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

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==Solution==
 
==Solution==
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, it follows that the number of people in the room must be divisible by <math>5</math> and <math>4</math>. Since we are trying to find the minimum <math>x</math>, we should also use the smallest possible value for the number of people in the room. Similarly, as we are trying to minimize <math>x</math>, we can assume that everyone in the room must be wearing either a hat or a glove or both. That is, there are no people present who wearing neither of the two items. 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.
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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 <math>LCM(4,5)=20</math>. Since we are trying to find the minimum <math>x</math>, we must use the smallest possible value for the number of people in the room. Similarly, as we are trying to minimize <math>x</math>, we can assume that everyone in the room must be wearing either a hat or a glove or both. That is, there are no people present who wearing neither of the two items. 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.
  
 
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 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>. Since we know that this equals <math>20</math>, it follows that <math>23-x = 20</math>, which implies that <math>x=3</math>. Thus, <math>\boxed{\textbf{(A)}\ 3}</math> is the correct answer.
 
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 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>. Since we know that this equals <math>20</math>, it follows that <math>23-x = 20</math>, which implies that <math>x=3</math>. Thus, <math>\boxed{\textbf{(A)}\ 3}</math> is the correct answer.

Revision as of 15:59, 4 March 2012

Problem 20

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 $LCM(4,5)=20$. Since we are trying to find the minimum $x$, we must use the smallest possible value for the number of people in the room. Similarly, as we are trying to minimize $x$, we can assume that everyone in the room must be wearing either a hat or a glove or both. That is, there are no people present who wearing neither of the two items. Thus, we can say that there are $20$ people in the room, all of which are wearing at least a hat or a glove.

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 Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is $8+15-x = 23-x$. Since we know that this equals $20$, it follows that $23-x = 20$, which implies that $x=3$. Thus, $\boxed{\textbf{(A)}\ 3}$ is the correct answer.

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions