sets · applications of set theory 1. definition of set a well-defined collection of objects, is...

The set theory was developed by German Mathematician Georg Cantor(1845-1918). He f irst encountered sets while working on ‘problems ontrigonometric series’. This concept is used in every branch of Mathematics i.e.,Geometry, Algebra, etc.

In our daily life, while performing our regular work, we often come across avariety of things that occur in groups.

e.g., Team of cricket players, group of tall boys, group of teachers, etc.

The words used above like team, group, etc., convey the idea of certaincollections.

Well-defined Collection of Objects

If any given collection of objects is in such a way that it is possible to tellwithout any doubt whether a given object belongs to this collection or not,then such a collection of objects is called a well-defined collection of objects.e.g., ‘The rivers of India’ is a well-defined collection. Since, we can say that theriver Nile does not belong to this collection. On the other hand, the riverGanga does belong to this collection.

Difference between Not Well-defined Collection and Well-defined Collection

Not Well-defined Collection Well-defined Collection

A group of intelligent students. A group of students scoring more than 95% marks ofyour school.

A group of most talented writersof India.

A group of odd natural numbers less than 25.

Group of pretty girls. Group of girls of class XI of your school.

Here, a group of intelligent students, a group of most talented writers of India,group of pretty girls are not well-defined collections, because we can notdecide whether a given particular object belongs to the given collection or not.

Sets

1

1 Sets Their Typesand

3 Sets and Their Types

3 Subsets of Set

3 Venn Diagrams and Operations

on Sets

3 Applications of Set Theory

1

Definition of Set

A well-defined collection of objects, is called a set.Sets are usually denoted by the capital letters A, B, C, X, Yand Z etc.

The elements of a set are represented by small letters a, b, c,x, y and z etc.

If a is an element of a set A, then we say that a belongs to A.The word belongs to denoted by the Greek symbol∈ (epsilon).

Thus, in notation form, a belongs to set A is written asa A∈ and b does not belongs to set A is written as b A∉ .e.g.,

(i) 6 ∈ N N, being the set of natural numbers and0 ∉ N .

(ii) 36 ∈ P P, being the set of perfect square numbers, so5 ∉ P.

* Some examples of sets used particularly in Mathematics are

N → The set of natural numbers. {1, 2, 3, …}

Z → The set of all integers. {..., − − −3 2 1 0 1 2 3, , , , , , , ... }

Q → The set of all rational numbers.

... , , , , , , , ...− − −

3

4

2

3

1

2

1

2

2

3

3

4

R → The set of real numbers.

(rational and irrational numbers)

Z+→ The set of positive integers. {1, 2, 3, … }

Q+→ The set of positive rational numbers.

1

2

2

3

3

4

4

5, , , , ...

R+→ The set of positive real numbers.

(positive rational and irrational numbers)

Note Objects, elements and members of a set are synonymous terms.

Example 1. Which of the following are sets? Justify youranswer.

(i) The collection of all the months of a year beginning withthe letter J.

(ii) The collection of ten most talented writers of India.

(iii) A collection of novels written by the writer Munshi PremChand.

(iv) A collection of most dangerous animals of the world.

Solution

(i) We are sure that members of this collection are January, Juneand July.

So, this collection is well-defined. Hence, it is a set.

(ii) A writer may be most talented for one person and may not befor other. Therefore, we cannot definitely decide which writerwill be there in the collection.So, this collection is not well-defined. Hence, it is not a set.

(iii) Here, we can definitely decide whether a given novelbelongs to this collection or not.

So, this collection is well-defined. Hence, it is a set.

(iv) The term most dangerous is not a well-defined term. Ananimal may be most dangerous for one person and maynot be for the other.

So, it is not well-defined. Hence, it is not a set.

Representation of Sets

Sets are generally represented by following two ways

1. Roster or Tabular Form or Listing Method

2. Set-builder Form or Rule Method

1. Roster or Tabular Form or Listing Method

In this form, all the elements of a set are listed, theelements are being separated by commas and areenclosed within curly braces { }.e.g.,

(i) The set of all natural numbers less than 10 isrepresented in roster form as {1, 2, 3, 4, 5, 6, 7,8, 9}.

(ii) The set of prime natural numbers is { , , , , }2 3 5 7 K .Here, three dots tell us that the list of primenatural numbers continue indefinitely.

Note (i) In roster form, order in which the elements are listed is notimportant.Hence, the set of natural numbers less than 10 can also bewritten as {2, 4, 1, 3, 5, 6, 8, 7, 9} instead of {1, 2, 3, 4, 5,6, 7, 8, 9}.

Here, the order of listing elements is not important.

(ii) In roster form, element is not repeated i.e., all the elementsare taken as distinct.Hence, the set of letters forming the word‘MISCELLANEOUS’ is {M, I, S, C, E, L, A, N, O, U}.

2. Set-builder Form or Rule Method

In this form, all the elements of set possess a singlecommon property p x( ) , which is not possessed by anyother element outside the set.In such a case, the set is described by { : ( )x p x holds}.e.g.,In the set {a e i o u, , , , }, all the elements possess acommon property, i.e., each of them is a vowel of theEnglish alphabet and no other letter possesses thisproperty.If we denote the set of vowels by V, then we write

V = {x x: is a vowel in the English alphabet}The above description of the set V is read as ‘‘theset of all x such that x is a vowel of the Englishalphabet.’’

2 Mathematics Class 11thll ne

2

Some examples are given below

(i) Set of all natural numbers less than 10,

A x x N x= ∈ <{ : , }10

(ii) Set of all real numbers,

R x x R= ∈{ : }

* Working Rule to Write the Set in the

Set-builder Form

To convert the given set in set-builder form, we use the followingsteps

Step I Describe the elements of the set by using a symbol x or anyother symbol y z, etc.

Step II Write the symbol colon ‘:’.

Step III After the sign of colon, write the characteristic propertypossessed by the elements of the set.

Step IV Enclose the whole description within braces i.e., { }.

Worked out Problem

Write the following set A = {14, 21, 28, 35, 42, ..., 98}

in set-builder form.

Step I Describe the elements of the set by using a symbol.Let x represent the elements of given set.

Step II Write the symbol colon.Write the symbol ‘:’ after x.

Step III Find the characteristic property possessed by theelements of the set.Given numbers are all natural numbers greater than 7 which aremultiples of 7 and less than 100.

Step IV Enclose the whole description within braces.Thus, A = {x : x is a set of natural numbers greater than 7 whichare multiples of 7 and less than 100}.Which is the required set- builder form of given set.

Representation of a Statement in Both Form

Statement Roster form Set- builder form

The set of allnatural numbersbetween 10and 14.

{11, 12, 13} { : , }x x N x∈ < <10 14

The set of allschools in Delhibeginning withletter D.

{DAV School, DPS School,Decent School, ...}

{x x: is name of schoolin Delhi beginning withletter D}

The set of alldistinct lettersused in the word‘Friend’.

{F, r, i, e, n, d} {x x: is the distinctletters used in the word‘Friend’}

Example 2. Describe the set

(i) “The set of all vowels in the word EQUATION” inRoster form.

(ii) “The set of reciprocals of natural numbers” inset-builder form.

Solution

(i) The word “EQUATION” has following vowels i.e., A, E,I, O and U.

Hence, the required set can be described in roster form as{A, E, I, O, U}.

(ii) Given set can be described in set-builder form as { :x xis a reciprocal of a natural number}

or x xn

n N: ,= ∈

1

Example 3. Describe the following set in Rosterform.

{x : x is positive integer and a divisor of 9}

Solution Here, x is a positive integer and a divisor of 9.

So, x can take values 1, 3, 9.

∴ {x : x is a positive integer and a divisor of 9} = { , , }1 3 9

Example 4. Write the set of all natural numbers x suchthat 4x + 9 < 50 in roster form.

/ Firstly, simplify the inequality and then listed all the naturalnumbers under given condition.

Solution We have, 4 9 50x + <⇒ 4 9 9 50 9x + − < −

[subtracting 9 from both sides]

⇒ 4 41x < ⇒ x < 41

4

∴ x <1 50.2

Since, x is a natural number, so x can take values 1, 2, 3,4, 5, 6, 7, 8, 9, 10.

∴ Required set = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Example 5. Describe the following set in set-builderform.

B = { , , , , , , , , , }53 59 61 67 71 73 79 83 89 97

/ Firstly, find the characteristic property possessed by theelements of the set, then write in set-builder form.

Solution The given set is

{53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

We observe that these numbers are all prime numbersbetween 50 and 100.

∴ Given set, B x x= { : is a prime number and

50 100< <x }

Sets 3ll ne

3

Types of Sets

1. Finite Set

A set, which is empty or consists of a definitenumber of elements, is called a finite set.e.g.,

(i) The set {1, 2, 3, 4} is a finite set, because itcontains a definite number of elementsi.e., only 4 elements.

(ii) The set of solutions of x 2 25= is a finiteset, because it contains a definite numberof elements i.e., 5 and − 5.

(iii) An empty set, which does not contain anyelement is also a finite set.

The number of distinct elements in a finite set Ais called cardinal number of set and it is denotedby n A( ).

e.g., If A = − −{ , , , , }3 1 8 10 13 , then n A( ) = 5.

2. Infinite Set

A set which consists of infinite number ofelements is called an infinite set.

When we represent an infinite set in the rosterform, it is not possible to write all the elementswithin braces { } because the number of elementsof such a set is not finite, so we write a fewelements which clearly indicate the structure ofthe set following by three dots.

e.g., Set of squares of natural numbers is aninfinite set, because such natural numbers areinfinite and it can be represented as{ , , , ...}4 9 16 25 .

Note All infinite sets cannot be described in the rosterform. For example, the set of real numbers cannot bedescribed in this form, because the elements of thisset do not follow any particular pattern.

3. Empty Set

A set, which does not contain any element, iscalled an empty set or null set or void set. It isdenoted by φor { }.

e.g., A x x= { : is a natural number less than 1}

We know that, there is no natural number lessthan one. Therefore, set A contains no elementand hence it is an empty set.

4. Singleton Set

A set, consisting of a single element, is called a singleton set.

e.g., (i) The sets {0}, {5}, {− 7} are singleton sets.

(ii) A x x x Z= + = ∈{ : , }8 0 is a singleton set, because thisset contains only one integer, namely − 8.

5. Equivalent SetsTwo sets A and B are equivalent, if their cardinal numbers aresame i.e., n A n B( ) ( )= .e.g., Let A a b c d= { , , , } and B = { , , , },1 2 3 4 then n A( ) = 4 andn B( ) = 4.

Therefore, A and B are equivalent sets.

6. Equal SetsTwo sets A and B are said to be equal, if they have exactly the sameelements and we write A B= . Otherwise, two sets are said to be

unequal and we write A B≠ .

e.g., Let A a b c d= { , , , } and B c d b a= { , , , }, then A B= , becauseeach element of A is in B and vice-versa.

Note A set does not change, if one or more elements of the set are repeated.e.g., The sets A ={ , , }1 4 5 and B ={ , , , , }1 1 4 5 5 are equal because elementsof A is in B and vice-versa. That’s why, we generally do not repeat anyelement in describing a set.

Example 6. Identify which of the following set is an emptyset, singleton set, infinite set or equal sets.

(i) A x x= { : is a girl being living on the Jupiter}

(ii) B = {x : x is a letter in the word ‘‘MARS’’}

(iii) C y y= { : is a letter in the word ‘‘ARMS’’}

(iv) D x x x Q= − = ∈{ : , }3 2 0

(v) E x x N= ∈{ : and x is an odd number}

Solution

(i) A x x= { : is a girl being living on the Jupiter}

We know that, there is no human being or any girl being living onthe Jupiter. Hence, A is an empty set.

(ii) B x x= { : is a letter in the word ‘‘MARS’’}

⇒ B = {M, A, R, S}

(iii) C y y= { : is a letter in the word ‘‘ARMS’’}

⇒ C = {A, R, M, S}

Here, we observe that the elements of sets B and C are exactlysame, hence these sets are equal.

(iv) D x x x Q= − = ∈{ : , }3 2 0

⇒ D =

2

3Q 3 2 0

2

3x x Q− = ⇒ = ∈

Hence, D is a singleton set.

(v) E x x N= ∈{ : and x is an odd number}Clearly, it is an infinite set because there are infinite naturalnumbers which are odd.


4

Example 7. Which of the following pairs of sets areequal? Justify your answer.

(i) A x x= { : is a letter of the word ‘‘LOYAL’’},

B x x= { : is a letter of the word ‘‘ALLOY’’}

(ii) A x x Z= ∈{ : and x2 8< },

B x x R= ∈{ : and x x2 4 3 0− + = }

/ Firstly, describe the given sets in the roster form and checkwhether they have exactly same elements.

