2008 AMC 8 Problems/Problem 11

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((5,0),5)); label("$x$",(2.3,0),S); label("$20$",(7,0),S); label("$26$",(-2,0),S); [/asy]

Let $x$ be the number of students with both a dog and a cat.

Therefore, we have

\[26+20-x = 39\] \[46-x = 39\] \[x = \boxed{\textbf{(A)} ~7}\].

~MrThinker

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

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

Invalid username
Login to AoPS