Chapter 1 - Sets

Exercise: 1.1

Page Number:1.2

Question 1:

Solution:

Well-defined collections are sets.

Example:
The collection of good teachers in a school is not a set, It is a collection.
Thus, we can say that every set is a collection, but every collection is not necessarily a set.
The collection of vowels in English alphabets is a set.

Question 2.

(i) A collection of all natural numbers less than 50.

(ii) The collection of good hocky players in India.
(iii) The collection of all girls in your class.
(iv) The collection of most talented writers of India.
(v) The collection of difficult topic in mathematics.
(vi) The collection of all months of a yesr beginning with the letter J
(vii) A collection of novels by MunshiPrem Chand.
(viii) The collection of all questions in this chapter.
(ix) A collection of most dangerous animals of the world.
(x) The collection of prime integers.

Solution:
(i) The collection of all natural numbers less than 50 is a set because it is well defined.
(ii) The collection of good hockey players is not a set because the goodness of a hockey
player is not defined here. So, it is not a set.
(iii) The collection of all girls in a class is a set, as it is well defined that all girls of the class
are being talked about.
(iv)The collection of the most talented writers of India is a set because it is well defined.
(v) The collection of difficult topics in mathematics is not a set because a topic can be easy
for one student while difficult for the other student.
(vi) The collection of all months of a year beginning with the letter J is a set given by
{January, June, July}
(vii) A collection of novels written by MunshiPrem Chand is a set because one can determine
whether the novel is written by MunshiPrem Chand or not.

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(Viii) The collection of all question in this chapter is a set because one can easily check
whether it is a question of the chapter or not.
(ix) A collection of most dangerous animals of the world is not a set because we cannot
decide whether the animal is dangerous or not.
(x) The collection of prime integers is set given by 2, 3, 5..

Question 3:

Solution:

i  4 

 ii   4 

 iii  12  

 iv  9 

v 0 

 vi   2 

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Exercise: 1.2

Page Number: 1.6

Question 1:

Solution: Roster form:

In this form, a set is defined by listing elements, separated by commas, within braces {}.
 i  a, b, c, d 
 ii  1, 2,3, 4
 iii  11,13,17,19
 iv  2, 4, 6,8,10,
 v 
 vi  2,3,5
 vii  17, 26,35, 44,53, 62, 71,80
 viii  T , R, I , G, O, N , M , E, Y 
 ix  B, E , T , R
Question 2:

Solution:

Set-builder form:
To describe a set, a variable x (each element of the set) is written inside braces. Then, after
putting a colon, the common property P  x  ) possessed by each element of the set is written
within braces.
(i) { x : x  N , x  7}
(ii) {x : x  1n, x  N }
(iii) {x : x  3n, n  Z }
(iv) {x : x  N ,9  x  16}
(v)  x : x  0
(vi) { x 2 : x  N ,1  n  10}
(vii) {x : x  2n, n  N }
(viii) {5n : n  N ,1  n  4}

Question 3:

Solution:
(i) A  0, 1, 2, 3
1 1 1 1
(ii) B  1, , , ,
3 5 7 9

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 (iii) C  0,1, 2, 3, 4
(iv) D   A, E , I , O, U 
(v) E={February, April, June, September, November}
(vi) F  M , I , S , P

Question 4:

Solution:

(i) implies (vi)

(ii) implies(v)
(iii) implies(i)
(iv) implies(iv)
(v) implies(iii)
(vi) implies(ii)

Question 5:

Solution:

The set of vowels in the English alphabet that precede q is a, e, i, o .

Question 6.
Solution:

The set of all positive integers whose cube is odd is {2n  1 : n  Z , n  0}

Question 7.

Solution:

The set-builder form of the set

. 12, 25,310, 417,526, 637, 750 . is:
 n 
 2 : n  N , n  7
 n  1 

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Exercise: 1.3

Page Number: 1.9

Question 1:

Solution:

(i) All natural numbers that end with 0 are even & divisible by 5. Therefore, the given set is
not an example of empty set.

(ii) 2 is an even prime number. Therefore, the given set is not an example of empty set.

(iii) There is no rational number whose square is 2 such that x 2  2 . Therefore, it is example
of empty set.

(iv) It is not possible that x  8 and, at the same time, x  12 . Therefore, it is an example of
empty set.

(v) There is no common point in two parallel lines. Therefore, it is an example of empty set.

Question 2.

Solution:

(i) There can be infinite concentric circles in a plane. Therefore, it is an infinite set.
(ii) There are 26 letters in the set of English alphabet. Therefore, it is a finite set.
(iii) {x  N : x  5}  6, 7,8,9,  . There will be infinite numbers. So, it an infinite set.
(iv) There are finite elements in the set {x  N : x  200}. Therefore, it is a finite set.
(v) In this set, x  Z , so there would be infinite elements in the set {x  Z : x  5}. Therefore, it
is an infinite set.
(vi) In this set, x  R. We know real numbers include all numbers, i.e., decimal numbers,
rational numbers and irrational numbers.
So, there would be infinite elements in the set {x  R : 0  x  1}. Therefore, it is an infinite
set.

Question 3.

Solution:

Two sets A & B are equal if every element of A is a member of B & every element of B is a
member of A.
(i) A  1, 2,3
(ii) B   x  R : x 2  2 x  1  0 set B would be 1 .
(iii) C  1, 2, 2,3

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It can be written as 1, 2, 3 because we do not repeat the elements while writing the elements
of a set.
C  1, 2,3
(iv) D   x  R : x3  6  2  11x  6  0 includes elements 1, 2, 3 .
D  1, 2,3
Hence, we can say that A  C  D .

