2008 AMC 8 Problems/Problem 11: Difference between revisions
No edit summary |
No edit summary |
||
| Line 7: | Line 7: | ||
\textbf{(D)}\ 39\qquad | \textbf{(D)}\ 39\qquad | ||
\textbf{(E)}\ 46</math> | \textbf{(E)}\ 46</math> | ||
==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>. | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=10|num-a=12}} | {{AMC8 box|year=2008|num-b=10|num-a=12}} | ||
Revision as of 22:13, 24 December 2012
Problem
Each of the 
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 
students have a cat. How many students have both a dog and a cat?
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 
.
See Also
| 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 | ||
