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For set A = {1, 3, 5}, B = {2, 4, 6} and C = {2, 4, 6, 8}

we have (ii) and (iii) universal set, because it contains all the elements of the given sets A, B and C.

Answer is (C)

Elements of set A = {1, 4, 9, 16, 25,…} are the square of 1, 2, 3, 4, 5,… respectives.

i. C = {b, d}, B = {c, d, e ,f} 

C ⊆ B 

False 

Since, all the elements of C are not present in B. 

ii. A = {a, b, c, d, e}, D = {a, e} 

A ⊆ D 

False 

Since, all the elements of A are not present in D. 

iii. D = {a, e}, B = {c, d, e, f} 

D ⊆ B 

False 

Since, all the elements of D are not present in B. 

iv. D = {a, e}, A = {a, b, c, d, e} 

D ⊆ A 

True

Since, all the elements of D are present in A. 

v. B = {c, d, e, f}, A = {a, b, c, d, e} 

B ⊆ A 

False

Since, all the elements of B are not present in A. 

vi. C = {b, d}, A = {a, b, c, d, e} 

C ⊆ A 

True 

Since, all the elements of C are present in A.

A = {x | x is prime number and 10 < x < 20} 

∴ A = {11, 13, 17, 19} 

B = {11, 13, 17, 19} 

∴ All the elements in set A and B are identical. 

∴ A and B are equal sets, i.e. A = B

i. Rhombus, square, parallelogram and rectangle all are quadrilaterals. 

∴ Y ⊆ X,S ⊆ X,T ⊆ X,V ⊆ X 

ii. Every square is a rhombus, parallelogram and rectangle. 

∴ S ⊆ Y, S ⊆ T, S ⊆ V 

iii. Every rhombus and rectangle is a parallelogram. 

∴ Y ⊆ T, V ⊆ T

i.  A = { 2, 4, 6, 8,….}

ii. 2 is the only even prime number 

∴ B = { 2 }

iii. C = {-1, -2, -3,….} 

iv. D = {sa, re, ga, ma, pa, dha, ni}

A = Even prime numbers 

Since 2 is the only even prime number, 

∴ A = {2} …(i) 

B= {x | 7x – 1 = 13} 

Here, 7x – 1 = 13 

∴ 7x = 14 

∴ x = 2 

∴ B = {2} …(ii) 

∴ The element in set A and B is identical. … [From (i) and (ii)] 

∴ A and B are equal sets.

Natural numbers = 1, 2, …, 30, 31, 32, 33, 34, 35, 36, … 

The elements of this set are 30, 31, 32, 33, 34 and 35 only 

So, A = {30, 31, 32, 33, 34, 35}

i. A = {2, 4, 6, 8,….} 

ii. 2 is the only even prime number 

∴ B = {2} 

iii. C = {-1, -2, -3,….} 

iv. D = {sa, re, ga, ma, pa, dha, ni}

Correct option is (D) {3, 5}

i. (4/3) is an element of set Q. 

ii. -2 is not an element of set N. 

iii. Set P is a set of all p’s such that p is an odd number.

(B) {1, 2, 3, 4, 5}

i. A = {x| x < 10, x is a natural number} 

∴ A = {1,2, 3,4, 5,6, 7, 8, 9} 

The number of elements in A are limited and can be counted. 

∴A is a finite set.

ii. B = (y | y < -1, y is an integer} 

∴ B = { …,-4, -3, -2} 

The number of elements in B are unlimited and uncountable. 

∴ B is an infinite set. 

iii. C = Set of students of class 9 from your school. 

The number of students in a class is limited and can be counted. 

∴ C is a finite set. 

iv. Set of people from your village. 

The number of people in a village is limited and can be counted. 

∴ Given set is a finite set. 

v. Set of apparatus in laboratory 

The number of apparatus in the laboratory are limited and can be counted. 

∴ Given set is a finite set. 

vi. Set of whole numbers 

The number of elements in the set of whole numbers are unlimited and uncountable. 

∴ Given set is an infinite set. 

vii. Set of rational number 

The number of elements in the set of rational numbers are unlimited and uncountable. 

∴ Given set is an infinite set.

(D) The answer is {3, 5}

(B) {3, 4, 5, 6}

(C) Finite set

(i) Null Set 

(a) A = {x : x is a natural number less than 1} 

(b) B = {x : x is a positive natural number which lies between 2 and 3} 

(ii) Finite set 

(a) A = {x : x² < 10, where x is a prime number} 

(b) B = {x : x, is any month of a year} 

(iii) Infinite set 

(a) P = {x : x = 2n, where n is a natural number} 

(b) Q = {x : x = p/q, where p and q are integer and q ≠ 0} 

(iv) Universal set 

(a) Set of integers for natural numbers. 

