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We have, N= {1,2, 3,4,…, 100}

(i) subset of N whose elements are even numbers = {2,4, 6, 8,…, 100}

(ii) subset of N whose elements are perfect square = {1,4, 9, 16, 25, 36,49, 64, 81, 100}

Given L,= {1,2, 3,4}, M= {3,4,5,6} and N= {1,3,5}

M⋃N= {1,3,4, 5,6}

L – (M⋃N) = {2}

Now, L-M= {1, 2} and L-N= {2,4}

{L-M) ⋂ {L-N)= {2}

A total number of subsets for any given ser is 2ⁿ. 

In every set there is only one improper set i.e. the set itself. 

∴ proper subsets would be 2ⁿ–1. 

Example : 

A = {1,2,3}There will be 2³ - 1 = 8 - 1 = 7 proper subsets.These are [ { ∅},{1},{2},{3},{1,2}, {1,3},{2,3}].

(i) False 

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

{c,d} ϵ A 

(ii) True 

{c,d} ϵ A 

(iii) True 

{c,d} is a subset of A. 

(iv) It is true that a belong to A. 

(v) False 

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

(vi) True 

(vii) False 

{a,b,e} ⊂ A this is the correct form. 

(viii) False 

{a,b,c} is not a subset of A 

(ix) False 

ϕ is a subset of A. 

ϕ ⊂ A. 

(x) False

(i) False 

1 is not an element of A. 

(ii) True 

{1,2,3} ϵ A. this is correct. 

(iii) True. 

{6, 7, 8} ϵ A. 

(iv) False 

2 is not an element of A, and ϕ is also not an element of A. Hence, false.

(v) False 

2 alone is not an element of A. Hence, false. 

(vi) True 

{1, 2, 3} is the set belonging to A. {2, {1}} does not belong to A. 

(vii) True 

{{2}}, {1}} does not belongs to A. Hence, true. 

(viii) False 

Φ does not belong to A.

There is only one element in set A. 

Now, 1 is not an element of B. 

Therefore, A ⊄ B.

Let A – Apple juice; O – Orange juice be the sets.

Given; n(U) = 400; n(A) = 100;

n(O) = 150; n(A ∩ O) = 75

n (neither apple juice nor orange juice)

= n(A’ ∩ O’) = n((A ∪ O)’)

= n(U) – n(A ∪ O)

= 400 – [n(A) + n(O) – n(A ∩ O)]

= 400 – [100 + 150 – 75] 

= 400 – 175 = 225.

Let U = set of students in a school.

A = set of students taking tea

B = set of students taking coffee

Then n(∪) = 600; n(A) = 150; n(B) = 225; n(A∩B) = 100 

∴ n(Au B) = n(A) + n(B)-n(A∩B) = 150 + 225-100 =275 

Now n(A’∩B’) = n(A∪B) 

= n(∪)-n(A∪5) = 600-275 = 325 

∴ 325 students were taking neither tea nor coffee.

(b) 2 and 3

1. A – (B ∪ C) = (A – B) ∪ (A – C) 

(A – B) ∪ (A – C) 

⇒ For all (x ∈ A and x ∉ B) or (x ∈ A and x ∉ C) 

⇒ For all (x ∈ A and x ∉ B and x ∉ C) 

⇒ x ∈ A – (B ∩ C ) 

Hence not correct. 

2. x ∈ (A – (A ∩ B)) ⇒ x ∈ A and x ∉ A ∩ B 

⇒ x ∈ A and x ∉ B 

⇒ x ∈ (A – B) 

3. x ∈ [(A ∩ B) ∪ (A – B)] 

⇒ (x ∈ A and x ∉ B) or (x ∈ A and x ∈ B′) 

⇒ x ∈ A and x ∈ B and x ∈ B′ 

⇒ x ∈ A and x ∈ B ∩ B′ 

⇒ x ∈ A and x ∈ ϕ 

⇒ x ∈ A ∪ ϕ 

⇒ x ∈ A 

Hence (1) is incorrect.

