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 10:34, 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.