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

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

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.

@@ Line 1: / Line 1: @@
-==Problem 20==
+==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?
@@ Line 5: / Line 5: @@
 ==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.
+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>. 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 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==
+https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
+==Video Solution by WhyMath==
+https://youtu.be/Ym28CYMKIW8
+~savannahsolver
 ==See Also==
-{{AMC8 box|year=2011|num-b=19|num-a=21}}
+{{AMC8 box|year=2010|num-b=19|num-a=21}}
+{{MAA Notice}}

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