# NCERT Solutions Class 11 Maths Chapter 1

## NCERT Solutions Class 11 Mathematics Chapter 1 – Sets

Students can start preparing for the upcoming exams with NCERT Solutions Class 11 Mathematics Chapter 1. It is made available for the students to get accustomed to the critical techniques to solve questions related to sets. Every student can get familiar with Chapter 1 Class 11 Mathematics concepts. With a considerable amount of time and practice, they will be able to perform better in the examination.

Sets are defined as the collection of objects. A set without any element is known as an empty set. Thus, the students will learn about definite elements and finite sets. They will understand how to calculate two sets and some properties of operation of the intersection. In addition, the students can also grasp their knowledge of the operations of unions and types of sets. All essential topics are covered in NCERT Solutions Class 11 Mathematics Chapter 1, such as powerset, subset, and singleton set.

Get access to NCERT Solutions and CBSE syllabus for both primary and secondary Classes from Extramarks website. Here are a few links for students to refer to primary Class solutions – NCERT solution Class 1, NCERT solution Class 2, NCERT solution Class 3, and NCERT solution Class 4.

### Key Topics Covered In NCERT Solution for Class 11 Mathematics Chapter 1

It is essential to understand sets, not only for Maths but also for other subjects. Every student will understand the elements of sets and the operations that can be performed on them. Regular practice of Chapter 1 Mathematics Class 11 will help students acquire the knowledge necessary to perform functions on set elements. These solutions include examples and sample problems to help the students understand the topic step-by-step.

Thus, the following are the key topics covered in the NCERT Solutions Class 11 Mathematics Chapter 1:

 Exercises Topics 1.1 Introduction to Sets 1.2 Types of Sets 1.3 Power Sets, Universal sets, and Subsets 1.4 Union and Intersection 1.5 Complementary Sets 1.6 Operations around Sets Other Miscellaneous

1.1 Introduction to Sets

The introduction to sets and how they are represented begins with sets. The roaster form is a tabular presentation of a set. A set builder form is another way to represent a set. Students will be able to practice the questions in this exercise and learn more about these forms.

1.2 Types of Sets

This section focuses on understanding null sets, finite and infinite sets, and equal sets and their applications. For example, null sets are set without any elements. They can be a set of all odd numbers divisible by 2.

• Empty Sets: A set that does not contain any elements is empty, void set, or null. It is denoted with {} or Φ.
• Singleton Set: A single-element set is referred to as a singleton setup.
• Finite and Infinite Sets: A set that contains a limited number of elements is called a finite or infinite set.
• Equal Sets: If every element in A is also present in B, or vice versa, then A and B can be considered equal. Two equal sets will contain precisely the same element.
• Equivalent Sets are two finite sets A or B that are equivalent if they have the same number of elements, i.e. n(A) = n(B)

1.3 Power Sets, Universal Sets, and Subsets

This section in NCERT Solutions Class 11 Mathematics Chapter 1 is about power sets, subsets, and universal sets. It deals with problems based on correctly expressing statements using the correct symbols. Set theory symbols play a crucial role in the expression of ideas about subsets and related terms. This exercise provides multiple examples that illustrate the use of symbols in sets.

• Power Set: The collection of all subsets in a set is known as the power set. It is indicated by P(A). If A has more elements than A, i.e. If n(A), then the number of factors in A = n, then P(A) = 2n
• Universal Set: The universal set is a set that includes all sets within a given context.
• Subset: If all elements of set A belong to set B, set A can be considered a subset. We write symbols

A ⊆ B, if x ∈ A ⇒ x ∈ B

1.4 Union and Intersection

The sum of all elements in a set is called the union. The symbol for the union of sets, ‘U’, is used.

All elements common to given sets at the intersection are collected, called’∩’ (Uniformity of Sets).

The intersection of two sets A or B is denoted A B, common to both A AND B.

Thus, A ∩ B = {x : x ∈ A and x ∈ B}

1.5 Complementary Sets

It covers problems that arise from the complement of sets. A set’s complement is the sum of all elements in a set, excluding those found in its given sets. A’ is the complement to a set. This part of the exercise in NCERT Solutions Class 11 Mathematics Chapter 1 consists of seven problems based on the laws of double complementation, laws for the empty set, and universal set.

Let U be the universal collection, and A be a subset. Then, the complement is the set of elements U that are not A.

Thus, A’ = U – A = {x : x ∈ U and x ∉ A}

1.6 Operations around Sets

The union of two sets A, B and B, is denoted A B. It is the set that contains all elements found in either A or B or both. Thus, A ∪ B = {x : x ∈ A or x ∈ B}.

1.7 Miscellaneous

These problems are based on sets and their representations. They also include practical problems on the union and the intersection of sets. The sums are complex but can be used to help students learn this lesson. NCERT Solutions Class 11 Mathematics Chapter 1 miscellaneous section offers various types of problems. It is based on the operations of sets, application and subtraction of sets.

Different Laws of Algebra of Sets in NCERT Solutions Class 11 Mathematics Chapter 1

1. Idempotent Laws: We have for any set A.

A ∪ A = A

A ∩ A = A

1. Identity Laws: We have for any set A.

A ∪ Φ = A

A ∩ U = A

1. CommutativeLaws: We have for any two sets A or B.

A ∪ B = B ∪ A

A ∩ B = B ∩ A

1. Associative Laws: We have A, C, and B for any three sets.

A ∪ (B ∪ C) = (A ∪ B) ∪ C

A ∩ (B ∩ C) = (A ∩ B) ∩ C

1. Distributive Laws: If A and B are three sets, then

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

1. De Morgan’s Laws If A or B are two sets, then

(A ∪ B)’ = A’ ∩ B’

(A ∩ B)’ = A’ ∪ B’

### NCERT Solutions Class 11 Mathematics Chapter 1 Exercise & Solutions

NCERT Solutions Class 11 Mathematics Chapter 1 elaborates on the most fundamental concepts based on sets, types, and applications. The sets are majorly used to represent the relations and functions. The students can study the Class 11 Mathematics Chapter 1 with the help of NCERT solutions. It will help them make a foundation for upcoming topics such as geometry, sequences, and probability.

With the help of NCERT Solutions Class 11 Mathematics Chapter 1, the students can get well-versed with the topics and their applications in real-life situations. In the solutions guide for Class 11 Mathematics, we have covered a list of formulas, the union of sets, the intersection of sets, and the Venn diagram. Thus, the students will clearly understand all the terms and operations of settings. Further, it will also encourage them to enhance their speed while problem-solving .

The students can click on the below links to refer exercise specific questions and solutions for NCERT Solutions Class 11 Mathematics Chapter 1:

• Class 11 Maths Chapter 1 Ex 1.1 – 6 Questions
• Class 11 Maths Chapter 1 Ex 1.2 – 6 Questions
• Class 11 Maths Chapter 1 Ex 1.3 – 9 Questions
• Class 11 Maths Chapter 1 Ex 1.4 – 12 Questions
• Class 11 Maths Chapter 1 Ex 1.5 – 7 Questions
• Class 11 Maths Chapter 1 Ex 1.6 – 8 Questions
• Class 11 Maths Chapter 1 Miscellaneous Ex – 16 Questions

Apart from the exercise and solutions, the students can also refer to the NCERT solutions on our website.

• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions for Class 9
• NCERT Solutions for Class 10
• NCERT Solutions for Class 11
• NCERT Solutions for Class 12

NCERT Exemplar Class 11 Mathematics PDF

NCERT Exemplar acts as the best study material to test the knowledge on the topic and know your grey areas.  r. It includes short-form questions, long-form questions, and multiple-choice questions. Further, the Exemplar contains various examples which have detailed solutions. Practising with the exemplar can help students strengthen their concepts on Sets-Chapter 1 Class 11.

Also, they can improve their knowledge and skills on the topic. The difficulty level varies from the topic given in a Chapter. For example, the difficulty level on Sets- Chapter 1 is moderate. So, it is suitable for Class 11, 12 and so on. Chapter 11 of NCERT Mathematics is on Sets. The Chapter defines Sets, Relations between Sets, a Product of Sets, Operations on Sets, and many more.

This Chapter deals with set theory in Mathematics which is a branch of algebra or Mathematical logic that deals with concepts related to the collection like union, intersection, complement, and so on. . This Chapter is necessary as it provides good practice to test how well you can apply your concepts on sets. The NCERT Exemplar gives a strong foundation in Mathematics..

To improve your score, the students can refer to NCERT Solutions Class 11 Mathematics Chapter 1 and other NCERT solutions. It is equally important to use the right reference materials from authentic, dependable, and reliable sources like Extramarks and be confident of the results.

