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Class 11 NCERT Solutions – Chapter 1 Sets – Exercise 1.6
  • Last Updated : 27 Nov, 2020

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.

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