Question 1. If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y).
Solution:
n (X) = 17
n (Y) = 23
n (X U Y) = 38
So we will write this as :
n (X U Y) = n (X) + n (Y) – n (X ∩ Y)
Putting values,
38 = 17 + 23 – n (X ∩ Y)
So,
n (X ∩ Y) = 40 – 38 = 2
∴ n (X ∩ Y) = 2
Question 2. If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?
Solution:
n (X U Y) = 18
n (X) = 8
n (Y) = 15
So we will write this as :
n (X U Y) = n (X) + n (Y) – n (X ∩ Y)
Putting values,
18 = 8 + 15 – n (X ∩ Y)
So,
n (X ∩ Y) = 23 – 18 = 5
∴ n (X ∩ Y) = 5
Question 3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Solution:
Let ‘A’ is the set of people who speak Hindi & ‘B’ is the set of people who speak English
Given,
n(A ∪ B) = 400
n(A) = 250
n(B) = 200
So we will write this as :
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Putting values,
400 = 250 + 200 – n(A ∩ B)
400 = 450 – n(A ∩ B)
So,
n(A ∩ B) = 450 – 400
n(A ∩ B) = 50
∴ 50 people can speak both Hindi & English.
Question 4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?
Solution:
Given, n(S) = 21
n(T) = 32
n(S ∩ T) = 11
So we will write this as :
n (S ∪ T) = n (S) + n (T) – n (S ∩ T)
Putting values,
n (S ∪ T) = 21 + 32 – 11
So,
n (S ∪ T) = 42
∴ the set (S ∪ T) has 42 elements.
Question 5. If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?
Solution:
Given, n(X) = 40
n(X ∪ Y) = 60
n(X ∩ Y) = 10
So we will write this as :
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
Putting values,
60 = 40 + n(Y) – 10
n(Y) = 60 – (40 – 10) = 30
∴ the set Y has 30 elements.
Question 6. In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?
Solution:
Let ‘A’ is the set of people who like coffee & ‘B’ is the set of people who like tea
Given,
n(C ∪ T) = 70
n(A) = 37
n(B) = 52
So we will write this as :
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Putting values,
70 = 37 + 52 – n(A ∩ B)
70 = 89 – n(A ∩ B)
So,
n(A ∩ B) = 89 – 70 = 19
∴ 19 people like both coffee and tea.
Question 7. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution:
Let ‘A’ is the set of people who like cricket & ‘B’ is the set of people who like tennis
Given,
n(A ∪ B) = 65
n(A) = 40
n(A ∩ B) = 10
So we will write this as :
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Putting values,
65 = 40 + n(B) – 10
65 = 30 + n(B)
So,
n(B) = 65 – 30 = 35
∴ 35 people like tennis.
And we can say,
(B – A) ∪ (B ∩ A) = B
So,
(B – A) ∩ (B ∩ A) = Φ
n (B) = n (B – A) + n (B ∩ A)
Putting values,
35 = n (B – A) + 10
n (B – A) = 35 – 10 = 25
∴ 25 people like only tennis.
Question 8. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?
Solution:
Let ‘A’ is the set of people in the committee who speak French & ‘B’ is the set of people in the committee who speak Spanish
Given,
n(A) = 50
n(B) = 20
n(B ∩ A) = 10
So we will write this as :
n(B ∪ A) = n(B) + n(A) – n(B ∩ A)
Putting values,
n(B ∪ A) = 20 + 50 – 10
n(B ∪ A) = 70 – 10
n(B ∪ A) = 60
∴ 60 people in the committee speak at least one of these two languages i.e French & Spanish.