Difference between revisions of "2011 AMC 8 Problems/Problem 6"
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There are <math>351</math> total adults, and <math>45</math> own a motorcycle. The number of adults that don't own a motorcycle is <math>351 - 45 = 306</math>. Since everyone owns a car or motorcycle and one who doesn't own a motorcycle owns a car, the answer is <math>\boxed{\textbf{(D)}\ 306}</math>. | There are <math>351</math> total adults, and <math>45</math> own a motorcycle. The number of adults that don't own a motorcycle is <math>351 - 45 = 306</math>. Since everyone owns a car or motorcycle and one who doesn't own a motorcycle owns a car, the answer is <math>\boxed{\textbf{(D)}\ 306}</math>. | ||
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| + | ==Solution 3== | ||
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| + | Note that since there are some adults that own both, we can eliminate answer choice <math>E</math>. It is fairly obvious that the answer must be in the 300 range, giving us <math>\boxed{\textbf{(D)}\ 306}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=5|num-a=7}} | {{AMC8 box|year=2011|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:42, 24 August 2020
Problem
In a town of 
adults, every adult owns a car, motorcycle, or both. If 
adults own cars and 
adults own motorcycles, how many of the car owners do not own a motorcycle?
Solution 1
By PIE, the number of adults who own both cars and motorcycles is 
Out of the car owners, 
of them own motorcycles and 
of them don't.
Solution 2
There are total adults, and own a motorcycle. The number of adults that don't own a motorcycle is 
. Since everyone owns a car or motorcycle and one who doesn't own a motorcycle owns a car, the answer is 
.
Solution 3
Note that since there are some adults that own both, we can eliminate answer choice 
. It is fairly obvious that the answer must be in the 300 range, giving us
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
