Difference between revisions of "2017 AMC 10B Problems/Problem 13"
I.j purple (talk | contribs) m (→Solution) |
(→Solution) |
||
| Line 10: | Line 10: | ||
Number of people in at least two sets is <math>\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9.</math> | Number of people 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 = \boxed{\textbf{(C) } 3}.</math> | So, <math>20 = (10 + 13 + 9) - (9 + 2x) + x,</math> which gives <math>x = \boxed{\textbf{(C) } 3}.</math> | ||
| + | |||
| + | ==Solution 1== | ||
| + | By System of Equations: | ||
| + | The total number of classes taken is <math>10 + </math>13 + <math>9 = </math>32. Each student is taking at least one class so let's subtract the <math>20 classes (</math>1 per each of the <math>20 students) from </math>32 classes to get <math>12. </math>12 classes is the total number of extra classes taken by the students who take <math>2 or </math>3 classes. | ||
| + | |||
| + | Now let's set up our system of equations: Let x be equal to the number of students taking <math>2 classes and let y be equal to the number of students taking </math>3 classes. | ||
| + | |||
| + | x + y = <math>9 | ||
| + | x + 2y = </math>12 | ||
| + | |||
| + | (Note: We know there are <math>9 total students taking either </math>2 or <math>3 classes and we already subtracted one class per each of the </math>20 students (the <math>9 students are included) from the total number of classes so it is only </math>1x and <math>2y.) | ||
| + | |||
| + | Solving for this system of equations we get, y = </math>3. Therefore the answer is \boxed{\textbf{(C) } 3}.$ | ||
==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 21:30, 1 January 2021
Contents
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), we have
Number of people in at least two sets is 
So, 
which gives 
Solution 1
By System of Equations:
The total number of classes taken is 
13 + 
32. Each student is taking at least one class so let's subtract the 
1 per each of the 
32 classes to get 
12 classes is the total number of extra classes taken by the students who take 
3 classes.
Now let's set up our system of equations: Let x be equal to the number of students taking 
3 classes.
x + y = 
12
(Note: We know there are 
2 or 
20 students (the 
1x and $2y.)
Solving for this system of equations we get, y =$ (Error compiling LaTeX. Unknown error_msg)3. Therefore the answer is \boxed{\textbf{(C) } 3}.$
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. 