### Key Features of NCERT Solutions Class 11 Mathematics Chapter 1

Every student wishes to attain good marks in the examination. Thus, to get the best score, refer to NCERT Solutions Class 11 Mathematics Chapter 1. It offers a complete solution for all the complex examples. The key features in the solutions include:

• In Chapter 11 Sets, the students will be able to apply the concepts required to solve many sets of related problems.
• The NCERT solutions help clear doubts and obstacles by explaining the Chapter’s complex concepts.
• The solutions are provided with detailed explanations and presented by subject matter experts.
• With the help of NCERT Solutions Class 11 Mathematics Chapter 1, the students can gain knowledge of the concepts related to sets.
• Class 11 Mathematics NCERT solutions Chapter 1 helps the student get a strong foundation for the operations of sets.

## NCERT Solutions Class 11 Maths Chapter 1 Sets

(i) The collection of all the months of a year beginning with the letter J.
(ii) The collection of ten most talented writers of India.
(iii) A team of eleven best-cricket batsmen of the world.
(iv) The collection of all boys in your class.
(v) The collection of all natural numbers less than 100.
(vi) A collection of novels written by the writer Munshi Prem Chand.
(vii) The collection of all even integers.
(viii) The collection of questions in this Chapter.
(ix) A collection of most dangerous animals of the world.

(i) Yes, the collection of all the months of a year beginning with the letter J is a set.
Set = {January, June, July}.

(ii) The collection of ten most talented writers of India is not a set. Since, there is no specific parameter to define a talented writer.

(iii) A team of eleven best-cricket batsmen of the world is not a set. Since, it is difficult to find the best-cricket batsman of the world.

(iv) The collection of all boys of in our class is a set.
Since, this set can be represented as
A = {x: x is a boy of our class.}

(v) Collection of all natural numbers less than 100 is a set. Because it is well define and can be represented as = {x: x is a natural number less than 100}

(vi) A collection of novels written by the writer Munshi Prem Chand, is a set. Since, it is a collection of well defined elements.

(vii) The collection of even integers is a set because it’s all elements are well defined.

(viii) The collection of questions in this Chapter is a set. Since, all the elements of set are well defined.

(ix) A collection of most dangerous animals is not a set as most dangerous animals are not well defined.

Q.2

$\begin{array}{l}\text{Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol}\in \text{or}\notin \\ \text{in the blank spaces:}\\ \text{(i) 5 . . . A (ii) 8 . . . A (iii) 0 . . . A}\\ \text{(iv) 4 . . . A (v) 2 . . . A (vi) 10 . . . A}\end{array}$

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }5\in \mathrm{A}\text{ }\left(\mathrm{ii}\right)\text{ }8\notin \mathrm{A}\text{ }\left(\mathrm{iii}\right)\text{ }0\notin \mathrm{A}\\ \left(\mathrm{iv}\right)\text{ }4\in \mathrm{A}\text{ \hspace{0.17em}}\left(\mathrm{v}\right)\text{ }2\in \mathrm{A}\text{ }\left(\mathrm{vi}\right)\text{ }10\notin \mathrm{A}\end{array}$

Q.3 Write the following sets in roster form:

(i) A = {x: x is an integer and –3 < x < 7}
(ii) B = {x: x is a natural number less than 6}
(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}
(iv) D = {x: x is a prime number which is divisor of 60}
(v) E = The set of all letters in the word TRIGONOMETRY
(vi) F = The set of all letters in the word BETTER

(i) A = {x: x is an integer and –3 < x < 7}
= {– 2, –1, 0, 1, 2, 3, 4, 5, 6}
(ii) B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5}
(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}
= {17, 26, 35, 44, 53, 62, 71, 80}
(iv) D = {x: x is a prime number which is divisor of 60}
= {2, 3, 5}
(v) E = The set of all letters in the word TRIGONOMETRY
= {T, R, I, G, O, N, M, E, Y}
(vi) F = The set of all letters in the word BETTER
= {B, E, T, R}

Q.4 Write the following sets in the set-builder form:

(i) (3, 6, 9, 12}
(ii) {2, 4, 8,16,32}
(iii) {5, 25, 125, 625}
(iv) {2, 4, 6, . . .}
(v) {1,4,9, . . .,100}

(i) (3, 6, 9, 12} = {x : 3n, n is a natural number and 1 ≤ n ≤ 4}.
(ii) {2, 4, 8, 16, 32} = {x: 2n, n is a natural number and 1 ≤ n ≤ 5}.
(iii) {5, 25, 125, 625 } = {x: 5n, n is a natural number and 1 ≤ n ≤ 4}.
(iv) {2, 4, 6, . . .}= {x: x is an even natural number}
(v) {1,4,9, . . .,100} = {x: x = n2, n is a natural number and 1 ≤ n ≤ 10 }

Q.5

$\begin{array}{l}\text{List all the elements of the following sets:}\\ \text{\hspace{0.17em}}\left(\text{i}\right)\text{\hspace{0.17em}A = {x : x is an odd natural number}}\\ \text{\hspace{0.17em}}\left(\text{ii}\right)\text{\hspace{0.17em}B = {x : x is an integer,\hspace{0.17em} –}\frac{\text{1}}{\text{2}}\text{< x <}\frac{\text{9}}{\text{2}}\text{}}\\ \left(\text{iii}\right){\text{\hspace{0.17em}C = {x : x is an integer, x}}^{\text{2}}\le \text{4}}\\ \left(\text{iv}\right)\text{\hspace{0.17em}D = {x : x is a letter in the word “LOYAL”}}\\ \left(\text{v}\right)\text{\hspace{0.17em}E = {x : x is a month of a year not having 31 days}}\\ \left(\text{vi}\right)\text{\hspace{0.17em}F = {x : x is a consonant in the English alphabet which precedes k}.}\end{array}$

(i) A = {1, 3, 5, 7, …}
(ii) B = {0, 1, 2, 3, 4}
(iii) C = {–2, –1, 0, 1, 2}
(iv) D = {L, O, Y, A }
(v) E = { February, April, June, September, November}
(vi) F = {b, c, d, f, g, h, j}

Q.6 Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(i) {1, 2, 3, 6} (a) {x: x is a prime number and a divisor of 6}
(ii) {2, 3} (b) {x: x is an odd natural number less than 10}
(iii) {M,A,T,H,E,I,C,S} (c) {x: x is natural number and divisor of 6}
(iv) {1, 3, 5, 7, 9} (d) {x: x is a letter of the word MATHEMATICS}.

(i) {1, 2, 3, 6} = (c) {x: x is natural number and divisor of 6}
(ii) {2, 3} = (a) {x: x is a prime number and a divisor of 6}
(iii) {M, A, T, H, E, I, C, S} = (d) {x: x is a letter of the word MATHEMATICS}
(iv) {1, 3, 5, 7, 9} = (b) {x: x is an odd natural number less than 10}

Q.7 Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) {x: x is a natural numbers, x < 5 and x > 7 }
(iv) {y : y is a point common to any two parallel lines}

(i) Set of odd natural numbers divisible by 2 is null set.
(ii) Set of even prime number is not a null set, because set of even prime number is {2}.
(iii) {x:x is a natural numbers, x < 5 and x > 7} = {} It is a null set.
(iv) It is an example of null set because there is no common point in two parallel lines.

Q.8 Which of the following sets are finite or infinite

(i) The set of months of a year
(ii) {1, 2, 3, . . .}
(iii) {1, 2, 3, . . .99, 100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99

(i) Set of months is a finite set because number of elements in this set are limited to count.
(ii) It is an example of infinite set.
(iii) It is an example of finite set.
(iv) It is an example of infinite set.
(v) It is an example of finite set.

Q.9 State whether each of the following set is finite or infinite:

(i) The set of lines which are parallel to the x-axis
(ii) The set of letters in the English alphabet
(iii) The set of numbers which are multiple of 5
(iv) The set of animals living on the earth
(v) The set of circles passing through the origin (0, 0)

(i) This is an example of an infinite set. Since there are uncountable parallel lines to the x-axis.
(ii) This is an example of a finite set because there are 26 English alphabets.
(iii) The set of numbers that are multiples of 5 is an infinite set because there are uncountable multiples of 5.
(iv) The number of animals living on the earth is countable. So, it is an example of a finite set.
(v) The number of circles passing through the origin (0, 0) are uncountable. So, it is an example of an infinite set.

Q.10 In the following, state whether A = B or not:

(i) A = {a, b, c, d}; B = { d, c, b, a }
(ii) A = {4, 8, 12, 16}; B = { 8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}; B = {x: x is positive even integer and x ≤ 10}
(iv) A = { x : x is a multiple of 10};
B = { 10, 15, 20, 25, 30, . . . }

(i) A = B
(ii) A ≠ B
(iii) A = B
(iv) A≠ B

Q.11 Are the following pair of sets equal? Give reasons.