Question 4.

Solution:

Here, A  B because every element of A is a member of B & every element of B is a member

of A.
But every element of C is not a member of A and B.
Also, every element of A and B is not a member of C.
Therefore, we can say that these sets are not equal.

Question 5.

Solution:

Two sets A and B are equivalent if their cardinal numbers are equal,
That is,
n  A  n  B  n  A
 3n  B 
 5n  C 
 3n  D 
5
Therefore, equivalent sets are (A and C) and (B and D).

Question 6.

Solution:
(i) A  2,3 , B  2, 3
A is not equal to B because every element of A is not a member of B and every element of B
is not a member of A.
(ii) A  W , O, L, F 
And, B   F , O, L, W 
Here, A  B because every element of A is a member of B and every element of B is a
member of A.

Question 7.

Solution:

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Equal sets:
(a) B and D, because every element of B is a member of D and every element of D is a
member of B.
(b) C and F, because every element of C is a member of F and every element of F is a
member of C
Equivalent sets:
(a) A, E and H
Since,
n  A  n  E   n  H   2
(b) B, D and G
Since,
n  B   n  D   n G   4
(c) C and F
Since,
n C   n  F   3

Question 8.

Solution:

A  1, 2
B  1, 2
C  3, 1
D  1, 3
E  1, 2, 1, 1  1, 2
F  1, 1, 3  1, 3
Therefore, A  B  E and C  D  F

Question 9.

Solution:

Letters required to spell CATARACT are {C, A, T, R}. Let this set be denoted as E.
E = {C, A, T, R}
Letters required to spell TRACT are {T, R, A, C}. Let this set be denoted as F.
F = {T, R, A, C}
The two sets E & F are equal because every element of E is a member of F & every element
of F is a member of E.

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Exercise: 1.4

Page Number: 1.16

Question 1.

Solution:

(i) False.
It is not necessary that for any two sets A & B, either A⊆B or B⊆A.It is not satisfactory
always.
Let: A  1, 2 & B  a, b, c
Here, neither A⊆B nor B⊆A.

(ii) False.

A  1, 0,1, 2,3 is a finite set that is a subset of infinite set Z.

(iii) True.

Every subset of a finite set is a finite set.

(iv) False.
.
ϕ does not have a proper subset.

(v) False.
a, b, a, b, a, b,... will be equal to a, b which is a finite set.
(vi) True.
a, b, c and 1, 2, 3 are equivalent sets because the number of elements in both the sets are
same.

(vii) False.
In the set A  1, 2 , subsets can be {ϕ}, {1} and {2}, which are finite

Question 2.

Solution:

(i) True

(ii) False
It should be written as a  b, c, a or a  b, c, a .

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(iii) False
It should be written as a  b, c, a or a  b, c, a
.
(iv) True

(v) False
The element of the set  x : x  8  8 is {0}. Therefore, it is not an empty or null set.

Question 3.

Solution:
We have:
A   x : x satisfies x 2  8 x  12  0.  2, 6
B  2, 4, 6
C  2, 4, 6, 8,
D  6
Therefore, we can say that D  A  B  C .

Question 4.

Solution:

(i) True
m
A rational number is any , where m and n are any integers ( n  0 ). Any integer can be put
n
into that form by setting n = 1. Therefore, the set of all integers is contained in the set of all
rational numbers.

(ii) True
All crows are birds. Therefore, the set of all crows is contained in the set of all birds.

(iii) False
Every square can be a rectangle, but every rectangle cannot be a square.

(iv) True
Every real number can be written in the (a + bi) form. Thus, we can say that the set of all real
numbers is contained in the set of all complex numbers.

(v) False
P = {a}
B = {{a}} = {P}
P≠{P}

(vi) True
We have:
A = {x:x is a letter of the word LITTLE} = {L, I, T, E}
B = {x:x is a letter of the word TITLE} = {T, I, L, E}

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Sets A & B are equal because every element of A is a member of B & every element of B is a
member of A.

Question 5.

Solution:

Here, (viii) is correct.

The correct forms of each of the incorrect statements are:
(i) a∈a, b, c
(ii) a⊂a, b, c
(iii) {a}∈{a, b}
(iv) {a}⊂a, b
(v) b, c∈a,b, c
(vi) {a,b}⊄{a,{b,c}}
(vii) ϕ⊂a, b
(ix) x:x+3=3≠ϕ

Question 6.

Solution:
A = {a, b, {c, d}, e}
(i) False
The correct statement would be {c, d}⊂A.
(ii) True
(iii) True
(iv) True
(v) False
The correct statement would be {a}⊂ A or a ∈ A.
(vi) True
(vii) False
The correct statement would be a, b, e⊂A.
(viii) False
The correct statement would be {a, b, c} ⊄ A.
(ix) False
A null set is a subset of every set. Therefore, the correct statement would be ϕ⊂A.
(x) False
ϕ is an empty set; in other words, this set has no element. It is denoted by ϕ. Therefore, the
correct statement would be ϕ⊂A.

Question 7.

Solution:
(i) False
If it could be 1∉A , then it would be true .
(ii) False
The correct form would be 1, 2, 3∈A or {1, 2, 3}⊂A.
(iii) True
(iv) True

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(v) False
A null set is a subset of every set. Therefore, the correct form would be ϕ ⊂ A.
(vi) True

Question 8.

