Difference between revisions of "2008 AMC 8 Problems/Problem 11"
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\textbf{(E)}\ 46</math> | \textbf{(E)}\ 46</math> | ||
| − | ==Solution== | + | ==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 <math>20+26-39 = \boxed{\textbf{(A)}\ 7}</math>. | 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>. | ||
| + | |||
| + | ==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 <math>x</math> be the number of students with both a dog and a cat. | ||
| + | |||
| + | Therefore, we have | ||
| + | |||
| + | <cmath>26+20-x = 39</cmath> | ||
| + | <cmath>46-x = 39</cmath> | ||
| + | <cmath>x = \boxed{\textbf{(A)} ~7}</cmath>. | ||
| + | |||
| + | ~MrThinker | ||
==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}} | ||
Latest revision as of 09:35, 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("$x$",(2.3,0),S); label("$20$",(7,0),S); label("$26$",(-2,0),S); [/asy]](http://latex.artofproblemsolving.com/0/2/7/0278976caa54753192763fd786ae0de7a7db44a1.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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
