Difference between revisions of "2017 AMC 10B Problems/Problem 13"
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==Solution 2 (Subtraction)== | ==Solution 2 (Subtraction)== | ||
The total number of classes taken is <math>10 + 13 + 9 = 32</math>. Each student is taking at least one class so let's subtract the <math>20</math> classes ( <math>1</math> per each of the <math>20</math> students) from <math>32</math> classes to get <math>12</math>. <math>12</math> classes is the total number of extra classes taken by the students who take <math>2</math> or <math>3</math> classes. Since we know that there are <math>9</math> students taking at least <math>2</math> classes, there must be <math>12 - 9 = \boxed{\textbf{(C) } 3}</math> students that are taking all <math>3</math> classes. | The total number of classes taken is <math>10 + 13 + 9 = 32</math>. Each student is taking at least one class so let's subtract the <math>20</math> classes ( <math>1</math> per each of the <math>20</math> students) from <math>32</math> classes to get <math>12</math>. <math>12</math> classes is the total number of extra classes taken by the students who take <math>2</math> or <math>3</math> classes. Since we know that there are <math>9</math> students taking at least <math>2</math> classes, there must be <math>12 - 9 = \boxed{\textbf{(C) } 3}</math> students that are taking all <math>3</math> classes. | ||
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+ | ==Solution 3 (Algebra)== | ||
+ | Assume there are <math>a</math> students taking one class, <math>b</math> students taking two classes, ad <math>c</math> students taking three classes. Because there are <math>20</math> students total, <math>a+b+c = 20</math>. Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, <math>a+2b+3c = 10+13+9 = 32</math>. There are <math>9</math> students taking two or three classes, so <math>b+c = 9</math>. Solving this system of equations gives us <math>c=\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}} |
Latest revision as of 23:45, 19 January 2021
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 1
By PIE (Property of Inclusion/Exclusion), we have
Number of people in at least two sets is So, which gives
Solution 2 (Subtraction)
The total number of classes taken is . Each student is taking at least one class so let's subtract the classes ( per each of the students) from classes to get . classes is the total number of extra classes taken by the students who take or classes. Since we know that there are students taking at least classes, there must be students that are taking all classes.
Solution 3 (Algebra)
Assume there are students taking one class, students taking two classes, ad students taking three classes. Because there are students total, . Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, . There are students taking two or three classes, so . Solving this system of equations gives us .
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 |
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