Interactive ISC Mathematics Class 11 Solutions Chapter 1 Sets MCQs engage students in active learning and exploration.

## ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets MCQs

Choose the correct answer from the given four options in questions (1 to 29) :

Question 1.

Which of the following collection of objects is not a set?

(a) The collection of all even integers.

(b) The collection of all months of a year beginning with letter J.

(c) The collection of most talented writers of India.

(d) The collection of all prime numbers less than 20.

Answer:

(c) The collection of most talented writers of India.

The collection of most talented writers of India is not a well defined collection. Since according to some one, one witer is talented and according to some other the same writer is non-talented so the word talented is vague.

∴ Thus collection does not represent a set.

Question 2.

If A = {1, 2, 3, 4, 5}, then which of the following is not true?

(a) 0 ≠ A

(b) 3 ∈ A

(c) {3} ∈ A

(d) {3} ⊂ A

Answer:

(c) {3} ∈ A

Clearly 0, 3 are members of A so both are belongs to set A and {3} is a subset of A but {3} ∉ A.

Question 3.

Which of the following is a null set?

(a) {x : x ∈ N, 2x – 1 = 3}

(b) {x : x ∈ N, x^{2} < 20}

(c) (x : x ∈ N, x is a factor of 128}

(d) {x : x ∈ I, x ≤ 7}

Answer:

(c) (x : x ∈ N, x is a factor of 128}

For option (a) ;

2x – 1 = 3

⇒ 2x = 4

⇒ x = 2

∴ {x : x ∈ N, 2x – 1 = 3} = {2) which is not an empty set.

For option (b) ;

x^{2} < 20 and x ∈ N

∴ x = 1 ;

∵ 1^{2} < 20 ;

x = 3

∵ 3^{2} < 20 ;

x = 2

∵ 2^{2} < 20 ;

x = 4

∵ 4^{2} < 20

∴ {x : x ∈ N ; x^{2} < 20} = {1, 2, 3, 4}

For option (d) :

3x + 7 = 1

⇒ x = – 2 {x : x ∈ N ; 3x + 7 = 1}

= {- 2} which is not a null set

For option (c) ;

since there is only one even prime number, which is 2 and their is no even prime number which is > 2.

∴ given set is a null set.

Question 4.

Which of the following is a finite set?

(a) {x : x = 2n, n° ∈ N}

(b) {x : x is a prime number}

(c) {x : x ∈ N, x is a factor of 128}

(d) {x : x ∈ I, x ≤ 7}

Answer:

(c) {x : x ∈ N, x is a factor of 128}

xisafactorof 128 and x ∈ N

∴ x = 1, 2, 4, 8, 16, 32, 64, 128

Thus {x : x ∈ N, x is a factor of 128} = {1, 2, 4, 8, 16, 32, 64, 128}

which contains 8 elements and hence it is a finite set.

Question 5.

Given set A = {1, 3, 5, 7, 9}, B = {0, 2, 4, 6} and C = {7, 8, 9}. Which of the following may be taken as universal set for all the three sets A, B and C?

(a) {0, 1, 2, 3, 4, 5, 6, 7, 8}

(b) {1, 2, 3, 4, 5, 6, 7, 8, 9}

(c) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(d) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Answer:

(d) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Now, universal set contains all elements of A three sets A, B and C

∴ U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

So universal set contain some more elements other than elements of A, B and C.

Question 6.

Number of proper subsets of a set containing 4 elements is

(a) 4^{2}

(b) 4^{2} – 1

(c) 2^{4}

(d) 2^{4} – 1

Answer:

(d) 2^{4} – 1

We know that, the number of proper subsets of set containing n elements = 2^{n} – 1

∴ No. of subsets of a set containing 4 elements = 2^{4} – 1 = 15.

Question 7.

Which of the following is not correct?

(a) N ⊂ R

(b) N ⊂ Q

(c) Q ⊂ R

(d) N ⊂ T

Answer:

(d) N ⊂ T

Since every natural number is a real number

as well as rational number

∴ N ⊂ R and N ⊂ Q both are true.

Also rational and irrational numbers combine to form real number

∴ Q ⊂ R is correct

Question 8.

