Difference between revisions of "2017 AMC 10B Problems/Problem 13"

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==Problem==
 
==Problem==
There are <math>20</math> students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are 10 students taking yoga, 13 taking bridge, and 9 taking painting. There are 9 students taking at least two classes. How many students are taking all three classes?
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There are <math>20</math> students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are <math>10</math> students taking yoga, <math>13</math> taking bridge, and <math>9</math> taking painting. There are <math>9</math> students taking at least two classes. How many students are taking all three classes?
  
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>
  
==Solution==
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==Solution 1==
Placeholder
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By PIE (Property of Inclusion/Exclusion), we have
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<math>|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|.</math>
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Number of people in at least two sets is <math>\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9.</math>
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So, <math>20 = (10 + 13 + 9) - (9 + 2x) + x,</math> which gives <math>x = \boxed{\textbf{(C) } 3}.</math>
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==Solution 2 (Subtraction)==
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The total number of classes taken is <math>10 + 13 + 9 = 32</math>. Each student is taking at least one class so let's subtract the <math>20</math> classes ( <math>1</math> per each of the <math>20</math> students) from <math>32</math> classes to get <math>12</math>. <math>12</math> classes is the total number of extra classes taken by the students who take <math>2</math> or <math>3</math> classes. Since we know that there are <math>9</math> students taking at least <math>2</math> classes, there must be <math>12 - 9 = \boxed{\textbf{(C) } 3}</math> students that are taking all <math>3</math> classes.
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==Solution 3 (Algebra)==
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Assume there are <math>a</math> students taking one class, <math>b</math> students taking two classes, ad <math>c</math> students taking three classes. Because there are <math>20</math> students total, <math>a+b+c = 20</math>. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, <math>a+2b+3c = 10+13+9 = 32</math>. There are <math>9</math> students taking two or three classes, so <math>b+c = 9</math>. Solving this system of equations gives us <math>c=\boxed{\textbf{(C) 3}}</math>.
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2017|ab=B|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:45, 19 January 2021

Problem

There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution 1

By PIE (Property of Inclusion/Exclusion), we have

$|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|.$ Number of people in at least two sets is $\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9.$ So, $20 = (10 + 13 + 9) - (9 + 2x) + x,$ which gives $x = \boxed{\textbf{(C) } 3}.$

Solution 2 (Subtraction)

The total number of classes taken is $10 + 13 + 9 = 32$. Each student is taking at least one class so let's subtract the $20$ classes ( $1$ per each of the $20$ students) from $32$ classes to get $12$. $12$ classes is the total number of extra classes taken by the students who take $2$ or $3$ classes. Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \boxed{\textbf{(C) } 3}$ students that are taking all $3$ classes.

Solution 3 (Algebra)

Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\boxed{\textbf{(C) 3}}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 10 Problems and Solutions

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