(i) A = {2, 3},
B = {x: x is solution of x2 + 5x + 6 = 0}

(ii) A = {x: x is a letter in the word FOLLOW}
B = {y: y is a letter in the word WOLF}

(i) A ≠ B
Since, x2 + 5x + 6 = 0
(x + 2)(x + 3) = 0
x = –2, – 3
So, B = {–2, – 3}

(ii) A = {F, O, L, W} and B = {W, O, L, F}
So, A = B.

Q.12 From the sets given below, select equal sets:

A = {2, 4, 8, 12},
B = {1, 2, 3, 4},
C = {4, 8, 12, 14},
D = {3, 1, 4, 2}
E = {–1, 1},
F = {0, a},
G = {1, –1},
H = {0, 1}

B = D [because each is {1, 2, 3, 4}]
E = G [because each is {–1, 1}]

Q.13 Make correct statements by filling in symbols ⊂ or in the blank spaces:

$\begin{array}{l}\left(\text{i}\right)\text{ }\left\{2,3,4\right\}...\left\{1,2,3,4,5\right\}\\ \left(\mathrm{ii}\right)\text{ }\left\{\text{a},\text{b},\text{c}\right\}...\left\{\text{b},\text{c},\text{d}\right\}\\ \left(\mathrm{iii}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{student}\text{ }\mathrm{of}\\ \mathrm{Class}\text{ }\mathrm{XI}\text{ }\mathrm{of}\text{ }\mathrm{your}\text{ }\mathrm{school}\end{array}\right\}...\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{student}\text{ }\mathrm{of}\text{ }\mathrm{your}\\ \mathrm{school}\end{array}\right\}\\ \left(\mathrm{iv}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{circle}\\ \mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{plane}\end{array}\right\}...\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{circle}\text{ }\mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{same}\\ \mathrm{plane}\text{ }\mathrm{with}\text{ }\mathrm{radius}\text{ }1\text{ }\mathrm{unit}\end{array}\right\}\\ \left(\text{v}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{triangle}\\ \mathrm{in}\text{ }\mathrm{a}\text{ }\mathrm{plane}\end{array}\right\}...\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{rectangle}\\ \mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{plane}\end{array}\right\}\\ \left(\mathrm{vi}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{an}\text{ }\mathrm{equilateral}\\ \mathrm{triangle}\text{ }\mathrm{in}\text{ }\mathrm{a}\text{ }\mathrm{plane}\end{array}\right\}...\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{triangle}\text{ }\mathrm{in}\\ \mathrm{the}\text{ }\mathrm{same}\text{ }\mathrm{plane}\end{array}\right\}\\ \left(\mathrm{vii}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{an}\text{ }\mathrm{even}\\ \mathrm{natural}\text{ }\mathrm{number}\end{array}\right\}...\left\{\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{an}\text{ }\mathrm{integer}\right\}\end{array}$

$\begin{array}{l}\left(\text{i}\right)\text{ }\left\{2,3,4\right\}\subset \left\{1,2,3,4,5\right\}\\ \left(\mathrm{ii}\right)\text{ }\left\{\text{a},\text{b},\text{c}\right\}\not\subset \left\{\text{b},\text{c},\text{d}\right\}\\ \left(\mathrm{iii}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{student}\text{ }\mathrm{of}\text{ }\mathrm{Class}\\ \mathrm{XI}\text{ }\mathrm{of}\text{ }\mathrm{your}\text{ }\mathrm{school}\end{array}\right\}\subset \left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{student}\text{ }\mathrm{of}\text{ }\mathrm{your}\\ \mathrm{school}\end{array}\right\}\\ \left(\mathrm{iv}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{circle}\\ \mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{plane}\end{array}\right\}\not\subset \left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{circle}\text{ }\mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{same}\\ \mathrm{plane}\text{ }\mathrm{with}\text{ }\mathrm{radius}\text{ }1\text{ }\mathrm{unit}\end{array}\right\}\\ \left(\text{v}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{triangle}\\ \mathrm{in}\text{ }\mathrm{a}\text{ }\mathrm{plane}\end{array}\right\}\not\subset \left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{rectangle}\\ \mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{plane}\end{array}\right\}\\ \left(\mathrm{vi}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{an}\text{ }\mathrm{equilateral}\\ \mathrm{triangle}\text{ }\mathrm{in}\text{ }\mathrm{a}\text{ }\mathrm{plane}\end{array}\right\}\subset \left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{a}\text{ }\mathrm{triangle}\text{ }\mathrm{in}\\ \mathrm{the}\text{ }\mathrm{same}\text{ }\mathrm{plane}\end{array}\right\}\\ \left(\mathrm{vii}\right)\text{ }\left\{\begin{array}{l}\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{an}\text{ }\mathrm{even}\\ \mathrm{natural}\text{ }\mathrm{number}\end{array}\right\}\subset \left\{\text{x}:\mathrm{x}\text{ }\mathrm{is}\text{ }\mathrm{an}\text{ integer}\right\}\end{array}$

Q.14

$\begin{array}{l}\text{Examine whether the following statements are true or false:}\\ \left(\text{i}\right)\text{{a,b}}\not\subset \text{{b,c, a}}\\ \left(\text{ii}\right)\text{{a,e}}\subset \text{{x : x is a vowel in the English alphabet}}\\ \left(\text{iii}\right)\text{}\left\{\text{1, 2, 3}\right\}\subset \text{{ 1, 3,5 }}\\ \left(\text{iv}\right)\text{{a}}\subset \text{{a,b, c}}\\ \left(\text{v}\right)\text{{a}}\in \text{{a,b, c}}\\ \left(\text{vi}\right)\text{}\left\{\begin{array}{l}\text{x : x is an even natural}\\ \text{number less than 6}\end{array}\right\}\subset \left\{\begin{array}{l}\text{x : x is a natural number}\\ \text{which divides 36}\end{array}\right\}\end{array}$

$\begin{array}{l}\left(\text{i}\right)\text{{a,b}}\not\subset \text{{b,c, a}}\to \text{False}\\ \left(\text{ii}\right)\text{{a,e}}\subset \text{{x : x is a vowel in the English alphabet}}\to \text{True}\\ \left(\text{iii}\right)\text{}\left\{\text{1, 2, 3}\right\}\subset \text{{ 1, 3,5 }}\to \text{False}\\ \left(\text{iv}\right)\text{{a}}\subset \text{{a,b, c}}\to \text{True}\\ \left(\text{v}\right)\text{{a}}\in \text{{a,b, c}}\to \text{False}\\ \left(\text{vi}\right)\text{}\left\{\begin{array}{l}\text{x : x is an even natural}\\ \text{number less than 6}\end{array}\right\}\subset \left\{\begin{array}{l}\text{x : x is a natural number}\\ \text{which divides 36}\end{array}\right\}\text{}\to \text{True}\end{array}$

Q.15 Let A = {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why?

(i) {3, 4} ⊂ A
(ii) {3, 4} ∈ A
(iii) {{3, 4}} ⊂ A
(iv) 1 ∈ A
(v) 1 ⊂ A
(vi) {1, 2, 5} ⊂ A
(vii) {1, 2, 5} ∈ A
(viii) {1, 2, 3} ⊂ A
(ix) Φ ∈ A
(x) Φ ⊂ A
(xi) {Φ} ⊂ A

(i) {3, 4}⊂ A —-Incorrect statement.
Correct statement is {3, 4} ∈ A

(ii) {3, 4} ∈ A —- Correct statement.

(iii) {{3, 4}} ⊂ A—- Correct statement.

(iv) 1 ∈ A —- Correct statement.

(v) 1 ⊂ A —-Incorrect statement.
Correct statement is 1 ∈ A

(vi) {1, 2, 5} ⊂ A —- Correct statement.

(vii) {1, 2, 5} ∈ A —- Incorrect statement.
Correct statement is {1, 2, 5} ⊂ A

(viii) {1, 2, 3} ⊂ A —- Incorrect statement.
Because 3 is not individual element of set A.

(ix) φ ∈ A—- Incorrect statement.
Since, φ is a subset of set A.

(x) φ ⊂ A —- Correct statement.

(xi) {φ} ⊂ A —- Incorrect statement.
Since, φ is not an element of set A.

Q.16 How many elements has P(A), if A = Φ?

P(A) is called power set of set A. It is the collection of subsets of set A. Since, there is no element in set A, so there will be one subset of Φ.
Thus, P(A) = 1.

