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**OBJECTIVE TYPE QUESTIONS**

**Q.29. Suppose A _{1}, A_{2}, …, A_{30} are thirty sets each having 5 elements and B_{1}, B_{2}, …, B_{n} are**

Number of elements in A

But each element is repeated 10 times

∴ n(S) = 30 × 5/10 = 150/10 = 15 .........(i)

Number of elements in B

But each element is repeated 09 times

∴ n(S) = 3n/9 = n/3 ...........(ii)

From(i) and (ii) we get

n/3 = 15

⇒ n = 15 × 3 = 45

Hence, the correct option is (c).

We have the number of subsets of set containing n elements = 2

So, according to the question, we have

2

⇒ 2

∴ 2

⇒ n = 4 and 2

⇒ n = 4 and m – n = 3

⇒ m – 4 = 3

⇒ m = 7

Hence,the correct option (b).

we know that: (A ∩ B)’ = A’ ∪ B’ [De Morgan's law]

∴ (A ∩ B’)’ ∪ (B ∩ C) = [A’ ∪ (B’)’] ∪ (B ∩ C)

= (A’ ∪ B) ∪ (B ∩ C) [∵ (B’)’ = B]

= A’ ∪ B

Hence, the correct option is (b)

We know that rectangles, rhombus and square in a plane is a parallelogram but trapezium is not a parallelogram.

∴ F

Hence,the correct option is (d).

The given conditions of the question may be represented by the following Venn diagram. From the given Venn diagram, we clearly conclude that

S ∪ T ∪ C = S

Hence, the correct option is (c).

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

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

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

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

Let OABC be a rectangle whose sides a and b are along the positive direction of X

and Y respectively.

∴ Clearly,

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

Hence, the correct option is (d).

Total number of students = 60

⇒ n(U) = 60

Number of students who play cricket = 25

⇒ n(C) = 25

Number of students who play tennis = 20

⇒ n(T) = 20

Number of students who play cricket and tennis both = 10

⇒ n(C ∩ T) = 10

∴ n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

= 25 + 20 – 10 = 45 – 10 = 35

∴ n(C’ ∩ T’) = n(U) – n(C ∪ T)

= 60 – 35 = 25

Hence, the correct option is (b).

(a) 210

(b) 290

(c) 180

(d) 260

Total number of persons in a town = 840

⇒ n(U) = 840

Number of persons who read Hindi = 450

⇒ n(H) = 450

Number of persons who read English = 300

⇒ n(E) = 300

Number of persons who read both = 200

⇒ n(H ∩ E) = 200

∴ n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

= 450 + 300 – 200 = 550

n(H’ ∩ E’) = n(U) – n(H ∪ E)

= 840 – 550 = 290

Hence, the correct option is (b).

Given that: X = {8

And Y = {49n - 49 | n ∈ N} = {0, 49, 98, ...}

Here, it is clear that every element belonging to X is also present in Y.

∴ X ⊂ Y. Hence, the correct option is (a).

Let p% of the people watch a channel and q% of the people watch another channel

∵ n(p ∩ q) = x% and n(p ∪ q) ≤ 100

So, n(p ∪ q) ≥ n(p) + n(q) – n(p ∩ q)

100 ≥ 63 + 76 – x

100 ≥ 139 – x

⇒ x ≥ 139 – 100

⇒ x ≥ 39

Now n(p) = 63

∴ n(p ∩ q) ≤ n(p)

⇒ x ≤ 63

So 39 ≤ x ≤ 63.

Hence, the correct option is (c).

(a) A ∩ B = A

(b) A ∩ B = B

(c) A ∩ B = φ

(d) A ∪ B = A

Give that: A = {(x, y) | y = 1/x, 0 ≠ x ∈ R},

and B = {(x, y)|y = – x, x ∈ R}

It is very clear that y = 1/x and y = – x

∵ 1/x ≠ -x

∴ A ∩ B = ϕ

Hence, the correct option is (c).

Given that: A ∩ (A ∪ B)

Let x ∈ A ∩ (A ∪ B)

⇒ x ∈ A and x ∈ (A ∪ B)

⇒ x ∈ A and (x ∈ A or x ∈ B)

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

⇒ x ∈ A or x ∈ (A ∩ B)

⇒ x ∈ A.

Hence the correct option is (a)

Given that:

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

B = {2, 4, …, 18}

U= N = {1, 2, 3, 4, 5, …}

A’ ∪ (A ∪ B) ∩ B’ = A’ ∪ [(A ∩ B’) ∪ (B ∩ B’)]

= A’ ∪ (A ∩ B’) ∪ Φ [∵ A ∩ A’ = Φ]

= A’ ∪ (A ∩ B’)

= (A’ ∪ A) ∩ (A’ ∪ B’)

= N ∪ (A’ ∪ B’) [∵ A’ ∪ A = N]

= A’ ∪ B’

= (A ∩ B)’ = (Φ)’ = N [∵ A ∩ B = Φ]

Hence, the correct option is (b).

Given that: S ={x|x is a positive multiple of 3 < 100}

∴ S = {3, 6, 9, 12, 15, 18,…, 99}

⇒ n(S) = 33

T = (x | x is a prime number < 20}

∴ T = {2, 3, 5, 7, 11, 13, 17, 19}

⇒ n(T) = 8

So, n(S) + n(T) = 33 + 8 = 41

Hence, the correct option is (b).

