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1. (-4, 6]

2. [0, 7)

3. [3, 4]

Let H = set of students, who know Hindi 

E = set of students, who know English 

Then n(H) = 100; n(E) = 50 and n(H∩E) = 25. 

∴ n(H∪E) = n(H) + n(E)-n(H∩E) 

= 100 + 50 - 25 = 125 

Hence, 125 students are there in the group.

1. Given; n(U) = 200; n(A) = 120;

n(B) = 50; n(A∩B) = 30

n (Chemical A but not chemical B)

= n(A ∩ B’) = n(A) – n(A ∩ B) = 120 – 30 = 90

2. n (Chemical B but not chemical A)

= n(A’ ∩ B) = n(B) – n(A ∩ B) = 50 – 30 = 20

3. n (Chemical A or chemical B)

= n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

= 120 + 50 – 30 = 140.

Let U = set of surveyed students 

A = set of students taking apple juice 

B = set of students taking orange juice. 

Then n(∪) = 400; n(A) = 100; n(B) = 150; and n(A∩B) = 75 

Now n(A’∩B’) = n(A∪B)’ 

= n(∪) – n(A∪B) 

= n(∪)-n(A)-n(B) + n(A∩S) 

= 400-100-150 + 75 = 225 

∴ 225 students were taking neither apple juice nor orange juice.

(i) The subsets of {A} are ϕ and {A} 

(ii) The subsets of {a, b} are ϕ, {a}, {b}, and {a, b}. 

(iii) The subsets of {–2, 3} are ϕ, {-2}, {3}, and {-2, 3}. 

(iv) The subsets of {–1,0,1} are ϕ ,{-1},{0},{1},{-1,0},{0,1},{-1,1},{–1, 0, 1} 

(v)ϕ has only one subset ϕ. 

(vi) let x = {3} 

Then, F = {2,x} 

The subsets of {2,x} are ϕ , {2},{x},{2,x} 

i.e ϕ , {2},{{3}},{2,{3}} 

(vii) let x = {5,6} 

Then, G = {3,4,x} 

The subsets of {3,4,x } are ϕ ,{3},{4},{x},{3,x},{4,x},{3,4}, {3,4,x}

i.e. ϕ ,{3},{4},{{5,6}},{3,{5,6}},{4,{5,6}},{3,4},{3,4,{5,6}}

(i) {1},{2},ϕ 

(ii) {1},{2},{3},{1,2},{2,3},{1,3},ϕ. 

(iii) ϕ ,{1}.

(i) Subsets of given set are {a}, ϕ. 

(ii) Subsets of given set are {0},{1},{0,1},ϕ. 

(iii) Subsets of given set are {a},{b},{c},{a,b},{b,c},{a,c},{a,b,c},ϕ. 

(iv) Subsets of given set are {1},{{1}},{1,{1}},ϕ. 

(v) Subsets of given set are{ϕ},ϕ.

(i) True 

Φ is a member of set A. Hence, true. 

(ii) True 

{ Φ} is a member of set A. Hence, true. 

(iii) False 

1 alone is not a member of A. Hence, false. 

(iv) False 

We can see that 2 is a member of set A, {2, Φ} is not. Hence, false 

(v) True 

2 is a member of set A. Hence, true. 

(vi) True 

{1} is not a member of set A. 

(vii) True 

Neither {2} and nor {1} is a member of set A. Hence, true. 

(viii) True 

All three are members of set A. Hence, true. 

(ix) False 

{{ϕ}} is not a member of set A. Hence, false

A ⊂ B 

Explanation: we have, A = {2,4,6,8,...} and B = {. . ., –3, –2, –1, 0, 1, 2, 3, . . .}. 

since, even natural numbers are also integers, we observe that elements of A belongs to B. 

Thus , A ⊂ B.

(i) Given: a is an element of set A 

this means a ∈ A 

(ii) Given: b is not an element of A 

this means b ∉ A 

(iii) Given: A is an empty set 

and B is a non – empty set 

this means A = ɸ 

and B ≠ ɸ 

(iv) Given: In set A, total number of elements is 6 

This means, |A| = 6

(v) Given: 0 is a whole number but not a natural number 

This means 0 ∈ W but 0 ∉ N

A ⊂ B 

Explanation: A is a null set. Since, ϕ is a subset of every set therefore A ⊂ B.

