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$",(2.3,1.5),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

2008 AMC 8 (Problems • Answer Key • Resources)
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.

Page

Toolbox

Search

2008 AMC 8 Problems/Problem 11

Contents

Problem

Solution 1

Solution 2 (Venn Diagram)

See Also