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Let P(A) = P(B) 

To show: 

A = B 

Let x ∈ A A ∈ P(A) = P(B) 

∴ x ∈ C, for some C ∈ P(B) 

Now, C ⊂ B ∴ x ∈ B ∴ A ⊂ B 

Similarly, B ⊂ A ∴ A = B 

False 

Let   A = {0, 1} and B = {1, 2}  7 

∴ A ∪ B = {0, 1, 2} 

P(A) = {Φ, {0}, {1},{0, 1}}

P(B) = {Φ, {1}, {2}, {1, 2}}

 P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}} 

P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}} 

∴ P(A) ∪ P(B) ≠ P(A ∪ B) 

Calculation:

Given: The four sets have 150, 180, 210 and 240 elements respectively

n(A) = 150

n(B) = 180

n(C) = 210

n(D) = 240

Each pair of sets has 15 elements

n(A ∩ B) = 15

n(A ∩ C) = 15

n(A ∩ D) = 15

n(B ∩ C) = 15

n(B ∩ D) = 15

n(C ∩ D) = 15

Each triple of sets has 3 elements

n(A ∩ B ∩ C) = 3

n(A ∩ B ∩ D) = 3

n(A ∩ C ∩ D) = 3

n(B ∩ C ∩ D) = 3

A ∩ B ∩ C ∩ D = ϕ 

n(A ∩ B ∩ C ∩ D) = 0

Now, number of elements in the union of 4 sets A, B, C and D

n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D) - n(A ∩ B) - n(A ∩ C) - n(A ∩ D) - n(B ∩ C) - n(B ∩ D) - n(C ∩ D) + n(A ∩ B ∩ C) + n(A ∩ B ∩ D) + n(A ∩ C ∩ D) + n(B ∩ C ∩ D) - n(A ∩ B ∩ C ∩ D)

⇒ 150 + 180 + 210 + 240 - 6 × 15 + 4 × 3 - 0

∴ The required number of elements is 702.

Concept:

N denote the set of natural numbers which is an infinite set.

Let A and B be any two sets

A ∪ B = {x: x ∈ A, x ∈ B}

A' = {x: x \(\rm ∈ N , x \notin A\)}

A ∩ B = {x: x \(\rm ∈ A \;\; and \;\; x ∈ B\)}

A proper subset of a set A is a subset of A that is not equal to A

Calculations:

Let N denote the set of natural numbers

⇒ N = {1, 2, 3, 4,....}

A = {n2 : n ∈ N}

⇒ A = {1, 4, 9, 16, ....}

B = {n3 : n ∈ N}.

⇒ B = {1, 8, 27, 64, ....}

Consider the statement "A ∪ B = N"

⇒A ∪ B = {1, 4, 8, 9, 16, 27,....} ≠  N

Hence, the statement A ∪ B = N is not true.

Consider the statement "The complement of (A ∪ B) is an infinite set"

We know that  The complement of a set, denoted A', is the set of all elements in the given universal set U that are not in A.

⇒ (A ∪ B)' = U - (A ∪ B) = {2, 3, 5, 6,....} = infinite set

⇒ (A ∪ B)' = Infinite set

Hence, the statement "The complement of (A ∪ B) is an infinite set" is true.

Consider, the statement "A ∩ B must be a finite set"

A ∩ B  = {1}

Hence, the statement "A ∩ B must be a finite set" is true.

Consider, the statement "A ∩ B must be a proper subset of {m6 : m ∈ N}"

Let S = {m6 : m ∈ N}

S = {1, 64, ....}

and A ∩ B  = {1}

A ∩ B must be a proper subset of {m6 : m ∈ N}

Hence, the statement "A ∩ B must be a proper subset of {m6 : m ∈ N}" is True.

Hence,  if N denote the set of natural numbers and A = {n2 : n ∈ N} and B = {n3 : n ∈ N}. then the statement (2), (3) and (4) are correct.

Concept:

Let A and B denote two sets of elements.

• n(A) and n(B) are the numbers of elements present in set A and B respectively.
• n(A ⋃ B) is the total number of elements present in either set A or B.
• n(A ⋂ B) is the number of elements present in both sets A and B.
• n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B)

Calculations:

Let A be the set of students who read "Hitavad" and B the set of students who read "Hindustan".