Let x ∈ X ∩ (X ∪ Y)′

⇒ x ∈ X ∩ (X’ ∩ Y’)

⇒ x ∈ (X ∩ X’) ∩ (X ∩ Y’)

⇒ x ∈ Φ ∩ (X ∩ Y’) [∵A ∩ A’ = Φ]

⇒ x ∈ Φ

Hence, the correct option is (c).

The set {x ∈ R : 1 ≤ x < 2} can be written as [1, 2)

Hence, the filler is [1, 2).

Here A = Φ

∴ n(A)=0

∵ n[P(A)]= 2

= 2

Hence, the filler is 1.

Since A ⊂ B

∴ n(A ∪ B) = n(B)

Hence, the filler is n(B).

From the Venn diagram it is clear that

A – B = A ∩ B’.

Hence the filler is A ∩ B’.

Power set of A = P(A)

= {{1}, {2}, {1, 2}, Ф}

Hence, the filler is {{1}, {2}, {1, 2}, Ф }.

Given that: A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}

∴Universal set of all the given sets is

U = {0, 1, 2, 3, 4, 5, 6, 8}

Hence, the filler is {0, 1, 2, 3, 4, 5, 6, 8}.

Given that: U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 5} B = {2, 4, 6, 7} and C = {2, 3, 4, 8}

Hence, the filler is {1, 5, 9, 10}.

Hence, the filler is {1, 2, 3, 5, 6, 7, 9, 10}.

Q.51. For all sets A and B, A – (A ∩ B) is equal to _______.

⇒ A – (A ∩ B)= A ∩ B’

Hence, the filler is A ∩ B’.

= [(A ∩ B)’ ∩ A)’]’ [∵ A’ ∪ B’= (A ∩ B)’]

= [(A ∩ B)’]’ ∪ (A’)’ [∵(A’)’ = A]

= (A ∩ B) ∪ A

= A.

So (i) is match with (b).

**(ii)** [B’ ∪ (B’ – A)]’ = [B’ ∪ (B’ ∩ A’)]’ [∵ A – B = A ∩ B’]

= (B’)’ ∩ (B’ ∩ A’)’ [∵ A’ ∩ B’ = (A ∪ B)’]

= B ∩ (B ∪ A) = B

So (ii) is matched with (c).

**(iii)** (A – B) – (B – C)

= (A ∩ B’) – (B ∩ C’) [∵A – B = (A ∩ B’)]

= (A ∩ B’) ∩ (B ∩ C’)’

= (A ∩ B’) ∩ (B’ ∩ (C’)’) [(A ∩ B)’ = A’ ∪ B’]

= (A ∩ B’) ∩ (B’ ∩ C)

= [A ∩ (B’ ∪ C)] ∩ [B’ ∩ (B’ ∪ C)]

= [A ∩ (B’ ∪ C)] ∩ B’

= (A ∩ B’) ∩ (B’ ∪ C) ∩ B’

= (A ∩ B’) ∩ B’ = (A ∩ B’)= A – B

So, (iii) is matched with (a).

**(iv)** (A – B) ∩ (C – B) = (A ∩ B’) ∩ (C ∩ B’) [∵A – B= (A – B’)]

= (A ∩ C) ∩ B’

= (A ∩ C) – B [∵ A ∩ B’ = A – B]

Hence, (iv) is matched with (f).

**(v)** A × (B ∪ C) = (A × B) ∩ (A × C)

So, (v) is matched with (d).

**(vi)** A × (B ∪ C) = (A × B) ∪ (A × C)

So, (vi) is matched with (e).**State True or False for the following statements****Q.53. If A is any set, then A ⊂ A****Ans.**

Since every set is a subset of itself. So, it is ‘True’.**Q.54. Given that M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and if B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then B ⊄ M****Ans.**

M = {1, 2, 3, 4, 5, 6, 7, 8, 9}

and B = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Since M and B has same elements

∴ M = B, so B ⊄ M is False.**Q.55. The sets {1, 2, 3, 4} and {3, 4, 5, 6} are equal.****Ans.**

False, since the two sets do not contain the same elements.**Q.56. Q ∪ Z = Q, where Q is the set of rational numbers and Z is the set of integers.****Ans.**

True, since every integer is a rational number.

∴ Z ⊂ Q, so Q ∪ Z = Q.**Q.57. Let sets R and T be defined as****R = {x ∈ Z | x is divisible by 2}****T = {x ∈ Z | x is divisible by 6}. Then T ⊂ R****Ans.**

We can written the given sets is Roster form

R = {..., – 8, – 6, – 4, – 2, 0, 2, 4, 6, 8, ...}

and T = {..., – 18, – 12, – 6, 0, 6, 12, 18, ...}

Since every element of T is present in R.

So, T ⊂ R.

Hence, the statement is ‘True’.**Q.58. Given A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2}. Then A = B.****Ans.**

Here A = {0, 1, 2}, B is a set having all real numbers from 0 to 2.

So A ≠ B.

Hence, the given statement is ‘False’.

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