Solution

(i) Given, A x x= { : is a letter in the word ‘‘LOYAL’’}

= {L, O, Y, A} = {A, L, O, Y}

and B x x= { : is a letter of the word ‘‘ALLOY’’}

= {A, L, O, Y}

Here, we see that both sets have exactly the same elements.

∴ A B=

(ii) Given, A x x Z= ∈{ : and x 2 8< }

= − −{ , , , , }2 1 0 1 2

[Qx 2 8< ⇒ − < <2 2 2 2x and as x Z∈

∴ ∈ − −x { , , , , }2 1 0 1 2 ]

and B x x R= ∈{ : and x x2 4 3 0− + = } = { , }1 3

[Qx x2 4 3 0− + = ⇒ ( )( )x x− − =1 3 0 ⇒ x =1 3, ]

Here, we see that set A has 5 distinct elements and set B has2 distinct elements. So, they do not have same elements.

∴ A B≠

Sets 5ll ne

Very Short Answer Type Questions [1 Mark each]

1. If A = { , , , , , , , }1 3 5 7 9 11 13 15 , then insert the

appropriate symbol ∈ or ∉ in each of the followingblank spaces.(i) 1... A (ii) 6 ... A

(iii) 9 ... A (iv) 14 ... A

Sol. Given set is A = { , , , , , , , }1 3 5 7 9 11 13 15

(i) 1 ∈ A

(ii) 6 ∉ A

(iii) 9 ∈ A

(iv) 14 ∉ A [1]

2. Write {x x: is an integer and − ≤ <3 7x } in roster

form.

Sol. { , , , , , , , , , }− − −3 2 1 0 1 2 3 4 5 6 [1]

3. Write set A = { , , , , }3 6 9 12 15 in set-builder form.

Sol. A = {x x: is a natural number multiple of 3 and x <18}. [1]

4. Write set A={1, 4, 9, ..., 100} in set-builder form.

Sol. A x x n n N n= = ∈ <{ : , }2 11and [1]

5. Write set D =

1

2

2

5

3

10

4

17

5

26

6

37

7

50, , , , , , in

set-builder form.

Sol. D x xn

nn N n= =

+∈ ≤

: ,2 1

7and [1]

6. Use listing method to express the set

A = {x x n: ,= 3 n N∈ and x < 80}.

Sol. A = { , , , }1 8 27 64 [1]

7. List the elements of the set

A x x x= − < <

: is an integer,1

2

9

2

Sol. A = { , , , , }0 1 2 3 4 [1]

8. Express the set

D x xn

nn N= = −

+∈

: ,2

2

1

1and n < 4

in roster form.

Sol. D =

03

5

4

5, , [1]

9. Describe

(i) “The set of vowels in the wordMATHEMATICS” in roster form.

(ii) “The set of all odd natural numbers’’ in setbuilder form.

Sol. (i) The word “MATHEMATICS” has following vowelsi.e., A, E, I.

Hence, the required set can be described in roster formas {A, E, I}. [1/2]

(ii) An odd natural number can be written in the form( )2 1n − .

Hence, given set can be described in set-builder form as

{ : , }x x n n N= − ∈2 1 [1/2]

EXAM

5

Short Answer Type Questions [4 Marks each]

10. Which of the following sets are empty?

(i) A = {x x N: ∈ and x ≤ 1}

(ii) B = {x x x N: ,3 1 0+ = ∈ }

(iii) C = {x x:2 3< < , x N∈ }

/ A set which does not contain any element is called theempty set.

Sol. (i) A = {x x N: ∈ and x ≤1}

Since, x ≤1i.e., x <1and x =1, which is a natural number.

∴ A = { }1

So, this is not an empty set. [1]

(ii) B = {x x x N: ,3 1 0+ = ∈ }

Since, 3 1 0x + =

⇒ x N= − ∉1

3

Thus, set B does not contain any element.

Hence, set B is an empty set. [ ]112

(iii) C = {x x: 2 3< < , x N∈ }

Since, there is no natural number between 2 and 3.

Hence, set C is an empty set. [ ]112

11. Let T xx

x

x

x= +

−− = −

−

:5

75

4 40

13.

Is T an empty set? Justify your answer.

Sol. The given set T is not an empty set.

Justification

Qx

x

x

x

+−

− = −−

5

75

4 40

13

⇒ x

x

x

x

+−

− = −−

5

7

5

1

4 40

13[1]

⇒ x x

x

x

x

+ − +−

= −−

5 5 35

7

4 40

13

⇒ 40 4

7

4 40

13

−−

= −−

x

x

x

x[1]

⇒ − −−

− −−

=( )4 40

7

4 40

130

x

x

x

x

⇒ ( )( ) ( )

4 401

13

1

70x

x x−

−+

−

= [1]

⇒ ( )( ) ( )

4 406

13 70x

x x−

− −

=

⇒ 4 40 0x − =∴ x =10

Hence, T is not an empty set. [1]

12. Which of the following sets are finite and whichare infinite?

(i) The set of the months of a year.

(ii) {1, 2, 3, ...}

(iii) The set of prime numbers less than 99.

Sol. (i) It is a finite set, as there are 12 members in the set whichare months of the year. [1]

(ii) It is an infinite set, since there are infinite natural numbers.

[ ]112

(iii) It is a finite set, because the set is {2, 3, 5, 7, ..., 97}. [ ]112

13. From the following sets, select equal sets.

A = { , , , }2 4 6 8 , B = { , , , , }1 2 3 4 5 ,

C = −{ , , , }2 4 6 8 , D = { , , , , }2 3 5 4 1 , E = { , , , }8 6 2 4

/ Two sets are said to be equal, if they have exactly the sameelements.

Sol. A = { , , , }2 4 6 8 and E = { , , , }8 6 2 4

Since, each element of set A presents in set E and vice-versa.[1]

Therefore, A and E are equal sets. [1]

Similarly, B = { , , , , }1 2 3 4 5 and D = { , , , , }2 3 5 4 1Since, each element of set B presents in set D and vice-versa.

[1]

Therefore, B and D are also equal sets. [1]

14. Are the following pair of sets equal? Give reason.

(i) A = { , }2 3 and B = {x x: is a solution of

x x2 5 6 0+ + = }

(ii) A = {x x: is a letter in the word ‘‘FOLLOW’’}

and B = {y y: is a letter in the word ‘‘WOLF’’}

/ Firstly, convert the given sets in roster form and thencheck whether they have exactly the same elements.

Sol. (i) Here, A = { , }2 3

and B = {x x: is a solution of x x2 5 6 0+ + = }First, we find the solution of x x2 5 6 0+ + = .

Now, x x x2 3 2 6 0+ + + =⇒ x x x( ) ( )+ + + =3 2 3 0⇒ ( )( )x x+ + =2 3 0 ⇒ x = − −2 3,∴ B = − −{ , }2 3 [1]

Since, the elements of A and B are not same, thereforeA B≠ . [1]

(ii) Here, A = {x x: is a letter of the word ‘‘FOLLOW’’}= {F, O, L, W}

and B y y= { : is a letter of the word ‘‘WOLF’’}

= {W, O, L, F} [1]

Since, every element of A is in B and every element of Bis in A i.e., both have exactly same elements.

∴ A B= [1]


6

1 Mark Questions

1. Write set A = {x x: is an integer and x2 20< } in the roster form.

2. Write set A = {x x: is a planet} in the roster form.

3. Which of the following is/are the examples of an empty set?

(i) Set of all even prime numbers.

(ii) {x x: is a point common to any two parallel lines}

4. Write the following sets in the roster form.

(i) B x x x x R= = ∈{ : , }2

(ii) C = {x x: is a positive factor of a prime number p}

5. Write the following sets in the set-builder form.

(i) { , , , }5 25 125 625

(ii) { , , , , }2 4 6 8 10

4 Marks Questions

6. From the sets given below, pair the equivalent sets.

A = { , , },1 2 3 B t p q r s= { , , , , }, C = { , , }α β γ and D a e i o u= { , , , , }

7. From the sets given below, select empty set, singleton set, infinite set and equal sets.

(i) A = {x x: < 1and x > 3} (ii) B x x x R= − = ∈{ : , }3 1 0

(iii) C = {x x N: ∈ and x is a prime number} (iv) D = { , , , , }2 4 6 8 10

(v) E = {x : x is positive even integer and x ≤ 10}

8. From the sets given below, select equal sets and equivalent sets.

A a= { , }0 , B = { , , , }1 2 3 4 , C = { , , }4 8 12 ,

D = { , , , }3 1 2 4 , E = { , }1 0 , F = { , , }8 4 12 ,

G = { , , , }1 5 7 11 and H a b= { , }

9. State which of the following statements are true and which are false? Justify your answer.

(i) 35 ∈ {x x: has exactly four positive factors}

(ii) 128 ∈ {y : the sum of all the positive factors of y is 2y}

(iii) 3 5 24 3 2∉ − +{ :x x x x − + =112 6 0x }

(iv) 496 ∉ {y : the sum of all the positive factors of y is 2y}

ll ne

YOURSELFTRY

7

SubsetLet A and B be two sets. If every element of A is anelement of B, then A is called a subset of B.

If A is a subset of B, then we write A B⊂ , which isread as ‘‘A is a subset of B ” or A is contained in B.

In other words, A B⊆ , if whenever a A∈ , thena B∈ . It is often convenient to use the symbol‘‘⇒ ’’ which means implies. Using this symbol, wecan write the definition of subset as follows

A B⊂ , if x A∈⇒ x B∈The above statement is read asA is subset of B, if x is an element of A, then it impliesthat x is also an element of B.

If A is not a subset of B, then we write A B/⊆ .

If A B⊂ , then B is called a superset of A written as

B A⊃ .

e.g., Consider the sets A and B, where set A denotesthe set of all students in your class, B denotes theset of all students in your school. We observe that,every element of A is also an element of B.Therefore, we can say that A is subset of B i.e.,A B⊂ .

If it happens for both sets A and B, i.e., everyelement of A is in B and every element of B is in A,then in this case, A and B are same sets. Thus wehave,

A B⊆ and B A⊆ ⇔ =A B, where ‘‘⇔’’ is asymbol for two ways implications and is usuallyread as if and only if (iff ).

Note (i) { } { , , }1 1 2 3⊂(ii) 2 1 2 3⊂ { , , }, which is not possible.

Proper Subset

If A B⊆ and A B≠ , then A is called a propersubset of B, written as A B⊂ and B is called propersuperset of A.

e.g., Let A = {x x: is an even natural number}

and B = { :x x is a natural number}

Then, A = {2, 4, 6, 8,...}

and B = {1, 2, 3, 4, 5,...}

∴ A B⊂

❖ Some Important Results1. Every set is a subset of itself.

Proof Let A be any set. Then, each element of A is clearly in A. Hence,

A A⊆ .

2. The empty set φis a subset of every set.

Proof Let Abe any set and φbe the empty set. To show that φ ⊆ A, i.e., to

show that every element of φis an element of Aalso. But we know that

empty set φcontains no element. So, every element of φis in A. Hence,

φ ⊆ A.

3. A set itself and an empty set are always subsets of every set and set itself

is called improper subset of the set.

4. The total number of subsets and proper subset of a finite set containing n

elements is 2n and 2 1n − , respectively.

5. If A B⊆ andB C⊆ , then A C⊆ .

6. A B= , if and only if A B⊆ andB A⊆ .

Example 1. 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) φ ⊂ A (ix) φ ∈ A

(x) { }φ ⊂ A

Solution We have, A = {1, 2, {3, 4}, 5}

(i) Since, {3, 4} is a member of set A.

∴ {3, 4} ∈ AHence, {3, 4} ⊂ A is incorrect.

(ii) Since, {3, 4} is a member of set A.

Hence, {3, 4} ∈ A is correct.

(iii) Since, {3, 4} is a member of set A.

So, {{ , }}3 4 is a subset of A.

Hence, {{ , }}3 4 ⊂ A is correct.

(iv) Since, 1 is a member of A.

Hence, 1 ∈ A is correct.

(v) Since, 1 is a member of set A.

Hence, 1 ⊂ A is incorrect.

(vi) Since, 1, 2, 5 are members of set A.

So, {1, 2, 5} is a subset of set A.

Hence, {1, 2, 5} ⊂ A is correct.

(vii) Since, 1, 2 and 5 are members of set A.

So, {1, 2, 5} is a subset of A.

Hence, {1, 2, 5} ∈ A is incorrect.

(viii) Since, φis subset of every set. Hence, φ ⊂ A is correct.

(ix) Since, φis not a member of set A.

Hence, φ ∈ A is incorrect.

(x) Since, φis not a member of set A.

Hence, { }φ ⊂ A is incorrect.

ll ne

2 Subsets of Set

8

Subsets of the Set of RealNumbers

Rational numbers and irrational numbers taken together,are known as real numbers. Thus, every real number iseither a rational or an irrational number. The set of realnumbers is denoted by R.There are many important subsets of set of real numbers whichare given below

1. Natural Numbers

The numbers being used in counting as 1, 2, 3, 4,...,called natural numbers.