Q.17 Write down all the subsets of the following sets

(i) {a}
(ii) {a, b}
(iii) {1, 2, 3}
(iv) Φ

(i) Subsets of {a} are: {}, {a}.
(ii) Subsets of {a, b} are: {}, {a}, {b}, {a, b}.
(iii) Subsets of {1, 2, 3} are: {}, {1}, {2}, {3},{1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
(iv) Subset of Φ is Φ only.

Q.18 Write the following intervals in set-builder form:

(i) (– 3, 0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [–23, 5)

 S. No. interval Set builder form of intervals (i) (– 3 , 0) {x: x ∈ R, – 3 < x < 0} (ii) [6, 12] {x: x ∈ R, 6 ≤ x ≤ 12} (iii) (6, 12] {x: x ∈ R, 6 < x ≤ 12} (iv) [–23, 5) {x: x ∈ R, – 23 ≤ x < 5}

Q.19 Write the following as intervals:

(i) {x: x ∈ R, – 4 < x ≤ 6}
(ii) {x: x ∈ R, – 12 < x < –10}
(iii) {x: x ∈ R, 0 ≤ x < 7}
(iv) {x: x ∈ R, 3 ≤ x ≤ 4}

 S. No. Set builder form of interval Intervals (i) {x: x ∈ R, – 4 < x ≤ 6} (– 4, 6] (ii) {x: x ∈ R, – 12 < x < –10} (– 12, – 10) (iii) {x: x ∈ R, 0 ≤ x < 7} [0, 7) (iv) {x: x ∈ R, 3 ≤ x ≤ 4} [3, 4]

Q.20 What universal set(s) would you propose for each of the following:

(i) The set of right triangles.
(ii) The set of isosceles triangles.

(i) Universal set for the set of right triangles is set of triangles or set of polygons.
(ii) Universal set for the set of isosceles triangles is the set of triangles or set of polygons.

Q.21 Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for all the three sets A, B and C

(i) {0, 1, 2, 3, 4, 5, 6}
(ii) Φ
(iii) {0,1,2,3,4,5,6,7,8,9,10}
(iv) {1, 2, 3, 4, 5, 6, 7, 8}

(i) Since, elements of sets A and B are in {0, 1, 2, 3, 4, 5, 6} but elements of set C are not contained in {0, 1, 2, 3, 4, 5, 6}. So, {0, 1, 2, 3, 4, 5, 6} is not a universal set (s).
(ii) Since, elements of sets A, B and C are not contained in Φ. So, it is not a universal set.
(iii) Since, all the elements of sets A, B and C are in given set. So, {0,1,2,3,4,5,6,7,8,9,10} is a universal set (s) for all the three sets A, B and C.
(iv) Since, elements of set C are not in given set {1, 2, 3, 4, 5, 6, 7, 8}, so it is not a universal set(s) of all the three sets A, B and C.

Q.22 Let A = {a, b}, B = {a, b, c}. Is A ⊂ B?
What is A ∪ B?

Since, all the elements of set A belongs to the set B,
So, A ⊂ B.
A ∪ B = {a, b, c} = B

Q.23 If A and B are two sets such that A ⊂ B, then what is A ∪ B?

Since, A ⊂ B so, all the elements of set A belong to set B, then A ∪ B = B.

Q.24 If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find

(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D

(i) A ∪ B = {1, 2, 3, 4, 5, 6}
(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) B ∪ C = {3, 4, 5, 6, 7, 8}
(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

Q.25 If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find

(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
(xi) C – D
(xii) D – C

(i) A – B = {3, 6, 9, 15, 18, 21}
(ii) A – C = {3, 9, 15, 18, 21}
(iii) A – D = {3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 16, 20}
(v) C – A = {2, 4, 8, 10, 14, 16}
(vi) D – A = {5, 10, 20}
(vii) B – C = {20}
(viii) B – D = {4, 8, 12, 16}
(ix) C – B = {2, 6, 10, 14}
(x) D – B = {5, 15, 20}
(xi) C – D = {2, 4, 6, 8, 12, 14, 16}
(xii) D – C = {5, 15, 20}

Q.26 If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

Since, R = set of real numbers Q = set of rational numbers Therefore,
R – Q = Set of irrational numbers

Q.27 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9},
A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find

(i) A′
(ii) B′
(iii) (A ∪ C)′
(iv) (A ∪ B)′
(v) (A′)′
(vi) (B – C)’

(i) A’ = U – A = {5, 6, 7, 8, 9}
(ii) B’ = U – B = {1, 3, 5, 7, 9}
(iii) A ∪ C = {1, 2, 3, 4, 5, 6}
(A ∪ C)’ = U – (A ∪ C)
= {7, 8, 9}
(iv) A ∪ B = {1, 2, 3, 4, 6, 8}
(A ∪ B)’ = U – (A∪B)
= {5, 7, 9}
(v) (A’)’ = U – (A’)
= {1, 2, 3, 4}
(vi) B – C = {2, 8}
(B – C)’ = U – (B – C)
= {1, 3, 4, 5, 6, 7, 9}

Q.28 If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}

(i) A = {a, b, c}
A’ = U – A
= {d, e, f, g, h}

(ii) B = {d, e, f, g}
B’ = U – B
= {a, b, c, h}

(iii) C = {a, c, e, g}
C’ = U – C
= {b, d, f, h}

(iv) D = {f, g, h, a}
D’ = U – D
= {b, c, d, e}

Q.29 Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(i) {x: x is an even natural number}
(ii) {x: x is an odd natural number}
(iii) {x: x is a positive multiple of 3}
(iv) {x: x is a prime number}
(v) {x: x is a natural number divisible by 3 and 5}
(vi) {x: x is a perfect square}
(vii) {x: x is a perfect cube}
(viii) {x: x + 5 = 8}
(ix) {x: 2x + 5 = 9}
(x) { x : x ≥ 7 }
(xi) { x : x ∈ N and 2x + 1 > 10 }

Here, U = N = {1, 2, 3, 4, 5, …}

(i) Let A = {x: x is an even natural number}
A’ = U – A
= {1, 3, 5, 7, …}
= {x: x is an odd natural number}

(ii) Let B = {x: x is an odd natural number}
B’ = U – B
= {2, 4, 6, 8, …}
= {x : x is an even natural number}

(iii) Let C = {x: x is a positive multiple of 3}
C’ = U – C
= {1, 3, 4, 5, 7, …}
= {x : x is a natural number but not a multiple of 3}

(iv) Let D = {x: x is a prime number}
D’ = U – D
= {x : x is composite number and 1}

(v) Let E = {x: x is a natural number divisible by 3 and 5}
E’ = U – E
= {x: x is a natural number neither divisible by 3 nor by 5}

(vi) Let F = {x: x is a perfect square}
F’ = U – F
= {x : x is not a perfect square}

(vii) Let G = {x: x is a perfect cube}
G’ = U – G
= {x : x is not a perfect cube}

(viii) Let H = {x: x + 5 = 8}
= {3}
H’ = U – H
= {x : x is a natural number except 3}

(ix) Let I = {x: 2x + 5 = 9}
= {2}
I’ = U – I
= {x : x is a natural number except 2}

(x) Let J = { x:x ≥ 7}
J’ = U – J
= {1, 2, 3, 4, 5, 6}

(xi) Let K = {x : x ∈ N and 2x + 1 > 10}
= { x : x ∈ N and x > 9/2}
= {5, 6, 7, 8, …}
K’ = U – K
= {1, 2, 3, 4}

Q.30 If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that

(i) (A ∪ B)′ = A′ ∩ B′
(ii) (A ∩ B)′ = A′ ∪ B′

A ∪ B = {2, 3, 4, 5, 6, 7, 8}
A ∩ B = {2}
(i) L.H.S.:
(A ∪ B)′ = U – (A ∪ B)
= {1, 9}
R.H.S.:
A′ ∩ B′ =(U – A) ∩ ( U – B)
= {1, 3, 5, 7, 9}∩{1, 4, 6, 8, 9}
= {1, 9}
Thus,
(A ∪ B)′ = A′ ∩ B′

(ii) L.H.S.:
(A ∩ B)′ = U – (A ∩ B)
= {1, 3, 4, 5, 6, 7, 8, 9}
R.H.S.:
A′ ∪ B′=(U – A) ∪ ( U – B)
= {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}
= {1, 3, 4, 5, 6, 7, 8, 9}
Thus,
(A ∩ B)’ = A′ ∪ B′

Q.31 Draw appropriate Venn diagram for each of the following :

(i) (A ∪ B)’, (ii) A′ ∩ B′,
(iii) (A ∩ B)′, (iv) A′ ∪ B′

(i) (A ∪ B)’ (ii) A′ ∩ B′=(A ∪ B)’ [By De Morgan’s Law] (iii) (A ∩ B)′ (iv) A′ ∪ B′ = (A ∩ B)′ [By De Morgan’s Law] Q.32 Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A′?

