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

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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>
 
<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>
 
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>
 
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>
So, <math>20 = (10 + 13 + 9) - (9 - 2x) + x,</math> which gives <math>x = \boxed{\textbf{(C) } 3}.</math>
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So, <math>20 = (10 + 13 + 9) - (9 + 2x) + x,</math> which gives <math>x = \boxed{\textbf{(C) } 3}.</math>
  
 
==Solution 2 (Subtraction)==
 
==Solution 2 (Subtraction)==

Revision as of 13:02, 9 January 2022

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 among the 20 students 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)

Total class count is 32. 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
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 AMC 10 Problems and Solutions

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