The set of natural numbers is denoted by N.

N = {1, 2, 3, 4, 5, 6, ...}

2. Whole Numbers

The natural numbers along with number 0 (zero) formthe set of whole numbers i.e., 0, 1, 2, 3, ... , are wholenumbers. The set of whole numbers is denoted by W.

W = {0, 1, 2, 3, ...}Set of natural numbers is the proper subset of the set ofwhole numbers.

⇒ N W⊂

3. Integers

The natural numbers, their negatives and zero make theset of integers and it is denoted by Z.

Z = − − − − −{..., , , , , , , , , , , ...}5 4 3 2 1 0 1 2 3 4

Set of whole numbers is the proper subset of integers.

⇒ W Z⊂

4. Rational Numbers

A number of the formp

q, where p and q both are integers

and q ≠ 0, is called a rational number (division by 0 is not

permissible).

The set of rational numbers is generally denoted by Q.

Qp

qp q Z q= ∈ ≠

: , and 0

All the whole numbers are also rational numbers, sincethey can be represented as the ratio.

e.g., 33

1

6

2

18

6= , , , etc.

The set of integers is the proper subset of the set ofrational numbers i.e., Z Q⊂ and N Z Q⊂ ⊂ .

5. Irrational Numbers

A number which cannot be written in the form p/q, wherep and q both are integers and q ≠ 0, is called an irrationalnumber i.e., a number which is not rational is called anirrational number.

The set of irrational numbers is denoted by T.

T x x R x Q= ∈ ∉{ : }and

e.g., 0.535335333..., 2, 3 are irrational numbers.

Above subsets can be represented diagramatically as givenbelow

Intervals as Subsets of R

Let a, b belongs to R and a b< . Then, the set of realnumbers { : }x a x b< < is called an open interval and is

denoted by ( , )a b .

All the real numbers between a and b belongs to the openinterval (a, b) but a and b do not belong to this set(interval).

The interval which contains the first and last members ofthe set, is called closed interval and it is denoted by [a, b].

Here, [ , ] { : }a b x a x b= ≤ ≤

Semi-open or Semi-closed IntervalSome intervals are closed at one end and open at theother, such intervals are called semi-closed or semi-openintervals.

[ , ) { : }a b x a x b= ≤ < is an open interval from a to bwhich includes a but excludes b.

( , ] { : }a b x a x b= < ≤ is an open interval from a to bwhich excludes a but includes b.

e.g.,

(i) (2, 8) is a subset of ( , )−1 11 .

(ii) [4, 6) is a subset of [4, 6].

(iii) The set [ , )0 ∞ defines the set of non-negative realnumbers.

(iv) The set ( , )− ∞ 0 defines the set of negative realnumbers.

(v) ( , )−∞ ∞ is the set of real numbers.

Sets 9ll ne

Real number

Natural

number

Whole

number

Non-negative

rational

number

Irrational

number

Negative rational

number

9

On real line, we can draw the interval, which is shown by the dark portion onthe number line

Length of an interval The number ( )b a− is called the length of

any of the intervals ( , )a b , [ , ]a b , [ , )a b or ( , ]a b .

Example 2. (a) Write the following as intervals

(i) { : , }x x R x∈ − < ≤5 6 (ii) { : , }x x R x∈ − < < −11 9

(b) Write the following as intervals and also represent on the numberline.

(i){ : , }x x R x∈ ≤ <2 8 (ii) { : , }x x R x∈ ≤ ≤5 6

/ If an inequality is of the form ≤ or ≥, then we use the symbol of closedinterval, otherwise we use the symbol of open interval.

Solution

(a) (i) { : , }x x R x∈ − < ≤5 6 is the set that does not contain −5 butcontains 6. So, it can be written as an interval whose first end point isopen and last end point is closed i.e., ( , ]−5 6 .

(ii) { : , }x x R x∈ − < < −11 9 is the set that neither contains−11nor −9,so it can be represented as open interval i.e., ( , )− −11 9 .

(b) (i) { : , }x x R x∈ ≤ <2 8 is the set that contains 2 but not contain 8.So, itcan be represented as an interval whose first end point is closed andthe other end point is open i.e., [2, 8).

On the real line [2, 8) may be graphed as shown in figure givenbelow

The dark portion on the number line is the required set.

(ii) { : , }x x R x∈ ≤ ≤5 6 is the set which contains 5 and 6 both. So, it isequivalent to a closed interval i.e., [5, 6].

On the real line [5, 6] may be graphed as shown in the figure givenbelow

The dark portion on the number line is the required set.

Universal SetIf there are some sets under consideration, then there happens to be aset which is a superset of each one of the given sets. Such a set isknown as the universal set and it is denoted by U.

e.g.,

(i) Let A = {2, 4, 6}, B = {1, 3, 5} andC = {0, 7}

Then,U = {0, 1, 2, 3, 4, 5, 6, 7} is an universal set.

(ii) For the set of all integers, the universal set can be the set ofrational numbers or the set of real numbers.

Example 3. What universal sets would youpropose for each of the following?

(i) The set of right triangles.

(ii) The set of isosceles triangles.

Solution

(i) The universal set for the set of right trianglesis set of triangles.

(ii) The universal set for the set of isoscelestriangle is set of equilateral triangles or set oftriangles.

Power SetThe collection of all subsets of a set A, iscalled the power set of A and it is denoted byP(A). In P(A), every element is a set.e.g., Let A = {3, 4, 5}

Then, P A( ) = {φ, {3}, {4}, {5}, {3, 4}, {3, 5},

{4, 5}, {3, 4, 5}}

Note (i) If set A has n elements, then the total numberof elements in its power set is 2n .

(ii) If A is an empty set φ, then P A( ) has just oneelement i.e., P A( ) { }= φ .

Properties of Power Sets

(i) Each element of a power set is a set.

(ii) If A B⊆ , then P A P B( ) ( )⊆ .

(iii) P A P B P A B( ) ( ) ( )∩ = ∩(iv) P A B P A P B( ) ( ) ( )∪ ≠ ∪

* Method to Write the Power Set of a

Given Set

Let a set having n elements is given, then forwriting its power set, we use the following steps

Step I Write all the possible subsets having singleelement of given set.

Step II Write all the possible subsets having twoelements at a time of given set.

Step III Write all the possible subsets having threeelements at a time of given set. Repeat thisprocess for writing all possible subsetshaving n elements at a time, as the given sethas n elements.

Step IV Form a set, with the help of the subsetsobtained from steps I, II and III andelement φ.This set will give the power set of given set.


a b

( )a, b

a b

[ ]a, b

a b

[ )a, b

a b

( ]a, b

0 2 8

0 5 6

10

Worked out Problem

If A ={ , , }1 2 3 then find the power set of A.

Step I Write all the possible subsets having singleelement of given set.

All possible subsets of a given set having singleelement are { }, { }, { }1 2 3 .

Step II Write all the possible subsets of a given sethaving two elements at a time.

All possible subsets of a given set having twoelements at a time are {1, 2}, {2, 3}, {3, 1}.

Step III Write all the possible subsets of a given sethaving three elements at a time.

All possible subsets of a given set having threeelements at a time is {1, 2, 3}.

Step IV Form a set with the help of the subsetsobtained from steps I, II and III and element φ.Hence, the required power set is

{{1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3}, φ}.

Example 4. Let A = { , , , }1 2 3 4 , B = { , , }1 2 3 and C = { , }2 4 .Find all sets X satisfying each pair of conditions

(i) X B⊆ and X C/⊆ (ii) X B⊆ , X B≠ and X C/⊆Sol. Given, A = { , , , }1 2 3 4 , B = { , , }1 2 3 and C = { , }2 4

Now, P A( ) { , { },= φ 1 { }, { },2 3 { },4 { , },1 2 { , }1 3 , { , }1 4 ,{ , }2 3 , { , }2 4 , { , }, { , , }3 4 1 2 3 , { , , }1 2 4 , { , , }1 3 4 , { , , }2 3 4 , { , , , }1 2 3 4 }

...(i)

P B( ) { , { }, { }, { }, { , }= φ 1 2 3 1 2 , { , }1 3 , { , }2 3 , { , , }1 2 3 } ...(ii)

and P C( ) { , { }, { }, { , }}= φ 2 4 2 4 ...(iii)

(i) Now, X B⊆ and X C/⊆⇒ X P B∈ ( )and X P C∉ ( )

From Eqs. (ii) and (iii), we get

X = { }, { }, { , }, { , }, { , }1 3 1 2 1 3 2 3 , { , , }1 2 3

(ii) Now, X B⊆ , X B≠ and X C/⊆⇒ X P B∈ ( ), X B≠ , X P C∉ ( )

From Eqs. (ii) and (iii), we get

X = { }, { }, { , }, { , }, { , }1 3 1 2 1 3 2 3

Here, X does not contain B = { , , }1 2 3 as X B≠ .

Sets 11ll ne


1. Consider the following sets φ, A = {1, 2}

and B = {1, 4, 8}

Insert the following symbols ⊂ or ⊄between each of the following pair of sets.

(i) φ... B (ii) A B...

Sol. (i) Since, null set is proper subset of every set.

∴ φ ⊂ B [1/2]

(ii) Given, A = {1, 2} and B = { , , }1 4 8 . Since,element 2 ∉ B.

∴ A B⊄ [1/2]

2. Prove that A ⊆ φ implies A = φ.

Sol. Given, A ⊆ φ ...(i)

But φ ⊆ A [empty set is a subset of each set] …(ii)[1/2]

From Eqs. (i) and (ii), we get

A = φ [1/2]

3. Write the following subset of R as interval.Also find the length of interval andrepresent on number line.

{ : , }x x R x∈ − ≤ ≤ −12 10

/ If inequalities are of the form ≥ or ≤, then use the symbol of closedinterval and then find the length of the interval, which is equal to thedifference of its extreme values.

Sol. { : , } [ , ]x x R x∈ − ≤ ≤ − = − −12 10 12 10

and length of interval = − − − =10 12 2( )

On the real line set [− −12 10, ] may be graphed as shown in figuregiven below

The dark portion on the number line is the required set. [1]

4. Let A = {a, b, {c, d}, e}. Which of the following

statements is/are true?

(i) {c, d} ∈ A (ii) {{c, d }} ⊂ A

Sol. Given, A = {a, b, {c, d}, e}

(i) Since a, b, {c, d} and e are elements of A.

∴ {c, d} ∈ A

Hence, it is a true statement. [1/2]

(ii) As{ , }c d A∈ and{{ , }}c d represents a set, which is a subset of A.

∴ {{c, d}} ⊂ A

Hence, it is a true statement. [1/2]

EXAM

0–12 –10

11

5. Write down the subsets of the following sets.

(i) {1, 2, 3} (ii) {φ}

Sol. (i) The subsets of {1, 2, 3} are

φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}. [1/2]

(ii) Clearly, { }φ is the power set of empty set φ. Now, itssubsets are φand {φ}. [1/2]

6. Let A = {1, 3, 5} and B = {x x: is an odd natural

number < 6}.

(i) Is A B⊆ ? (ii) Is A B= ?

Sol. Given, A = {1, 3, 5}

and B = {x x: is an odd natural number < 6}⇒ B = {1, 3, 5}

(i) Yes, here A B⊆ ,because all elements of set A arepresent in set B. [1/2]

(ii) Yes, here, B A⊆ , because all elements of set B arepresent in set A.

Hence, A B= , because both sets contain equal and sameelements. [1/2]

7. Write down the power set of the following sets.

(i) B = {0, 1, 3} (ii) C = {1, {2}}

Sol. (i) Given, B = {0, 1, 3}

∴ P B( ) = {φ, {0}, {1}, {3}, {0, 1}, {0, 3}, {1, 3}, {0, 1, 3}}

[1/2]

(ii) Given, C = { , { }}1 2

P C( ) = {φ, {1}, {{2}}, {1, {2}}} [1/2]

8. In each of the following, determine whether thestatement is true or false. If it is true, then prove it.If it is false, then give an example.

(i) If x A∈ and A B∈ , then x B∈ .

(ii) If A B⊂ and B C∈ , then A C∈ .

Sol. (i) False,Let A = { }2 , B = {{2}, 3}

Clearly, 2 ∈ A and A B∈ , but 2 ∉ B.

So, x A∈ and A B∈ need not imply that x B∈ . [1/2]

(ii) False,Let A = { }2 , B = { , }2 3 and C = {{2, 3}, 4}

Clearly, A B⊂ and B C∈ , but A C∉ .

Thus, A B⊂ and B C∈ need not imply that A C∈ .[1/2]

9. Given, the sets A = {1, 3, 5}, B = {2, 4, 6} and

C = {0, 2, 4, 6, 8}. Which of the following may beconsidered as universal set(s) for all three sets A,B and C?

