Exercise 1.6 of **NCERT Solutions for Class 11 Maths Chapter 1 â€“ 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 BYJUâ€™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 BYJUâ€™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 â€“ Sets Exercise 1.6

### Class 11 Maths Chapter 1- Sets Exercise 1.6 Solutions

**1. If X and Y are two sets, such that n(X) = 17,Â n(Y) = 23 andÂ n(XÂ âˆªÂ Y) = 38, findÂ n(XÂ âˆ© 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) â€“ n (X âˆ© Y)

Substituting the values,

38 = 17 + 23 â€“ n (X âˆ© Y)

By further calculation,

n (X âˆ© Y) = 40 â€“ 38 = 2

So, we get

n (X âˆ© Y) = 2

**2. If X and Y are two sets, such that X âˆªY has 18 elements, X has 8 elements, and Y has 15 elements, how many elements does X âˆ© 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) â€“ n (X âˆ© Y)

Substituting the values,

18 = 8 + 15 â€“ n (X âˆ© Y)

By further calculation,

n (X âˆ© Y) = 23 â€“ 18 = 5

So, we get

n (X âˆ© 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*(HÂ âˆªÂ E) = 400

*n*(H) = 250

*n*(E) = 200

It can be written as

*n*(HÂ âˆªÂ E) =Â *n*(H) +Â *n*(E) â€“Â *n*(HÂ âˆ©Â E)

By substituting the values,

400 = 250 + 200 â€“Â *n*(HÂ âˆ©Â E)

By further calculation,

400 = 450 â€“Â *n*(HÂ âˆ©Â E)

So, we get

*n*(HÂ âˆ©Â E) = 450 â€“ 400

*n*(HÂ âˆ©Â 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 SÂ âˆ©Â T has 11 elements, how many elements does SÂ âˆªÂ T have?**

**Solution:**

We know that

*n*(S) = 21

*n*(T) = 32

*n*(SÂ âˆ©Â T) = 11

It can be written as

*n*Â (SÂ âˆªÂ T) =Â *n*Â (S) +Â *n*Â (T) â€“Â *n*Â (SÂ âˆ©Â T)

Substituting the values,

*n*Â (SÂ âˆªÂ T) = 21 + 32 â€“ 11

So, we get

*n*Â (SÂ âˆªÂ T)= 42

Therefore, the set (SÂ âˆªÂ T) has 42 elements.

**5. If X and Y are two sets such that X has 40 elements, X âˆªY has 60 elements, and X âˆ©Y has 10 elements, how many elements does Y have?**

**Solution:**

We know that

*n*(X) = 40

*n*(XÂ âˆªÂ Y) = 60

*n*(XÂ âˆ©Â Y) = 10

It can be written as

*n*(XÂ âˆªÂ Y) =Â *n*(X) +Â *n*(Y) â€“Â *n*(XÂ âˆ©Â Y)

By substituting the values,

60 = 40 +Â *n*(Y) â€“ 10

On further calculation,

*n*(Y) = 60 â€“ (40 â€“ 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*(CÂ âˆªÂ T) = 70

*n*(C) = 37

*n*(T) = 52

It is given that

*n*(CÂ âˆªÂ T) =Â *n*(C) +Â *n*(T) â€“Â *n*(CÂ âˆ©Â T)

Substituting the values,

70 = 37 + 52 â€“Â *n*(CÂ âˆ©Â T)

By further calculation,

70 = 89 â€“Â *n*(CÂ âˆ©Â T)

So, we get

*n*(CÂ âˆ©Â T) = 89 â€“ 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*(CÂ âˆªÂ T) = 65

*n*(C) = 40

*n*(CÂ âˆ©Â T) = 10

It can be written as

*n*(CÂ âˆªÂ T) =Â *n*(C) +*Â n*(T) â€“Â *n*(CÂ âˆ©Â T)

Substituting the values,

65 = 40 +Â *n*(T) â€“ 10

By further calculation,

65 = 30 +Â *n*(T)

So, we get

*n*(T) = 65 â€“ 30 = 35

Hence, 35 people like tennis.

We know that

(T â€“ C)Â âˆªÂ (TÂ âˆ©Â C) = T

So, we get,

(T â€“ C)Â âˆ©Â (TÂ âˆ©Â C) =Â Î¦

Here,

*n*Â (T) =Â *n*Â (T â€“ C) +Â *n*Â (TÂ âˆ©Â C)

Substituting the values,

35 =Â *n*Â (T â€“ C) + 10

By further calculation,

*n*Â (T â€“ C) = 35 â€“ 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*(SÂ âˆ©Â F) = 10

It can be written as

*n*(SÂ âˆªÂ F) =Â *n*(S) +Â *n*(F) â€“Â *n*(SÂ âˆ©Â F)

By substituting the values,

*n*(SÂ âˆªÂ F) = 20 + 50 â€“ 10

By further calculation,

*n*(SÂ âˆªÂ F) = 70 â€“ 10

*n*(SÂ âˆªÂ 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 â€“ 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 â€“ Sets here.

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