Multiple Choice Questions on Sets Class 11 OP Malhotra Exe-1M ISC Maths Solutions Ch-1 Latest editions. In this article you would learn to solve MCQs on Sets. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE  for detail information about ISC Board Class-11

Multiple Choice Questions on Sets Class 11 OP Malhotra Exe-1M ISC Maths Solutions Ch-1

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-1	Sets
Writer	OP Malhotra
Exe-1(M)	Multiple Choice Questions.

Multiple Choice Questions on Sets

Class 11 OP Malhotra Multiple Choice Questions Solutions

Que-1: If A = {1, 2, 3, 4, 5}, and B = {2, 4, 6} then A − B is equal to

(a) {1, 3, 5, 6}
(b) {0, 1, 3, 5, 6}
(c) {1, 3, 5}
(d) {2, 4}

Sol: (c) {1, 3, 5}
A − B means elements of A that are not in B.
A = {1, 2, 3, 4, 5}
B = {2, 4, 6}
Remove common elements (2, 4) from A:
A − B = {1, 3, 5}.

Que- 2: If A = {x | x ∈ N, x is a prime number less than 12} and B = {x | x ∈ N, x is a factor of 10}, then A ∩ B is equal to

(a) {2}
(b) {2, 5}
(c) {2, 5, 10}
(d) {1, 2, 5, 10}

Sol: (b) {2, 5}
A = Prime numbers less than 12 = {2, 3, 5, 7, 11}
B = Factors of 10 = {1, 2, 5, 10}
A ∩ B = Common elements = {2, 5}

Que-3:  Two sets A and B have 27 and 33 elements respectively. If (A ∪ B) contain 48 elements, then how many elements are there in (A ∩ B)?

(a) 6
(b) 9
(c) 12
(d) 15

Sol: (c) 12
We know the formula:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Substitute the given values:
48 = 27 + 33 − n(A ∩ B)
48 = 60 − n(A ∩ B)
n(A ∩ B) = 60 − 48
n(A ∩ B) = 12

Que-4: A set T is given as T = {x ∈ R / (x − 3)² = (x + 4)², x < 2}. Which of these would be true about the set T?

(a) T is the empty set
(b) T is a singleton set
(c) x = 3 is a member of T
(d) x = −4 is a member of T

Sol: (a) T is the empty set
(x − 3)² = (x + 4)²
Taking square root on both sides:
x − 3 = ±(x + 4)
Case 1: x − 3 = x + 4 → -3 = 4 (Not possible)
Case 2: x − 3 = −(x + 4)
x − 3 = −x − 4
2x = -1
x = -1/2
Check condition: x < 2 (True)
T = { -1/2 }
So, T is a singleton set.

Que-5: Write the set builder form of A = {−1, 1}

(a) A = {x : x is an integer}
(b) A = {x : x is a root of the equation x² + 1 = 0}
(c) A = {x : x is a real number}
(d) A = {x : x is a root of the equation x² = 1}

Sol: (d) A = {x : x is a root of the equation x² = 1}
Given: A = {−1, 1}
(a) A = { x : x is an integer } (Incorrect)
(b) A = { x : x is a root of x² + 1 = 0 } (Incorrect, roots are ±i)
(c) A = { x : x is a real number } (Incorrect)
(d) A = { x : x is a root of x² = 1 } (Correct)

Que-6: If a set A has 4 elements, then the total number of proper subsets of set A is

(a) 16
(b) 14
(c) 15
(d) 17

Sol: (c) 15
Number of all subsets of a set with n elements = 2ⁿ
Here n = 4
Total subsets = 2⁴ = 16
Proper subsets = Total subsets − 1 (excluding the set itself)
Proper subsets = 16 − 1 = 15

Que-7: If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, then A ∪ (B ∩ C) is equal to

(a) {1, 2}
(b) Φ
(c) {4, 5}
(d) {1, 2, 3, 4}

Sol: (d) {1, 2, 3, 4}
B ∩ C = {4}
A ∪ (B ∩ C) = {1, 2, 3} ∪ {4}
= {1, 2, 3, 4}

Que-8: The shaded region in Fig. 1.66 represents

(a) A ∩ B
(b) A ∪ B
(c) B − A
(d) A − B
(e) (A − B) ∪ (B − A)

Sol: (e) (A − B) ∪ (B − A)
The shaded part shows the region which belongs to A only and B only, but not their common part.
So it represents elements in A but not in B, and in B but not in A.
Hence, the required region is:
(A − B) ∪ (B − A)

Que-9: If A = {x : x is a square of a natural number and x < 100} and B is a set of even natural numbers, then what is the cardinality of A ∩ B?

(a) 4
(b) 5
(c) 9
(d) None of these 

Sol: (a) 4
A = perfect squares less than 100 = {1, 4, 9, 16, 25, 36, 49, 64, 81}
B = even natural numbers = {2, 4, 6, 8, 10, …}
A ∩ B = even perfect squares = {4, 16, 36, 64}
Number of elements in A ∩ B = 4

Que-10: Let S = set of points inside the square, T = set of points inside the triangle and C = 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:  (c) S ∪ T ∪ C = S
Since both the triangle (T) and circle (C) lie completely inside the square (S), we have:
T ⊂ S and C ⊂ S
Also, triangle and circle intersect ⇒ T ∩ C ≠ φ
Now,
S ∪ T ∪ C = S
because union of subsets with the main set gives the main set itself.

