11 / 15

CBSE Class 11 Mathematics

NCERT Exemplar Solutions

Chapter 1

Sets

OBJECTIVE TYPE QUESTIONS

Choose the correct answer out of the given four options in each of the Exercises from 29

to 43 (M.C.Q.)

29. Suppose A1, A2, …, A30 are thirty sets each having 5 elements and B1, B2, …, Bn are n

sets each with 3 elements. Let and each element of S belongs to

exactly 10 of the Ai’s and exactly 9 of the Bj’s then n is equal to

(A). 15 (B). 30 (C). 45 (D) none

Sol. Number of elements in A1 A2 A3 … A30 = 30×5 = 150 (when repetition is not

allowed)

But each element is repeated 10 times

∴ ................................(i)

Number of elements in B1 B2 B3 … Bn = 3n (when repetition is not allowed)

But each element is repeated 09 times

∴ .........................................(ii)

From(i) and (ii) we get

Hence, the correct option is (c).

30. Two finite sets have m and n elements. The number of subsets of the first set is 112

http

s://w

ww.scho

olcon

nects

.in/

22 / 15

more than that of the second set. The values of m and n are, respectively,

(A) 4, 7 (B) 7, 4 (C) 4, 4 (D) 7, 7

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

So, according to the question, we have

2m – 2n = 112

⇒ 2n.(2m – n – 1) = 24.7

∴ 2n = 24 and 2m – n – 1 = 7

⇒ n = 4 and 2m – n = 1 + 7 = 8 = 23

⇒ n = 4 and m – n = 3 ⇒ m – 4 = 3 ⇒ m = 7

Hence, the correct option (B).

31. The set (A B’)’ (B C) is equal to

(A) A′ B C

(B) A′ B

(C) A′ C′

(D) A′ B

Sol. 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)

http

s://w

ww.scho

olcon

nects

.in/

33 / 15

32. Let F1 be the set of parallelograms, F2 the set of rectangles, F3 the set of rhombuses,

F4 the set of squares and F5 the set of trapeziums in a plane. Then F1 may be equal to

(A) F2 F3

(B) F3 F4

(C) F2 F5

(D) F2 F3 F4 F1

Sol. We know that rectangles, rhombus and square in a plane is a parallelogram but

trapezium is not a parallelogram.

∴ F1 = F2 F3 F4 F1

Hence, the correct option is (D).

33. Let S = set of points inside the square, T = the 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

Sol. 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

http

s://w

ww.scho

olcon

nects

.in/

44 / 15

Hence, the correct option is (C).

34. Let R be 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}

Sol. 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).

35. In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10

http

s://w

ww.scho

olcon

nects

.in/

55 / 15

students play both the games. Then, the number of students who play neither is

(A) 0

(B) 25

(C) 35

(D) 45

Sol. 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).

36. In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read

both. Then the number of persons who read neither is

(A) 210

(B) 290

(C) 180

(D) 260

http

s://w

ww.scho

olcon

nects

.in/

66 / 15

Sol. 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).

37. If X = {18n - 7n-1| n∈ N} and Y= {49n-49 | n ∈ N}, then

(a) X ⊂ Y

(b) Y ⊂ X

(c) X = Y

(d) X Y = Φ

Sol. Given that: X = {8n – 7n – 1 |n ∈ N} = {0, 49, 490, …}

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).

38. A survey shows that 63% of the people watch a News Channel whereas 76% watch

another channel. If x% of the people watch both channel, then

(A) x = 35 (B) x = 63 (C) 39 ≤ x ≤ 63 (D) x = 39

http

s://w

ww.scho

olcon

nects

.in/

77 / 15

Sol. 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).

39. If sets A and B are defined as

then

(a) A B = A

(b) A B = B

(c) A B = Φ

(d) A B = A

Sol. Given that:

and

It is very clear that and y = - x

http

s://w

ww.scho

olcon

nects

.in/

88 / 15

∵

∴ A B = Φ

Hence, the correct option is (c).

