Difference between revisions of "2017 AMC 10B Problems/Problem 13"
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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>. | ||
| + | ==Solution 4 ( A Combination of the Principle of Inclusion/Exclusion and Algebra)== | ||
| + | Let us assign the following variables: | ||
| + | <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. | ||
| + | |||
| + | |||
| + | \documentclass{article} | ||
| + | |||
| + | \usepackage{tikz} | ||
| + | \usetikzlibrary{shapes,backgrounds} | ||
| + | \begin{document} | ||
| + | |||
| + | \def\firstcircle{(0,0) circle (1.5cm)} | ||
| + | \def\secondcircle{(60:2cm) circle (1.5cm)} | ||
| + | \def\thirdcircle{(0:2cm) circle (1.5cm)} | ||
| + | \begin{tikzpicture} | ||
| + | \begin{scope}[shift={(6cm,0cm)}] | ||
| + | \begin{scope}[even odd rule]% first circle without the second | ||
| + | \clip \thirdcircle (-3, -3) rectangle (3,3); | ||
| + | \clip \secondcircle (-3, -3) rectangle (3,3); | ||
| + | \fill[yellow] \firstcircle; | ||
| + | |||
| + | |||
| + | \end{scope} | ||
| + | \draw \firstcircle node {<math>X</math>}; | ||
| + | \draw \secondcircle node {<math>Y</math>}; | ||
| + | \draw \thirdcircle node {<math>Z</math>}; | ||
| + | \end{scope} | ||
| + | \end{tikzpicture} | ||
==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 21:12, 4 July 2022
Contents
Problem
There are 
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 
students taking yoga, 
taking bridge, and 
taking painting. There are students taking at least two classes. How many students are taking all three classes?
Solution 1
By PIE (Property of Inclusion/Exclusion), we have
Number of people in at least two sets is 
So, 
which gives 
Solution 2 (Subtraction)
The total number of classes taken among the 20 students is 
. Each student is taking at least one class so let's subtract the classes ( 
per each of the students) from 
classes to get 
. classes is the total number of extra classes taken by the students who take 
or 
classes. Since we know that there are students taking at least classes, there must be 
students that are taking all classes.
Solution 3 (Algebra)
Total class count is 32. Assume there are 
students taking one class, 
students taking two classes, ad 
students taking three classes. Because there are students total, 
. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, 
. There are students taking two or three classes, so 
. Solving this system of equations gives us 
.
Solution 4 ( A Combination of the Principle of Inclusion/Exclusion and Algebra)
Let us assign the following variables:
which designates the number of people taking exactly Bridge and Yoga.
which designates the number of people taking exactly Bridge and Painting.
\documentclass{article}
\usepackage{tikz} \usetikzlibrary{shapes,backgrounds} \begin{document}
\def\firstcircle{(0,0) circle (1.5cm)} \def\secondcircle{(60:2cm) circle (1.5cm)} \def\thirdcircle{(0:2cm) circle (1.5cm)} \begin{tikzpicture} \begin{scope}[shift={(6cm,0cm)}]
\begin{scope}[even odd rule]% first circle without the second
\clip \thirdcircle (-3, -3) rectangle (3,3);
\clip \secondcircle (-3, -3) rectangle (3,3);
\fill[yellow] \firstcircle;
\end{scope}
\draw \firstcircle node {
};
\draw \secondcircle node {
};
\draw \thirdcircle node {
};
\end{scope}
\end{tikzpicture}
};
\draw \secondcircle node {
};
\draw \thirdcircle node {
};
\end{scope}
\end{tikzpicture}
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. 