Here, U = set of all triangles in a plane,
A = set of all triangles with at least one angle different from 60°
A’ = set of all triangles with each angle 60° i.e., set of all equilateral triangles.

Q.33 If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38, find n ( X ∩ Y ).

$\begin{array}{l}\mathrm{We}\text{have,\hspace{0.17em}\hspace{0.17em}n}\left(\text{X}\right)=\text{17}\mathrm{}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}\left(\text{Y}\right)=\text{23}\\ \text{n\hspace{0.17em}}\left(\text{X}\cup \text{Y}\right)\text{}=\text{38}\\ \text{Since,}\\ \text{\hspace{0.17em}n\hspace{0.17em}}\left(\text{X}\cap \text{Y}\right)=\mathrm{n}\left(\mathrm{X}\right)+\mathrm{n}\left(\mathrm{Y}\right)-\mathrm{n}\left(\mathrm{X}\cap \mathrm{Y}\right)\\ \text{}=17+23-38\\ \text{}=40-38\\ \text{}=2\\ \therefore \text{\hspace{0.17em}n\hspace{0.17em}}\left(\text{X}\cap \text{Y}\right)=2\end{array}$

Q.34 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?

$\begin{array}{l}\mathrm{We}\text{have, n}\left(\mathrm{X}\cup \mathrm{Y}\right)=18,\text{\hspace{0.17em}}\mathrm{n}\left(\mathrm{X}\right)=8\text{and n}\left(\mathrm{Y}\right)=15\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}n}\left(\mathrm{X}\cap \mathrm{Y}\right)=\mathrm{n}\left(\mathrm{X}\right)+\text{n}\left(\mathrm{Y}\right)-\text{n}\left(\mathrm{X}\cup \mathrm{Y}\right)\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}\left(\mathrm{X}\cap \mathrm{Y}\right)=8+15-18\\ =5\\ \mathrm{Thus},\text{the number of elements in the set}\mathrm{X}\cap \mathrm{Y}\text{}\mathrm{is}\text{5.}\end{array}$

Q.35 In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{H}=\mathrm{Set}\text{of Hindi speaking people}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{E}=\mathrm{Set}\text{of English speaking people}\\ \mathrm{Given}:\\ \mathrm{n}\left(\mathrm{H}\right)=250,\text{\hspace{0.17em}}\mathrm{n}\left(\mathrm{E}\right)=200\text{and n}\left(\mathrm{H}\cup \mathrm{E}\right)=400\\ \mathrm{n}\left(\mathrm{H}\cap \mathrm{E}\right)=\mathrm{Number}\text{of people who speak Hindi and English}\\ \mathrm{Then},\text{\hspace{0.17em}n}\left(\mathrm{H}\cap \mathrm{E}\right)=\mathrm{n}\left(\mathrm{H}\right)+\mathrm{n}\left(\mathrm{E}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{E}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=250+200-400\\ \text{\hspace{0.17em} \hspace{0.17em}}=50\\ \mathrm{Thu}\text{s, the number of people who speak Hindi and English}\\ \text{both is 50.}\end{array}$

Q.36 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?

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{n}\left(\mathrm{S}\right)=21,\text{n}\left(\mathrm{T}\right)=32\text{and n}\left(\mathrm{S}\cap \mathrm{T}\right)=11\\ \mathrm{Let}n\left(\mathrm{S}\cup \mathrm{T}\right)=\mathrm{x}\\ \mathrm{Since},\text{n}\left(\mathrm{S}\cup \mathrm{T}\right)=\mathrm{n}\left(\mathrm{S}\right)+\mathrm{n}\left(\mathrm{T}\right)-\mathrm{n}\left(\mathrm{S}\cap \mathrm{T}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\text{\hspace{0.17em}}21+32-11\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=42\\ \text{Thus, the number of elements in}\left(\mathrm{S}\cup \mathrm{T}\right)\text{is 42.}\end{array}$

Q.37 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?

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{n}\left(\mathrm{X}\right)=40,\text{n}\left(\mathrm{X}\cup \mathrm{Y}\right)=60\text{and n}\left(\mathrm{X}\cap \mathrm{Y}\right)=10\\ \mathrm{To}\text{find: n}\left(\mathrm{Y}\right)\\ \mathrm{Since},\text{n}\left(\mathrm{x}\cup \mathrm{Y}\right)=\mathrm{n}\left(\mathrm{X}\right)+\mathrm{n}\left(\mathrm{Y}\right)-\mathrm{n}\left(\mathrm{X}\cap \mathrm{Y}\right)\\ \mathrm{then},\text{\hspace{0.17em}\hspace{0.17em}​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 60}=40+\mathrm{n}\left(\mathrm{Y}\right)-10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}60-30=\mathrm{n}\left(\mathrm{Y}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}30=\mathrm{n}\left(\mathrm{Y}\right)\\ \mathrm{Thus},\text{set Y has 30 elements.}\end{array}$

Q.38 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?

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}C}=\text{Set of people who like coffee}\\ \text{\hspace{0.17em}\hspace{0.17em}T}=\text{Set of people who like tea}\\ \text{and\hspace{0.17em}\hspace{0.17em}}\mathrm{C}\cap \mathrm{T}=\mathrm{Set}\text{of people who like coffee and tea}\\ \text{Here, n}\left(\mathrm{C}\right)=37,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}\left(\mathrm{T}\right)=52\text{and n}\left(\mathrm{C}\cup \mathrm{T}\right)=70\\ \mathrm{Since},\text{\hspace{0.17em}}\mathrm{n}\left(\mathrm{C}\cap \mathrm{T}\right)=\mathrm{n}\left(\mathrm{C}\right)+\mathrm{n}\left(\mathrm{T}\right)-\mathrm{n}\left(\mathrm{C}\cup \mathrm{T}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=37+52-70\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=89-70\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=19\\ \mathrm{Thus},\text{​ 19 people like both coffee and tea.}\end{array}$

Q.39 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?

$\begin{array}{l}\mathrm{C}=\mathrm{Set}\text{of people who like cricket}\\ \text{T}=\mathrm{Set}\text{of people who like tennis}\\ \text{C}\cap \text{T}=\mathrm{Set}\text{of people who like cricket and tennis}\\ \text{n}\left(\mathrm{C}\right)=40,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}\left(\mathrm{C}\cup \mathrm{T}\right)=65,\text{\hspace{0.17em}}\mathrm{n}\left(\text{C}\cap \text{T}\right)=10\\ \mathrm{n}\left(\mathrm{C}\cup \mathrm{T}\right)=\mathrm{n}\left(\mathrm{C}\right)+\mathrm{n}\left(\mathrm{T}\right)-\mathrm{n}\left(\text{C}\cap \text{T}\right)\\ \text{\hspace{0.17em}}65 =40+\mathrm{n}\left(\mathrm{T}\right)-10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}\left(\mathrm{T}\right)=65-30\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=35\\ \mathrm{Thus},\text{35 people like tennis.}\end{array}$

Q.40 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?

$\begin{array}{l}\mathrm{Let}\text{F}=\mathrm{Set}\text{of people who speak french}\\ \text{S}=\mathrm{Set}\text{of people who speak Spanish}\\ \text{F}\cap \text{S}=\mathrm{Set}\text{of people who speak\hspace{0.17em}}\mathrm{French}\text{and Spanish}\\ \mathrm{F}\cup \mathrm{S}=\mathrm{Set}\text{of people who speak\hspace{0.17em}}\mathrm{French}\text{or Spanish}\\ \text{n}\left(\mathrm{F}\right)=50,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}\left(\mathrm{S}\right)=20\text{and n}\left(\mathrm{F}\cap \mathrm{S}\right)=10\\ \mathrm{n}\left(\mathrm{F}\cup \mathrm{S}\right)=\mathrm{n}\left(\mathrm{F}\right)+\mathrm{n}\left(\mathrm{S}\right)-\mathrm{n}\left(\mathrm{F}\cap \mathrm{S}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50+20-10\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\\ \mathrm{Thus},\text{60 people speak at least one of these two languages.}\end{array}$

Q.41 Decide, among the following sets, which sets are subsets of one and another:

A = {x: x ∈ R and x satisfy x2 – 8x + 12 = 0},
B = {2, 4, 6}, C = {2, 4, 6, 8, . . .}, D = { 6 }.