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

Sol. We know that, universal set for sets A, B and C is supersetof A, B andC i.e., universal set contains all elements of A, Band C.

(i) φcannot be considered as universal set. [1/2]

(ii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for thegiven sets A, B and C as all the elements of sets A, B andC are in this set. [1/2]



10. If A = {3, {4, 5}, 6}, then find which of the following

statements are true.

(i) { , }4 5 ⊂ A

(ii) { , }4 5 ∈ A

(iii) φ ⊂ A

(iv) { , }3 6 ⊂ A

Sol. Given, A = {3, {4, 5}, 6}

Here, 3, {4, 5}, 6 all are elements of A.

(i) { , }4 5 ⊂ A, which is not true.

Since, element of any set is not a subset of any set andhere { , }4 5 is an element of A. [1]

(ii) { , }4 5 ∈ A, is a true statement. [1]

(iii) It is always true that φ ⊂ A. [1]

(iv) {3, 6} makes a set, so it is a subset of A i.e., { , }3 6 ⊂ A.[1]

11. Let A and B be two sets. Then, prove thatA B= ⇔ A B⊆ and B A⊆ .

Sol. Given A B=To prove A B⊆ and B A⊆Proof By definition of equal sets, every element of A is in Band every element of B is in A. [1]

∴ A B⊆ and B A⊆Thus, A B= ⇒ A B⊆ and B A⊆Again, let A ⊆ B and B ⊆ A [1]

By the definition of a subset, if A B⊆ then it follows thatevery element of A is in B and if B A⊆ then it follows thatevery element of B is in A.

∴ A B= [1]

Hence, A B= ⇔ A B⊆ and B A⊆ [1]

Hence proved.

12

12. Examine whether the following statements aretrue or false.

(i) {a, b} ⊄ {b, c, a}

(ii) {a, e} ⊂ {x x: is a vowel in the Englishalphabet}

(iii) {a} ∈ {a, b, c}

(iv) {0,1,2,3,4,5,6,7,8,9,10} is the universal set forthe sets {1, 3, 5} and {2, 4, 6}.

Sol. (i) Since, the elements of the set {a, b} are also present in theset {b, c, a}.So, { , } { , , }a b b c a⊂ [1]

⇒ {a, b} ⊄ {b, c, a} is false.

(ii) Vowels in the English alphabets are a, e, i, o, u.

∴ {a, e} ⊂ {x x: is a vowel in English alphabet}is true.[1]

(iii) False, since a ∈ {a, b, c} and not {a}. [1]

(iv) True, since all elements of both sets {1, 3, 5} and

{2, 4, 6} are present in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Hence, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universalset for given two sets. [1]

13. Show that n P P P{ [ ( ( ))]}φ =4.

Sol. We have P ( ) { }φ = φ∴ P P( ( )) { , { }}φ = φ φ [1]

⇒ P P P[ ( ))] { , { }, {{ }}, { , { }}}φ = φ φ φ φ φ [1]

Hence, number of elements in P P P[ ( ( ))]φ is 4. [1]

i.e., n P P P{ [ ( ( ))]}φ =4 [1]

Hence proved.

1 Mark Questions

Directions (Q.Nos. 1 to 3) Determine whether the statement is true or false.

1. If A = {3, 6, 7}, B = {2, 3, 7, 8, 10}, then A B⊂ .

2. If A = {x x x: 2 4 21 0+ − = , x N∈ } and B = −{ , }7 3 , then A B⊆ .

3. If A B= ={ , , }, { , , }1 7 9 2 4 6 and C = { , , }0 2 4 , then U = { , , , , , , , , , }0 1 2 3 4 5 6 7 8 9 is the universal set for all three

sets.

4. Let A = {4, 5, {6}, 7}, then which of the following statements are incorrect and why?

(i) { }6 ∈ A (ii) { }6 ⊂ A

5. Find the power set of a set A = {0, 1, 2}.

6. If set A = {1, 3, 5}, then find the number of elements in P P A{ ( )}.

7. If A x x n n= = ={ : , , , }2 1 2 3 , then find the number of proper subsets.

4 Marks Questions

8. Write the following intervals in the set-builder form.

(i) (−6, 0) (ii) [3, 21)

(iii) [ , ]2 21 (iv) ( , ]− 23 5

9. Write the following as intervals.

(i) { : , }x x R x∈ − < ≤3 7 (ii) { : , }x x R x∈ − < < −11 7

(iii) { : , }x x R x∈ ≤ <0 11 (iv) { : , }x x R x∈ ≤ ≤2 9

Sets 13ll ne

YOURSELFTRY

13

Venn DiagramsVenn diagrams are named after the English logician,John Venn (1834-1883). Venn diagrams represent mostof the relationship between sets. These diagrams consistof rectangles and closed curves usually circles.

In Venn diagrams, the universal set is represented by arectangular region and its subset are represented bycircle or a closed geometrical figure inside the universalset. Also, an element of a set is represented by a pointwithin the circle of set.

e.g., IfU = { , , , , ..., }1 2 3 4 10 and A = {1, 2, 3},

then its Venn diagram is as shown in the figure

Example 1. Draw the Venn diagrams to illustrate thefollowing relationship among sets E, M and U,where E is the set of students studying English in aschool, M is the set of students studyingMathematics in the same school and U is the set ofall students in that school.(i) All the students who study Mathematics also study

English, but some students who study English do notstudy Mathematics.

(ii) Not all students study Mathematics, but everystudent studying English studies Mathematics.

/ Firstly, make a relation between sets under given condition.Then, it is easy to draw a Venn diagram.

Solution

Given, E = Set of students studying English

M = Set of students studying Mathematics

U = Set of all students

(i) Since, all of the students who study Mathematics alsostudy English, but some students who study English donot study Mathematics.

∴ M E U⊂ ⊂Through, Venn diagram we represent it as

(ii) Since, every student studying English studiesMathematics.Hence, E M U⊂ ⊂Through, Venn diagram we represent it as

Operations on SetsThere are some operations which when performed ontwo sets give rise to another set. Here, we will definecertain operations on set and examine their properties.

1. Union of Sets

Let A and B be any two sets. The union of A and B is theset of all those elements which belong to either in A or inB or in both. It is denoted by A B∪ and read as A unionB. The symbol ‘∪ ’ is used to denote the union.

∴ A B x x A x B∪ = ∈ ∈{ : }ore.g., Let A = { , }2 3 and B = { , , }3 4 5 ,

then A B∪ = { , , , }2 3 4 5

The union of sets A and B is represented by the followingVenn diagram.

The shaded portion represents A B∪ .

Some Properties of Union of Sets

(i) A B B A∪ = ∪ [commutative law]

(ii) ( ) ( )A B C A B C∪ ∪ = ∪ ∪ [associative law]

(iii) A A∪ =φ[law of identity element, φ is the identity of ∪ ]

(iv) A A A∪ = [idempotent law]

(v) U A U∪ = [law of U]

ll ne

3 Venn Diagrams Operations on Setsand

76

5

410

8

9

32

1

A

U

M

EU

E

M

U

UA

B

UA B

C

14

Example 2. Find the union of each of the followingpairs of sets.

(i) A a e i o u B a c d= ={ , , , , }, { , , }

(ii) A B= ={ , , }, { , , }1 3 5 2 4 6

(iii) A x x= { : is a natural number and 1 5< ≤x }

and B x x= { : is a natural number and 5 10< ≤x }

/ Firstly, convert the given set in roster form, if it is not givenin that. Then union of two sets, is the set which consists of allthose elements which are either in A or in B.

Solution

(i) A a, e, i, o, u , B a, c, d= ={ } { }

⇒ A B a c d e i o u∪ = { , , , , , , }

(ii) A B= ={ , , }, { , , }1 3 5 2 4 6

⇒ A B∪ = { , , , , , }1 2 3 4 5 6

(iii) A x x= { : is a natural number and 1 5< ≤x }

⇒ A = { , , , }2 3 4 5

B x x= { : is a natural number and5 10< ≤x }

⇒ B = { , , , , }6 7 8 9 10

∴ A B∪ = ∪{ , , , } { , , , , }2 3 4 5 6 7 8 9 10

= { , , , , , , , , }2 3 4 5 6 7 8 9 10

2. Intersection of Sets

Let A and B be any two sets. The intersection of A and Bis the set of all those elements which belong to both A andB. It is denoted by A B∩ and read as A intersection B.The symbol ‘∩’ is used to denote the intersection.

∴ A B x x A x B∩ = ∈ ∈{ : }and

The intersection of sets A and B is represented by thefollowing Venn diagram.

The shaded portion represents A B∩ .

e.g., Let A = { , , , }2 3 4 5 and B = { , , , }1 3 6 4

Then, A B∩ = { , }3 4

Some Properties of Intersection of Sets

(i) A B B A∩ = ∩ [commutative law]

(ii) ( ) ( )A B C A B C∩ ∩ = ∩ ∩ [associative law]

(iii) φ φ,∩ = ∩ =A U A A [law of φand U]

(iv) A A A∩ = [idempotent law]

(v) If A B, andC are any three sets, then(a) A B C A B A C∩ ∪ = ∩ ∪ ∩( ) ( ) ( )

[distributive law i.e., ∩ distributes over ∪ ]

This can be shown with the Venn diagram as given below

For LHS

For RHS

(b) A B C A B A C∪ ∩ = ∪ ∩ ∪( ) ( ) ( )

[∪ distributes over∩]

Example 3.

(i) If A = { , , , , }1 3 5 7 9 and B = { , , , , }2 3 6 8 9 , then find

A B∩ .

(ii) If A e f g= { , , } and B = φ, then find A B∩ .

(iii) If A x x n n Z= = ∈{ : , }3 and

B x x n n Z= = ∈{ : , }4 , then find A B∩ .

Solution

(i) Given, A = { , , , , }1 3 5 7 9

and B = { , , , , }2 3 6 8 9

⇒ A B∩ = { , }3 9

[Q3 and 9 are only elements which are common]

(ii) Given, A e f g= { , , } and B = φ⇒ A B∩ = φ [since, there is no common element]

Sets 15ll ne

U

AB

UA B

C

UA B

C

UA B

C

(i) Shaded portion

represents ( )B C∪(ii) Shaded portion

represents A B C∩ ∪( )

UA B

C

UA B

C

(iii) Shaded portion

represents ( )A B∩(iv) Shaded portion

represents ( )A C∩

UA B

C

(v) Shaded portion represents

( ) ( )A B A C∩ ∪ ∩

15

(iii) Let x A B∈ ∩⇒ x A∈ and x B∈⇒ x is a multiple of 3 and x is a multiple of 4.⇒ x is a multiple of 3 and 4 both.⇒ x is a multiple of 12.⇒ x n n Z= ∈12 ,Hence, A B x x n n Z∩ = = ∈{ : , }12

3. Difference of Sets

Let A and B be any two sets. The difference of sets A andB in this order is the set of all those elements of A whichdo not belong to B. It is denoted by A B− and read as Aminus B.The symbol ‘−’ is used to denote the differenceof sets.

∴ A B x x A x B− = ∈ ∉{ : and }

Similarly, B A x x B x A− = ∈ ∉{ : and }

The difference of two sets A and B can be represented bythe following Venn diagram

The shaded portion represents the difference of two setsA and B.

e.g., Let A = { , , , , }1 2 3 4 5

and B = { , , , }3 5 7 9

Then, A B− = { , , }1 2 4

and B A− = { , }7 9

Some Properties of Intersection of Sets(i) A B A B− = ∩ ′ (ii) B A B A− = ∩ ′

(iii) A B A− ⊆ (iv) B A B− ⊆

Example 4.

(i) If X = {a, b, c, d, e, f}and Y = { f, b, d, g, h, k}, then find X – Y and Y – X.

(ii) If A = { , , , , }1 2 3 4 5 and B = { , , }2 4 6 , then find

A B− and B A− .

Solution

(i) Given, X a b c d e f= { , , , , , } and Y f b d g h k= { , , , , , }

∴ X Y a b c d e f f b d g h k− = −{ , , , , , } { , , , , , } = { , , }a c e

[only those elements of X which do not belong to Y ]

andY X f b d g h k a b c d e f− = −{ , , , , , } { , , , , , }

= { , , }g h k

[only those elements of Y which do not belong to X ]

(ii) Given, A = { , , , , }1 2 3 4 5

and B = { , , }2 4 6

∴ A B− = −{ , , , , } { , , }1 2 3 4 5 2 4 6

⇒ A B− = { , , }1 3 5

[only those elements of A which do not belong to B]

and B A− = −{ , , } { , , , , }2 4 6 1 2 3 4 5 = { }6

[only those elements of B which do not belong to A]

4. Symmetric Difference of Two Sets

Let A and B be any two sets. The symmetric difference of Aand B is the set ( ) ( )A B B A− ∪ − . It is denoted by A∆Band read as A symmetric difference B. The symbol ‘∆’ isused to denote the symmetric difference.

