2017 AMC 10B Problems/Problem 13

Revision as of 18:59, 17 January 2021 by Maxliu2k (talk | contribs) (Solution 2 (Subtraction))

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.

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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