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{A}=\left\{\mathrm{x}:\mathrm{x}\in \mathrm{R}\text{}\mathrm{and}\text{}\mathrm{x}\text{}\mathrm{satisfy}\text{}{\mathrm{x}}^{2}–8\mathrm{x}+12=0\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}\left\{\mathrm{x}:\mathrm{x}\in \mathrm{R}\text{}\mathrm{and}\text{}\mathrm{x}\text{}\mathrm{satisfy}\text{}\left(\mathrm{x}-2\right)\left(\mathrm{x}-6\right)=0\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{2,\text{\hspace{0.17em}}6\right\}\\ \mathrm{}\mathrm{B}=\left\{2,\text{}4,\text{}6\right\},\\ \mathrm{C}=\left\{2,\text{}4,\text{}6,\text{}8,\text{}.\text{}.\text{}.\right\},\\ \mathrm{D}=\left\{\text{}6\text{}\right\}\\ \mathrm{Since},\text{elements of set A belongs to sets B and C.}\\ \mathrm{So},\text{A}\subset \text{B,​\hspace{0.17em}A}\subset \text{C.}\\ \mathrm{Since},\text{elements of set B belongs to set C.}\\ \mathrm{So},\text{B}\subset \text{C.}\\ \mathrm{Since},\text{elements of set D belongs to sets A, B and C.}\\ \mathrm{So},\text{D}\subset \text{A,​\hspace{0.17em}D}\subset \text{B and D}\subset \text{C.}\\ \text{Thus, A}\subset \text{B,​\hspace{0.17em}A}\subset \text{C},\text{\hspace{0.17em}B}\subset \text{C},\text{D}\subset \text{A,​\hspace{0.17em}D}\subset \text{B and D}\subset \text{C.}\end{array}$

Q.42 In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

$\begin{array}{l}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{ }\mathbf{If}\mathbf{ }\mathbf{x}\mathbf{\in }\mathbf{A}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{A}\mathbf{\in }\mathbf{B}\mathbf{,}\mathbf{ }\mathbf{then}\mathbf{ }\mathbf{x}\mathbf{\in }\mathbf{B}\\ \mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{ }\mathbf{If}\mathbf{ }\mathbf{A}\mathbf{\subset }\mathbf{B}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{B}\mathbf{\in }\mathbf{C}\mathbf{,}\mathbf{ }\mathbf{then}\mathbf{ }\mathbf{A}\mathbf{\in }\mathbf{C}\\ \mathbf{\left(}\mathbf{iii}\mathbf{\right)}\mathbf{ }\mathbf{If}\mathbf{ }\mathbf{A}\mathbf{\subset }\mathbf{B}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{B}\mathbf{\subset }\mathbf{C}\mathbf{,}\mathbf{ }\mathbf{then}\mathbf{ }\mathbf{A}\mathbf{\subset }\mathbf{C}\\ \mathbf{\left(}\mathbf{iv}\mathbf{\right)}\mathbf{ }\mathbf{If}\mathbf{ }\mathbf{A}\mathbf{\not\subset }\mathbf{B}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{B}\mathbf{\not\subset }\mathbf{C}\mathbf{,}\mathbf{ }\mathbf{then}\mathbf{ }\mathbf{A}\mathbf{\not\subset }\mathbf{C}\\ \mathbf{\left(}\mathbf{v}\mathbf{\right)}\mathbf{ }\mathbf{If}\mathbf{ }\mathbf{x}\mathbf{\in }\mathbf{A}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{A}\mathbf{\not\subset }\mathbf{B}\mathbf{,}\mathbf{ }\mathbf{then}\mathbf{ }\mathbf{x}\mathbf{\in }\mathbf{B}\\ \mathbf{\left(}\mathbf{vi}\mathbf{\right)}\mathbf{ }\mathbf{If}\mathbf{ }\mathbf{A}\mathbf{\subset }\mathbf{B}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{x}\mathbf{\notin }\mathbf{B}\mathbf{,}\mathbf{ }\mathbf{then}\mathbf{ }\mathbf{x}\mathbf{\notin }\mathbf{A}\end{array}$

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}}\mathrm{False},\\ \text{Since, x in the element of set A and A is the element}\\ \text{of set B.So, x can not be an element of set B.}\\ \text{Example: A}=\left\{4,\text{}2,\text{}6\right\},\text{B}=\left\{1,\left\{4,\text{}2,\text{}6\right\},\text{\hspace{0.17em}\hspace{0.17em}}5,\text{\hspace{0.17em}}7\right\}\\ \text{Here, 2}\in \text{A but 2}\notin \text{C.}\\ \left(\mathrm{ii}\right)\mathrm{False},\\ \mathrm{Since},\text{set A is subset of set B and set B is an element of set C.}\\ \text{So, A can not be an element of set C.}\\ \text{Example: A}=\left\{\mathrm{a},\text{b, c}\right\},\text{​ B}=\left\{\mathrm{a},\text{b, c, d, e}\right\}\text{}\\ \text{}\mathrm{and}\text{C}=\left\{\mathrm{f},\text{\hspace{0.17em}}\left\{\mathrm{a},\text{b, c, d, e}\right\},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{k},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{m}\right\}\\ \text{Here, A}\subset \text{B and B}\in \text{C but A}\notin \text{C}\\ \left(\mathrm{iii}\right)\mathrm{True},\\ \mathrm{Since},\text{​\hspace{0.17em}set A is subset of set B and set B is subset of set C then}\\ \text{set A is also a subset of set C.}\\ \text{Example:\hspace{0.17em}}\mathrm{Let}\text{A}=\left\{2,\text{4}\right\},\text{B}=\left\{2,\text{4, 6, 8}\right\},\text{\hspace{0.17em}\hspace{0.17em}C}=\left\{2,\text{4, 6, 8, 10, 11}\right\}\\ ⇒\mathrm{A}\subset \mathrm{BandB}\subset \mathrm{CandA}\subset \mathrm{C}.\\ \left(\mathrm{iv}\right)\mathrm{False},\text{}\\ \text{Since, set A is not a subset of set B and set B is not a subset}\\ \text{of C. So, A may be or may not be a subset of C.}\\ \text{Example:}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}A}=\left\{1,2,3\right\},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=\left\{4,5,6\right\}\mathrm{C}=\left\{1,2,3,5,7,8\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{A}\not\subset \mathrm{BandB}\not\subset \mathrm{C},\mathrm{butA}\subset \mathrm{C}\\ \left(\mathrm{v}\right)\text{\hspace{0.17em}}\mathrm{False},\\ \text{Example:\hspace{0.17em}}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=\left\{1,2,3,4,5\right\},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=\left\{1,2,3,4,6\right\}\\ \mathrm{Here},\text{2}\in \text{A and A}\not\subset \text{\hspace{0.17em}}\mathrm{B}\text{but 2}\in \text{B.}\\ \left(\mathrm{vi}\right)\text{\hspace{0.17em}}\mathrm{True},\\ \text{Example:\hspace{0.17em}}\mathrm{Let}\text{A}=\text{\hspace{0.17em}}\left\{4,8,6\right\}\text{, B}=\left\{1,2,4,6,8\right\}\\ \mathrm{if}\text{9}\notin \text{B, then 9}\notin \text{A.}\end{array}$

Q.43 Assume that P (A) = P (B). Show that A = B

$\begin{array}{l}\mathrm{Let}\text{x be any element of A and X be a subset of A such that x}\in \mathrm{X}.\\ \mathrm{So},\text{X}\subset \mathrm{A}⇒\mathrm{X}\in \mathrm{P}\left(\mathrm{A}\right)\\ \text{\hspace{0.17em}}⇒\mathrm{X}\in \mathrm{P}\left(\mathrm{B}\right) \left[\because \mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{B}\right)\right]\\ \text{\hspace{0.17em}}⇒\mathrm{X}\subset \mathrm{B}\\ \text{\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{B}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\in \mathrm{A}⇒\mathrm{x}\in \mathrm{B}\\ \mathrm{Therefore},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\subset \mathrm{B}...\left(\mathrm{i}\right)\end{array}$ $\begin{array}{l}\mathrm{Now},\text{let y}\in \mathrm{B}\text{and Y}\subset \mathrm{B}\text{as y}\in \mathrm{Y}.\text{\hspace{0.17em}}\mathrm{Then},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Y}\subset \mathrm{B}⇒\mathrm{Y}\in \mathrm{P}\left(\mathrm{B}\right)\\ \text{\hspace{0.17em}}⇒\mathrm{Y}\in \mathrm{P}\left(\mathrm{A}\right) \left[\because \mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{B}\right)\right]\\ \text{\hspace{0.17em}}⇒\mathrm{Y}\subset \mathrm{A}\\ \text{\hspace{0.17em}}⇒\mathrm{y}\in \mathrm{A}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}\in \mathrm{B}⇒\mathrm{y}\in \mathrm{A}\\ \mathrm{Therefore},\text{B}\subset \mathrm{A}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right),\text{we have}\\ \text{A}=\text{B}\end{array}$

Q.44 Using properties of sets, show that

(i) A ∪ ( A ∩ B ) = A
(ii) A ∩ ( A ∪ B ) = A.