∴ A B A B B A∆ = − ∪ −( ) ( )

= ∈ ∈ ∉ ∩{ : }x x A x B x A Bor but

The symmetric difference of sets A and B is representedby the following Venn diagram

Hence, the shaded portion represents the symmetricdifference of sets A and B.e.g., Let A = { , , , }1 2 3 4 and B = { , , , }3 4 5 6Now, A B− = { , }1 2

B A− = { , }5 6∴ A B A B B A∆ = − ∪ −( ) ( ) = { , , , }1 2 5 6

* Method of Finding Symmetric Difference of Two

Sets

If two sets A and B (say) are given, then we find their symmetricdifference with the help of following steps

Step I Write the given sets in tabular form (if not given intabular form) and assume them A and B (say).

Step II Find the difference of sets A and B i.e., A B− .

Step III Find the difference of sets B and A i.e., B A− .

Step IV Find the union of sets obtained from steps II and III,which will give the required symmetric difference.

Worked out Problem

Find the symmetric difference of setsA ={ , , , , }1 3 5 6 7 and B ={ , , , }3 7 8 9 .

Step I Write the given sets in tabular formGiven sets are A = { , , , , }1 3 5 6 7 and B = { , , , }3 7 8 9 which arein tabular form.

Step II Find the difference of sets A and B i.e., A B− .We know that, difference of sets A and B is the set ofthose elements of A, which are not present in B.

So, A B− = −{ , , , , } { , , , }1 3 5 6 7 3 7 8 9 = { , , }1 5 6


U

A B

U

A B

A – B

B – A

UA B

A – B B – A

16

Step III Find the difference of sets B and A i.e., B A− .We know that, difference of sets B and A is the set ofthose elements of B, which are not present in A.So, B A− = −{ , , , } { , , , , }3 7 8 9 1 3 5 6 7 = { , }8 9

Step IV Find the union of sets obtained from steps II and III,which will give the required symmetric difference.

∴ Required symmetric difference,

A B A B B A∆ = − ∪ −( ) ( ) = ∪{ , , } { , }1 5 6 8 9 = { , , , , }1 5 6 8 9

Example 5. If A = { , , , , }1 3 5 7 9 and

B = { , , , , }2 3 5 7 11 ,

then find A B∆ .

/ Firstly, determine the sets A B− and B A− and then find theunion of these sets.

Solution We know that, A B∆ is the symmetric difference ofsets A and B.

∴ A B A B B A∆ = − ∪ −( ) ( )

Given, A = { , , , , },1 3 5 7 9

and B = { , , , , }2 3 5 7 11

Then, A B− = { , }1 9

and B A− = { , }2 11

∴ A B A B B A∆ = − ∪ −( ) ( )

= ∪{ , } { , }1 9 2 11 = { , , , }1 2 9 11

Disjoint Sets

Two sets A and B are said to be disjoint sets, if they haveno common element i.e., A B∩ = φ.

The disjoint of two sets A and B can be represented bythe following Venn diagram

Hence, separation of sets represents the two disjoint sets.

e.g., Let A = { , , }2 4 6 and B = { , , }1 3 5

Then, A B∩ = φ. Hence, A and B are disjoint sets.

Note The sets A B− , A B∩ and B A− are mutually disjoint sets i.e.,the intersection of any of these two sets is an empty set.

Example 6. Which of the following pairs of sets aredisjoint?

(i) A = { , , , , , }1 2 3 4 5 6

and B x x= { : is a natural number and 4 6≤ ≤x }

(ii) A x x= { : is the boys of your school}

B x x= { : is the girls of your school}

/ Firstly, convert all the sets in roster form, if it is not given inthat. Then use the condition for disjoint sets i.e., A B∩ = φ.

Solution

(i) Given, A = { , , , , , }1 2 3 4 5 6

and B = { , , }4 5 6

∴ A B∩ = ∩{ , , , , , } { , , }1 2 3 4 5 6 4 5 6

= ≠{ , , }4 5 6 φHence, this pair of sets is not disjoint.

(ii) Here, A b b= { , , ... }1 2 and B g g= { , , ... }1 2 ,

where b b1 2, , ... , are the boys and g g1 2, ,…, are the girlsof school.

Clearly, A B∩ = φHence, this pair of set is disjoint set.

Complement of a SetLet U be the universal set and A be any subset of U,then complement of A with respect to U is the set of allthose elements of U which are not in A.It is denoted by A or A ′ and read as‘A complement’.

Thus, A x x U= ∈{ : and x A∉ }

or A U A′ = −The complement of set A is represented by the followingVenn diagram

Hence, the shaded portion represents the complementof set A.

e.g., Let U = { , , , , , }1 2 3 4 5 6

and A = { , }2 4

Then, A U A′ = −= −{ , , , , , } { , }1 2 3 4 5 6 2 4

= { , , , }1 3 5 6

Note If A is a subset of the universal set U, then its complement A′ isalso a subset of U.

Some Properties of Complement of Sets

(i) ( )A A′ ′ = [law of double complementation]

(ii) (a) A A U∪ ′ =(b) A A∩ ′ = φ [complement law]

(iii) (a) φ′ =U

(b) U ′ = φ [laws of empty set and universal set]

Sets 17ll ne

A

UA′

UA B

17

(iv) (a) ( )A B A B∪ ′ = ′∩ ′(b) ( )A B A B∩ ′ = ′ ∪ ′ [De-Morgan’s law]

i.e., the complement of the union of two sets is theintersection of their complements [figure (a)] and thecomplement of the intersection of two sets is the union oftheir complements [figure (b)].

All the operations between two sets can be represented ina single Venn diagram, as given below

Example 7. Let U {1, 2, 3, 4, 5, 6, 7, 8, 9},=A B= ={ , , , } { , , , }2 4 6 8 2 3 5 7and .

Verify that

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

/ Use the formulae,

( ) ( ),A B U A B A U A∪ ′ = − ∪ ′ = − and B U B′ = − and thensimplify it.

Solution

(i) We have, U = { , , , , , , , , }1 2 3 4 5 6 7 8 9

A

B

==

{ , , , }

{ , , , }

2 4 6 8

2 3 5 7

∴ A B∪ = ∪ 2,{ , , , } { , , }2 4 6 8 3 5 7

⇒ A B∪ = { , , , , , , }2 3 4 5 6 7 8

Now,( ) ( )A B U A B∪ ′ = − ∪= { , , , , , , , , }1 2 3 4 5 6 7 8 9 − { , , , , , , }2 3 4 5 6 7 8

= { , }1 9 ...(i)

Now, A U A′ = −= −{ , , , , , , , , } { , , , }1 2 3 4 5 6 7 8 9 2 4 6 8

= { , , , , }1 3 5 7 9

and B U B′ = −= { , , , , , , , , }1 2 3 4 5 6 7 8 9 −{ , , , }2 3 5 7

= { , , , , }1 4 6 8 9

∴ A B′∩ ′ = ∩{ , , , , } { , , , , }1 3 5 7 9 1 4 6 8 9

={ , }1 9 ...(ii)


( )A B A B∪ ′ = ′ ∩ ′ Hence verified.

(ii) Here, A B∩ = ∩{ , , , } { , , , }2 4 6 8 2 3 5 7 = { }2

∴ ( ) ( )A B U A B∩ ′ = − ∩= −{ , , , , , , , , } { }1 2 3 4 5 6 7 8 9 2

⇒ ( ) { , , , , , , , }A B∩ ′ = 1 3 4 5 6 7 8 9 …(i)

Now, A B′∪ ′ = ∪{ , , , , } { , , , , }1 3 5 7 9 1 4 6 8 9

= { , , , , , , , }1 3 4 5 6 7 8 9 …(ii)


( )A B A B∩ ′ = ′∪ ′ Hence verified.

Example 8. If

A B= ={ , , , , }, { , , , , , , }2 4 6 8 10 1 2 3 4 5 6 7 ,

C = { , , , }2 6 7 10 and

U = { , , , , , , , , , }1 2 3 4 5 6 7 8 9 10 , then verify that

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

(iii) A B C A B C∩ ∩ = ∩ ∩( ) ( )

(iv) A B C A B A C∪ ∩ = ∪ ∩ ∪( ) ( ) ( )

(v) A B C A B A C∩ ∪ = ∩ ∪ ∩( ) ( ) ( )

Solution Given, A B{2, 4, 6, 8, 10}, {1, 2, 3, 4, 5, 6, 7}= = ,

C = { , , , }2 6 7 10 andU = { , , , , , , , , , }1 2 3 4 5 6 7 8 9 10

(i) B C∪ = ∪{ , , , , , , } { , , , }1 2 3 4 5 6 7 2 6 7 10

= { , , , , , , , }1 2 3 4 5 6 7 10

∴ ( ) ( )B C U B C∪ ′ = − ∪= { , , , , , , , , , }1 2 3 4 5 6 7 8 9 10 − { , , , , , , , }1 2 3 4 5 6 7 10

= { , }8 9 ...(i)

Similarly, B U B′ = − = { , , }8 9 10

and C U C′ = − = { , , , , , }1 3 4 5 8 9

∴ B C′∩ ′ = { , }8 9 ... (ii)


( )B C B C∪ ′ = ′∩ ′ Hence verified.

(ii) Here, A = { , , , , }2 4 6 8 10

Now, B C∪ = { , , , , , , , }1 2 3 4 5 6 7 10

∴ A B C∪ ∪ =( ) { , , , , , , , , }1 2 3 4 5 6 7 8 10 ...(iii)

Now, A B∪ = { , , , , , , , , }1 2 3 4 5 6 7 8 10

∴ ( A B C∪ ∪ =) { , , , , , , , , }1 2 3 4 5 6 7 8 10 ∪ { , , , }2 6 7 10

⇒ ( ) { , , , , , , , , }A B C∪ ∪ = 1 2 3 4 5 6 7 8 10 ...(iv)

From Eqs. (iii) and (iv), we get

A B C A B C∪ ∪ = ∪ ∪( ) ( ) Hence verified.


A B

A B∆

( )A B∪ ′

A B∩B A−

A B–

A B∪(Shaded Part)(Portion shaded by

vertical lines in

circle )B

(Portion shaded by horizontal

line between and )A B

(Without shaded

part)

(Portion shaded by vertical Line

in both circles)

(Portion shaded

by vertical lines

in circle )A

U

U

A B

(a)

U

(b)

A B

18

(iii) Here, B C∩ = ∩{ , , , , , , } { , , , }1 2 3 4 5 6 7 2 6 7 10

= { , , }2 6 7

A B C∩ ∩ = ∩( ) { , , , , } { , , }2 4 6 8 10 2 6 7

= { , }2 6 ...(v)

Now, A ∩ = ∩B { , , , , } { , , , , , , }2 4 6 8 10 1 2 3 4 5 6 7

= { , , }2 4 6

∴ ( ) { , , } { , , , }A B C∩ ∩ = ∩2 4 6 2 6 7 10

= { , }2 6 ...(vi)

From Eqs. (v) and (vi), we get

A B C A B C∩ ∩ = ∩ ∩( ) ( )

Hence verified.

(iv) Here, B C∩ = ∩{ , , , , , , } { , , , }1 2 3 4 5 6 7 2 6 7 10

= { , , }2 6 7

A B C∪ ∩ = ∪( ) { , , , , } { , , }2 4 6 8 10 2 6 7

= { , , , , , }2 4 6 7 8 10 ... (vii)

Now, ( ) { , , , , } { , , , , , , }A B∪ = ∪2 4 6 8 10 1 2 3 4 5 6 7

= { , , , , , , , , }1 2 3 4 5 6 7 8 10

and ( ) { , , , , } { , , , }A C∪ = ∪2 4 6 8 10 2 6 7 10

= { , , , , , }2 4 6 7 8 10

∴ ( ) ( ) { , , , , , , , , }A B A C∪ ∩ ∪ = 1 2 3 4 5 6 7 8 10

∩ { , , , , , , }2 4 6 7 8 10

= { , , , , , }2 4 6 7 8 10 ...(viii)

From Eqs. (vii) and (viii), we get

A B C A B A C∪ ∩ = ∪ ∩ ∪( ) ( ) ( )

Hence verified.

(v) Now, B C∪ = ∪{ , , , , , , } { , , , }1 2 3 4 5 6 7 2 6 7 10

= { , , , , , , , }1 2 3 4 5 6 7 10

A B C∩ ∪ =( ) { , , , , }2 4 6 8 10 ∩ { , , , , , , , }1 2 3 4 5 6 7 10

⇒ A B C∩ ∪ =( ) { , , , }2 4 6 10 ...(ix)

Now, ( ) { , , , , } { , , , , , , }A B∩ = ∩2 4 6 8 10 1 2 3 4 5 6 7

= { , , }2 4 6

and ( ) { , , , , } { , , , }A C∩ = ∩2 4 6 8 10 2 6 7 10

= { , , }2 6 10

∴ ( ) ( ) { , , } { , , }A B A C∩ ∪ ∩ = ∪2 4 6 2 6 10

= { , , , }2 4 6 10 ...(x)

From Eqs. (ix) and (x), we get

A B C A B A C∩ ∪ = ∩ ∪ ∩( ) ( ) ( )

Hence verified.