Solution :
(i) True
(ii) True
(iii) False
The correct form would be 1⊂A.
(iv) True
(v) False
The correct form would be 1∈A.
(vi) True
(vii) True
(viii) True
(ix) True

Question 9.

Solution:
(i) ϕ ,{a}
(ii) (ii) ϕ,{0},{1},{0,1}
(iii) (iii) ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}
(iv) (iv) ϕ,{1},{{1}},{1,{1}}
(v) (v) ϕ,{ϕ}

Question 10.

Solution:
(i) {1}, {2}
(ii) {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}
(iii) No proper subsets are there in this set.

Question 11.

Solution:
We know that the total number of subsets of a finite set consisting of n elements is 2n.
Therefore, the total number of proper subsets of a set consisting of n elements is 2n-1.

Question 12.

Solution:
To prove: A    A  
Proof:
Let: A  
If A is a subset of an empty set, then A is the empty set.
∴ A=ϕ

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Now, let A=ϕ.
This means that A is an empty set.
We know that every set is a subset of itself.
∴ A
Thus, we have:
A  A

Question 13.

Solution:
Let x  A  x  B  A  B  x  C  B  C  x  A  x  C  A  C …(1)
It is given that,C⊆A …(2)
From 1 and 2, we have A  C

Question 14.

Solution:
Given: A   This means P  A     .
Hence, P  A  would have one element.

Question 15.

Solution:
(i) The set of all triangles in a plane.
(ii) The set of all triangles in a plane

Question 16.

Solution:

Given:
X  8n  7 n  1: n  N and Y  49  n  1 : n  N
To prove:
X⊆Y
Let : xn  8n  7 n  1, n  N
 x1  8  7  1  0
For any n2, we have :
n
xn  8n  7 n  1  1  7   7 n  1
2 3
 xn  nC0  nC1  7   nC2  7   nC3  7   nCn 7 n  7 n  1
 xn  1  7 n  nC0 7 2  nC0 73   7 n  7 n  1
[ nC0  1 and nC1  n]
 xn  7 2  C  C  7   C  7    C 7 
n
2
n
3
n
4
2 n
n
n2

 49  C  C  7   C  7   C 7 
n n n 2 n n2
 xn 2 3 4 n

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Thus, xn is some positive integral multiple of 49 for all n⩾2. X consists of all those positive

 2
integral multiples of 49 that are of the form 49 nC2  nC3  7   nC4  7   nCn  7 
n2

along with zero. Y  {49  n  1 : n  N } implies that it consists of all integral multiples of 49
along with zero.∴ X⊆Y

Exercise: 1.5

Page Number:1.21

Question 1:

Solution:

From the Venn diagrams given below, we can clearly say that if A and B are two sets such

that A  B then

(i) Form the given Venn diagram, we can see that A  B  A

(ii) Form the given Venn diagram, we can see that A  B  B

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Question 2

Solution:

Given:
A  1, 2, 3, 4,5 , B  4,5, 6, 7,8 , C  7,8,9,10,11 and D  10,11,12,13,14

(i) A  B  1, 2,3, 4,5, 6, 7,8

(ii) A  C  1, 2, 3, 4, 5, 7,8,9,10,11
(iii) B  C  4,5, 6, 7,8, 9,10,11
(iv) B  D  4, 5, 6, 7,8,10,11,12,13,14
(v) A  B  C  1, 2,3, 4,5, 6, 7,8, 9,10,11

(vi) A  B  D  1, 2,3, 4,5, 6, 7,8,10,11,12,13,14

(vii) B  C  D  4, 5, 6, 7,8,9,10,11,12,13,14
(viii) A  B  C  4,5
(ix) A  B  B  C  
(x) A  D  B  C  4,5,10,11

Question 3

Solution:

A=x:x∈N={1,2,3,…}B=x:x-2n, n∈N={2,4,6,8,…} C=x:x=2n-1, n∈N={1,3,5,7,…}

D = {x:x is a prime natural number.} = {2, 3, 5, 7,…}
(i) A∩B = B
(ii) A∩C = C
(iii) A∩D = D
(iv) B∩C = ϕ
(v) B∩D = {2}
(vi) C∩D = D-{2}

Question 4

Solution:

Given:
A = {3, 6, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5,
10, 15, 20}
(i) A-B = {3, 6, 15, 18, 21}
(ii) A-C = {3, 15, 18, 21}

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(iii) A-D = {3, 6, 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}

Question 5

Solution:

Given:
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}
(i) A’ = {5, 6, 7, 8, 9}
(ii) B’ = {1, 3, 5, 7, 9}
(iii) A∩C’ = {1, 2, 5, 6, 7, 8, 9}
(iv) A∪B’ = {5, 7, 9}
(v) A” = {1, 2, 3, 4} = A
(vi) B-C’ = {1, 3, 4, 5, 6, 7, 9}

Question 6

Solution:

Given:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}
We have to verify:

(i) A∪B’=A’∩B’
LHS
A∪B ={2,3,4,5,6,7,8}A∪B ‘={1,9}

RHS
A’={1,3,5,7,9}B’={1,4,6,8,9}A’∩B’={1,9}

LHS = RHS
Hence proved.

