2017 AMC 10B Problems/Problem 13

Revision as of 22:30, 1 January 2021 by Niranjana (talk | contribs) (Solution)

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

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 1

By System of Equations: 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.

Now let's set up our system of equations: Let x be equal to the number of students taking $2 classes and let y be equal to the number of students taking$3 classes.

x + y = $9 x + 2y =$12

(Note: We know there are $9 total students taking either$2 or $3 classes and we already subtracted one class per each of the$20 students (the $9 students are included) from the total number of classes so it is only$1x and $2y.)

Solving for this system of equations we get, y =$ (Error compiling LaTeX. Unknown error_msg)3. Therefore the answer is \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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