(b) A – (B ∪ C) = A ∩ (B ∪ C)′

A. A ∪ (B – C) = A ∪ (B ∩ C′) 

B. A – (B ∪ C) = A ∩ (B ∪ C)′ = A ∩ (B′ ∪ C′) 

C. A – (B ∩ C) = A ∩ (B ∩ C)′ = A ∩ (B′ ∪ C′)

D. A ∩ (B – C) = A ∩ (B ∩ C′) 

Hence B is the correct option.

Let A be the set of people who read newspaper H.

Let B be the set of people who read newspaper T.

Let C be the set of people who read newspaper I.

Given n(A) = 25, n(B) = 26, and n(C) = 26

n(A∩C) = 9, n(A∩B) = 11, and (B∩C) = 8

n(A∩B∩C) = 3

Let U be the set of people who took part in the survey.

n(A∪B∪C) = n(A) + n(B) + n(C) − n(A∩B) − n(B∩C) − n(C∩A) + n(A∩B∩C)

= 25 + 26 + 26 − 11 − 8 − 9 + 3

= 52

Hence, 52 people read at least one of the newspaper.

(c) (A × B) ∪ (B × A) = (A × B) ∪ (B × C)

(A × B) = { (1,1), (1,2), (2,1), (2,2), (3,1), (3,2)} 

(B × A) = { (1,1), (1,2), (1,3), (2,1), (2,2), (2,3)} 

(B × C) = { (1,2), (1,3), (2,2), (2,3)} 

(C × B) = { (2,1), (2,2), (3,1), (3,2)} 

(A × C) = { (1,2), (1,3), (2,2), (2,3), (3,2) (3,3)} 

(C × A) = { (2,1), (2,2), (2,3), (3,1), (3,2) (3,3)} 

∴ (A × B) ∪ (B × A) = {(1,1), (1,2), (2,1), (2,2), (3,1), (3,2) (1,3), (2,3)} 

and (A × B) ∪ (B × C) = { (1,1), (1,2), (2,1), (2,2), (3,1), (3,2), (1,3), (2,3)} 

∴ (A × B) ∪ (B × A) = (A × B) ∪ (B × C)

(c) n (A – B) = n (A) – n (A ∩ B)

n (A – B) = No. of these elements of A which are not common in A and B 

= n (A) – n (A ∩ B)

a. All elements of set A are not present in set B, C and D.

 ∴ A ⊆ B, 

 ∴ A ⊆ C,

 ∴ A ⊆ D 

 ∴ Statement (a) is false. 

b. All elements of set B are not present in set A, C and D. 

 ∴ B ⊆ A, 

 ∴ B ⊆ C,

 ∴ B ⊆ D 

 ∴ Statement (b) is false.

Let C = set of people who like cricket 

T = set of people who like tennis. 

Then n(C ∪ T) = 65; n(C) = 40; n(C ∩ T) = 10

∴ n(T) = n{C∪T) + n(C∩T)-n(C)

= 65 + 10-40 = 35

Number of people who like tennis only

n(T-C) = n(T)-n(C ∩ T)

= 35-10 = 25

Let F – French; S – Spanish be the sets.

Given; n(F) = 50; n(S) = 20;w(F ∩ S) = 10

n (speaks at least one of these two languages)

= n(F ∪ S) = n(F) + n(S) – n(F ∩ S)

= 50 + 20 – 10 = 60.

U = {1, 3, 9, 11, 13, 18, 19},

B= {3, 9, 11, 13} ….(i)

∴ B’= {1, 18, 19} 

(B’)’= {3, 9, 11, 13} ….(ii) 

∴ (B’)’ = B … [From (i) and (ii)] 

∴ Complement of a complement is the given set itself.

