8. A set is a collection of well defined objects.
Definition
9. A set is a collection of well defined objects.
Definition
NOTE
(a) Sets are generally denoted by capital letters
like A, B, X, Y etc.
Elements of the set are denoted by small letters
like a, b, x, y etc.
(a) If a is an element of the set A, then it’s written as a ∈ A.
(b) Elements are not repeated in the set.
10. Description
Sets are described in two ways.
(i) Roster form (ii) Set Builder form
Elements are listed using
commas and brackets
For example,
{a, e, i, o, u}
Elements are told by telling
the property or rule they
follow
For example,
{x : x is a vowel}
11. Match the following
(a) {1, 2, 4, 8} (i) {x : x is a prime divisor of 12}
(b) {2, 3} (ii) {x : x ∈ N and x is a divisor of 8}
(c) {A, C, E, H, I, M, S, T} (iii) {x : x is a letter of the word MATHEMATICS}
(iv) {x : x ∈ N and x < 9}
(v) {x : x is a letter of the word ALCHEMIST}
15. Given set is A = {1, 2, 4, 8, 16, 32, 64}
By observation, we can say
A = {2n - 1 : n ≤ 7 and n ∈ N}
Alternatively, we can say
A = {2n : n ≤ 6 and n ∈ W}
Solution:
19. Null set or Empty set or Void set
A set which does not contain any element is called Empty set or Null
set or Void set. It’s denoted by 𝜙 or { }.
20. Singleton Set
A set containing single element is called singleton set.
Remark
Curly brackets ie { . } are use to write singletons.
Eg. {z} , {0} , {𝜙} , { {1,4,5} } are singleton sets.
21. Remark
Finite Set
A set which is either empty or has finite number of elements is called
a finite set.
Number of elements in a finite set A is called the order or the
cardinality of the set A, generally denoted by o(A) or n(A).
22. If A = {1, 2, {3, 4} }, then n(A) = _____.
A B C D
2 3 4 None of these
Recall
Eg. {z}, {0}, {𝜙}, { {3,4,5} } are singleton sets.
23. If A = {1, 2, {3, 4} }, then n(A) = _____.
A B C D
2 3 4 None of these
24. Given A = {1, 2, {3, 4} }
Here, {3, 4} is an element of A
So, n(A) = 3
Solution:
25. Infinite Set
A set having infinite number of elements is called infinite set.
Eg. (i) {1, 2, 3, 4, ...},
(ii) {..., -2, -1, 0, 1, 2, ...} etc.
26. Equal Set
Two sets are said to be equal if they have exactly the same elements.
Eg. If A = {0, 1, 2} , B = {2, 1, 0} , C = {0, 1, 1, 2}, then A = B = C.
27. Equivalent Set
Two finite sets are said to be equivalent if their cardinalities are equal.
Eg. {1, 2, 3} is equivalent to {dog, cat, parrot}.
29. Set A is said to be a subset of set B if every element of A is
also an element of B. It is generally denoted as A ⊂ B.
Ex. If A = {1, 2, 3} and B = {1, 2, 3, 4}, then A ⊂ B.
Definition
30. NOTE
(a) Every set is a subset of itself.
(b) ɸ is a subset of all sets.
(c) A ⊂ B and B ⊂ A ⇒ A = B .
33. Let A = {1, 2}
The subsets of A are: 𝜙 , {1}, {2}, {1,2}
Solution:
34. Let B = {a, b, c} and C = {b, d}. Find all sets X such that:
X ⊂ B and X ⊂ C
35. Solution:
We have,
Subsets of B are : ϕ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}
and
Subsets of C are : ϕ, {b}, {d}, {b, d}
Now, X ⊂ B and X ⊂ C
⇒ X ∈ {ϕ, {b}}
37. The collection of all the subsets of a set A is called the power
set of A, denoted by P(A).
Definition
38. Write down the power sets of the following sets.
(a) A = {1, 2} (b) A = {1, 2, 3}
39. (a) Given A = {1, 2}
Power set of A = {ɸ, {1}, {2}, {1, 2}}
(b) Given A = {1, 2, 3}
P(A) = {ɸ, {1}, {2}, {3}, {1, 2} {2, 3}, {1, 3}, {1, 2, 3}}
Solution:
41. Solution:
Let B = {b}. Then, A = {a, B}
∴ P(A) = {ϕ, {a}, {B}, {a, B}} = {ϕ, {a}, {{b}}, {a, {b}}}.
42. Result
If n(A) = p, then the number of subsets of A is 2p.
Clearly, same will be n(P(A)).
43. Set A has m elements and Set B has n elements. If the total number
of subsets of A is 112 more than the total number of subsets of B,
then the value of mn is_____.
