Difference between revisions of "2008 AMC 8 Problems/Problem 11"
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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 20+26-39 = A | + | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:22, 8 August 2021
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