Sets 19ll ne


1. If A = { , , , , }3 5 7 9 11 , B = { , , , }7 9 11 13 and

C = { , , }11 13 15 , then find

(i) A B∩ (ii) B C∩Sol. We have, A = { , , , , }3 5 7 9 11

B = { , , , }7 9 11 13 and C = { , ,11 13 15}

(i) A B∩ = ∩{ , , , , } { , , , }3 5 7 9 11 7 9 11 13

= { , , }7 9 11 [1/2]

(ii) B C∩ = ∩{ , , , } { , , }7 9 11 13 11 13 15

= { , }11 13 [1/2]

2. Find A B− and B A− when A = { , , , , , }1 2 3 4 5 6 and

B = { , , , , }2 4 6 8 10 .

Sol. Now, A B− = −{ , , , , , } { , , , , }1 2 3 4 5 6 2 4 6 8 10 = { , , }1 3 5

[1/2]

and B A− = −{ , , , , } { , , , , , }2 4 6 8 10 1 2 3 4 5 6

= { , }8 10 [1/2]

3. State whether each of the following statement istrue or false.

(i) A = { , , , }2 4 6 8 and B = { , , }1 3 5 are disjoint sets.

(ii) A a e i o u= { , , , , } and B a b c d= { , , , } are disjointsets.

Sol. (i) We have, A = { , , , }2 4 6 8 and B = { , , }1 3 5

Now, A B∩ = ∩{ , , , } { , , }2 4 6 8 1 3 5

⇒ A B∩ = φTherefore, A and B are disjoint sets.

Hence, given statement is true. [1/2]

(ii) We have, A a e i o u= { , , , , } and B a b c d= { , , , }

Now, A B a∩ = { }

i.e., A B∩ ≠ φTherefore, A and B are not disjoint sets.

Hence, given statement is false. [1/2]

4. Prove that( ) ( )A B B C∩ ′ ′ ∪ ∩ = A B′∪ .

Sol. LHS =( ) ( )A B B C∩ ′ ′ ∪ ∩= { ( ) } ( )A B B C′∪ ′ ′ ∪ ∩ [by De-Morgan’s law]

= ′∪ ∪ ∩( ) ( )A B B C [Q( )′ ′ =B B]

= ′∪ ∪ ∩ ′ ∪ ∪(( ) ) (( ) )A B B A B C

= ′∪ ∪ ∩ ′ ∪ ∪( ( )) ( )A B B A B C

= ′ ∪ ∩ ′ ∪ ∪( ) ( )A B A B C

= ′∪( )A B = RHS [1]

Hence proved.

EXAM

19

5. Let S = Set of points inside the square, T = Set of

points inside the triangle and C = Set of pointsinside the circle. If the triangle and circleintersect each other and are contained in asquare. Then, prove that S T C S∪ ∪ = , by Venndiagram.

Sol. Given, S = Set of points inside the square

T = Set of points inside the triangle

C = Set of points inside the circle

According to the given condition, the Venn diagram is givenbelow

It is clear from the Venn diagram that, S T C S∪ ∪ = [1]

Hence proved.



6. If U a b c d e f g h= { , , , , , , , }, then find the

complement 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= { , , }

Sol. We have, U a b c d e f g h= { , , , , , , , }

(i) Complement of A is A ′.∴ A U A′ = − = −{ , , , , , , , } { , , }a b c d e f g h a b c

= { , , , , }d e f g h [1]

(ii) Complement of B is B ′.∴ B U B′ = − = −{ , , , , , , , } { , , , }a b c d e f g h d e f g

= { , , , }a b c h [1]

(iii) Complement of C is C ′.∴ C U C′ = − = −{ , , , , , , , } { , , , }a b c d e f g h a c e g

= { , , , }b d f h [1]

(iv) Complement of D is D ′.∴ D U D′ = − = −{ , , , , , , , } { , , }a b c d e f g h f g h

= { , , , , }a b c d e [1]

7. Find A B∆ , if

(i) A = { , , }1 3 4 and B = { , , , }2 5 9 11 .

(ii) A = { , , , , }1 3 6 11 12 and B = { , }1 6 .

Sol. (i) We have,

A = { , , }1 3 4

and B = { , , , }2 5 9 11

Then, ( ) { , , } { , , , }A B− = −1 3 4 2 5 9 11

= { , , }1 3 4

and ( ) { , , , } { , , }B A− = −2 5 9 11 1 3 4

= { , , , }2 5 9 11 [1]

∴ A B A B B A∆ = − ∪ −( ) ( )

= ∪{ , , } { , , , }1 3 4 2 5 9 11

= { , , , , , , }1 2 3 4 5 9 11 [1]

(ii) We have,

A = { , , , , }1 3 6 11 12

and B = { , }1 6

Then, ( ) { , , , , } { , }A B− = −1 3 6 11 12 1 6

= { , , }3 11 12

and ( ) { , } { , , , , }B A− = −1 6 1 3 6 11 12 = φ [1]

∴ A B A B B A∆ = − ∪ −( ) ( )

= ∪{ , , }3 11 12 φ ={ , , }3 11 12 [1]

8. If U = { , , , , , , , , , }2 3 4 5 6 7 8 9 10 11 ,

A = { , , }2 4 7 , B = { , , , , }3 5 7 9 11 and

C = { , , , , }7 8 9 10 11 , then compute

(i) ( ) ( )A U B C∩ ∩ ∪ (ii) C B−(iii) B C− (iv) ( )B C− ′

Sol. Given,

U = { , , , , , , , , , }2 3 4 5 6 7 8 9 10 11 ,

A = { , , }2 4 7 , B = { , , , , }3 5 7 9 11 and C = { , , , , }7 8 9 10 11

(i) ( ) { , , }A U∩ = 2 4 7 ∩ { , , , , , , , , , }2 3 4 5 6 7 8 9 10 11

= { , , }2 4 7

B C∪ = ∪{ , , , , } { , , , , }3 5 7 9 11 7 8 9 10 11

= { , , , , , , }3 5 7 8 9 10 11

∴ ( ) ( ) { , , }A U B C∩ ∩ ∪ = 2 4 7 ∩ { , , , , , , }3 5 7 8 9 10 11

= { }7 [1]

(ii) C B− = −{ , , , , } { , , , , }7 8 9 10 11 3 5 7 9 11

= { , }8 10 [1]

(iii) B C− = −{ , , , , } { , , , , }3 5 7 9 11 7 8 9 10 11

= { , }3 5 [1]

(iv) ( ) ( )B C U B C− ′ = − −= −{ , , , , , , , , , } { , }2 3 4 5 6 7 8 9 10 11 3 5

= { , , , , , , , }2 4 6 7 8 9 10 11 [1]

9. Verify( )A B A B∩ ′ = ′ ∪ ′ , where A = { , , , }3 4 5 6 and

B = { , , , }3 6 7 8 are subsets of

U = { , , , , , , , }1 2 3 4 5 6 7 8 .

Sol. We have, A = { , , , }3 4 5 6 , B = { , , , }3 6 7 8

and U = { , , , , , , , }1 2 3 4 5 6 7 8

Now, A B∩ = ∩{ , , , } { , , , }3 4 5 6 3 6 7 8

= { , }3 6 [1]

ST

C

20

∴ ( )A B∩ ′ = − ∩U A B( ) = −{ , , , , , , , } { , }1 2 3 4 5 6 7 8 3 6

⇒ ( )A B∩ ′ = { , , , , , }1 2 4 5 7 8 ... (i) [1]

Also, A ′ = −U A

= −{ , , , , , , , } { , , , }1 2 3 4 5 6 7 8 3 4 5 6

= { , , , }1 2 7 8 [1]

and B U B′ = −= −{ , , , , , , , } { , , , }1 2 3 4 5 6 7 8 3 6 7 8

= { , , , }1 2 4 5 [1]

∴ A B′∪ ′ = ∪{ , , } { , , , }1 7 8 1 2 4 5

= { , , , , , }1 2 4 5 7 8 ... (ii)


( )A B A B∩ ′ = ′ ∪ ′ [1]

Hence verified.

10. Let A and B be any two sets. Using properties ofsets prove that

(i) ( )A B B A B− ∪ = ∪ (ii) ( )A B A A− ∪ =(iii) ( )A B B− ∩ = φ (iv) ( )A B A A B− ∩ = ∩ ′

Sol. (i) ( ) ( )A B B A B B− ∪ = ∩ ′ ∪ [ ]Q A B A B− = ∩ ′= ∪ ∩ ′∪( ) ( )A B B B

[since, ∪ is distributes over ∩]

= ∪ ∩( )A B U [ ]QB B U′ ∪ == ∪A B [1]

(ii) ( )A B A A− ∪ = [ ]Q A B A− ⊆ [1]

(iii) ( ) ( )A B B A B B− ∩ = ∩ ′ ∩ [ ]Q A B A B− = ∩ ′= ∩ ′∩A B B( ) [since, ∩ is associative]

= ∩A φ = φ [ ]QB B′∩ = φ [1]

(iv) ( )A B A A B− ∩ = − = ∩ ′A B [1]

Hence proved.

11. For any two sets A and B, prove by usingproperties of sets that

( ) ( ) ( ) ( )A B A B A B B A∪ − ∩ = − ∪ −Sol. LHS =( ) ( )A B A B∪ − ∩

= ∪ ∩ ∩ ′( ) ( )A B A B [ ]Q X Y X Y− = ∩ ′= ∩ ′∪ ′X A B( ), where X A B= ∪

[ ( ) ]Q A B A B∩ ′ = ′∪ ′ [1]

= ∩ ′ ∪ ∩ ′( ) ( )X A X B [∩ distributes over ∪ ]

= ∩ ′ ∪ ∩ ′( ) ( )B A A B [1]

Q X A A B A A A B A

B A B A

∩ ′ = ∪ ∩ ′ = ∩ ′ ∪ ∩ ′= ∪ ∩ ′ = ∩ ′

( ) ( ) ( )

( )φSimilarly,

X B A B B A B B B

A B A B

∩ ′ = ∪ ∩ ′ = ∩ ′ ∪ ∩ ′= ∩ ′ ∪ = ∩ ′

( ) ( ) ( )

( ) ( )φ

[1]

= ∩ ′ ∪ ∩ ′( ) ( )A B B A = − ∪ −( ) ( )A B B A

[Q A B A B B A B A− = ∩ ′ − = ∩ ′and ] [1]

Hence proved.

12. Prove that A B C A B A C− ∪ = − ∩ −( ) ( ) ( ).

Sol. Let x be any element of A B C− ∪( ).

Then, x A B C∈ − ∪( )

⇒ x A∈ and x B C∉ ∪⇒ x A∈ and (x B∉ and x C∉ ) [1/2]

⇒ (x A∈ and x B∉ ) and (x A∈ and x C∉ )

⇒ x A B∈ −( ) and x A C∈ −

⇒ x A B A C∈ − ∩ −( ) ( )

∴ A B C A B A C− ∪ ⊆ − ∩ −( ) ( ) ( ) ...(i) [1]

Now, let x be any element of ( ) ( )A B A C− ∩ − .

Then, x A B A C∈ − ∩ −( ) ( )

⇒ x A B∈ −( )

and x A C∈ −( ) [1/2]

⇒ (x A∈ and x B∉ )

and (x A∈ and x C∉ )

⇒ x A∈ (and x B x C∉ ∉and )

⇒ x A x B C∈ ∉ ∪and

⇒ x A B C∈ − ∪( ( ))

∴ ( ) ( ) ( )A B A C A B C− ∩ − ⊆ − ∪ ...(ii) [1]


A B C A B A C− ∪ = − ∩ −( ) ( ) ( ) [1]

Hence proved.

13. Suppose A A A1 2 30, , ..., are thirty sets each

having 5 elements and B B Bn1 2, , ..., are n setseach having 3 elements.

Let U Ui

ij

n

jA B= =

= =1

30

1

5and each element of S

belongs to exactly 10 of Ai ’s and exactly 9 of Bj ’s.Then, find the value of n.