JEE Main 2020
44. Set A has m elements and Set B has n elements. If the total number
of subsets of A is 112 more than the total number of subsets of B,
then the value of mn is_____.
JEE Main 2020
Ans: 28
45. Solution:
Number of subsets of A = 2m
Number of subsets of B = 2n
Given = 2m − 2n = 12
∴ (m, n) = (7, 4)
⇒ mn = 28
47. To understand operations on sets, first we need to know the term
“universal set”.
Operations on Sets
48. Universal Set
If we talk about any particular context in sets then the set which is
superset of all possible sets in the given context is called universal set.
(Generally, it’s given in the question).
49. Universal Set
If we talk about any particular context in sets then the set which is
superset of all possible sets in the given context is called universal set.
(Generally, it’s given in the question).
Venn Diagram
Ex. A = {1, 2, 3}
B = {3, 4, 5}
U = {1, 2, 3, 4, 5, 6}
50. Following are the operations on sets that we are now going to learn.
● Union of two Sets
● Intersection of two Sets
● Difference of two Sets
● Symmetric Difference of two Sets
● Complement of a Set
51. Union of two Sets
A ∪ B = {x : x ∈ A or x ∈ B}
Ex. {1, 2, 3} ∪ {2, 3, 4} = {1, 2, 3, 4}
A B
52. Intersection of two Sets
A ∩ B = {x : x ∈ A and x ∈ B simultaneously}
Ex. {1, 2, 3} ∩ {2, 3, 4} = {2, 3}
A B
53. A
B
C
D
{3, 6, 9}
{5, 10, 15, 20, ...}
{15, 30, 45, …}
None of these
Let A = {x: x is a multiple of 3} and B = {x: x is a multiple of 5}. Then
A ∩ B is given by
54. A
B
C
D
{3, 6, 9}
{5, 10, 15, 20, ...}
{15, 30, 45, …}
None of these
Let A = {x: x is a multiple of 3} and B = {x: x is a multiple of 5}. Then
A ∩ B is given by
55. Solution:
Since x ∈ A ∩ B ⇔ x ∈ A and x ∈ B ⇔ x is a multiple of 3
and x is a multiple of 5 ⇔ x is a multiple of 15.
Hence A ∩ B = {x |x| is a multiple of 15} = {15, 30, 45, …}
Hence (C) is the correct answer.
56. NOTE
(a) If C ⊂ B, then
(i) C ∪ B = B
(ii) C ∩ B = C
57. NOTE
(a) If C ⊂ B, then
(i) C ∪ B = B
(ii) C ∩ B = C
(b) C ∩ (B ∪ A) = (C ∩ B) ∪ (C ∩ A); C ∪ (B ∩ A)
= (C ∪ B) ∩ (C ∪ A)
58. NOTE
(a) If C ⊂ B, then
(i) C ∪ B = B
(ii) C ∩ B = C
(b) C ∩ (B ∪ A) = (C ∩ B) ∪ (C ∩ A); C ∪ (B ∩ A)
= (C ∪ B) ∩ (C ∪ A)
(c) If C ∩ B = ɸ, then C and B are called disjoint sets.
59. A
Ac
Complement of a Sets
Ac or A’ = {x : x ∉ A and x ∈ U}
Eg.
If U = {1, 2, 3, 4, 5, 6}, then complement of
{1, 2, 3, 4} is {5, 6}.
74. Result
(a) n(A) = x1 + x2
(b) n(only A) = x1
(c) n(B) = x2 + x3
(d) n(only B) = x3
(e) n(none of A or B) = x4
B
A
x1
x4
x2
x3
75. In a school, every teacher teaches either maths or physics. If 15
teach maths, 24 teach physics, while 6 teach both, then find
(a) the number of teachers.
(b) the number of teachers who teach physics only.
76. In a school, every teacher teaches either maths or physics. If 15
teach maths, 24 teach physics, while 6 teach both, then find
(a) the number of teachers.
77. Let A be the set of teacher who teaches Maths
B be the set of teacher who teaches Physics
Then n(A) = 15, n(B) = 24
n(A ∩ B) = 6
∴ Total number of teachers
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
= 15 + 24 – 6
= 33
A B
Solution:
78. In a school, every teacher teaches either maths or physics. If 15
teach maths, 24 teach physics, while 6 teach both, then find
(b) the number of teachers who teach physics only.