A ⊄ B 

Explanation: A ⊄ B since 0∈A and 0∉B.

A ⊄ B 

Explanation: we have, A = { -1,1} and B={1} 

Since, -1∈A and -1∉B thus A ⊄ B .

A ⊄ B 

Explanation: A ⊄ B since 3∈A and 3∉B.

(i) True 

A rational number is represented by the form p/q where p and q are integers and (q not equal to 0) keeping q =1 we can place any number as p. Which then will be an integer. 

(ii) True 

All crows are birds, so they are contained in the set of all birds. 

(iii) False 

Every square can be a rectangle, but every rectangle cannot be a square. 

(iv) False 

Every square can be a rectangle, but every rectangle cannot be a square. 

(v) False 

P = {a} 

B = {{a}} 

But {a} = P 

B = {P} 

Hence they are not equal. 

(vi) True 

A = For “LITTLE” 

A = {L,I,T,E} = {E,I,L,T} 

B = For “TITLE” 

B = {T,I,L,E} = {E,I,L,T}

(i) True

Since, 1 belongs to the given set {1, 2, 3} as it is present in it.

(ii) False

Here, a is an element and not a subset of a set {b, c, a}

(iii) False

Here, {a} is a subset of set {b, c, a} and not an element.

(iv) True

Since, we do not have repeat same elements in a given set.

(v) False

Given as, x + 8 = 8

i.e. x = 0

Therefore, the given set is a single ton set {0}. Here, it is not a null set.

(i) True

Since, a rational number is represented by the form p/q where p and q are integers and (q not equal to 0) keeping q = 1 we can place any number as p. Which then will be an integer.

(ii) True

Crows are also birds, Therefore, they are contained in the set of all birds.

(iii) False

Since, very square can be a rectangle, but every rectangle cannot be a square.

(iv) False

P = {a}

B = {{a}}

But {a} = P

B = {P}

Thus, they are not equal.

(v) True

A = For the “LITTLE”

A = {L, I, T, E} = {E, I, L, T}

B = For the “TITLE”

B = {T, I, L, E} = {E, I, L, T}

∴ A = B

A = x² – 8x + 12=0 

= (x – 6)(x – 2) = 0 

= x = 2 or x = 6A = {2,6} 

B = {2,4,6} 

C = {2,4,6,8} 

D = {6} 

So we can say 

D⊂A⊂B⊂C

A = x² – 8x + 12=0

= (x – 6) (x – 2) =0

= x = 2 or x = 6

A = {2, 6}

B = {2, 4, 6}

C = {2, 4, 6, 8}

D = {6}

Therefore we can say

D ⊂ A ⊂ B ⊂ C

(i) False: For example, let A = {1, 2} and

B = (2, 3, {1, 2}}. As 1∈ A and A∈ B but 1∉ B 

(ii) False: For example, let A = {1, 2}, B = {1, 2, 3} and C = {3, 4,{1, 2, 3}}

Now clearly, A ⊂ B and B⊂C but A ∉ C, 

(iii) True: Given A ⊂ B and 

Now we have to prove A ⊂ C 

∴ x ∈ C ∵ B ∈ C. 

∀ X ∈ A ⇒X ∈ C∴ A ∈ C. 

(iv) False: For example, let 

A = (1, 2), B = {2, 3),C = (1, 2, 4) 

Clearly A⊄B and B⊄C but A ⊂ C 

(v) False: For example, let A=(l,2), B={2, 3, 4, 5} 

Now 1 ∈ A and A ∉ B but 1 ∉ B 

(vi) True: Suppose x∈ A then X∈ B A⊂ B.

Given: n(A) = 8, n(B) = 11, n(A ∪ B) = 14 

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

⇒ 14 = 8 + 11 – n(A ∩ B) 

⇒ 14 = 19 – n(A ∩ B) 

⇒ n(A ∩ B) = 19 – 14 

⇒ n(A ∩ B) = 5 

Hence n(A ∩ B) = 5