40. Let A and B are two sets then A (A B) equals to

(a) A

(b) B

(c) Φ

(d) A B

Sol. 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)

41. If A = {1, 3, 5, 7, 9, 11, 13, 15, 17} B = {2, 4, ..., 18} and N the set of natural numbers is

the universal set, then A′ (A B) B′ is

(A) Φ

(B) N

(C) A

http

s://w

ww.scho

olcon

nects

.in/

99 / 15

(D) B

Sol. 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).

42. Let S = {x | x is a positive multiple of 3 less than 100} P = {x | x is a prime number

less than 20}. Then n(S) + n(T) is

(a) 34

(b) 41

(c) 33

(d) 30

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

∴ S = {3, 6, 9, 12, 15, 18, …, 99}

http

s://w

ww.scho

olcon

nects

.in/

1010 / 15

⇒ 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).

43. If X and Y are two sets and X′ denotes the complement of X, then X

equal to

(X

Y)′ is

(A) X

(B) Y

(C) Φ

(D) X Y

Sol. 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).

Fill in the Blanks in Each of the Exercises 44 to 52.

44. The set {x ∈ R : 1 ≤ x < 2} can be written as .

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

Hence, the filler is [1, 2).

http

s://w

ww.scho

olcon

nects

.in/

1111 / 15

45. When A = Φ then number of elements is P(A) is .

Sol. Here A = Φ ∴ n(A) = 0

∵ n[P(A)] = 2n(A) = 20 = 1

Hence, the filler is 1.

46. If A and B are finite sets such that A ⊂ B, then n(A B) = .

Sol. Since A ⊂ B ∴ n(A B) = n(B)

Hence, the filler is n(B).

47. If A and B are any two sets, then A – B is equal to .

Sol. From the Venn diagram it is clear that

A – B = A B’.

Hence the filler is A B’.

48. Power set of the set A = {1, 2} is .

Sol. Power set of A = p(A) =

Hence, the filler is

49. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Then the universal set of

http

s://w

ww.scho

olcon

nects

.in/

1212 / 15

all the three sets A, B and C can be .

Sol. 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}

50. If 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}. Then

(i) (B C)′ is .

(ii)(C – A)′ is .

Sol. 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}

(i) (B C)’ = {2, 3, 4, 6, 7, 8}’ = {1, 5, 9, 10}

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

(ii) (C-A)’ = {4, 8}’= {1, 2, 3, 5, 6, 7, 9, 10}

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

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

Sol. Since A – B = A B’

⇒ A – (A B) = A B’

http

s://w

ww.scho

olcon

nects

.in/

1313 / 15

Hence, the filler is A B’.

52. Match the following sets for all sets A, B and C

(i) ((A’ B’) – A)’ (a) A – B

(ii) [B’ (B’ – A)]’ (b) A

(iii) (A – B) – (B – C) (c) B

(iv) (A – B) (C – B) (d) (A×B) (A×C)

(v) A × (B C) (e) (A × B) (A×C)

(vi) A × (B C) (f) (A C) – B

Sol. (i) ((A’ B’) – A)’ = [(A’ B’) A)’]’ [∵ A – B = 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)

http

s://w

ww.scho

olcon

nects

.in/

1414 / 15

= [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 of False for the statements in Each of the Exercises 53 to 58.

53. If A is any set, then A ⊂ A.

Sol. Since every set is a subset of itself. So, it is ‘True’.

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.

Sol. 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’.

http

s://w

ww.scho

olcon

nects

.in/

1515 / 15

55. The sets {1, 2, 3, 4} and {3, 4, 5, 6} are equal.

Sol. False, since the two sets do not contain the same elements.

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.

57. Let sets R = T be defined as

R = {x ∈ Z| x is divisible by 2}

T = (x ∈ Z| x is divisible by 6}. Then T ⊂ R.

Sol. 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’.

58. Given A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2} then A = B.

Sol. 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’.

http

s://w

ww.scho

olcon

nects

.in/