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, the number of people in the room must be a multiple | + | 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.) |
| − | 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> | + | 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>. |
==Video by MathTalks== | ==Video by MathTalks== | ||
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https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be | https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be | ||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/Ym28CYMKIW8 | ||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=19|num-a=21}} | {{AMC8 box|year=2010|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:47, 23 October 2024
Problem
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 4 and 5. Since we are trying to find the minimum , we must use the least common multiple. 
. Thus, we can say that there are 
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 
people wearing gloves and 
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 
, where is the number wearing both. Since everyone in the room is wearing at least one item (see above), 
, and so 
.
Video by MathTalks
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
Video Solution by WhyMath
~savannahsolver
See Also
| 2010 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
