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

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==Solution 2 (Subtraction)==
 
==Solution 2 (Subtraction)==
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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 = <math>\boxed{\textbf{(C) } 3}</math> students that are taking all 3 classes.
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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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Solving for this system of equations we get, y = 3. Therefore the <math>y = \boxed{\textbf{(C) } 3}.</math>
 
  
 
==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 17:59, 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 (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 (Problems • Answer Key • Resources)
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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