Difference between revisions of "2017 AMC 10B Problems/Problem 13"
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==Solution== | ==Solution== | ||
By PIE (Property of Inclusion/Exclusion), 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>. | ||
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| + | <math>|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|.</math> | ||
| + | Number of ppl in at least two sets is <math>\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9</math> | ||
| + | So, <math>20 = (10 + 13 + 9) - (9 + 2x) + x,</math> which gives <math>x = 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 23:15, 28 November 2019
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 
.
Number of ppl in at least two sets is 
So, 
which gives 
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. 