79. Let x1 = num of teachers who teach physics only
Let x2 = num of teachers who teach maths only
Let x3 = num of teachers who teach both maths and physics
According to the question,
x1 + x3 = 24 and x2 + x3 = 15 and x3 = 6
Solving, we get x1 = 24 - 6 = 18
x1 x3 x2
Solution:
80. In a class 60% passed their Physics examination and 58% passed
in Mathematics. Atleast what percentage of students of this class
passed both their physics and Mathematics examination?
A B C D
18% 17% 16% 2%
81. In a class 60% passed their Physics examination and 58% passed
in Mathematics. Atleast what percentage of students passed both
their physics and Mathematics examination?
A B C D
18% 17% 16% 2%
82. In a survey of 600 students in a school, 150 students were found to
be drinking Tea and 225 drinking Coffee, 100 were drinking both
Tea and Coffee. Find how many students were drinking neither Tea
nor Coffee.
83. In a survey of 600 students in a school, 150 students were found to
be drinking Tea and 225 drinking Coffee, 100 were drinking both
Tea and Coffee. Find how many students were drinking neither Tea
nor Coffee.
Ans: 325
84. Solution:
Given,
Total number of students = 600
Number of students who were drinking Tea = n(T) = 150
Number of students who were drinking Coffee = n(C) = 225
Number of students who were drinking both Tea and Coffee = n(T ∩ C) = 100
n(T ∪ C) = n(T) + n(C) - n(T ∩ C)
= 150 + 225 - 100
= 375 - 100
= 275
Hence, the number of students who are drinking neither Tea nor Coffee
= 600 - 275 = 325
85. Let X = {n ∈ N : 1 ≤ n ≤ 50}. If A = {n ∈ X : n is a multiple of 2} and
B = {n ∈ X : n is a multiple of 7}, then the number of elements in the
smallest subset of X containing both A and B is______.
JEE Main 2020
86. Let X = {n ∈ N : 1 ≤ n ≤ 50}. If A = {n ∈ X : n is a multiple of 2} and
B = {n ∈ X : n is a multiple of 7}, then the number of elements in the
smallest subset of X containing both A and B is______.
JEE Main 2020
Ans: 29
87. Solution:
∵ X = {1, 2, 3, 4, …, 50}
A = {2, 4, 6, 8, …, 50}
B = {7, 14, 21, 28, 35, 42, 49}
Here n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 29
∴ Number of elements in smallest subset of X
containing both A and B is 29.
88. In a school, if 15 teachers teach maths and 25 teachers teach
physics, and if total number of teachers is 30, then try to
observe that minimum value of n(M ∩ P) is 10.
Observation
89. In a school, if 15 teachers teach maths and 25 teachers teach
physics, and if total number of teachers is 50, then try to
observe that minimum value of n(M ∩ P) is 0.
Observation
90. Remark
(a) If n(A) + n(B) ≥ n(U) , then minimum value of n(A ∩ B) is given
by n(A) + n(B) − n(U)
(b) If n(A) + n(B) < n(U), then minimum n(A ∩ B) = 0
91. If n(A) = 5 and n(B) = 10 , then find maximum and minimum
possible values of n(A ∩ B) if
(a) n(U) = 8 (b) n(U) = 20
92. (a) Given, n(A) = 5 and n(B) = 10
Here, maximum possible value of n(A ∩ B) = 5
Now, as n(U) = 8 and n(A) + n(B) = 15
Thus, minimum value of n(A ∩ B) = 7
Solution:
93. (b) Given, n(A) = 5 and n(B) = 10
Here, maximum possible value of n(A ∩ B) = 5
Now, as n(U) = 20 and n(A) + n(B) = 15
Thus, minimum value of n(A ∩ B) = 0
Solution:
94. If A and B be two sets containing 3 and 6 element respectively,
what can be the minimum number of elements in A ∪ B? Find also,
the maximum number of elements in A ∪ B.
95. If A and B be two sets containing 3 and 6 element respectively,
what can be the minimum number of elements in A ∪ B? Find also,
the maximum number of elements in A ∪ B.
Ans: 6
96. A B C D
There are 25 students. Every student reads 10 newspapers and
each newspaper is read by (i.e., shared by) 5 students. Find the
number of newspapers.
97. A B C D
There are 25 students. Every student reads 10 newspapers and
each newspaper is read by (i.e., shared by) 5 students. Find the
number of newspapers.
98. Since every student reads 10 newspaper
Thus, total number of newspaper read = 10 × 25 = 250
But each news paper is shared by 5 students
Thus actual number of newspapers =
Solution:
99. Suppose A1, A2, ….., A30 are 30 sets, each with 5 elements and
B1, B2, …., Bn are n sets, each with 3 elements.
and each element of S belongs to exactly 10 of Ai‘s and 9 of Bj’s.
Find ‘n’.