From the given information:

n(A ⋃ B) = 50 - 10 = 40.

n(A) = 30.

n(B) = 35.

By using n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B), we get:

40 = 30 + 35 - n(A ⋂ B)

⇒ n(A ⋂ B) = 25.

∴ The number of students who read both "Hitavad" and "Hindustan" newspapers is n(A ⋂ B) = 25.

Concept:

The complement of an event:

The complement of an event is the subset of outcomes in the sample space that are not in the event.

The probability of the complement of an event is one minus the probability of the event.

P (A') = 1 - P (A)

For two events A and B we have

1. P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

2. De morgan's Law P (A ∪ B)' = P (A' ∩ B')

3. \(P({A'\over B'})={{P(A'\cap B')\over P(B')}}={{P(A\cup B)'\over P(B')}}\)

Calculation:

Given:

\(P(A) = \dfrac{1}{2}, P(B) = \dfrac{1}{4}, P(A \cap B) = \dfrac{1}{5}\)

we have,

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

\(P(A ∪ B) = {1\over 2}+{1\over4}-{1\over5}={11\over 20}\)

The complement of events:

\(P(B')=1-P(B)=1-{1\over 4}={3\over 4}\)

\(P(A\cup B)'=1-P(A\cup B)=1-{11\over 20}={9\over 20}\)

Now,

\(P({A'\over B'})={{P(A'\cap B')\over P(B')}}={{P(A\cup B)'\over P(B')}}\)

\(P({A'\over B'})={{{9\over20}\over {3\over 4}}}={9\over 20}\times {4\over 3}={3\over 5}=0.6\)

Concept:

Set theory:

• A ∪ B means set of all the values in the set A and B.
• A ∩ B is the set of common elements of A and B.
• Subset (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from.

Calculation

Given: B = {2x + 1 : x ∈ N}

So, Set B = {3, 5, 7, 9, 11, 13, ......} (all the odd positive integers excluding 1)

Given: A = {x : x is square of natural numbers ≤ 8}

So, Set A = {1, 4, 9, 16, 25, 36, 49, 64}

∴ A ∩ B = {9, 25, 49}

Concept:

If U is a universal set and A be any subset of U then the complement of A is the set of all members of the universal set U which are not the elements of A.

It is denoted by A'

A' = U - A

Calculation:

U = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

A = {3, 9, 15}

A' = U - A = {1, 5, 7, 11, 13, 17, 19}

Let U be the set of all students in the group. 

Let E be the set of all students who know English. 

Let H be the set of all students who know Hindi. 

∴ H ∪ E = U 

Accordingly, 

n(H) = 100 and n(E) = 50 

n( H U E ) = n(H) + n(E) – n(H ∩ E)

 = 100 + 50 – 25 = 125 

Hence, 

there are 125 students in the group. 

A = {x: x ∈ R and x satisfies x2 – 8x + 12 = 0} 

2 and 6 are the only solutions of

 x2 – 8x + 12 = 0. 

∴ A = {2, 6} B = {2, 4, 6},

 C = {2, 4, 6, 8 …}, 

D = {6} 

∴ D ⊂ A ⊂ B ⊂ C 

Hence, 

A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C 

Concept:

Set theory:

• A ∪ B means set of all the values in the set A and B.
• A ∩ B is the set of common elements of A and B.
• Subset (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from.

Calculation:

Set A ∪ B = {1, 2, 3, a, b, 4, 7, 9} 

Set C = {1, 3, a, b, 2}

As all the elements of C are present in A ∪ B 

∴ C ⊂ (A∪B)

Concept:

Let A and B are two sets, having a number of elements n(A) and n(B) respectively.

Then, the number of distinct relation = \(\rm 2^{n(A)\times n(B)}\)

Calculation:

Here,  P = {1, 2, 3} and Q = {4, 5, 6, 7}

n(P) = 3

n(Q) = 4

∴ Number of distinct relation = \(\rm 2^{n(A)\times n(B)}\)

\(\rm 2^{n(P)\times n(Q)} \\=2^{3\times 4}\)

= 2¹²

= 4096

Hence, option (4) is correct.