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

(→‎Solution 4 ( A Combination of the Principle of Inclusion/Exclusion and Algebra))
(→‎Solution 4 ( A Combination of the Principle of Inclusion/Exclusion and Algebra))
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Total class count is 32. Assume there are <math>a</math> students taking one class, <math>b</math> students taking two classes, ad <math>c</math> students taking three classes. Because there are <math>20</math> students total, <math>a+b+c = 20</math>. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, <math>a+2b+3c = 10+13+9 = 32</math>. There are <math>9</math> students taking two or three classes, so <math>b+c = 9</math>. Solving this system of equations gives us <math>c=\boxed{\textbf{(C) 3}}</math>.
 
Total class count is 32. Assume there are <math>a</math> students taking one class, <math>b</math> students taking two classes, ad <math>c</math> students taking three classes. Because there are <math>20</math> students total, <math>a+b+c = 20</math>. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, <math>a+2b+3c = 10+13+9 = 32</math>. There are <math>9</math> students taking two or three classes, so <math>b+c = 9</math>. Solving this system of equations gives us <math>c=\boxed{\textbf{(C) 3}}</math>.
  
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==Solution 4 ( A Combination of the Principle of Inclusion/Exclusion and Algebra)==
+
==Solution 4 ( A Combination of the Venn Diagram and Algebra)==
 
Let us assign the following variables:
 
Let us assign the following variables:
 
<math>a</math> which designates the number of people taking exactly Bridge and Yoga.
 
<math>a</math> which designates the number of people taking exactly Bridge and Yoga.
 
<math>b</math> which designates the number of people taking exactly Bridge and Painting.
 
<math>b</math> which designates the number of people taking exactly Bridge and Painting.
 +
<math>c</math> which designates the number of people that took all <math>3</math> classes or what we want to find.
 +
<math>d</math> which designates the number of people taking exactly Yoga and Painting.
  
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+
Note: The Venn Diagram is linked here: https://artofproblemsolving.com/wiki/index.php/File:IMG_20220704_192122676_2.jpg#filelinks
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<math>\documentclass{article}
 
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\usepackage{tikz}
 
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\usetikzlibrary{shapes,backgrounds}
 
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\begin{document}
 
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\def\firstcircle{(0,0) circle (1.5cm)}
 
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\def\secondcircle{(60:2cm) circle (1.5cm)}
 
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\def\thirdcircle{(0:2cm) circle (1.5cm)}
 
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\begin{tikzpicture}
 
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\begin{scope}[shift={(6cm,0cm)}]
 
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        \begin{scope}[even odd rule]% first circle without the second
 
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            \clip \thirdcircle (-3, -3) rectangle (3,3);
 
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            \clip \secondcircle (-3, -3) rectangle (3,3);
 
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        \fill[yellow] \firstcircle;
 
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        \end{scope}
 
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        \draw \firstcircle node {</math>X<math>};
 
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        \draw \secondcircle node {</math>Y<math>};
 
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        \draw \thirdcircle node {</math>Z<math>};
 
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    \end{scope}
 
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  \end{tikzpicture}</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 21:27, 4 July 2022

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 among the 20 students 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.

Solution 3 (Algebra)

Total class count is 32. Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$. There are $9$ students taking two or three classes, so $b+c = 9$. Solving this system of equations gives us $c=\boxed{\textbf{(C) 3}}$.

Solution 4 ( A Combination of the Venn Diagram and Algebra)

Let us assign the following variables: $a$ which designates the number of people taking exactly Bridge and Yoga. $b$ which designates the number of people taking exactly Bridge and Painting. $c$ which designates the number of people that took all $3$ classes or what we want to find. $d$ which designates the number of people taking exactly Yoga and Painting.

Note: The Venn Diagram is linked here: https://artofproblemsolving.com/wiki/index.php/File:IMG_20220704_192122676_2.jpg#filelinks

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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