NCERT Solutions for Class 11 Maths Chapter 1- Sets Exercise 1.6

The Exercise 1.6 of NCERT Solutions for Class 11 Maths Chapter 1- Sets is based on the Practical Problems on Union and Intersection of Two Sets. Solving the problems in this exercise will help you in understanding the problem solving method of questions related to Union and Intersection of Two Sets. The NCERT Solutions for Class 11 provided at BYJU’S follows the syllabus and guidelines prescribed by the CBSE.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 1- Sets Exercise 1.6

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Class 11 Maths Chapter 1- Sets exercise 1.6 Solutioins

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:

Given

n (X) = 17

n (Y) = 23

n (X U Y) = 38

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

38 = 17 + 23 – n (X ∩ Y)

By further calculation

n (X ∩ Y) = 40 – 38 = 2

So we get

n (X ∩ Y) = 2

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:

Given

n (X U Y) = 18

n (X) = 8

n (Y) = 15

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

18 = 8 + 15 – n (X ∩ Y)

By further calculation

n (X ∩ Y) = 23 – 18 = 5

So we get

n (X ∩ Y) = 5

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:

Consider H as the set of people who speak Hindi

E as the set of people who speak English

We know that

n(H ∪ E) = 400

n(H) = 250 

n(E) = 200

It can be written as

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

By substituting the values

400 = 250 + 200 – n(H ∩ E)

By further calculation

400 = 450 – n(H ∩ E)

So we get

n(H ∩ E) = 450 – 400

n(H ∩ E) = 50

Therefore, 50 people can speak both Hindi and English.

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:

We know that

n(S) = 21

n(T) = 32 

n(S ∩ T) = 11

It can be written as

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

Substituting the values

n (S ∪ T) = 21 + 32 – 11

So we get

n (S ∪ T)= 42

Therefore, the set (S ∪ T) has 42 elements.

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:

We know that

n(X) = 40

n(X ∪ Y) = 60

n(X ∩ Y) = 10

It can be written as

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

By substituting the values

60 = 40 + n(Y) – 10

On further calculation

n(Y) = 60 – (40 – 10) = 30

Therefore, the set Y has 30 elements.

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:

Consider C as the set of people who like coffee

T as the set of people who like tea

n(C ∪ T) = 70

n(C) = 37 

n(T) = 52

It is given that

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values

70 = 37 + 52 – n(C ∩ T)

By further calculation

70 = 89 – n(C ∩ T)

So we get

n(C ∩ T) = 89 – 70 = 19

Therefore, 19 people like both coffee and tea.

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:

Consider C as the set of people who like cricket

T as the set of people who like tennis

n(C ∪ T) = 65 

n(C) = 40 

n(C ∩ T) = 10

It can be written as

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values

65 = 40 + n(T) – 10

By further calculation

65 = 30 + n(T)

So we get

n(T) = 65 – 30 = 35

Hence, 35 people like tennis.

We know that,

(T – C) ∪ (T ∩ C) = T

So we get,

(T – C) ∩ (T ∩ C) = Φ

Here

n (T) = n (T – C) + n (T ∩ C)

Substituting the values

35 = n (T – C) + 10

By further calculation

n (T – C) = 35 – 10 = 25

Therefore, 25 people like only tennis.

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:

Consider F as the set of people in the committee who speak French

S as the set of people in the committee who speak Spanish

n(F) = 50

n(S) = 20 

n(S ∩ F) = 10

It can be written as

n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

By substituting the values

n(S ∪ F) = 20 + 50 – 10

By further calculation

n(S ∪ F) = 70 – 10

n(S ∪ F) = 60

Therefore, 60 people in the committee speak at least one of the two languages.


Access other exercise solutions of Class 11 Maths Chapter 1- Sets

Exercise 1.1 Solutions 6 Questions

Exercise 1.2 Solutions 6 Questions

Exercise 1.3 Solutions 9 Questions

Exercise 1.4 Solutions 12 Questions

Exercise 1.5 Solutions 7 Questions

Miscellaneous Exercise On Chapter 1 Solutions 16 Questions

Download All Questions of NCERT Solutions for Class 11 Maths Chapter 1- Sets here.

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