On real axis if A = [1, 5] and B = [3, 9], then A – B is

(a) (5, 9)

(b) (1, 3)

(c) [5, 9)

(d) [1, 3)

Answer:

(d) [1, 3)

A = [1, 5]

and B = [3, 9]

∴ A – B = A ∩ B^{c}

= A – A ∩ B

= [1, 3)

Question 9.

If n (A – B) = 10, n (B – A) = 23, n (A ∪ B) = 50, then n (A ∩ B) is

(a) 7

(b) 17

(c) 27

(d) 33

Answer:

(b) 17

∴ n (A ∪ B) = n (A – B) + n (A ∩ B) + n (B – A)

⇒ 50 = 10 + n (A ∩ B) + 23

⇒ n (A ∩ B) = 50 – 33 = 17

Question 10.

Two finite sets A and B are such that A ⊂ B. then which of the following is not correct ?

(a) A ∪ B = B

(b) A ∩ B = A

(c) A – B = Φ

(d) B – A = Φ

Answer:

(d) B – A = Φ

When A ⊂ B

∴ A ∪ B = B

and A ∩ B = A

A – B = A – A ∩ B

= A – A = Φ

and B – A = B – B

= B – A ≠ Φ

Question 11.

Two finite sets have m and n elements respectively. The total number of subsets of the first set is 192 more than the total number of subsets of the second set. The values of in and n respectively are

(a) 7, 6

(b) 8, 6

(c) 8, 5

(d) 9, 7

Answer:

(b) 8, 6

The total number of subsets of set containing m elements = 2^{m}

and total no. of subsets of set containing n elements = 2^{n}

Then according to given condition.

2^{m} = 2^{n} + 192

⇒ 2^{m} – 2^{n} = 192

⇒ 2^{n} (2^{m – n} – 1) = 2^{6} × 3^{1}

2^{n} (2^{m-n} – 1) = 2^{6} × (2^{2} – 1)

∴ n = 6

and m – n = 2

⇒ m = 8

Question 12.

For any two sets A and B, A ∩ (A ∪ B) is equal to

(a) A

(b) B

(c) Φ

(d) A ∩ B

Solution:

(a) A

A ∩ (A ∪ B) = A

[∵ A ⊂ A ∪ B]

Question 13.

The symmetric difference of A = {0, 1, 2) and B {2, 3, 4} is

(a) {0, 1}

(b) {3, 4}

(c) {0, 1, 3, 4}

(d) {0, 1, 2, 3, 4}

Solution:

(c) {0, 1, 3, 4}

A = {0, 1, 2)

and B = {2, 3, 4}

∴ A – B = {0, 1} and B – A = {3, 4}

Thus, A ∆ B = (A – B) ∪ (B – A)

= {0, 1} ∪ {3, 4}

= {0, 1, 3, 4}

Question 14.

The symmetric difference of sets A and B is equal to

(a) (A – B) ∪ A

(b) (B – A) ∪ B

(c) (A ∪ B) – (A ∩ B)

(d) (A ∪ B) (A ∩ B)

Solution:

(c) (A ∪ B) – (A ∩ B)

A ∆ B = (A – B) ∪ (B – A)

= (A ∩ B’) ∪ (B ∩ A’)

= {(A ∪ B) ∩ (B’ ∪ B)} ∩ {(A ∪ A’) ∩ (B’ ∪ A’)}

= {(A ∪ B) ∩ U} ∩ {U ∩ (B’ ∪ A’)}

= (A ∪ B) ∩ (B’ ∪ A’)

= (A ∪ B) ∩ (B ∩ A)’ [Demorgan’s law]

= (A ∪ B) – (B ∩ A)

Question 15.

