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

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So, <math>20 = (10 + 13 + 9) - (9 + 2x) + x,</math> which gives <math>x = \boxed{\textbf{(C) } 3}.</math>
 
So, <math>20 = (10 + 13 + 9) - (9 + 2x) + x,</math> which gives <math>x = \boxed{\textbf{(C) } 3}.</math>
  
==Solution 1 (System of Equations) ==
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==Solution 2 (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.  
 
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.  
  

Revision as of 18:52, 17 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 (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 + 2 y = 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 1 x and 2 y.)

Solving for this system of equations we get, y = 3. Therefore the $y = \boxed{\textbf{(C) } 3}.$

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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