Difference between revisions of "2008 AMC 8 Problems/Problem 11"

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\textbf{(E)}\ 46</math>
 
\textbf{(E)}\ 46</math>
  
==Solution==
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==Solution 1==
 
The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is <math>20+26-39 = \boxed{\textbf{(A)}\ 7}</math>.
 
The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is <math>20+26-39 = \boxed{\textbf{(A)}\ 7}</math>.
  
 +
==Solution 2 (Venn Diagram)==
 +
We create a diagram:
 +
<asy>
 +
draw(circle((0,0),5));
 +
draw(circle((4,0),5));
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</asy>
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=10|num-a=12}}
 
{{AMC8 box|year=2008|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:29, 4 September 2022

Problem

Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$

Solution 1

The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \boxed{\textbf{(A)}\ 7}$.

Solution 2 (Venn Diagram)

We create a diagram: [asy] draw(circle((0,0),5)); draw(circle((4,0),5)); [/asy]

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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