(b) Set of real numbers for natural numbers, integers and rational numbers.

Disjoint sets have their intersections as Φ. 

(i) A = {3, 4, 5, 6} and B = {2, 5, 7, 9} Are pairs of disjoint sets. 

(ii) C = {1, 2, 3, 4, 5} and D = {6, 7, 9, 11} Are pairs of disjoint sets. 

(iii) E = {x : x ϵ N, x is even and x < 8} = {2, 4, 6} and 

F = {x : x = 3n, n ϵ N, and x < 4} = {3, 6, 9} Are not pairs of disjoint sets. 

(iv) G = {x : x ϵ N, x is even} and H {x : x ϵ N, x is prime} 

∵ 2 is an even prime number; their intersection is not Φ 

Are not pairs of disjoint sets. 

(v) J = {x : x ϵ N, x is even} and K = {x : x ϵ N, x is odd} 

∵ there is no number which is both odd and even. 

∴ J and K are pairs of disjoint sets.

Let A and B be any two disjoint sets, 

i.e., A and B have no elements in common. 

∴ A ∩ B is a null set. (∵ A ∩ B is the set of all elements which are common to both A and B)

Consider the disjoint sets 

A = {1, 2, 3, 4} and B = {a, b, c}

i) Rule: If two sets are disjoint their intersection is null set. 

= {2, 3, 4, 5} n {3, 6} = { 3 } ≠ φ 

∴ Given statement is False.

ii) Given sets are 

{a, e, i, o, u} and {a, b, c, d} 

= {a, e, i, o, u} ∩ {a, b, c, d} 

= { a } ≠ φ 

∴ Given statement is False. 

iii) Given sets are 

{2, 6, 10, 14} and {3, 7, 11, 15} 

= {2, 6, 10, 14} ∩ {3, 7, 11, 15} 

= { } 

∴ Given statement is True. 

iv) Given sets are 

{2, 6, 10} and {3, 7, 11} 

= {2, 6, 10} ∩ {3, 7, 11} = { } 

∴ Given statement is True.

If A and B are disjoint then A ∩ B is a null set. 

∴ n(A ∩ B) = 0 and it gives us n(A ∪ B) = n (A) + n(B).

Let x ϵ A and y ϵ B 

A – B = The set of values of A that are not in B. 

A ∩ B = The set containing common values of A and B 

B – A = The set of values of B that are not in A. 

Two sets X and Y are called disjoint if, 

X ∩ Y = ϕ 

(A – B) ∩ (A ∩ B) = ((A – B) ∩ A) ∪ ((A – B) ∩B) 

(A – B) ∩ (A ∩ B) = ϕ ∪ ϕ 

(A – B) ∩ (A ∩ B) = ϕ 

Similarly, 

(B – A) ∩ (A ∩ B) = ((B – A) ∩ A) ∪ ((B – A) ∩B) 

(B – A) ∩ (A ∩ B) = ϕ 

Hence, 

the three sets are pair wise disjoint.

For “CATRACTR” 

Letters in word are 

{C, A, T, R} ={A, C, R, T} 

For “TRACT” 

Letters in word are 

{T, R, A, C} = {A, C, R, T} 

As we see letters need to spell cataract is equal to set of letters need to spell tract. 

Hence Proved.

A = {1, 2}

B = {1, 2}

C = {3, 1}

D = {1, 3} (Here, the odd natural numbers less than 5 are 1 and 3)

E = {1, 2} (Here, repetition is not allowed)

F = {1, 3} (Here, repetition is not allowed)

∴ Sets A, B and E are equal.

Hence, C, D and F are equal.

For the A

The letters in word reap

A ={R, E, A, P} = {A, E, P, R}

For the B

The letters in word paper

B = {P, A, E, R} = {A, E, P, R}

For the C

The letters in word rope

C = {R, O, P, E} = {E, O, P, R}.

Set A = Set B

Because every element of set A is present in set B

But Set C is not equal to either of them because all elements are not present.

A = x is a natural number. And x is less than 3 

So all natural numbers less than 3 constitute set A. 

{1, 2} 

A ={1, 2} 

B = {1, 2} 

C = {1, 3} 

D = x is a natural number. And x is less than 5 and is odd. 

So all odd natural numbers less than 5 constitute set D. 

{1, 3} 

D = {1, 3} 

E ={1, 2, 1, 1} 

We don’t repeat same elements in a set. 

∴ E = {1, 2} 

F ={1, 1, 3} 

We don’t repeat same elements in a set. 