(ii) A∩B’=A’∪B’

LHS
A∩B={2}A∩B’={1,3,4,5,6,7,8,9}

RHS
A’={1,3,5,7,9}B’={1,4,6,8,9}A’∪B’={1,3,4,5,6,7,8,9}

LHS = RHS
Hence proved.

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Exercise: 1.6

Page Number:1.27

Question 1.

Solution :
We have to find the smallest set A such that A  1, 2  1, 2,3, 5,9 . The union of the two
sets A & B is the set of all those elements that belong to A or to B or to both A & B .
Thus, A must be 3, 5, 9 .

Question 2.
i  A  B  C  A  B  A  C
 ii  A  B  C  A  B  A  C
(iii) A  B  C  A  B  A  C
 iv  A  B  C  A  B  A  C
v A  B  C  A  B  A  C
 vi  A  BC  A  BA  C

Solution :
Given:
A  1, 2, 4,5 , B  2, 3, 5, 6 and C  4, 5, 6, 7 We have to verify the following identities:
i  A  B  C  A  B  A  C
LHS
 B  C   5, 6 A   B  C   1, 2, 4,5, 6
RHS
A  B  1, 2, 3, 4,5, 6 A  C  1, 2, 4,5, 6, 7 A  B  A  C  1, 2, 4, 5, 6
LHS = RHS
 A B C  A B  AC
 ii  A  B  C  A  B  A  C
LHS
( B  C )  2,3, 4,5, 6, 7 A  ( B  C )  2, 4,5
RHS
A  B  2, 5 A  C  4,5 A  B  A  C  2, 4,5
LHS = RHS
 A B C  A B  AC
(iii) A  B  C  A  B  A  C
LHS
 B  C   2,3 A   B  C   2
RHS
 A  B   2,5 A  C   4,5 A  B    A  C   2

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LHS = RHS
 A B C  A B  AC
 iv  A  B  C  A  B  A  C
LHS
( B  C )  2,3, 4,5, 6, 7 A  ( B  C )  1
RHS
 A  B   1, 4 A  C   1, 2 A  B    A  C   1
LHS = RHS
 A B C  A B  AC
v A  B  C  A  B  A  C
LHS
 B  C   5, 6 A   B  C   1, 2, 4
RHS
 A  B   1, 4 A  C   1, 2 A  B    A  C   1, 2, 4
LHS = RHS
 A B C  A B  AC
 vi  A  BC  A  BA  C
LHS
 BC    B  C    C  B  B  C   2,3 C  B   4,7 B  C    C  B   2,3, 4,7
  BC   2,3, 4,7 A   BC   2, 4
RHS
 A  B   2, 5 A  C   4, 5 A  B    A  C    A  B    A  C    A  C    A  B   A  B 
  A  C   2 A  C    A  B   4 A  B    A  C   { A  C    A  B }  2, 4
  A  B    A  C   2, 4

LHS = RHS
 A  BC  A  BA  C

Question 3.

Solution :
Given:
U = {2, 3, 5, 7, 9}
A = {3, 7}
B = {2, 5, 7, 9}
To prove :
(i)  A  B   A  B
(ii) A∩B’=A’∪B’
Proof :
(i) LHS:
(A∪B)={2,3,5,7,9}(A∪B)’=ϕ
RHS:
A’={2,5,9}B’={3}A’∩B’=ϕLHS=RHS

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∴ A∪B’=A’∩B’
(ii) LHS:
(A∩B)={7}(A∩B)’={2,3,5,9}
RHS:
A’={2,5,9}B’={3}A’∪B’={2,3,5,9}
LHS = RHS
∴ A∩B’=A’∪B’

Question 4.

Solution:
 i  For all x  B
 x  Aor x  B
 x A B  Difinition of union of sets 
 B  A B
 ii  For all x  A  B
 x  A and x  B  Difinition of union of sets 
 x A

Question 5.

Solution :

We have that the following statements are equivalent:

(i) A⊂B
(ii) A-B=ϕ
(iii) A∪B=B
(iv) A∩B=A
Proof:
Let A⊂BLet x be an arbitrary element of (A-B).
Now, x   A  B   x  A & x  B (Which is contradictory) Also,∵A⊂B⇒A-B⊆ϕ …(1) We
know that null sets are the subsets of every set.
∴ϕ⊆ A-B …(2)From (1) & (2), we get,(A-B)=ϕ∴(i)=(ii)
Now, we have,(A-B)=ϕ
That means that there is no element in A that does not belong to B.
Now, A∪B=B∴(ii)=(iii) We haveA∪B=B⇒A⊂B⇒A∩B=A∴(iii)=(iv)We have, A∩B=A
It should be possible if A⊂B.
Now,A⊂B∴ (iv)=(i)We have,(i)=(ii)=(iii)=(iv)
Therefore, we can say that all statements are equivalent.

Question 6.

Solution :
(i) Let A  2, 4, 5, 6 , B  6, 7, 8, 9 and C  6, 10, 11, 12,13
So, A∩B=6 and A∩C=6

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Hence, A  B  A  C but B  C
(ii) Let z  C  B …(1)
 z  C and z  B  z  C and z  A A  B  z  C  A … (2)From (1) and (2),
we get C  B  C  A

Question 7.

Solution :
(i)

LHS  A  A  B
 LHS  A  A  A  B
 LHS  A  A  B
 A  A B
 LHS  A  RHS

(ii)

LHS  A  A  B
 LHS  A  A  A  B
 LHS  A  A  B
 LHS  A  RHS

Question 8.