Que-11: What does the shaded region in Fig. 1.67 represent?

(a) (A ∩ B) ∩ C
(b) (A ∪ B) ∩ C
(c) (A ∪ B) − C
(d) None of these

Sol: (b) (A ∪ B) ∩ C
From the figure, the shaded portion lies in the region common to set C and also includes the combined area of sets A and B.
So the shaded region = (A ∪ B) ∩ C

Que-12: A, B and C are three sets and U is the universal set such that n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B) = 100, then what is the value of n(A ∪ B)?

(a) 100
(b) 200
(c) 300
(d) 400

Sol: (c) 300
We use the formula:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
= 200 + 300 − 100
= 400

Que-13: The set (A ∩ B)’ ∪ (B ∩ C)’ is equal to

(a) A’ ∪ B ∪ C
(b) A’ ∪ B
(c) A’ ∪ C’
(d) A’ ∩ B

Sol: (b) A’ ∪ B
Using De Morgan’s Law:
(A ∩ B)’ = A’ ∪ B’
(B ∩ C)’ = B’ ∪ C’
So,
(A ∩ B)’ ∪ (B ∩ C)’
= (A’ ∪ B’) ∪ (B’ ∪ C’)
= A’ ∪ B’ ∪ C’
= A’ ∪ (B’ ∪ C’)
= A’ ∪ (B ∩ C)’

Que-14: If X = {8n − 7 | n ∈ N} and Y = {49(n − 1) | n ∈ N}, then

(a) X ⊂ Y
(b) Y ⊂ X
(c) X = Y
(d) X ∩ Y = φ

Sol: (a) X ⊂ Y
First write few elements:
X = {8n − 7} = {1, 9, 17, 25, 33, 41, 49, 57, …}
Y = {49(n − 1)} = {0, 49, 98, 147, …}
Common element in both sets = 49
So, X ∩ Y ≠ φ and X ≠ Y
Also, not every element of X is in Y and not every element of Y is in X

Que-15: Let U = {x ∈ N : 1 ≤ x ≤ 10} be the universal set, N being the set of natural numbers. If A = {1, 2, 3, 4} and B = {2, 3, 6, 10}, then what is the complement of (A − B)?

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

Sol: (c) {1, 2, 3, 4}
A − B = {1, 4}
Complement of (A − B) = U − {1, 4}
U = {1,2,3,4,5,6,7,8,9,10}
⇒ U − {1,4} = {2,3,5,6,7,8,9,10}

Que-16: Two sets have m and n elements, the number of subsets of the first set is 112 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: (b) 7, 4
Number of subsets of a set with k elements = 2^k
Given: 2^m − 2ⁿ = 112
112 = 128 − 16 = 2⁷ − 2⁴
So, m = 7 and n = 4

Que-17: If A = {x, y} then the power set of A is

(a) {x, y}
(b) {φ, x, y}
(c) {φ, {x}, {y}}
(d) {φ, {x}, {y}, {x, y}}

Sol: (d) {φ, {x}, {y}, {x, y}}
Set A = {x, y} has 2 elements.
Power set of a set with n elements has 2ⁿ subsets.
So total subsets = 2² = 4
Subsets are:
φ (empty set)
{x}
{y}
{x, y}

Que-18: In a class of 60 students, if 25 students play cricket, 20 students play tennis and 10 students play both the games, then the number of students who play neither is

(a) 45
(b) 0
(c) 25
(d) 35

Sol: (c) 25
Using formula:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
= 25 + 20 − 10 = 35
Students who play at least one game = 35
Students who play neither = Total − (at least one)
= 60 − 35 = 25

Que-19: If A and B are two sets, then (A ∪ B)’ ∪ (A’ ∩ B) is

(a) A’
(b) A
(c) B’
(d) None of these

Sol: (a) A’
(A ∪ B)’ = A’ ∩ B’    (De Morgan’s Law)
So, (A ∪ B)’ ∪ (A’ ∩ B)
= (A’ ∩ B’) ∪ (A’ ∩ B)
Take A’ common:
= A’ ∩ (B’ ∪ B)
But (B’ ∪ B) = Universal Set (U)
Therefore,
= A’ ∩ U = A’

Que-20: A and B are subsets of universal set U such that n(U) = 800, n(A) = 300, n(B) = 400, and n(A ∩ B) = 100. The number of elements in the set A ∪ B is   

(a) 100
(b) 200
(c) 300
(d) 400

Sol:  (b) 200
We use the formula:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Substitute values:
n(A ∪ B) = 300 + 400 − 100 = 600

–: End Sets Class 11 OP Malhotra Exe-1M ISC Maths Ch-1 Latest editions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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