Sol. If elements are not repeated, then number of elementsA A A A1 2 3 30∪ ∪ ∪ ∪K is 30 5× but each element isused 10 times, so

n S( ) = × =30 5

1015 ... (i) [1]

Similarly, if elements in B B Bn1 2, , ..., are not repeated,then total number of elements is 3n, but each element isrepeated 9 times, so

n Sn

( ) = 3

9[1]

⇒ 153

9= n

[from Eq. (i)]

∴ n = × =15 9

345 [2]

Sets 21ll ne

21

1 Mark Questions

1. Let A = { , , , }2 4 6 8 and B = { , , , }6 8 10 12 . Find A B∪ .

2. Let A a e i o u= { , , , , } and B a i u= { , , }. Show that A B A∪ = .

3. If A = { , , , , }1 3 5 7 9 and B = { , , , , }2 3 5 7 11 , then find A B− .

4. Let U = { , , , , , , , , }1 2 3 4 5 6 7 8 9 , A = { , , , }1 2 3 4 and B = { , , , }2 4 6 8 . Find ( )A B∪ ′ .

5. If A and B are two sets, then verify A A B A∩ ∪ =( ) .

6. If U = { , , , , , , , , , }1 2 3 4 5 6 7 8 9 10 , A = { , , , }1 2 3 5 , B = { , , , }2 4 6 7 and C = { , , , }2 3 4 8 , then find ( )B C∪ ′ .

4 Marks Questions

7. If A = { , , , , }2 4 6 8 10 , B = { , , , , , , }1 2 3 4 5 6 7 and C = { , , , }2 6 7 10 , then verify that

(i) A B C A B A C− ∪ = − ∩ −( ) ( ) ( )

(ii) A B C A B A C− ∩ = − ∪ −( ) ( ) ( )

8. If U = { , , , , , , , , , }1 2 3 4 5 6 7 8 9 10 , A = { , , }1 3 4 and B = { , }5 6 , then verify that

A B A B B A− = ∩ ′ = ′ − ′

9. If U = { , , , , , , , }1 2 3 4 5 6 7 8 , A = { , , , }1 2 3 4 , B = { , , }3 4 6 and C = { , , , }5 6 7 8 , then verify that

(i) A B C A B A C∪ ∩ = ∪ ∩ ∪( ) ( ) ( )

(ii) ( )A B A B∩ ′ = ′∪ ′

10. If A = { , , , , , , , , }1 3 5 7 9 11 13 15 17 , B = { , , ..., }2 4 18 and N, the set of natural numbers is the universal set, then

prove that A A B B N′ ∪ ∪ ∩ ′ ={( ) } .

11. (i) If L M= ={ , , , }, { , , , }1 2 3 4 3 4 5 6 and N = { , , }1 3 5 , then verify that L M N L M L N− ∪ = − ∩ −( ) ( ) ( )

(ii) If A and B are subsets of the universal set U, then show that A A B⊆ ∪ .

12. If X and Y are two sets and X′ denotes the complement of X, then verify that X X Y∩ ∪ ′ =( ) φ.

13. A, B and C are subsets of universal set U. If A = { , , , , , }2 4 6 8 12 20 , B = { , , , , }3 6 9 12 15 , C = { , , }5 10 15 and U is

the set of all whole numbers, then draw a Venn diagram showing the relation of U, A, B and C.

14. Determine, whether each of the following statement is true or false. Justify your answer.

(i) For all sets A and B, ( ) ( )A B A B A− ∪ ∩ = .

(ii) For all sets A, B and C, if A C⊂ and B C⊂ , then A B C∪ ⊂ .

ll ne

YOURSELFTRY

22

ll ne

In this topic, we discuss some practical problemsrelated to our daily life, which are based on union andintersection of two sets.

For solving these types of problem, we have followingformulae.

If A and B are two finite sets, then two cases arise

Case I If A and B are disjoint sets, then there is nocommon element in A and B i.e., A B∩ = φ.Then,

n A B n A n B( ) ( ) ( )∪ = +

Case II If A and B are not disjoint sets, then there arecommon elements in A and B. Then,

n A B n A n B n A B( ) ( ) ( ) ( )∪ = + − ∩

It is called addition theorem for two sets.

We know that, sets A B− , A B∩ and B A− aredisjoint and their union is A B∪ . Then,

(i) n A B n A B n B A( ) ( ) ( )∪ = − + − + ∩n A B( )

(ii) n A n A B n A B( ) ( ) ( )= − + ∩(iii) n B n B A n A B( ) ( ) ( )= − + ∩

Example 1. If X and Y are two sets such that X has40 elements, X Y∪ has 60 elements and X Y∩has 10 elements, then how many elements does Yhave?

/ Use the formula, n X Y n X nY n X Y( ) ( ) ( ) ( )∪ = + − ∩and simplify it.

Solution Given, n X( ) = 40,

n X Y( )∪ = 60,

n X Y( )∩ = 10,

Using the formula,

n X Y n X n Y n X Y( ) ( ) ( ) ( )∪ = + − ∩ , we get

60 40 10= + −n Y( )

⇒ 60 30= +n Y( )

⇒ n Y( ) = −60 30

∴ n Y( ) = 30

Hence, total number of elements inY have 30.

❖ Important Results

1. n X Y n X Y n X Y n X C( ) ( ) ( ) ( )∪ = ∩ ′ + ′∩ + ∩

2. n A B n A B B A( ) [( ) ( )]∆ = − ∪ − = − + −n A B n B A( ) ( )

[since, ( )A B− and ( )B A− are disjoint sets]

= − ∩ + − ∩n A n A B n B n A B( ) ( ) ( ) ( )

= + − ∩n A n B n A B( ) ( ) ( )2

3. n A B n A B( ) [( ) ]′∪ ′ = ∩ ′ = − ∩n U n A B( ) ( )

4. n A B n A B( ) [( ) ]′∩ ′ = ∪ ′ = − ∪n U n A B( ) ( )

5. If A B, and C are finite sets, then

n A B C n A B C( )∪ ∪ = ∪ ∪[( ) ]= + + − ∩( ) ( ) ( ) ( )n A n B n C n A B

− ∩ − ∩( ) ( )n B C n A C + ∩ ∩( )n A B C

6. Let M = Set of students who have Mathematics.

P = Set of students who have Physics.

C = Set of students who have Chemistry.

On applying the different operations on these three sets, we get

following important results.

(i) ′ =M Set of students who have no Mathematics.

(ii) P ′= Set of students who have no Physics.

(iii) C ′= Set of students who have no Chemistry.

(iv) M P∪ = Set of students who have atleast one subject

Mathematics or Physics.

(v) P C∪ = Set of students who have atleast one subject

Physics or Chemistry.

(vi) C M∪ = Set of students who have atleast one subject

Chemistry or Mathematics.

(vii) M P∩ = Set of students who have both subjects

Mathematics and Physics.

(viii) P C∩ = Set of students who have both subjects Physics

and Chemistry.

(ix) C M∩ = Set of students who have both subjects Chemistry

and Mathematics.

(x) ( )M P C∩ ∩ = Set of students who have all the subjects.

(xi) ( )M P C∪ ∪ = Set of all students who have three subjects.

4 Applications of Set Theory

UA B

UA B

A B∩ B – AA – B

23

Solution of Word ProblemsWord problems related to sets can be solved by following twomethods

1. By Using Operations

In this method, we use the operations related to setaccording to the given information.This method can beunderstand with the help of an worked out problemwhich is given below.

Worked out Problem

Each student in a class of 40 students study atleastone of the subjects English, Mathematics andEconomics. 16 students study English, 22

Economics and 26 Mathematics, 5 study Englishand Economics, 14 Mathematics and Economicsand 2 English, Economics and Mathematics. Findthe number of students who study English andMathematics.

Step I Consider the set which represent students in givensubjects.

Let A, B and C denote the set of students who study English,Economics and Mathematics, respectively.

Step II Find the relation between number of students of eachsubject given in question.

Given, total number of students, n A B C( )∪ ∪ = 40

Number of students who study English, n A( ) = 16

Number of students who study Economics, n B( ) = 22

Number of students who study Mathematics, n C( ) = 26

Number of students who study English and Economics,

n A B( )∩ = 5

Number of students who study Mathematics andEconomics, n B C( )∩ = 14

and number of students who study all subjects,

n A B C( )∩ ∩ = 2

Step III Use the suitable formula, for three sets. Then,substitute all the given values and simplify it.

We know that,

n A B C( )∪ ∪ = + +n A n B n C( ) ( ) ( )

− ∩ ) − ∩n A B n B C( ( )− ∩ + ∩ ∩n A C n A B C( ) ( )On putting the values from step II, we get

40 16 22 26 5= + + − − − ∩ +14 2n C A( )

⇒ 40 66 19= − − ∩n C A( )

⇒ n C A( )∩ = −47 40

= 7

Hence, number of students who study English andMathematics are 7.

Example 2. In a class of 60 students, 25 studentsplay Cricket, 20 students play Tennis and10 students play both the games. Then, find thenumber of students who play neither games.

Solution Let C and T denote the students who play Cricket,and Tennis, respectively and U denotes the total number ofstudents in a class.

Then, n C n T( ) , ( )= =25 20 and n U( ) = 60

We know that,

n C T n C n T n C T( ) ( ) ( ) ( )∪ = + − ∩On putting the values, we get

n C T( )∪ = + −25 20 10 = − =45 10 35

∴ Number of students who play neither games,

n C T n C T( ) ( )′∩ ′ = ∪ ′ [by De-Morgan’s Law]

= − ∪n U n C T( ) ( ) = − =60 35 25

Hence, 25 students play neither games.

2. By Using Venn Diagram

Sometimes a word problem contains three sets and manyinformation related to these sets but not exactinformation related to such set separately. Then, thatproblem become difficult to solve by previous method.Such problems can be solved by considering a variable oneach part of the venn diagram containing three sets andthen make some equations, which are given below

(i) Union of three sets

a + b + c + d + e + f + g = Constant

(ii) For Set A a + b + d + e = Constant

(iii) For Set B b + c + e + f = Constant

(iv) For Set C d + e + f + g = Constant

(v) Intersection of two sets at a time

(a) b + e = Constant (b) d + e = Constant

(c) e + f = ConstantNow, solve these equations one by one by using givenvalues and get separate values of each set.


BA

C

cba

de

f

g

U

24

Example 3. In a survey of 25 students, it was found

that 15 had taken Maths, 12 had taken Physics,

11 had taken Chemistry, 5 had taken Maths and

Chemistry, 9 had taken Maths and Physics, 4 had

taken Physics and Chemistry and 3 had taken all

the three subjects. Find the number of students that

had taken

(i) only Chemistry.

(ii) only Maths.

(iii) only Physics.

(iv) Physics and Chemistry but not Maths.

(v) Maths and Physics but not Chemistry.

(vi) only one of the subject.

(vii) atleast one of the three subjects.

(viii) None of the subjects.

Solution Let M, P and C denote the set of students who hadtaken Mathematics, Physics and Chemistry, respectively.

Let in Venn diagram, a, b, c, d, e, f and g denote the numberof elements in respective regions.

From the Venn diagram, we get

n M a b d e( ) = + + +n P b c e f( ) = + + +n C d e f g( ) = + + +

n M P b e( )∩ = +n P C e f( )∩ = +

n M C d e( )∩ = +and n M P C e( )∩ ∩ =It is given that, number of students who had taken all threesubjects,

n M P C( )∩ ∩ = 3

⇒ e = 3

∴ Number of students who had taken Maths and Physics,

n M P( )∩ = 9

⇒ b e+ =9

⇒ b + =3 9 [Qe = 3]

⇒ b = 6

∴ Number of students who had taken Physics andChemistry,

n P C( )∩ = 4

⇒ e f+ = 4

⇒ 3 4+ =f [Qe = 3]

⇒ f =1

Number of students who had taken Maths and Chemistry,n M C( )∩ = 5

⇒ d e+ =5

⇒ d + =3 5 [Qe = 3]

⇒ d = 2

Number of students who had taken Maths,

n M( ) =15

⇒ a b d e+ + + =15

⇒ a + + + =6 2 3 15 [ , , ]Qb d e= = =6 2 3

⇒ a = −15 11 ⇒ a = 4

Number of students who had taken Physics.n P( ) =12

⇒ b c e f+ + + =12

⇒ 6 3 1 12+ + + =c [ , , ]Qb e f= = =6 3 1

⇒ c = − ⇒12 10 c = 2

Number of students who had taken Chemistry,n C( ) =11

⇒ d e f g+ + + =11

⇒ 2 3 1 11+ + + =g [Qd e f= = =2 3 1, , ]

⇒ g g= − ⇒ =11 6 5

Now, the number of students in various cases are as follows

(i) Only Chemistry, g = 5

(ii) Only Maths, a = 4

(iii) Only Physics, c = 2

(iv) Physics and Chemistry but not Maths, f =1

(v) Maths and Physics but not Chemistry, b = 6

(vi) Only one of the subject, a c g+ + = + + =4 2 5 11

(vii) Atleast one of the three subjects

a b c d e f g+ + + + + + = + + + + + + =4 6 2 2 3 1 5 23

(viii) None of the subjects,

25 − + + + + + +( )a b c d e f g = −25 23 = 2

Sets 25ll ne

PM

C

cba

de

f

g

U

25


1. If X and Y are two sets such thatn(X) , n(Y)= =17 23 and n X Y( )∪ = 38, then findn X Y( )∩ .

/ We will use the addition theorem,n A B n A n B n A B( ) ( ) ( ) ( )∪ = + − ∩ , to calculate anyone ofthe four values, if three of them are given.