100. Suppose A1, A2, ….., A30 are 30 sets, each with 5 elements and
B1, B2, …., Bn are n sets, each with 3 elements.
and each element of S belongs to exactly 10 of Ai‘s and 9 of Bj’s.
Find ‘n’.
Ans: 45
103. Remark
(a) n(A) = x1 + x4 + x6 + x7
(b) n(only A) = x1
(c) n(only A and B) = x1 + x4 + x2
(d) n(only one of A, B or C) = x1 + x2 + x3
(e) n(exactly two of A, B and C) = x4 + x5 + x6
(f) n(A, B or C) = x1 + x2 + x3 + x4 + x5 + x6 + x7
B
A
x1 x4 x2
x5
x6 x7
x3
C
Cardinality based Problems
104. Out of the members of three athletic teams in a school, 21 are in
the basketball team, 26 in hockey team and 29 in the football team,
14 play hockey and basketball, 15 play hockey and football, 12 play
football and basketball and 8 play all the games. The total number
of members is
A B C D
42 43 45 None
Recall
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C)
+ n(A ∩ B ∩ C)
105. A B C D
42 43 45 None
Out of the members of three athletic teams in a school, 21 are in
the basketball team, 26 in hockey team and 29 in the football team,
14 play hockey and basketball, 15 play hockey and football, 12 play
football and basketball and 8 play all the games. The total number
of members is
106. Let B, H, F be the sets of the three teams respectively
So n(B) = 21, n(H) = 26, n(F) = 29,
n(H ∩ B) = 14, n(H ∩ F) = 15, n(F ∩ B) = 12,
n(B ∩ H ∩ F) = 8 and
n(B ∪ H ∪ F) = n(B) + n(H) + n(F) - n(H ∩ B) - n(H ∩ F)
- n(F ∩ B) + n(B ∩ H ∩ F)
= 21 + 26 + 29 - 14 - 15 - 12 + 8 = 43
Solution:
107. A college awarded 38 medals in Football, 15 in Basketball and 20 to
Cricket. If these medals went to a total of 58 men and only three
men got medals in all the three sports, how many received medals
in exactly two of the three sports?
108. A college awarded 38 medals in Football, 15 in Basketball and 20 to
Cricket. If these medals went to a total of 58 men and only three
men got medals in all the three sports, how many received medals
in exactly two of the three sports?
Ans: 9
109. Solution: Let F denote the set of men who received medals in football, B the set of
men who received medals in Basketball and C the set of men who
received medals in Cricket. It is given that n(F) = 38, n(B) = 15, n(C) = 20,
n(F ∪ B ∪ C) = 58 and n(F ∩ B ∩ C) = 3.
Now,
n(F ∪ B ∪ C) = n(F) + n(B) + n(C) − n(F ∩ B) − n(B ∩ C) − n(F ∩ C) +
n(F ∩ B ∩ C)
⇒ 58 = 38 + 15 + 20 − n(F ∩ B) − n(B ∩ C) − n(F ∩ C) + 3
⇒ n(F ∩ B) + n(B ∩ C) + n(F ∩ C) = 76 − 58 = 18
Number of men who received medals in exactly two of the three sports
= n(F ∩ B) + n(B ∩ C) + n(F ∩ C) − 3n(F ∩ B ∩ C) = 18 − 3 × 3 = 9
Thus, 9 men received medals in exactly two of the three sports.
110. Out of 280 students in class XII of a school, 135 play Hockey, 1110
play football, 80 play volleyball, 35 these play hockey and football,
30 play volleyball and hockey, 20 play football and volleyball. Also,
each students plays at least one of the three games. How many
students play all the three games?
HW
111. Out of 280 students in class XII of a school, 135 play Hockey, 1110
play football, 80 play volleyball, 35 these play hockey and football,
30 play volleyball and hockey, 20 play football and volleyball. Also,
each students plays at least one of the three games. How many
students play all the three games?
Ans: 40
112. Solution:
Let H, F and V be the sets of students who play hockey,
football and volleyball respectively. Let x be the number of
students who play all the three games. It is given that 35
students play hockey and football. So, number of students
who play hockey and football only is (35 − x).
Similarly, the number of students playing various games are
written in the regions representing them in Fig
113. Solution:
It is given that each student plays at least
one of the three games.
∴ n(H ∪ F ∪ V) = 280
⇒ (70 + x) + (35 − x) + (30 − x) + x + (20 − x)
+ (55 + x) + (30 + x)
= 280 ⇒ 240 + x = 280 ⇒ x = 40
Hence, 40 students play all the three
games.
H F
V
70 + x 35 − x
55 + x
20 − x
30 − x
30 + x
x
