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True

According to the question,

There are three sets A, B and C

To check: if A ⊂ B, then A ∪ C ⊂ B ∪ C is true or false

Let x ∈ A ∪ C

⇒ x ∈ A or x ∈ C

⇒ x ∈ B or x ∈ C {∵ A ⊂ B}

⇒ x ∈ B ∪ C

⇒ A ∪ C ⊂ B ∪ C

Hence, the given statement “for all sets A, B and C, if A ⊂ B, then A ∪ C ⊂ B ∪ C” is true

True

According to the question,

There are three sets A, B and C

To check: if A ⊂ C and B ⊂ C, then A ∪ B ⊂ C is true or false

Let x ∈ A ∪ B

⇒ x ∈ A or x ∈ C

⇒ x ∈ C or x ∈ C {∵ A ⊂ C and B ⊂ C}

⇒ x ∈ C

⇒ A ∪ B ⊂ C

Hence, the given statement “for all sets A, B and C, if A ⊂ C and B ⊂ C, then A ∪ B ⊂ C” is true

According to the question,

There are two sets A and B

To prove: A ∪ (B – A) = A ∪ B

L.H.S = A ∪ (B – A)

Since, A – B = A ∩ B’, we get,

= A ∪ (B ∩ A’)

Since, distributive property of set ⇒ (A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C), we get,

= (A ∪ B) ∩ (A ∪ A’)

Since, A ∪ A’ = U, we get,

= (A ∪ B) ∩ U

= A ∪ B

= R.H.S

Hence Proved

The power set of set A is a collection of all subsets of A. 

Here A = {ϕ} 

Hence the subset of A will only be a null set ϕ 

Hence P(A) = {ϕ}

According to the question,

There are two sets A and B

To prove: A – (A – B) = A ∩ B

L.H.S = A – (A – B)

Since, A – B = A ∩ B’, we get,

= A – (A ∩ B’)

= A ∩ (A ∩ B’)’

Since, (A ∩ B)’ = A’ ∪ B’, we get,

= A ∩ [A’ ∪ (B’)’]

Since, (B’)’ = B, we get,

= A ∩ (A’ ∪ B)

Since, distributive property of set ⇒ (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C), we get,

= (A ∩ A’) ∪ (A ∩ B)

Since, A ∩ A’ = Φ, we get,

= Φ ∪ (A ∩ B)

= A ∩ B

= R.H.S

Hence Proved

According to the question,

There are two sets A and B

To prove: A – (A ∩ B) = A – B

L.H.S = A – (A ∩ B)

Since, A – B = A ∩ B’, we get,

= A ∩ (A ∩ B)’

= A ∩ (A ∩ B’)’

Since, (A ∩ B)’ = A’ ∪ B’, we get,

= A ∩ (A’ ∪ B’)

Since, Distributive property of set ⇒ (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C), we get,

= (A ∩ A’) ∪ (A ∩ B’)

Since, A ∩ A’ = Φ, we get,

= Φ ∪ (A ∩ B’)

= A ∩ B’

Since, A – B = A ∩ B’, we get,

= A – B

= R.H.S

Hence Proved

According to the question,

There are two sets A and B

To prove: (A ∪ B) – B = A – B

L.H.S = (A ∪ B) – B

Since, A – B = A ∩ B’, we get,

= (A ∪ B) ∩ B’

Since, Distributive property of set: (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C), we get,

= (A ∩ B’) ∪ (B ∩ B’)

Since, A ∩ A’ = Φ, we get,

= (A ∩ B’) ∪ Φ

= A ∩ B’

Since, A – B = A ∩ B’, we get,

= A – B

= R.H.S

Hence Proved

According to the question,

A, B and C are three given sets

To prove: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Let x ∈ A ∩ (B ∪ C)

⇒ x ∈ A and x ∈ (B ∪ C)

⇒ x ∈ A and (x ∈ B or x ∈ C)

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

⇒ x ∈ A ∩ B or x ∈ A ∩ C

⇒ x ∈ (A ∩ B) ∪ ( A ∩ C)

⇒ A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ ( A ∩ C) …(i)

Let y ∈ (A ∩ B) ∪ (A ∩ C)

⇒ y ∈ A ∩ B or x ∈ A ∩ C

⇒ (y ∈ A and y ∈ B) or (y ∈ A and y ∈ C)

⇒ y ∈ A and (y ∈ B or y ∈ C)

⇒ y ∈ A and y ∈ (B ∪ C)

⇒ y ∈ A ∩ (B ∪ C)

⇒ (A ∩ B) ∪ (A ∩ C) ⊂ A ∩ (B ∪ C) …(ii)

We know that:

P ⊂ Q and Q ⊂ P ⇒ P = Q

From equations (i) and (ii), we have,

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Hence Proved

(i) in English and Mathematics but not in Science

Answers is 2

(ii) in Mathematics and Science but not in English

Answers is 3

(iii) in Mathematics only

Answers is 3

(iv) in more than one subject only

Answers is 9

(a) The number of families which buy newspaper A only.

Answers is 3300

(b) The number of families which buy none of A, B and C

Answers is 4000

Given n(5) = 21; n(T) = 32; n(S∩T) = 11

n(S∪T) = n(S) + n(T) – n(S∩T)

= 21 + 32-11 =42.

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

But n(X∪Y) = n(X) + n(Y) – n(X ∩Y) n(X∩Y)

= 8 + 15-18 = 5.