Solution :
Let us consider the following sets,
A = {5, 6,10 }
B = {6,8,9}
C = {9,10,11}
Clearly, A∩B=6B∩C=9, A∩C=10 and A∩B∩C = ϕIt means that, A∩B,B∩C and A∩C are
non emptysetsand A∩B∩C = ϕ

Question 9.
Solution :

Let a  A  a  B  A  B   .
 a  B’
Thus, a  A and a  B’  A  B’

Question 10.

Solution :
(i) A-B and A∩BLet a  A  B  a  A and a  B  a  A  B
Hence, A-B and A∩B are disjoint sets.

(ii) B-A and A∩BLet a  B  A  a  B and a  A  a  A  B

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Hence, B-A and A∩B are disjoint sets.

(iii) A-B and B-A A-B=x:x∈A and x∉B B-A=x:x∈B and x∉A
Hence, A-B and B-A are disjoint sets.

Question 11.

Solution:

LHS  A  B  A  B’
 LHS  A  B  A  A  B  B’
 LHS  A  B  A  A  B  B’
 LHS  A  A  B  B’
 LHS  A  A  B’  B  B’ B  B  
 LHS  A  A  B’
 LHS  A  RHS

Question 12.

Solution :

(i)

Let a  A
 a U
 a  A’  B U  A’  B
 a  B  a  A’
Hence, A  B.

(ii)

Let a  A
 a  A’
 a  B’ B’  A’
 aB
Hence, A  B.

Question 13.

Solution :

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False,
Let X  P  A   P  B 
 X  P  A  or X  P  B 
 X  A or X  B
 X  A B
 X  P  A  B
 P  A  P  B   P  A  B 
Again, let X  P  A  B  But X  P  A  or x  P  B 
For example let A  2,5 and B  1, 3, 4 and take X  1, 2,3, 4
So, X  P  A  P  B 
Thus, P  A  B  is not necessarily a subset of P  A   P  B  .

Question 14.

Solution :
(i)
RHS   A  B    A  B 
 RHS   A  B    A  B ’
 RHS   A  B    A  A    B  B ’
 RHS  A   A  B ’   B  B ’
 RHS  A   A  B ’  U
 RHS  A   A  B ’
 RHS  A  LHS

(ii)

LHS  A   B  A 
 LHS  A   B  A ’
 LHS   A  B    A  A’
 LHS   A  B   U
 LHS  A  B  RHS

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Question 15.
20
It is given that each set X contains 5 elements and  X r  S .
r 1

 n  S   20  5  100

But, it is given that each element of S belong to exactly 10 of the X r ' s.

100
Number of distinct elements in S   10...... 1
10
It is also given that each set Ycontains 2 elements and
n

 Y = S.
r 1
r

 n  S   n  2  2n
Also, each element of S belong to eactly 4 of Yr 's.
2n
.
4
Number of distinct elements in S  4

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Exercise: 1.7

Page Number:1.34

Question1.

Solution:
LHS  A  B   A  B 

 C  D   C  D   A’  B   B  A ’  B  A
  C  D ’  C  D
RHS  B  A

So, LHS  RHS

Question 2.

Solution:

(i)
LHS  A   A’  B 
  A  A’   A  B 
    A  B
 A  B  RHS
Hence proved.

(ii)
LHS  A   A  B 
 A   A  B’
 A   A  B ' '
 A   A’   B ' '
 A   A’  B 
  A  A’   A  B 
    A  B
  A  B
 RHS

Hence proved

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(iii)
LHS  A   A  B ’
 A   A’  B’
  A  A’   A  B’
    A  B’
   RHS
Hence proved.

(iv)

LHS  A  A  B 
  A   A  B  '   A  B   A
  A   A  B  '   A  B   A '
  A   A ' B '    A  B   A '
  A  A '    A  B '    A  A '   B  A ' 
  A  B '    
  A  B '
 A B
 RHS

Question 3.

Solution:

Let a  C  B

 a  C and a  B
 a  C and a  A  A  B
 a C  A

Hence, C  B  C  A

Question 4.

Solution:
(i)

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  A  B  B   A  B  B '  X  Y  X  Y '
  A  B '   B  B '
  A  B '  
 A B'
 A B

(ii)
A   A  B  A   A  B '  X  Y  X  Y '
 A   A ' B '
  A  A '   A  B ' 
    A  B '
 A B'
 A B
(iii)

A   B  A  A   B  A '  X  Y  X  Y '
  A  B    A  A '  Distributive law
  A  B U   is the universal set 

(iv)

 A  B    A  B    A  B '   A  B   X  Y  X  Y '
.  A   B ' B   Distributive law 
 A U
A

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Exercise: 1.8

Page Number:1.46

Question 1:

Solution:
Consider,
n( A  B )  n  A   n  B   n  A  B 
 50  28  32  n  A  B 
 n  A  B   60  50  10

Question 2:

Solution:
Consider:
n  P   40
n  P  Q   60
n  P  Q   10
To find:n(Q)
n( P  Q )  n  P   n  Q   n  P  Q 
 60  40  n  Q   10
 n  Q   30

Question 3.

Solution:
Let A be the number of teachers who teach mathematics &B be the number of teachers who
teach physics.
Consider:
n  A   12
n  A  B   20
n  A  B  4
To find: n  B  consider:
n  A  B   n  A  n  B   n  A  B 
20  12  n  B   4
n  B   20  8  12
Therefore, 12 teachers teach physics.