The symmetric difference of sets A and B is not equal to

(a) (A – B) ∩ (B – A)

(b) (B – A) ∪ B

(c) (A ∪ B) – (A ∩ B)

d) ((A ∪ B) – B) ∪ ((A ∪ B) – A)

Solution:

(a) (A – B) ∩ (B – A)

A ∆ B = (A – B) ∪ (B – A)

Thus A ∆ B ≠ (A – B) ∩ (B – A)

For option (d) ;

= ((A ∪ B) – B) ∪ ((A ∪ B) – A)

= ((A ∪ B) ∩ B’) ∪ ((A ∪ B) ∩ A’)

= {(A ∩ B’) ∪ (B ∩ B’)) ∪ {(A ∩ A’) ∪ (B ‘ ∩ A’)}

= {(A ∩ B’) ∪ Φ) ∪ {Φ ∪ (B ∩ A’))

= (A ∩ B’) ∪ (B ∩ A’)

= (A – B) ∪ (B – A)

= A ∆ B

Question 16.

For any two sets X and Y, X ∩ (X ∪ Y)’ is equal to

(a) X

(b) Y

(c) Φ

(d) X ∩ Y

Solution:

(c) Φ

X ∩ (X ∪ Y)’ = X ∩ (X’ ∩ Y’) [Demorgan’s law]

= (X ∩ X’) ∩ Y’

= Φ ∩ Y’ = Φ

Question 17.

For any two sets A and B, ((A’ ∪ B’) – A is equal to

(a) A

(b) B

(c) Φ

(d) A ∩ B

Answer:

(a) A

((A’ ∪ B’) – A)’ = {(A’ ∪ B’) ∩ A’)’

= (A’ ∪ B’)’ ∪ (A’)’ [Demorgan’s law]

= [(A’)’ ∩ (B’)’} ∪ A

= (A ∩ B) ∪ A = A

[∵ A ∩ B ⊂ A]

Question 18.

For any two sets A and B, (B’ ∪ (B’ – A)]’ is equal to

(a) A

(b) B

(c) Φ

(d) A ∪ B

Answer:

(b) B

[B’ ∪ (B’ – A)]’ = [B’ ∪ (B’ ∩ A’)]’

= (B’)’ ∩ [B’ ∩ A’)’ [Demorgan’s law]

= B ∩ [(B’)’ ∪ (A’)’]

= B ∩ (B ∪ A) = B

[∵ B ⊂ B ∪ A]

Question 19.

For any re sets A, B and C, (A – B) ∩ (C – B) is equal to

(a) A – (B ∩ C)

(b) (A – C) ∩ B

(c) (A ∩ C) – B

(d) (A – B) ∩ C

Answer:

(c) (A ∩ C) – B

(A – B) ∩ (C – B) = (A ∩ B’) ∩ (C ∩ B’)

= (B’ ∩ A) ∩ (B’ ∩ C)

= B’ ∩ (A ∩ C)

= (A ∩ C) ∩ B’

= A ∩ C – B

Question 20.

Let A and B are two disjoint sets and ξ be the universal set, then A’ ∪ ((A ∪ B) ∩ B’) is equal to

(a) Φ

(b) ξ

(c) A

(d) B

Solution:

(b) ξ

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

= A’ ∪ ((A ∩ B’) ∪ Φ)

= A’ ∪ (A ∩ B’)

= (A’ ∪ A) ∩ (A’ ∪ B’)

= U ∩ (A’ ∪ B’)

= A’ ∪ B’

= (A ∩ B)’

[using Demorgan’s law]

= Φ’ = U = ξ

[∵ A and B disjoint sets = A ∩ B = Φ]

Question 21.

Let S = set of points inside the square, T = set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then

(a) S ∩ T ∩ C = Φ

(b) S ∪ T ∪ C = C

(c) S ∪ T ∪ C = S

(d) S ∪ T = S ∩ C

Answer:

(c) S ∪ T ∪ C = S

Therefore S ∪ T ∪ C = S

Question 22.

Let R be the set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of x-axis and y-axis. Then

(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}

Answer:

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

∴ R= {(x, y) ; 0 < x < a ; 0 < y < b}

Question 23.

In a class of 80 students, 39 students play football and 45 students play cricket and 15 students play both the games. Then the number of students who play neither is

(a) 11

(b) 14

(c) 16

(d) 18

Answer:

(a) 11

F : set of students who play football

C : set of students who play cricket

Then n (F) = 39 ;

n (C) = 45 ;

n (F ∩ C) = 15

and n (ξ) = 80

∴ n (F ∪ C) = n (F) + n (C) – n (F ∩ C)

= 39 + 45 – 15 = 69

Thus. No. of students who play neither = n (F’ ∩ C’)

= n {(F ∪ C)’}

= n (ξ) – n (F ∪ C)

= 80 – 69 = 11

Question 24.

