Difference between revisions of "2008 AMC 8 Problems/Problem 11"
(→Solution 2 (Venn Diagram)) |
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Line 15: | Line 15: | ||
<asy> | <asy> | ||
draw(circle((0,0),5)); | draw(circle((0,0),5)); | ||
− | draw(circle(( | + | draw(circle((5,0),5)); |
+ | label("$x$",(2.3,1.5),S); | ||
+ | label("$20$",(7,0),S); | ||
+ | label("$26$",(-2,0),S); | ||
</asy> | </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> | ||
+ | <math></math>x = \boxed{\textbf{(A)} ~7}$. | ||
+ | |||
+ | ~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}} |
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:
Let be the number of students with both a dog and a cat.
Therefore, we have
$$ (Error compiling LaTeX. Unknown error_msg)x = \boxed{\textbf{(A)} ~7}$.
~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.