$\begin{array}{l}\left(\mathrm{i}\right) \mathrm{A}\cup \left(\mathrm{A}\cap \mathrm{B}\right)=\left(\mathrm{A}\cup \mathrm{A}\right)\cap \left(\mathrm{A}\cup \mathrm{B}\right)\\ =\mathrm{A}\cap \mathrm{U}\\ =\mathrm{A}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \left(\mathrm{ii}\right) \mathrm{A}\cap \left(\mathrm{A}\cup \mathrm{B}\right)=\left(\mathrm{A}\cap \mathrm{A}\right)\cup \left(\mathrm{A}\cap \mathrm{B}\right)\\ =\mathrm{A}\cup \mathrm{\varphi }\\ =\mathrm{A}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.18 In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}n}\left(\text{C}\right)=\mathrm{Number}\text{\hspace{0.17em}of\hspace{0.17em}students\hspace{0.17em}taking\hspace{0.17em}coffee}\\ =225\\ \text{n}\left(\mathrm{T}\right)=\mathrm{Number}\text{\hspace{0.17em}of\hspace{0.17em}students\hspace{0.17em}taking\hspace{0.17em}tea}\\ =150\\ \text{n}\left(\mathrm{C}\cap \mathrm{T}\right)=\mathrm{Number}\text{\hspace{0.17em}of\hspace{0.17em}students taking coffee and tea both}\\ =100\\ \text{n}\left(\mathrm{C}\cup \mathrm{T}\right)=\mathrm{Number}\text{\hspace{0.17em}of\hspace{0.17em}students\hspace{0.17em}taking coffee or tea}\\ \text{Then,}\\ \text{\hspace{0.17em}n}\left(\mathrm{C}\cup \mathrm{T}\right)=\mathrm{n}\left(\mathrm{C}\right)+\mathrm{n}\left(\mathrm{T}\right)-\mathrm{n}\left(\mathrm{C}\cap \mathrm{T}\right)\\ =225+150-100\\ =275\\ \mathrm{Number}\text{\hspace{0.17em}of\hspace{0.17em}students who were taking neither tea nor coffee}\\ =600-275\\ =325\end{array}$

Q.45 In a group of students, 100 students know Hindi, 50 know English and 25 know both.
Each of the students knows either Hindi or English. How many students are there in the group?

$\begin{array}{l}\mathrm{Let}\text{n}\left(\mathrm{H}\right)=\mathrm{Number}\text{of students who know Hindi}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100\\ \mathrm{n}\left(\mathrm{E}\right)=\mathrm{Number}\text{of students who know English}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50\\ \mathrm{n}\left(\mathrm{H}\cap \mathrm{E}\right)=\mathrm{Number}\text{of students who know Hindi and English}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25\\ \mathrm{Number}\text{of students in who know only Hindi}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}\left(\mathrm{H}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{E}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100-25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=75\\ \mathrm{Number}\text{of students in who know only English}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}\left(\mathrm{E}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{E}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50-25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25\\ \mathrm{n}\left(\mathrm{HUE}\right)=\mathrm{Number}\text{of students who know Hindi and English}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}\left(\mathrm{H}\right)+\mathrm{n}\left(\mathrm{E}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{E}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100+50-25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=125\\ \mathrm{Thus},\text{total number of students in the group is 125.}\end{array}$

Q.46 In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers.
Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.

$\begin{array}{l}\mathrm{Let} \mathrm{n}\left(\mathrm{H}\right)=\mathrm{Number}\text{of people who read newspaper H}\\ =25\\ \mathrm{n}\left(\mathrm{T}\right)=\mathrm{Number}\text{of people who read newspaper T}\\ =26\\ \mathrm{n}\left(\mathrm{I}\right)=\mathrm{Number}\text{of people who read newspaper I}\\ =26\\ \mathrm{n}\left(\mathrm{H}\cap \mathrm{I}\right)=\mathrm{Number}\text{of people who read newspaper H and I}\\ =9\\ \mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\right)=\mathrm{Number}\text{of people who read newspaper H and T\hspace{0.17em}}=11\\ \mathrm{n}\left(\mathrm{T}\cap \mathrm{I}\right)=\mathrm{Number}\text{of people who read newspaper T and I\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\\ \mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\cap \mathrm{I}\right)=\mathrm{Number}\text{of people who read newspaper H, T and I\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ \left(\mathrm{i}\right)\\ \mathrm{n}\left(\mathrm{H}\cup \mathrm{T}\cup \mathrm{I}\right)=\mathrm{n}\left(\mathrm{H}\right)+\mathrm{n}\left(\mathrm{T}\right)+\mathrm{n}\left(\mathrm{I}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\right)-\mathrm{n}\left(\mathrm{T}\cap \mathrm{I}\right)-\mathrm{n}\left(\mathrm{I}\cap \mathrm{H}\right)\\ +\text{\hspace{0.17em}}\mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\cap \mathrm{I}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25+26+26-9-11-8+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=52\\ \mathrm{Thus},\text{number of people who read at least one of the news paper.}\\ \left(\mathrm{ii}\right)\text{n}\left(\mathrm{only}\text{\hspace{0.17em}}\mathrm{H}\right)=\mathrm{n}\left(\mathrm{H}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{I}\right)+\mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\cap \mathrm{I}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25-9-11+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}\left(\mathrm{only}\text{\hspace{0.17em}}\mathrm{T}\right)=\mathrm{n}\left(\mathrm{T}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\right)-\mathrm{n}\left(\mathrm{T}\cap \mathrm{I}\right)+\mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\cap \mathrm{I}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=26-11-8+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}\left(\mathrm{only}\text{\hspace{0.17em}}\mathrm{I}\right)=\mathrm{n}\left(\mathrm{I}\right)-\mathrm{n}\left(\mathrm{H}\cap \mathrm{I}\right)-\mathrm{n}\left(\mathrm{T}\cap \mathrm{I}\right)+\mathrm{n}\left(\mathrm{H}\cap \mathrm{T}\cap \mathrm{I}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=26-9-8+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\\ \mathrm{The}\text{number of people who read exactly one newspaper}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+10+12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=30\end{array}$

Q.47 In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

Here,
n(A) = Number of people who like product A = 21
n(B) = Number of people who like product B =26
n(C) = Number of people who like product C = 29
n(A ∩ B) = Number of people who like products A and B = 14
n(B ∩ C) = Number of people who like products B and C = 14
n(C ∩ A) = Number of people who like products C and A = 12
n(A ∩ B ∩ C) = Number of people who like products A, B and C = 8
n(Only C) = n(C) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C) = 29−14−12+8 = 11
Thus, 11 people like only product C.

Q.48 Find the union of each of the following pairs of sets:

(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = [ a, e, i, o, u} B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
B = {x: x is a natural number and 6 < x < 10}

(v) A = {1, 2, 3}, B = Φ

(i) X = {1, 3, 5} Y = {1, 2, 3}
X ∪ Y = {1, 2, 3, 5}
(ii) A = [a, e, i, o, u} B = {a, b, c}
A ∪ B = {a, b, c, e, i, o, u}
(iii )A = {x: x is a natural number and multiple of 3}
= {3, 6, 9, 12, 15, … }
B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5}
A∪B = {1, 2, 3, 4, 5, 6, 9, 12, 15, …}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
= {2, 3, 4, 5, 6}
B = {x: x is a natural number and 6 < x < 10}
= {5, 6, 7, 8, 9}
A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
(v) A = {1, 2, 3}, B = Φ
A ∪ B = {1, 2, 3} = A

Q.49 Find the intersection of each pair of sets of

(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = [ a, e, i, o, u} B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
B = {x: x is a natural number and 6 < x < 10}
(v) A = {1, 2, 3}, B = Φ

(i) X ∩ Y = {1, 3}
(ii) A ∩ B = {a}
(iii) A = {x: x is a natural number and multiple of 3}
= {3, 6, 9, 12, …}
B = {x: x is a natural number less than 6}
= {1, 2, 3, 4, 5}
A ∩ B = {3}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
= {2, 3, 4, 5, 6}
B = {x: x is a natural number and 6 < x < 10}
= {7, 8, 9}
A ∩ B = {} = Φ
(v) A ∩ B = {} = Φ

Q.50 If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find

(i) A ∩ B
(ii) B ∩ C
(iii) A ∩ C ∩ D
(iv) A ∩ C
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) (A ∩ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)

