Class 11 NCERT Solutions - Chapter 1 Sets - Exercise 1.6

Last Updated : 23 Jul, 2025

Chapter 1 of Class 11 NCERT Mathematics focuses on "Sets" a fundamental concept in mathematics that provides the basis for the various mathematical structures and operations. Understanding sets is crucial for grasping more advanced topics in algebra, calculus, and discrete mathematics. Exercise 1.6 in this chapter includes the problems that test and enhance students' skills in handling different types of sets and their operations. This exercise is essential for building a strong foundation in set theory which is often a critical component of exams.

Chapter Overview: Sets

In this chapter, students explore the concept of sets, learning about different methods of representing them, such as roster and set-builder forms. The chapter also covers various types of sets, including finite, infinite, empty, and universal sets. Understanding these concepts is essential for mastering set theory and its applications in mathematics. Exercise 1.6 is designed to provide practice in solving problems related to these concepts reinforcing students' understanding and application of set operations like union, intersection, and difference.

Note: Please note that Exercise 1.6 from Chapter 1, "Sets" in the NCERT Solutions, has been removed from the revised syllabus. As a result, this exercise will no longer be a part of your study curriculum.

Exercise 1.6: Importance and Focus

Exercise 1.6 is specifically designed to enhance students' understanding and application of set operations such as union, intersection, and difference. Although this exercise has been removed from the revised syllabus, practicing these problems is still valuable for building a strong foundation in set theory, which is a critical component of many competitive exams.

Importance in Exams

The concept of sets is not only fundamental to the understanding of mathematics but also plays a significant role in various competitive exams like JEE, NEET, and other entrance tests. Mastering this chapter helps students tackle more complex problems in algebra, calculus, and beyond, making it a vital part of their mathematical education.

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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Summary

Chapter 1 of the NCERT Class 11 Mathematics textbook covers the concept of Sets. In Exercise 1.6, students learn to solve problems involving the Venn diagram representation of sets and the algebraic representation of set operations. The exercises focus on understanding the relationships between sets, applying the properties of set operations, and solving complex problems involving the universal set and its subsets.

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