In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. The number of persons who read neither is

(a) 210

(b) 290

(c) 180

(d) 260

Answer:

(b) 290

Let us define the following sets as under:

H : set of persons who read Hindi

E : set of persons who read English

Then n (H) = 450 ;

n (E) = 300 ;

n (H ∩ E) = 200

∴ Required no. of persons who read neither = n (H’ ∩ E’)

= n (H ∪ E)’

= n (ξ) – n (H ∪ E)

= 840 – {n (H) + n (E)} – n (H ∩ E)}

= 840 – {450 + 300 – 200}

= 840 – 550 = 290

Question 25.

In a group of 70 people, 52 like soft drinks and 37 like tea and each person likes atleast one of the two drinks. Then the number of people who like both the drink

(a) 15

(b) 19

(c) 18

(d) 20

Answer:

(b) 19

Let S : set of peoples who like soft drinks

T : set of people, who like tea

Then n (S) = 52

and n (T) = 37

and n (S ∩ T) = 70

we know that,

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

⇒ 70 = 52 + 37 – n (S ∩ T)

⇒ n (S ∩ T) = 89 – 70 = 19.

Question 26.

A T.V. survey gives the following data for TV. watching 59 % of the people watch program A, 67 % of the people watch program B and x % of the people watch both the program, then

(a) x = 26

(b) x = 59

(c) 26 ≤ x ≤ 59

(d) x ≥ 59

Answer:

Let P: set of people watch program A

Q : set of people watch program B

Then n (P) = \(\frac{59}{100}\) ;

n (Q) = \(\frac{47}{100}\)

and n (P ∩ Q) = \(\frac{x}{100}\)

∴ n (P ∪ Q) = n (P) + n (Q) – n (P ∩ Q)

= \(\frac{59}{100}+\frac{67}{100}-\frac{x}{100}\)

A ∪ B ⊆ ξ

⇒ n (A ∪ B) ≤ n (ξ)

⇒ \(\frac{59}{100}+\frac{67}{100}-\frac{x}{100}\) ≤ 100 % = \(\frac{100}{100}\) = 1

⇒ 126 – x ≤ 100

⇒ 26 ≤ x ………………….(1)

Since A ∩ B ⊆ A

⇒ n (A ∩ B) ≤ n (A) = \(\frac{59}{300}\)

and A ∩ B ⊆ B

⇒ n (A ∩ B) ≤ n (B) = \(\frac{67}{100}\)

⇒ \(\frac{x}{100}\) ≤ min. \(\left\{\frac{59}{100}, \frac{67}{100}\right\}\)

⇒ \(\frac{x}{100} \leq \frac{59}{100}\)

⇒ x ≤ 59 ……………….(2)

From (1) and (2) ; we have

26 ≤ x ≤ 59

Question 27.

If A {(x, y) : y = \(\frac{1}{x}\) ≠ x ∈ R},

B = {(x, y) : y = – x, x ∈ R}, then

(a) A ∩ B = A

(b) A ∩ B = B

(c) A ∩ B = Φ

(d) A ∪ B = A

Answer:

(c) A ∩ B = Φ

Given A = {(x, y) ; y = \(\frac{1}{x}\) ; x ≠ 0 ∈ R}

and B = {(x, y) ; y – x ; x ∈ R}

∴ A ∩ B = {(x, y) ; y = \(\frac{1}{x}\) ; y = – x ; x ≠ 0 ∈ R}

Since y = \(\frac{1}{x}\) = – x

⇒ 1 + x^{2} = 0 which does not gives real values of x.

⇒ A ∩ B = Φ

Question 28.

If n (E) = 50, n (A) = 38, n (B) = 30, then the least value of n (A ∩ B) is

(a) 30

(b) 38

(c) 50

(d) 18

Answer:

(a) 30

Given n (ξ) = 50,

n (A) = 38,

n (B) = 30

∴ n (B) = 6

Thus n (A) – n (B) = 6 – 6 = 0