# 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

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