NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6

Exercise 1.6 of NCERT Solutions for Class 11 Maths Chapter 1 – Sets are based on the Practical Problems on Union and Intersection of Two Sets. Chapter 1 Sets of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. Solving the problems in this exercise will help students understand the problem-solving method of questions related to the Union and Intersection of Two Sets. The NCERT Solutions for Class 11 provided at BYJU’S follow the latest CBSE Syllabus and its guidelines.

The solutions are prepared with the aim of helping students understand the basic concepts effectively. Using the step-wise solutions given at BYJU’S, clearing the final exams will be an easy task. NCERT Class 11 Maths Solutions Chapter 1 Ex 1.6 can be downloaded in the form of a PDF using the link below.

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

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NCERT Solutions for Class 11 Maths Chapter 1 - Sets

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Class 11 Maths Chapter 1- Sets Exercise 1.6 Solutions

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 and

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, 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 and

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, and 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 and

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 and

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