(i) A ∩ B = {7, 9, 11}
(ii) B ∩ C = {11, 13}
(iii) A ∩C ∩ D = {} = Φ
(iv) A ∩ C = {11}
(v) B ∩ D = {} = Φ
(vi) A ∩ (B ∪C) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}
(vii) A ∩ D = {} = Φ
(viii) A ∩ (B ∪ D)
= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17}
= {7, 9, 11}
(ix) (A ∩ B ) ∩ ( B ∪ C )
= {7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}
(x) (A ∪ D) ∩ (B ∪ C)
= {3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11, 15}

Q.51 If A = {x: x is a natural number },
B = {x: x is an even natural number}

C = {x: x is an odd natural number} and
D = {x: x is a prime number}, find

(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D

(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D

A = {x: x is a natural number}
= {1, 2, 3, 4, 5, 6, 7, …}
B = {x: x is an even natural number}
= {2, 4, 6, 8, …}
C = {x: x is an odd natural number}
= {1, 3, 5, 7, …}
D = {x: x is a prime number}
= {2, 3, 5, 7,…}

(i) A∩B = {2, 4, 6, …} = B
(ii) A∩C = {1, 3, 5, 7, …} = C
(iii) A ∩ D = {2, 3, 5, 7,…} = D
(iv) B ∩ C = {} = Φ
(v) B ∩ D = {2}
(vi) C ∩ D = {3, 5, 7, …}
= {x: x is an odd prime number}

Q.52 Which of the following pairs of sets are disjoint

(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6 }
(ii) {a, e, i, o, u} and { c, d, e, f}
(iii) {x: x is an even integer} and {x: x is an odd integer}

(i) Let A = {1, 2, 3, 4} and
B = {x: x is a natural number and 4 ≤ x ≤ 6 } = {4, 5, 6}
A ∩ B = {4}
So, sets A and B are not disjoint sets.
(ii) Let P = {a, e, i, o, u} and Q = {c, d, e, f}
P ∩ Q = {e}
So, sets P and Q are disjoint sets.
(iii) Let C = {x: x is an even integer}
= {…, – 4, – 2, 2, 4, 6, … }
D = {x: x is an odd integer}
= {…, –5, –3, –1, 1, 3, 5,…}
C ∩ D = {} = Φ
So, sets C and D are disjoint sets.

Q.53 State whether each of the following statement is true or false. Justify your answer.

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, o, u } and { a, b, c, d }are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.

(i) False,
Since, {2, 3, 4, 5} ∩ {3, 6} = {3}
So, given sets are not disjoint sets.

(ii) False,
Since, {a, e, i, o, u} ∩{a, b, c, d}= {a}
So, given sets are not disjoint sets.

(iii) True,
Since, {2, 6, 10, 14} ∩ {3, 7, 11, 15} = {}
= Φ
So, given sets are disjoint sets.

(iv) True,
Since, {2, 6, 10} ∩ {3, 7, 11} = {}
= Φ
So, given sets are disjoint sets.

Q.54 Fill in the blanks to make each of the following a true statement:
(i) A ∪ A′ = . . .
(ii) Φ′ ∩ A = . . .
(iii) A ∩ A′ = . . .
(iv) U′ ∩ A = . . .

(i) A ∪ A′ = U
(ii) Φ′ ∩ A = A
(iii) A ∩ A′ = Φ
(iv) U′ ∩ A = Φ

Q.55 Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C

$\begin{array}{l}\mathrm{Given}:\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\cup \mathrm{B}=\mathrm{A}\cup \mathrm{C}\text{and A}\cap \text{B}=\text{A}\cap \text{C}\\ \text{To prove: B}=\text{C}\\ \text{Proof: Let x}\in \mathrm{B}⇒\mathrm{x}\in \mathrm{A}\cup \mathrm{B} \left[\because \mathrm{B}\subset \mathrm{A}\cup \mathrm{B}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{A}\cup \mathrm{C} \left[\because \mathrm{A}\cup \mathrm{B}=\mathrm{A}\cup \mathrm{C}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{A}\text{or}\mathrm{x}\in \mathrm{C}\\ \mathrm{Case}\text{\hspace{0.17em}}1:\mathrm{If}\text{x}\in \mathrm{A}\text{and also, x}\in \mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{A}\cap \mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{A}\cap \mathrm{C} \left[\because \text{A}\cap \text{B}=\text{A}\cap \text{C}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{A}\text{and}\mathrm{x}\in \mathrm{C}\\ \because \mathrm{x}\in \mathrm{C}\text{and x}\in \mathrm{B}⇒\mathrm{B}\subset \mathrm{C}...\left(\mathrm{i}\right)\\ \mathrm{Case}\text{\hspace{0.17em}}2:\mathrm{If}\text{x}\in \mathrm{C}\text{and also, x}\in \mathrm{A}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{A}\cap \mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{A}\cap \mathrm{B} \left[\because \text{A}\cap \text{B}=\text{A}\cap \text{C}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{x}\in \mathrm{A}\text{and}\mathrm{x}\in \mathrm{B}\\ \because \mathrm{x}\in \mathrm{B}\text{and x}\in \mathrm{C}⇒\mathrm{C}\subset \mathrm{B}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{case 1 and case 2, we have}\\ \text{B}=\text{C Hence proved.}\end{array}$

Q.56 Show that the following four conditions are equivalent:

(i) A ⊂ B
(ii) A – B = Φ
(iii) A ∪ B = B
(iv) A ∩ B = A

Taking (i) to prove (ii):
We have, A−B = { x ∈ A : x ∉ B }
Since, A⊂B.
Therefore, there is no element in A which does not belong to B.
∴ A−B = ∅
Hence, (i) ⇒ (ii).
Taking (ii) to prove (iii):
We have, A−B = ∅ ⇒ A ⊂ B ⇒ A ∪ B = B
Hence, (ii) ⇒ (iii)
Taking (iii) to prove (iv):
We have, A ∪ B = B ⇒ A ⊂ B ⇒ A ∩ B = A
∴ (iii) ⇒ (iv)
Taking (iv) to prove (i):
We have, A ∩ B = A ⇒ A ⊂ B
So,  (iv) ⇒ (i)
Therefore, (i) ⇔ (ii) ⇔ (iii) ⇔ (iv)

Q.57 Show that if A ⊂ B, then C – B ⊂ C – A.

Let x ∈ C−B ⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [ ∵ A ⊂ B ]
⇒ x ∈ C − A
Hence, C−B ⊂ C−A

Q.58 Is it true that for any sets A and B, P (A )∪ P (B) = P (A ∪ B)? Justify your answer.

False,
Let A = {1, 2, 3} and B = {3, 4}
then A∪B = {1, 2, 3, 4}
P(A) = {Φ, {1}, {2}, {3},{1,2},{2, 3},{3, 1},{1, 2, 3}}
P(B) = {Φ, {3}, {4}, {3,4}}
P(A∪B) = {Φ, {1}, {2}, {3}, {4}, {1,2}, {1, 3}, {2, 3},
{2, 4}, {3, 4},{4, 1}, {1, 2, 3}, {2, 3, 4},
{3, 4, 1}, {4, 1, 2}, {1, 2, 3, 4}}
And
P(A) ∪ P(B) = {Φ, {1}, {2}, {3}, {4}, {1,2},{2, 3},
{3, 1}, {3, 4}, {1, 2, 3}}
In this way we can see that
P(A) ∪ P(B) ≠ P(A∪B)

Q.59 Show that A ∩ B = A ∩ C need not imply B = C.

Let A = {1, 2, 3}, B = {1, 4} and C = {1, 5}
Then, A ∩ B = {1} and A ∩ C = {1}
Here, A ∩ B = A ∩ C = {1}
But, B ≠ C.

Q.60 Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A = B.

Given: A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X
Since,  A = A ∩ ( A ∪ X )
= A ∩ ( B ∪ X ) [ ∵ A ∪ X = B ∪ X ]
= ( A ∩ B ) ∪ ( A ∩ X )
= ( A ∩ B ) ∪ ϕ [ ∵ A ∩ X = ϕ ]
A = ( A ∩ B ) …(i)
And      B = B ∩ ( B ∪ X )
= B ∩ ( A ∪ X ) [ ∵ A ∪ X = B ∪ X ]
= ( B ∩ A ) ∪ ( B ∩ X )
= ( A ∩ B ) ∪ ϕ [ ∵ B ∩ X = ϕ ]
B = ( A ∩ B ) …(ii)
From (i) and (ii), we have A = B

Q.61 Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.

We can show these conditions on sets A, B and C by taking an example:
Let A = {1, 2, 3}, B = {1, 2, 6}, C = {6, 3, 7}.
Then, A ∩ B = {1, 2}, B ∩ C = {6} and C ∩ A = {3}
Therefore, A ∩ B, B ∩ C and C ∩ A are non-empty.
So, A ∩ B ∩ C = Φ.
Therefore, A = {1, 2, 3}, B = {1, 2, 6}, C = {6, 3, 7}.