Question 4.

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Solution:
Let A denote the set of the people who like tea & B denote the set of the people who like
coffee.
Consider:
n( A  B)  70
n  A   52
n  B   37
To find: n  A  B 
Consider:
n( A  B )  n  A   n  B  – n  A  B 
70  52  37 – n  A  B 
n  A  B   19
Therefore, 19 people like both tea & coffee.

Question 5.

Solution:
Consider:
n  A   20, n  A  B   42 and n  A  B   4
(i) Consider: n ( A  B )  n  A   n  B   n  A  B   42  20  n  B   4  n  B   26
(ii) n  A  B   n  A   n  A  B   n  A  B   20  4  16
(iii) Consider that sets follow the commutative property.
Therefore, n  A  B   n  B  A  n  B  A   n  B   n  B  A   n  B  A   26  4  22

Question 6.

Solution:
Consider:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}
We have to verify:
(i) A∪B’=A’∩B’
LHS
A∪B ={2,3,4,5,6,7,8}A∪B ‘={1,9}
RHS
A’={1,3,5,7,9}B’={1,4,6,8,9}A’∩B’={1,9}
LHS = RHS
Hence proved.
(ii) A∩B’=A’∪B’
LHS
A∩B={2}A∩B’={1,3,4,5,6,7,8,9}
RHS
A’={1,3,5,7,9}B’={1,4,6,8,9}A’∪B’={1,3,4,5,6,7,8,9}

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LHS = RHS
Hence proved.

Question 7.

Solution:
Let A & B denote the sets of the persons who like Hindi & English, respectively.
Consider:
n  A   750
n  B   460
n  A  B   950

(i) Consider:
n  A  B   n  A  n  B   n  A  B 
950  750  460  n  A  B 
n  A  B   260
Thus, 260 persons can speak both Hindi and English.

(ii)
n  A  B   n  A  n  A  B 
n  A  B   750  260  490
Thus, 490 persons can speak only Hindi.

(iii)
n  B  A  n  B   n  A  B 
n  B  A  460  260  200
Thus, 200 persons can speak only English.

Question 8.

Solution:
Let A&B denote the sets of the persons who drink tea & coffee, respectively.
Consider:
n  A  B   50
n  A   30
n  A  B   14
(i) n  A  B   n  A   n  A  B   14  30  n  A  B   n  A  B   16 Thus, 16 persons
drink tea and coffee both.
(ii) n  A  B   n  A   n  B   n  A  B   50  30  n  B   16  n  B   36
We have, n  B  A   n  B   n  A  B   n  B  A   36  16  20
Thus, 20 persons drink coffee but not tea.

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Question 9.

Solution:
Consider:
n  H   25
n T   26
n  I   26
nH  I   9
n  H  T   11
n T  I   8
nH T  I   3
(i) Consider:
n  H  T  I   n  H   n T   n  I   n  H  T   n  T  I   n  H  I   n  H  T  I 
n  H  T  I   25  26  26  11  8  9  3  52
Thus, 52 people can read at least one of the newspapers.
(ii) Now, we have to calculate the number of people who read exactly one newspaper.
We have:
n  H   n T   n  I   2n  H  T   2n T  I   2n  H  I   3n  H  T  I 
 25  26  26  22  16  18  9  30
Thus, 30 people can read exactly one newspaper.

Question 10.

Solution:
Let A, B&C be the sets of members in basketball team, hockey team & football team,
respectively.
Consider:
n  A   21
n  B   26
n  C   29
n  A  B   14
n  B  C   15
n  A  C   12
n A B C  8
Consider:
n  A  B  C   n  A  n  B   n  C   n  A  B   n  B  C   n  A  C   n  A  B  C 
 n  A  B  C   21  26  29  14  15  12  8  43
Therefore, there are 43 members in all teams.

Question 11.

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Solution:
Let A & B denote the sets of the persons who can speak Hindi & Bengali, respectively.
Consider:
n  A  B   1000
n  A   750
n  B   400
n  A  B   n  A  n  B   n  A  B 
 1000  750  400  n  A  B 
 n  A  B   150
Number of persons who can speak both Hindi and Bengali is n  A  B   150 Number of
persons who can speak only Hindi is n  A  B   n  A   n  A  B   750  150  600
Number of persons who can speak only Bengali is
n  B  A   n  B   n  A  B   400  150  250

Question 12.

Solution:
Let F, H B denote the sets of students who watch football, hockey and basketball,
respectively.
Also, let U be the universal set.
We have:
n  F   285,
n  H   195,
n  B   115,
n  F  B   45,
n  F  H   70 and n  H  B   50
Also, Consider:
n  F ’  H ’  B’  50
 n( F  H  B)’  50
 n U  – n( F  H  B)  50
 500 – n( F  H  B )  50
 n( F  H  B)  450
Number of students who watch all three games = n(F∩H∩B)
 n( F  H  B ) – n  F  – n  H  – n  B   n  F  B   n  F  H   n  H  B 
 450 – 285 – 195 – 115  45  70  50
 20
Number of students who watch exactly one of the three games
 n  F   n  H   n  B  – 2 n  F  B   n  F  H   n  H  B   3n  F  H  B 
 285  195  115 – 2  45  70  50   3  20   325

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Question 13.

