Difference between revisions of "2017 AMC 10B Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | By PIE, the answer is <math>10+13+9-9-20= \boxed{\textbf{(C) } 3}</math>. | + | By PIE (Property of Inclusion/Exclusion), the answer is <math>10+13+9-9-20= \boxed{\textbf{(C) } 3}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2017|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:42, 16 August 2017
Problem
There are 
students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are 
students taking yoga, 
taking bridge, and 
taking painting. There are students taking at least two classes. How many students are taking all three classes?
Solution
By PIE (Property of Inclusion/Exclusion), the answer is 
.
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
