Set theory: Application

Set theory: Application

Q1. Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).

Solution: 

 formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

 n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

                   = 20 + 28 - 36 = 12

                 

Q2. If n(A - B) = 18, n(A ∪ B) = 70 and n(A ∩ B) = 25, then find n(B).

Solution: 

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

                              70 = 18 + 25 + n(B - A)

                              70 = 43 + n(B - A)   = 27

                   
Now n(B) = n(A ∩ B) + n(B - A)

             = 25 + 27

             = 52

Q3. In a group of 60 people, 27 like cold drinks and 42 like hot drinks and each person likes at least one of the two drinks. How many like both coffee and tea?

Solution: 

Let A = Set of people who like cold drinks.

     B = Set of people who like hot drinks.


(A ∪ B) = 60            n(A) = 27       n(B) = 42 then;

Formula: n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

            = 27 + 42 - 60 =   9.0

          Therefore, 9 people like both tea and coffee. 

Q4. There are 35 students in art class and 57 students in dance class. Find the number of students who are either in art class or in dance class,  When two classes meet at different hours and 12 students are enrolled in both activities.

Solution: 

Let A be the set of students in art class. 
B be the set of students in dance class.

n(A) = 35,       n(B) = 57,       n(A ∩ B) = 12


(i) When 2 classes meet at different hours n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

                                                                    = 35 + 57 - 12  =   80

                                                                                 


  Q5. In a group of 100 persons, 72 people can speak English and 43 can speak French. How many can speak English only? How many can speak French only and how many can speak both English and French?

Solution: 

Let A be the set of people who speak English.

B be the set of people who speak French.

A - B be the set of people who speak English and not French.

B - A be the set of people who speak French and not English.

A ∩ B be the set of people who speak both French and English.

Given,

n(A) = 72       n(B) = 43       n(A ∪ B) = 100

Now, n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

                   = 72 + 43 - 100

                   = 115 - 100

                   = 15

Therefore, Number of persons who speak both French and English = 15

n(A) = n(A - B) + n(A ∩ B)

⇒ n(A - B) = n(A) - n(A ∩ B)

               = 72 - 15

               = 57

and n(B - A) = n(B) - n(A ∩ B)

                 = 43 - 15

                 = 28

Therefore, Number of people speaking English only = 57

Number of people speaking French only = 28

Q6.. Each student in a class of 40 plays at least one indoor game chess, carrom and scrabble. 18 play chess, 20 play scrabble and 27 play carrom. 7 play chess and scrabble, 12 play scrabble and carrom and 4 play chess, carrom and scrabble. Find the number of students who play (i) chess and carrom. (ii) chess, carrom but not scrabble.

Solution: 

Let A be the set of students who play chess

B be the set of students who play scrabble

C be the set of students who play carrom

Therefore, We are given n(A ∪ B ∪ C) = 40,

n(A) = 18,         n(B) = 20         n(C) = 27,

n(A ∩ B) = 7,     n(C ∩ B) = 12    n(A ∩ B ∩ C) = 4

We have

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

Therefore, 40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4

40 = 69 – 19 - n(C ∩ A)

40 = 50 - n(C ∩ A) n(C ∩ A) = 50 - 40

n(C ∩ A) = 10

Therefore, Number of students who play chess and carrom are 10.

Also, number of students who play chess, carrom and not scrabble.

= n(C ∩ A) - n(A ∩ B ∩ C)

= 10 – 4

= 6

Therefore, we learned how to solve different types of word problems on sets without using Venn diagram.