Solution:
Let A, B C be the sets of the persons who read magazines A, B and C, respectively. Also, let
U denote the universal set.
We have: n(U) = 100
n  A   28,
n  B   30,
n  C   42,
n  A  B   8,
n  A  C   10,
n  B  C   5 and n  A  B  C   3
Now,
Number of persons who read none of the three magazines = n(A’∩B’∩C’)
n( A  B  C )’  n U  – n( A  B  C )
 n U  – n  A   n  B   n  C  – n  A  B  – n  A  C  – n  B  C   n  A  B  C 
 100 –  28  30  42 – 8 – 10 – 5  3  20

Number of students who read magazine C only = n(C∩A’∩B’)

 n{C  ( A  B )’}
 n  C  – n C  ( A  B )
 n  C  – n{ C  A    C  B }  n  C  – n  C  A    C  B    A  B  C 
 42 – 10  5 – 3  30

Question 14.

Solution:
Let E, H and S be the sets of students who study English, Hindi and Sanskrit, respectively.
Also, let U be the universal set.
Now, we have:
n  E   26, n  S   48, n  E  S   8 and n  S  H   8
Also,
n  E  H ’  23
 n  E  – n  E  H   23
 26 – n  E  H   23
 nE  H   3
Therefore, the number of students studying English and Hindi is 3

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n  E  H ’  S’  18
 n  E  – n{E  ( H  S )’}  18
 26 – n{ E  H    E  S }  18
 26 – 3  8 – n  E  H  S   18
 nE  H  S   3
Also,
n  E’  H ’  S’  24
 n U  – n( E  H  S )  24
 n( E  H  S )  76

Therefore, Number of students studying Hindi is:

n( E  H  S ) – n  E  – n  S   n  E  H   n  E  S   n  S  H  – n  E  H  S 
 76 – 24 – 48  3  8  8 – 3  18

Question 15.
Solution:
Let P1, P2 and P3 denote the sets of persons liking products P1, P2 and P3, respectively.
Also, let U be the universal set.
Thus, we have: n(P1) = 21, n(P2) = 26 and n(P 3) = 29 And, n(P1  P2) = 14,
n(Pi  p3) = 12, n(P2  P3) = 14 and n(P1  P2  P3) = 8
Now, Number of people who like only product P3:


n  P3 P1' P2 '  n P3   P1 P2 
'

 n  P3   n  P3   P1 P2  
 n  P3   n  P3 P1    P3 P2   .
 n  P3    n  P3 P1   n  P3 P2   n  P1 P2 P3  
 29  12  14  8  11

Therefore, number of people like Product P3 is 11

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Exercise: Very short Question

Page number:1.49

Question 1.

Solution:
A set having n Elements has 2n Subsets or Elements.

Question 2.
Solution:
It is known that a set of n Elements has 2n Subsets or Elements.

A null set has no Element(s) in it.

∴ Number of Elements in the power set of null set  20  1

Question 3

Solution:
A  {x : x  N and x is a multiple of 3}
 3, 6, 9,12,15,18, 21, 24, 27,30, 33,36,39, 42, 45, 

B  {x : x  N and x is a multiple of 5}
 5,10,15, 20, 25,30, 35, 40, 45,

Therefore, it is clear that:

A  B  15, 30, 45,
 {x : x  N , where x is a multiple of 15}

Question 4.

Solution:
It is known that nA  B  nA  nB  nA  BnA  B is minimum when nA  B is maximum.
Therefore, nA  B  3
Thus , nA  B  nA  nB  nA  B  3  6  3  6

Question 5.

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Solution:
As given:
A  {x  C : x 2  1}
 A  1,1

Then,
B  {x  C : x 4  1}
 B   x 4  1  0
 B   x 2  1 2  1  0

Question 6.

Solution:
B’  A’  Not B – Not A  Nothing common in them  

Question 7.

Solution:
We know that nA  B  nA  nB  nA  BnA  B is maximum when nA  B is minimum.
Therefore , nA  B  0
Thus, nA  B  nA  nB  nA  B  4  7  0  11

Question 8.

Solution:
As we know:
A  x, y : y
 1x, 0  x  R
 1,1, 2,12,3,13, 4,14, 

Then,
B  x, y : y
  x, x  R
 1, 1, 2, 2,3, 3, 4, 4, 

Therefore, we get:
A B  

Question 9.

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Solution:
As we know:
A  0,1,1, e, 2, e2, 3, e3, 
B  0,1,1, e  1, 2, e  2,3, e  3, 
Therefore, It is clear that:
A  B  0,1

Question 10.

Solution:
As we know:
nA  20, nB  25 and nA  B  40

We know:
nA  B  nA  nB – nA  B
 nA  B  nA  nB  nA  B
 20  25 – 40
 5

Question 11

Solution:
nA  115,
nB  326 and
nA  B  47
Then,
nA – nA  B  nA  B
 115 – nA  B  47
 nA  B  68

Therefore, it is clear that:

nA  B  nA  nB – nA  B
 115  326 – 68
 373

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Exercise: 1.MCQ

Page Number :1.49

Question 1.

Solution:
(b) A
The Complement of the Complement of a set is the set itself.

Question 2.

Solution:
(c) A  B’

A  B belongs to those elements of A that do not belong to B.

 A  B  A  B’

Question 3.

Solution:
(d) 2 n
The total number of subsets of a finite set consisting of n elements is 2n.

Question 4

Solution:
(a) A
A  ( A  B)   A  A    A  B 
 A   A  B
A

Question 5.

