Difference between revisions of "2010 AMC 8 Problems/Problem 20"
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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 | + | 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, we can assume that there are no people present who are wearing neither of the two items since this would unnecessarily increase the number of people in the room. 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 15:07, 4 March 2012
Problem 20
In a room, 
of the people are wearing gloves, and 
of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
Solution
Let 
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 
. Since we are trying to find the minimum , we must use the smallest possible value for the number of people in the room. Similarly, we can assume that there are no people present who are wearing neither of the two items since this would unnecessarily increase the number of people in the room. Thus, we can say that there are 
people in the room, all of which are wearing at least a hat or a glove.
It follows that there are 
people wearing gloves and 
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 
. Since we know that this equals , it follows that 
, which implies that 
. Thus, 
is the correct answer.
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