∴ F = {1, 3} 

∴ A = {1, 2} 

B = {1, 2} 

C = {1, 3} 

D = {1, 3} 

E = {1, 2} 

F = {1, 3} 

Now, we can see clearly that set A, B, E are equal and set C, D, F are equal.

The equivalent set are different from equal sets, equivalent sets are those which have equal number of elements they do not have to be same.

A = {1, 2, 3}

The number of elements = 3

B = {t, p, q, r, s}

Here, the number of elements = 5

C = {α, β, γ}

Since, the number of elements = 3

D = {a, e, i, o, u}

Here, the number of elements = 5

∴ Set A is equivalent with Set C.

Hence, set B is equivalent with set D.

A = {0, a}

B = {1, 2, 3, 4}

C = {4, 8, 12}

D = {3, 1, 2, 4} = {1, 2, 3, 4}

E = {1, 0}

F = {8, 4, 12} = {4, 8, 12}

G = {1, 5, 7, 11}

H = {a, b}

For the equivalent sets:

i. A, E, H (all of them have exactly two elements in them)

ii. B, D, G (all of them have exactly four elements in them)

iii. C, F (all of them have exactly three elements in them)

For the equal sets:

i. B, D (all of them have exactly the same elements, therefore they are equal)

ii. C, F (all of them have exactly the same elements, therefore they are equal)

Note: Equivalent set are different from equal sets, Equivalent sets are those which have equal number of elements they do not have to be same.

 A = {1, 2, 3} 

Number of elements = 3 

B = {t, p, q, r, s} 

Number of elements = 5 

C = {α, β, γ} 

Number of elements = 3 

D = {a, e, I, o, u} 

Number of elements = 5 

Set A is equivalent with Set C and Set B is equivalent with Set D.

We first of all need to manipulate some of the sets 

D = {3, 1, 2, 4} = {1, 2, 3, 4, } 

F = {8, 4, 12} = {4, 8, 12} 

Equivalent sets: 

i. A, E, H (all of them have exactly two elements in them) 

ii. B, D, G (all of them have exactly four elements in them) 

iii. C, F (all of them have exactly three elements in them) 

Equal sets : 

i. B, D (all of them have exactly the same elements, so they are equal)

ii. C, F (all of them have exactly the same elements, so they are equal)

(i) A = {2, 3}

B = x² + 5x + 6 = 0

x² + 3x + 2x + 6 = 0

x(x+3) + 2(x+3) = 0

(x+3) (x+2) = 0

x = -2 and -3

= {–2, –3}

Here, A and B do not have exactly same elements thus they are not equal.

(ii) The every letter in WOLF

A = {W, O, L, F} = {F, L, O, W}

The every letter in FOLLOW

B = {F, O, L, W} = {F, L, O, W}

Here, A and B have same number of elements which are exactly same, thus they are equal sets.

Every letter in WOLF 

A = {W, O, L, F} = {F, L, O, W} 

Every letter in FOLLOW 

B = {F, O, L, W} = {F, L, O, W} 

As A and B have same number of elements which are exactly same hence they are equal sets.

A = {2, 3} 

B = {– 2, – 3} 

As A and B do not have exactly same elements hence they are not equal.

(i) A = B because A and B have same elements though in different order which is immaterial.

(ii) A≠B ∵ 12 ∈ A but 12 ∉ B 

(iii) A = B ∵ A and B have same elements 

(iv) A≠B ∵ 15 ∈ B 

(v) A = {2, 3, 5}, B = {2, 3, 5} ∴ A = B.

(i) x² + 5x + 6 = 0

⇒ (x + 3)(x + 2) = 0 

∴ x = -3,-2 

∴ B = {-3, -2}- But A = {2, 3} 

∴ A ≠ B 

(ii) Given sets are A = {F, O, L, W} 

B = {W,0,L,F} ∴ A = B 

∵ A and B have same elements.

Here, B = D and E = G.

A set A is said to be subset of set B if every element of A is also an element of B, and we write, A ⊆ B 

• If A ⊆ B then B is called a superset of A, and we write, B ⊇ A 
• If A ⊆ B and A ≠ B then A is called a proper subset of B.
• If there exists even a single element in A which is not in B, then A is not a subset of B and we write, A⊄B 
• Every set A is a subset of itself
• Empty set φ is a subset of every set.

(i) φ ⊂ B, as empty set is a subset of every set 

(ii) A ⊄ B, ∵ 3 ∈ A ⇒ but 3∉ B 

(iii) A⊂ C, as 1, 3 ∈ A ⇒1,3 ∈ C 

(iv) B ⊂ C, as 1, 5, 9∈ A ⇒ 1, 5, 9 ∈ C.