Solution:
(d) None of these
4 A
4  A
B A
Therefore, we can say that none of these options satisfy the given relation

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Question 6.

Solution:
(b) A  B  B  A

The symmetric difference of A and B is:-

 A  B    B  A

Question.7

Solution:
(b) 1, 2, 4,5

Here,
A  1, 2, 3 and B  3, 4, 5
The symmetric difference of A and B is : -
 A  B    B  A
Now, We have:
 A  B   1, 2 B  A
 4,5 A  B    B  A 
 1, 2, 4,5

Question 8.

Solution:
(c) A  B  A  B
A  B  B  A  A  B’  B  A’
 A  B  A’  B’  B  A’

Applying distribution law  A  B  A  A’  B’  B  B’  A’

Applying distribution law  A  B  U  U  B’  A’ A  A’  U

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  B’  B
 A  B  B’  A’ A  B  U
 A  B and U  B’  A’
 B’  A’
 A  B  A  B’ A  B’
 B’  A’
 A B  A B  A B
 A B  A B

Question 9.

Solution:
(c) A  B  A  B’

It includes all those elements of A, which do not belong to Complement of B, which is equal to
A∩B but not equal to A  B.
Therefore, (c) is false.

Question 10.

Solution:
(a) A  B  C  A  B  A  C
Consider x be any arbitrary element of A  B  C .

Therefore, we have,

x  A  B  C  x  A and x  B  C
 x  A and x  B and x  C
 x  A and x  B and x  A and x  C
 xA  B and x  A  C
 x A B – AC
 A B C  A B – AC

Similarly, A  B – A  C  A  B  C

Hence, A  B  C  A  B – A  C

Question 11.

Solution:

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(c)  4, 5 
A  x : x  R,
x  4 and B  x  R : x  5
A  B   4, 5

Question 12

Solution:
(c) 300
n  A’  B’  nA  B’
 nU – n( A  B )
 700 – 200  300  100
 300

Question 13

Solution:
We kNow:
nA  B  nA nB – nA  B
Now,
nA  B  nA  nB – n( A  B )
 16  14 – 25
5

Question 14

Solution:
(c) 31

The number of proper subsets of any set is given by the formula 2n-1, where n is the number of
elements in the set.
Here,
n5
∴ Number of proper subsets of A  25  1  31

Question 15.

Solution:
(c) x : x  x

Question 16.

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Solution:
(a) nA  nB
Two sets are disjoint if they do not have a common element in them, i.e., A  B  .
 nA  B  nA  nB

Question 17.

Solution:
(a) B  A
The union of two sets is a set of all those elements that belong to A or to B or to both A and B.
If A  B  A, then B  A .

Question 18.

Solution:
(d) 20
We have:
n  A  B   nA  nB – n( A  B )  70  60 – 110  20

Question 19.

Solution:
(d) A  Bc
A and B are two sets.
A  B is the common region in both the sets.
A  Bc is all the region in the universal set except A  B .
Now,
A  A  Bc  A  Bc

Question 20

Solution:
(b) A  B
A   x : x is a multiple of 3
A  3, 6,9,12,15,18, 21, 24, 27,30,33,36,39, 42, 45, 48, 
B   x : x is a multiple of 5.
B  5,10,15, 20, 25, 30, 35, 40, 45,50, 
Now, we have:
A  B  3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 
 A B

Question 21

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Solution:
(c) 60%

Suppose C and B represents the population travel by car and Bus respectively.
nC  B  nC  nB – nB  C  0.20  0.50 – 0.10  0.6 or 60%

Question 22

Solution:
(b) B  A
Only this case is possible.

Question 23.

Solution:
(b) 20

Question 24

Solution:
(c) 6, 4
A.T.Q.:

2m  1  48  2n  1
 2m  2n  48
 2m  2n  26  24
By comparing we get: m  6 and n  4

Question 25

Solution:
(c) 60

Consider M, P and C denote the sets of students who have opted for mathematics, physics, and
chemistry, respectively.
Here,
nM  100, nP  70 and n  C   40
Now,
nM  P  30, nM  C  28, nP  C  23 and nM  P  C  18

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Number of students who opted for only mathematics:
nM  P’  C’  nM  P  C’
 nM – nM  P  C
 nM – nM  P  M  C
 nM – nM  P  nM  C  nM  P  C
 100 – 30  28  18  60
Therefore, the number of students who opted for mathematics alone is 60

Question 26

Solution: The correct option is (b)

Question 27

Solution:
The correct option is (c)
2m  1  48  2n  1
 2m  2n  48
 2m  2n  26  24
By comparing we get :
m  6 and n  4
Question 28.

Solution: let M, P and C denote the sets of students who opted for mathematics, physics and
chemistry.
n  M   100, n  P   70 and n  C   40
Number of students who opted for only mathematics
n  M  P ' C '  n M   P  C  '
 n  M   n M   P  C 
 n  M   n  M  P    M  C 
 n  M   n  M  P   n  M  C   n  M  P  C 
 100   30  28  18 
 60

Question 29.

Solution:

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 A  B ' '  B  C 
  A '  B ' '   B, C   De Morgan law 
  A ' B    B  C 
  A ' C   B  Distributive law
Question 30

Solution: Since every rectangle, rhombus and square in a plane is parallel but trapezium,m is not
a parallelogram.
Thus, F1 is either F1 or F2 or F3 or F4
 F1  F1  F2 ,  F3  F4

Hence, the correct option is  d 

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