Counting calculus students. About A university offers 3 calculus classes: Math 2A, 2B and 2C. In both parts, you are given data about a group of students who have all taken at least one of the three classes. (a) Group A contains 157 students. Of these, 51 students in Group A have taken Math 2A, 80 have taken Math 2B, and 70 have takern Math 2C. 15 have taken both Math 2A and 2B, 20 have taken both Math 2A and 2C, and 13 have taken both Math 2B and 2C. How many students in Group A have taken all three classes? Solution (b) You are given the following data about Group B. 28 students in Group B have taken Math 2A, 28 have taken Math 2B, and 25 have taken Math 2C. 11 have taken both Math 2A and 2B, 9 have taken both Math 2A and 2C, and 10 have taken both Math 2B and 2C 3 have taken all three classes. How many students are in Group B? Feedback?

Question

Computers & Technology

Deja Pham

10 February, 12:29

Counting calculus students. About A university offers 3 calculus classes: Math 2A, 2B and 2C. In both parts, you are given data about a group of students who have all taken at least one of the three classes. (a) Group A contains 157 students. Of these, 51 students in Group A have taken Math 2A, 80 have taken Math 2B, and 70 have takern Math 2C. 15 have taken both Math 2A and 2B, 20 have taken both Math 2A and 2C, and 13 have taken both Math 2B and 2C. How many students in Group A have taken all three classes? Solution (b) You are given the following data about Group B. 28 students in Group B have taken Math 2A, 28 have taken Math 2B, and 25 have taken Math 2C. 11 have taken both Math 2A and 2B, 9 have taken both Math 2A and 2C, and 10 have taken both Math 2B and 2C 3 have taken all three classes. How many students are in Group B? Feedback?

+1

Answers (1)

Know the Answer?

Giada Mercado · Answer 1

Part (a) n (A∩B∩C) = 4

Part (b) n (A∪B∪C) = 54

Explanation:

n (A) = no. of students who took Math 2A

n (B) = no. of students who took Math 2B

n (C) = no. of students who took Math 2C

n (A∩B) = no. of students who took both Math 2A and 2B

n (A∩C) = no. of students who took both Math 2A and 2C

n (B∩C) = no. of students who took both Math 2B and 2C

n (A∩B∩C) = no. of students who took all three Math 2A, 2B and 2C

n (A∪B∪C) = no. of total students in a group

∩ represents Intersection and ∪ represents Union

Part (a)

n (A∪B∪C) = n (A) + n (B) + n (C) - n (A∩B) - n (A∩C) - n (B∩C) + n (A∩B∩C)

Where n (A∩B∩C) represents the number of students who took all three classes and n (A∪B∪C) represents the total number of students in group A

157 = 51 + 80 + 70 - 15 - 20 - 13 + n (A∩B∩C)

Re-arranging the equation to solve for n (A∩B∩C) since we want to find out those students who took all three classes

n (A∩B∩C) = 157 - 51 - 80 - 70 + 15 + 20 + 13

n (A∩B∩C) = 4

So there are 4 students in group A who took all three classes

Part (b)

n (A∪B∪C) = n (A) + n (B) + n (C) - n (A∩B) - n (A∩C) - n (B∩C) + n (A∩B∩C)

This time we are given n (A∩B∩C) students who took all three classes and want to find n (A∪B∪C) that is total number of students

n (A∪B∪C) = 28 + 28 + 25 - 11 - 9 - 10 + 3

n (A∪B∪C) = 54

So there are total 54 students in group B