Art of Problem Solving

2008 AMC 8 Problems/Problem 11: Difference between revisions

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==Solution==
==Solution==
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 20+26-39 = A:7


==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 16:15, 8 August 2021

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

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 = A:7

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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