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

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(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
 
By System of Equations:
 
By System of Equations:
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The total number of classes taken is <math>10 + </math>13 + <math>9 = </math>32 . Each student is taking at least one class so let's subtract the <math>20 classes (</math>1 per each of the <math>20 students) from </math>32 classes to get <math>12 . </math>12 classes is the total number of extra classes taken by the students who take <math>2 or </math>3 classes.  
+
The total number of classes taken is <math>10 + </math>13 + <math>9 = </math>32 . Each student is taking at least one class so let's subtract the <math>20 classes ( </math>1 per each of the <math>20 students) from </math>32 classes to get <math>12 . </math>12 classes is the total number of extra classes taken by the students who take <math>2 or </math>3 classes.  
  
 
Now let's set up our system of equations: Let x be equal to the number of students taking <math>2 classes and let y be equal to the number of students taking </math>3 classes.
 
Now let's set up our system of equations: Let x be equal to the number of students taking <math>2 classes and let y be equal to the number of students taking </math>3 classes.
  
 
x + y = <math>9
 
x + y = <math>9
−
x + 2y = </math>12
+
x + </math>2 y = <math>12
  
−
(Note: We know there are <math>9 total students taking either </math>2 or <math>3 classes and we already subtracted one class per each of the </math>20 students (the <math>9 students are included) from the total number of classes so it is only </math>1 x and <math>2 y.)
+
(Note: We know there are </math>9 total students taking either <math>2 or </math>3 classes and we already subtracted one class per each of the <math>20 students (the </math>9 students are included) from the total number of classes so it is only <math>1 x and </math>2 y.)
  
−
Solving for this system of equations we get, y = </math>3. Therefore the answer is C.
+
Solving for this system of equations we get, y = <math>3. Therefore the </math>y = \boxed{\textbf{(C) } 3}.$
  
 
==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 21:33, 1 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

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 +$2 y = $12

(Note: We know there are$ (Error compiling LaTeX. Unknown error_msg)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 (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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