8 box|year=2011|num-b=19|num-a=21}}
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{{AMC8 box|year=2010|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 20:47, 23 October 2024

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 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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