Difference between revisions of "2008 AMC 8 Problems/Problem 11"
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<cmath>26+20-x = 39</cmath> | <cmath>26+20-x = 39</cmath> | ||
<cmath>46-x = 39</cmath> | <cmath>46-x = 39</cmath> | ||
| − | < | + | <cmath>x = \boxed{\textbf{(A)} ~7}</cmath>. |
~MrThinker | ~MrThinker | ||
Revision as of 09:34, 4 September 2022
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 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 
.
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]](http://latex.artofproblemsolving.com/e/0/f/e0f9fceee4039e999d273310abcb3c09335affb2.png)
Let 
be the number of students with both a dog and a cat.
Therefore, we have
![\[26+20-x = 39\]](http://latex.artofproblemsolving.com/3/7/5/375e63084c9d655639e4cbc1c30f0eda7ec8a1cb.png)
![\[46-x = 39\]](http://latex.artofproblemsolving.com/6/2/c/62c38cca81668b09f0d7617cfe61f6f44c9d9d21.png)
![\[x = \boxed{\textbf{(A)} ~7}\]](http://latex.artofproblemsolving.com/d/4/b/d4b62274f2fb755a8ff15d331a1c70c6e2f0b1f7.png)
.
~MrThinker
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