Given n(X) = 17, n(Y) = 23, n(X∪Y) = 38

But n(X∪Y) = n(X) + n(Y)-n(X∩Y)∩(X∩Y)

= 17 + 23-38 = 2

D. R = {(x, y) : 0 < x < a, 0 < y < b}

Given: R be set of points inside a rectangle of sides a and b

Since, a, b > 1

a and b cannot be equal to 0

Therefore, R = {(x, y) : 0 < x < a, 0 < y < b}

(i) AUA'=U

(ii) Φ′ ∩ A = U ∩ A = A

∴ Φ′ ∩ A = A

(iii)A ∩ A′ = Φ

(iv)U′ ∩ A = Φ ∩ A = Φ

∴ U′ ∩ A = Φ

It is given that:

n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38

n(X ∩ Y) = ?

We know that:

n(X ∩ Y)=n(X)+n(Y)-n(X ∪ Y)

38=17+23-n(X ∪ Y)

n(X ∪ Y)=40-38=2

∴ n(X ∪ Y)=2

It is given that:

n(X ∪Y) = 18, n(X)=8, n(Y) = 15

n(X ïf ? Y) = ?

We know that:

n(X ∪Y)=n(X)+n(Y)-n(X∩Y)

18=8+15-n(X∩Y)

n(X∩Y)=5

A’ = ∪ - A = { 1,4,5,6}

B’ = U - B = { 1,2,6}

A’ ∩ B’ = {1,6}

A ∩ B = {2,3,4,5}

∴ (A ∪ B)’ = {1, 6} = A’ ∩ B’

A’ = U - A = {2, 4, 6, 8, 10}

Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A and is denoted by A’. 

Properties:

• A∪A’
• A∩A’=φ
• (A’)’=A
• (A∪B)’= A’∩B’
• (A∩B)’ -A’∪B’

Given: n(A) = 23, n(B) = 37, n(A – B) = 8 

Using the hint 

n(A) = n(A – B) + n(A ∩ B) 

⇒ 23 = 8 + n(A ∩ B) 

⇒ n(A ∩ B) = 23 – 8 

⇒ n(A ∩ B) = 15

Visualizing the hint given, 

We know that n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 

⇒ n(A ∪ B) = 23 + 37 – 15 

⇒ n(A ∪ B) = 45 

Hence n(A ∪ B) = 45

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

x² - 8x + 12 = 0

⇒(x - 6)(x - 2) = 0

⇒ x = 2, 6

∴ A = {2, 6}

∴ A ⊂ B and A ⊂ C; B ⊂ C, D ⊂ B and D ⊂ C.

Let x ∈ C - B ⇒ x ∈ C but x ∉ B 

⇒x ∈ C but x ∉ A ∵ A ⊂ B 

∴ x ∈ C- A Thus, C - B ⊂ C - A

(i) ⇔ (ii): 

A⊂B ⇔ All elements of A are in B ⇔ A – B = φ 

(ii) ⇔ (iii): 

A-B = φ⇔ All elements of A are in B ⇔ A∪B = B 

(iii) ⇔(iv) 

A∪B = B ⇔ All elements of A are in B ⇔ All elements of A are common in A and B. 

⇔ A∩B = A 

∴ All the four given conditions are equivalent.

(i) False

Since, {c, d} is not a subset of A but it belong to A.

Hence, {c, d} ∈ A

(ii) True

Thus, {c, d} ∈ A

(iii) True

Since, {c, d} is a subset of A.

(iv) Here, it is true that a belongs to A.

(v) False

Here, a is not a subset of A but it belongs to A

(vi) True

Thus, {a, b, e} is a subset of A.

(vii) False

Since, {a, b, e} does not belong to A, {a, b, e} ⊂ A this is the correct form.

(viii) False

Here, {a, b, c} is not a subset of A

(ix) False

Since, ϕ is a subset of A.

Hence, ϕ ⊂ A.

(x) False

Hence, {ϕ} is not subset of A, ϕ is a subset of A.

(i) True

Since, Φ belongs to set A. Thus, true.

(ii) True

Since, {Φ} is an element of set A. Thus, true.

(iii) False

Since, 1 is not an element of A. Thus, false.

(iv) True

Since, {2, Φ} is a subset of A. Thus, true.

(v) False

Since, 2 is not a subset of set A, it is an element of set A. Thus, false.

(vi) True

Since, {2, {1}} is not a subset of set A. Thus, true.

(vii) True

Neither {2} and nor {1} is a subset of set A. Thus, true.

(viii) True

Here, all three {ϕ, {ϕ}, {1, ϕ}} are subset of set A. Thus, true.

(ix) True

Since, {{ϕ}} is a subset of set A. Thus, true.

(i) This is not subset of given set but belongs to the given set.

Thus, the correct form would be

a ∈ {a,b,c}

(ii) In this {a} is subset of {a, b, c}

Thus, the correct form would be

{a} ⊂ {a, b, c}

(iii) ‘a’ is not the element of the set.

So, the correct form would be

{a} ∈ {{a}, b}

(iv) {a} is not a subset of given set.

So, the correct form would be

{a} ∈ {{a}, b}

(v) {b, c} is not a subset of given set. But it belongs to the given set.

Thus, the correct form would be

{b, c} ∈ {a,{b, c}}

(vi) {a, b} is not a subset of given set.

Thus, the correct form would be

{a, b}⊄{a,{b, c}}

(vii) ϕ does not belong to the given set but it is subset.

Thus, the correct form would be

ϕ ⊂ {a, b}

(viii) Since, it is the correct form. ϕ is subset of every set.

(ix) x + 3 = 3

x = 0 = {0}

It is not ϕ

Thus, the correct form would be

{x: x + 3 = 3} ≠ ϕ

i) This is null set. We know that there is no integer that lie between 2 and 3. 

ii) This is also a null set. We know that there is’ no natural number less than ‘1’.

iii) This is a null set. We know that odd numbers do not leave remainder zero when divided by 2.