Sol. Given, n X( ) =17

n Y( ) = 23

and n X Y( )∪ = 38

According to the addition theorem, we have

n X Y n X n Y n X Y( ) ( ) ( ) ( )∪ = + − ∩⇒ n X Y n X n Y n X Y( ) ( ) ( ) ( )∩ = + − ∪On putting the values, we get

n X Y( )∩ = + −17 23 38= −40 38

= 2 [1]

2. If X and Y are two sets such that X ∪ Y has18 elements. X has 8 elements and Y has15 elements, then how many elements does X ∪ Yhave?

Sol. Given, n X Y( )∪ = 18, n X n Y( ) ( )= =8 15and

According to the addition theorem, we have

n X Y n X n Y n X Y( ) ( ) ( ) ( )∪ = + − ∩⇒ 18 8 15= + − ∩n X Y( )

⇒ n X Y( )∩ = − =23 18 5 [1]

3. If n A , n , n( ) ( ) ( )= = =4 5 7B U and n A B( )∩ = 2,

then find the value of n (A B)∪ ′ .

Sol. Given, n A n B n U( ) , ( ) , ( )= = =4 5 7

and n A B( )∩ = 2

Q n A B n A n B n A B( ) ( ) ( ) ( )∪ = + − ∩∴ n A B( )∪ = + − =4 5 2 7

Now, n A B n U n A B( ) ( ) ( )∪ ′ = − ∪ = − =7 7 0 [1]

ll ne


4. In a town of 840 persons, 450 persons read Hindi,300 read English and 200 read both newspapers.Then, find the number of persons who readneither of the newspapers.

Sol. Let H and E denote Hindi and English newspapers.

Given, total number of persons, n U( ) = 840

Total number of persons who read Hindi newspaper,

n H( ) = 450

Total number of persons who read English newspaper,

n E( ) = 300

and n H E( )∩ = 200

We know that, n H E n H n E n H E( ) ( ) ( ) ( )∪ = + − ∩∴ n H E( )∪ = + −450 300 200

= − =750 200 550 [2]

∴ The number of persons who read neither of thenewspaper,

n H E n H E( ) ( )′ ∩ ′ = ∪ ′ [by De-Morgan’s law]

= − ∪ = − =n U n H E( ) ( ) 840 550 290

Hence, 290 persons read neither of the newspapers. [2]

5. In a survey of 400 movie viewers, 150 were listedas liking ‘Veer Zaara’, 100 were listed as liking‘Aitraaz’ and 75 were listed as both liking ‘Aitraaz’as well as ‘Veer Zaara’. Find how many peoplewere liking neither ‘Aitraaz’ nor ‘Veer Zaara’.

Sol. Let U denotes the set of movie viewers, V denotes the set ofpeople liking ‘Veer Zaara’ and A denotes the set of peopleliking ‘Aitraaz’.

Then, we have, n U n V n A( ) , ( ) , ( )= = =400 150 100

and n V A( )∩ = 75 [1]

Now, the number of people those were liking neither‘Aitraaz’ nor ‘Veer Zaara’,

n V A n V A( ) ( )′∩ ′ = ∪ ′ [by De-Morgan’s law]

= − ∪n U n V A( ) ( )

= − − + ∩n U n V n A n V A( ) ( ) ( ) ( )

[ ( ) ( ) ( ) ( )]Qn A B n A n B n A B∪ = + − ∩ [1]

= − − +400 150 100 75 = 225

Hence, 225 people were liking neither ‘Aitraaz’ nor ‘VeerZaara’. [2]

EXAM

26

6. There are 60 students in a Mathematics class and90 students in Physics class. Find the number ofstudents which are either in Physics class orMathematics class in the following cases

(i) Two classes meet at the same hour.

(ii) Two classes meet at different hours and30 students are enrolled in both the courses.

(iii) What value is shown here?

Sol. Let M be the set of students in Mathematics class and P bethe set of students in Physics class.

Given that, n M n P( ) ( )= =60 90and

(i) Two classes meet at same hour i.e.,M P∩ = φ

⇒ n M P( )∩ = 0 [1]

Now, n M P n M n P n M P( ) ( ) ( ) ( )∪ = + − ∩= + −60 90 0

=150

[1]

(ii) In this case, two classes meet at different hours and30 students are enrolled in both the courses.

∴ n M P( )∩ = 30

Now, n M P n M n P n M P( ) ( ) ( ) ( )∪ = + − ∩= + −60 90 30 = 120 [1]

(iii) Most of the students study Physics in comparison ofMathematics. [1]

7. In a group of 26 persons, 8 take tea but not coffee

and 16 take tea. How many persons take coffee

but not tea? What value is shown here?

Sol. Let T and C be the set of persons taking tea and coffee,respectively.

We have, n T C( )∪ = 26

n T C( )∩ ′ = 8

and n T( ) =16 [1]

Using the identity,

n T C n T n T C( ) ( ) ( )∩ ′ = − ∩⇒ 8 16= − ∩n T C( )

⇒ n T C( )∩ = −16 8 = 8 [1]

Using the identity,

n T C n T C n C T( ) ( ) ( )∪ = ∩ ′ + ∩ ′ + ∩n T C( )

⇒ 26 8 8= + ∩ ′ +n C T( )

⇒ n C T( )∩ ′ = −26 16 =10

Number of persons who take coffee but not tea,n C T( )∩ ′ =10 [1]

Here, we see that most people like coffee in comparison oftea. [1]

Sets 27ll ne

Long Answer Type Questions [6 Marks each]

8. In a survey, it is found that 21 people like productA, 26 people like product B and 29 people likeproduct C. If 14 people like products A and B, 12people like products C and A, 14 people likeproducts B and C, 8 people like all the threeproducts. Find how many like product C only.

Sol. Let A, B andC denote the set of people who liked productsA, B and C, respectively.

Let in Venn diagram a, , , , ,b c d e f and g denote thenumber of elements in respective regions.

Then, n A a b c d( ) = + + + =21 ...(i)

n B b c f g( ) = + + + =26 …(ii) [1]

n C c d e f( ) = + + + =29 …(iii)

n A B b c( )∩ = + =14 …(iv)

n C A c d( )∩ = + =12 …(v)

n B C c f( )∩ = + =14 …(vi)

and n A B C c( )∩ ∩ = = 8 …(vii) [1]

We have to find the number of people like product C onlyi.e., value of e.

On putting the value of c from Eq. (vii) in Eq. (iv), we get

b + =8 14

⇒ b = 6 [1]

Now, putting the value of c from Eq. (vii) in Eq. (v), weget

8 12+ =d

⇒ d = 4 [1]

UM

P

9060

UM P

6030 30

UT C

108 8

CA

B

eda

be

f

g

U

27

Now, putting the value of c from Eq. (vii) in Eq. (vi), weget

8 14+ =f ⇒ f = 6 [1]

On putting the values of c, d and f in Eq. (iii), we get

8 4 6 29+ + + =e

∴ e = −29 18 =11

Hence, number of people who like product C only is 11.

[1]

9. In a survey of 60 people, it was found that25 people read newspaper H, 26 readnewspaper T, 26 read newspaper Z, 9 read both Hand Z, 11 read both H and T, 8 read both T and Zand 3 read all three newspapers. Find

(i) the number of people who read atleast one ofthe newspaper.

(ii) the number of people who read exactly onenewspaper.

Sol. Given, set of people who read types of newspapers H, T andZ.

Let in Venn diagram a b c d e f, , , , , and g denote thenumber of elements in respective regions.

Here, n U a b c d( ) = + + + + + + =e f g 60 ...(i) [1]

n H a b c d( ) = + + + =25 ...(ii)

n T b c f g( ) = + + + =26 ...(iii)

n Z c d e f( ) = + + + =26 ...(iv)

n H Z c d( )∩ = + = 9 ...(v)

n H T b c( )∩ = + =11 ...(vi)

n T Z c f( )∩ = + = 8 ...(vii)

and n H T Z c( )∩ ∩ = = 3 ...(viii) [1]

On putting the value of c in Eq. (vii), we get

3 8 5+ = ⇒ =f f

On putting the value of c in Eq. (vi), we get

3 11 8+ = ⇒ =b b [1]

On putting the value of c in Eq. (v), we get

3 9 6+ = ⇒ =d d

On putting the values of c, d and f in Eq. (iv), we get

3 6 5 26+ + + =e

⇒ e = − =26 14 12 [1]

On putting the values of b, c and f in Eq. (iii), we get

8 3 5 26+ + + =g

⇒ g = − =26 16 10On putting the values of b, c and d in Eq. (ii), we get

a + + + =8 3 6 25

⇒ a = − =25 17 8 [1]

(i) Number of people who read atleast one of the threenewspapers,

A U( ) = + + + + + +a b c d e f g

= + + + + + +8 8 3 6 12 5 10 = 52

(ii) Number of people who read exactly one newspaper

= + +a e g = + +8 12 10 = 30 [1]

10. In a survey of 200 students of a school, it wasfound that 120 study Mathematics, 90 studyPhysics and 70 study Chemistry, 40 studyMathematics and Physics, 30 study Physics andChemistry, 50 study Chemistry and Mathematicsand 20 none of these subjects. Find the number ofstudents who study all the three subjects.

Sol. Let M, P and C denote the students studying Mathematics,Physics and Chemistry, respectively.

Then, we have,

n U n M n P n C( ) , ( ) , ( ) , ( ) ,= = = =200 120 90 70

n M P n P C n M C( ) , ( ) , ( )∩ = ∩ = ∩ =40 30 50 [1]

and n M P C( )∪ ∪ ′ = 20

Now, n M P C n U n M P C( ) ( ) ( )∪ ∪ ′ = − ∪ ∪∴ 20 200= − ∪ ∪n M P C( ) [1]

⇒ n M P C( )∪ ∪ = − =200 20 180 [1]

We know that, n M P C n M n P n C( ) ( ) ( ) ( )∪ ∪ = + +− ∩ − ∩ − ∩n M P n P C n C M( ) ( ) ( )

+ ∩ ∩n C M P( )

∴ 180 120 90 70 40 30 50= + + − − −+ ∩ ∩n C M P( )

⇒ 180 280 120= − + ∩ ∩n C M P( ) [1]

⇒ 180 120 280+ − = ∩ ∩n P C M( ) [1]

∴ n P C M( )∩ ∩ = − =300 280 20

Hence, 20 students study all the three subjects. [1]


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1 Mark Questions

1. If A and B are two sets, such that n A n B n A B( ) , ( ) ( )= = ∩ =28 32 10and , then find the value of n A B( )∪ .

2. If A and B are two sets, such that A B∪ has 18 elements, A has 8 elements and B has 15 elements, then how

many elements does A B∩ have?

4 Marks Questions

3. In a school, there are 20 teachers who teach Maths or Physics. Out of these, 12 teach Maths and 4 teach

Physics and Maths. How many teach Physics?

4. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How

many people speak atleast one of these two languages?

5. In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and

75 were listed as taking both apple as well as orange juice. Find how many sutdents were taking neither apple

juice nor orange juice.

6. In a group of 65 people, 40 like nutrition Indian food, 10 like both Indian nutrition food and fast food. How many

like only fast food? What is your opinion about this set of people?

6 Marks Questions

7. In a town of 10000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B,

10% families buy newspaper C, 5% families buy A and B, 3% families buy B and C and 4% families buy A and

C. If 2% families buy all the three newspapers, then find

(i) the number of families which buy newspaper A only.

(ii) the number of families which buy newspaper of A, B and C only.

8. In a group of 50 students, the number of students studying French, English and Sanskrit were found to be as

follows

French = 17, English = 13, Sanskrit = 15, French and English = 9, English and Sanskrit = 4, French and

Sanskrit = 5 and English, French and Sanskrit = 3.

Then, find the number of students who study

(i) French only (ii) English only (iii) French and English both

Which language is study by most of the students?

Answers for

Topic 1 Sets and Their Types

1. A = − − − −{ , , , , , , , , }4 3 2 1 0 1 2 3 4

2. A = {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune}

3. (ii)

4. (i) { , }0 1 (ii) { , }1 p

5. (i) { : , }x x n Nn= ∈ ≤ ≤5 4and 1 n

(ii) { : , }x x n n N= ∈ ≤ ≤2 and 1 n 5

6. A C B D, ; ,

7. (i) Empty set (ii) Singleton set (iii) Infinite set (iv) and (v) Equal sets

8. Equal sets-B D C F= =, and Equivalent sets-A E H B D G C F, , ; , , ; ,

9. (i) T (ii) F (iii) T (iv) F

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sets · applications of set theory 1. definition of set a well-defined collection of objects, is...

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