Here, n(A) = 15, n(A ∪ B) = 29, n(A ∩ B) = 7

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

∴ 29 = 15 + n(B) – 7 

∴ 29 – 15 + 7 = n(B) 

∴ n(B) = 21

(b) A × (B – C) = (A × B) – (A × C)

We known that the cartesian product of two sets is defined as 

X × Y = { (\(x\), y) : \(x\) ∈ \(x\) and \(x\) ∈ y} 

∴ A × (B – C) = (A × B) – (A × C)

(b) 2 only

1. A – ( B – C) = A – (B ∩ C′) 

= A ∩ (B ∩ C′)′ 

= A ∩ (B′ ∪ (C′)′) 

= A ∩ (B′ ∪ C) 

(Α – Β) ∪ C = (A ∩ B′) ∪ C 

Thus, A – ( B – C) ≠ (A – B) ∪ C 

2. A – ( B ∪ C) = A ∩ (B ∩ C)′ 

= A ∩ (B′ ∩ C′) 

(A – B) – C = (A ∩ B′) – C 

= A ∩ B′ ∩ C′ 

⇒ A – (B ∪ C) = (A – B) – C 

Associative property.

Here, we have,

The people who like oranges = 76%

The people who like bananas = 62%

Suppose people who like oranges be n (O)

Suppose people who like bananas be n (B)

The total number of people who like oranges or bananas = n (O ∪ B) = 100

The people who like both oranges and bananas = n (O ∩ B)

As we know,

n (O ∪ B) = n (O) + n (B) – n (O ∩ B)

By substituting the values we get

100 = 76 + 62 – n (O ∩ B)

100 = 138 – n (O ∩ B)

n (O ∩ B) = 38

Hence, people who like both oranges and banana is 38%.

(c) 259

Here, n (c) = 125, n (F) = 220, n (H) = 300 

n (H ∩ F) = 28 , n (C ∩ F) = 70, n (C ∩ H) = 32 and n ( C ∩ F ∩ H) = 26 

∴ Number of students who did not play any game 

= n (C′∩ F′∩ H′) 

= n ((C ∪ F ∪ H)′) 

= n ( ξ) – n (C ∪ F ∪ H) 

= n ( ξ) –[n (C) + n (F) + n (H) – n( C ∩ F) – n ( H ∩ F) – n ( C ∩ H) + n (C ∩ F ∩ H)] 

= 800–[ 125+220 + 300 – 70 –28–32– 26] 

= 800 – 541 = 259

According to the question,

There are three sets A, B and C

To show:

(A – B) ∩ (A – C) = A – (B ∪ C)

Let x ∈ (A – B) ∩ (A – C)

⇒ x ∈ (A – B) and x ∈ (A – C)

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

⇒ x ∈ A and (x ∉ B and x ∉ C)

⇒ x ∈ A and x ∉ (B ∪ C)

⇒ x ∈ A – (B ∪ C)

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

Let y ∈ A – (B ∪ C)

⇒ y ∈ A and y ∉ (B ∪ C)

⇒ y ∈ A and (y ∉ B and y ∉ C)

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

⇒ y ∈ (A – B) and y ∈ (A – C)

⇒ y ∈ (A – B) ∩ (A – C)

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

We know that,

If P ⊂ Q and Q ⊂ P

Then, P = Q

Therefore, from equations (i) and (ii),

A – (B ∪ C) = (A – B) ∩ (A – C)

True

According to the question,

There are two sets A and B

To check: (A – B) ∪ (A ∩ B) = A is true or false

L.H.S = (A – B) ∪ (A ∩ B)

Since, A – B = A ∩ B’,

We get,

= (A ∩ B’) ∪ (A ∩ B)

Using distributive property of set:

We get,

(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)

= A ∩ (B’ ∪ B)

= A ∩ U

= A

= R.H.S

Hence, the given statement “for all sets A and B, (A – B) ∪ (A ∩ B) = A” is true

A ∩ B means A intersection B. Common elements of A and B come in this group. 

Given 

A ⊂ B 

i.e. B is having all elements that is present in A. 

∴ A ∩ B = A. 

Example: 

A = {1,2}

B = {1,2,3,4}

A ∩ B = {1,2} = A

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 and x ∈ C

⇒ x ∈ B and 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.