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JEE • MATHEMATICS • SET THEORY

Set theory practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 151-180 of 273 questions

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151 Easy
151 of 273
Remembering +4 | -1
The number of elements of the power set of a set containing nn elements is
View answer and explanation
Correct option: B
2n2^n
Option B (2n2^n) is correct because the power set of a set with n elements contains 2n2^n elements.
152 Easy
152 of 273
Understanding +4 | -1
If A={1,2,5}A=\{1,2,5\} and B={3,4,5,9}B=\{3,4,5,9\}, then AΔBA \Delta B is equal to
View answer and explanation
Correct option: B
{1,2,3,4,9}\{1,2,3,4,9\}
Option B ({1,2,3,4,9}) is correct because the symmetric difference A Δ B consists of elements that are in A or B, but not in both.
153 Medium
153 of 273
Understanding +4 | -1
AA and BB are any two non-empty sets and AA is proper subset of BB. If n(A)=5,n\left( A \right) = 5, then find the minimum possible value of n(AΔB)n\left( {A\Delta B} \right)
View answer and explanation
Correct option: A
Is 1
Option A (Is 1) is correct because the minimum number of elements in B that are not in A is 1, given that A is a proper subset of B and n(A) = 5.
154 Easy
154 of 273
Understanding +4 | -1
Which one of the following is (A−B)∪(B−A)(A-B) \cup(B-A) ?
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Correct option: C
(A∪B)−(A∩B)(A \cup B)-(A \cap B)
Option C ((A∪B)−(A∩B)(A \cup B)-(A \cap B)) is correct because (A−B)∪(B−A)(A-B) \cup (B-A) represents the elements that are in A or B but not in both, which is the same as the union minus the intersection.
155 Easy
155 of 273
Application +4 | -1
In a rehabilitation programme, a group of 50 families were assured new houses and compensation by the government. Number of families who got both is equal to the number of families who got neither of the two. The number of families who got new houses is 6 greater than the number of families who got compensation. How many families got houses?
View answer and explanation
Correct option: B
28
Option B (28) is correct because it is the number of families who got new houses, derived from the given relationships between the families who received houses, compensation, or neither.
156 Easy
156 of 273
Understanding +4 | -1
If A={2,3,4,8,10},B={3,4,5,10,12}A=\{2,3,4,8,10\}, B=\{3,4,5,10,12\} and C={4,5,6,12,14}C=\{4, 5,6,12,14\}, then (A∩B)∪(A∩C)(A \cap B) \cup(A \cap C) is equal to
View answer and explanation
Correct option: A
{3,4,10}\{3,4,10\}
Option A ({3,4,10}\{3,4,10\}) is correct because it represents the union of the intersection of A and B, and A and C.
157 Easy
157 of 273
Application +4 | -1
The finite sets AA and BB have m and n elements respectively. if the total number of subsets of AA is 112 more than the total number of subsets of BB, then the volume of mm is
View answer and explanation
Correct option: A
7
The correct answer is A (7) because solving the equation 2^m - 2^n = 112 yields m = 7 and n = 4.
158 Easy
158 of 273
Understanding +4 | -1
If A={1,2,3}A=\{1,2,3\} and B={3,8}B=\{3,8\}, then (A∪B)×(A∩B)(A \cup B) \times(A \cap B) is
View answer and explanation
Correct option: B
{(1,3),(2,3),(3,3),(8,3)}\{(1,3),(2,3),(3,3),(8,3)\}
Option B ({ (1,3),(2,3),(3,3),(8,3)}) is correct because it represents the Cartesian product of (A ∪ B) and (A ∩ B).
159 Easy
159 of 273
Remembering +4 | -1
If a set A={a,b,c}A=\{a, b, c\} then the number of subsets of the set AA is
View answer and explanation
Correct option: A
8
Option A (8) is correct because a set with 3 elements has 23=82^3 = 8 subsets.
160 Easy
160 of 273
Application +4 | -1
Let AA and BB be two finite sets with mm and nn elements respectively. The total number of subsets of the set AA is 56 more than the total number of subsets of B.B. Then the distance of the point P(m,n)P(m, n) from the point Q(−2,−3)Q(-2,-3) is:
View answer and explanation
Correct option: C
10
Option C (10) is correct because solving the equation relating the number of subsets of A and B gives m=6 and n=3, and the distance between P(6,3) and Q(-2,-3) is 10.
161 Easy
161 of 273
Understanding +4 | -1
If AA and BB are any two sets, then A∪(A∩B)A \cup(A \cap B) is equal to:
View answer and explanation
Correct option: A
AA
Option A (AA) is correct because A∩BA \cap B is a subset of AA, so the union of AA and A∩BA \cap B is simply AA.
162 Easy
162 of 273
Understanding +4 | -1
If U={1,2,3}{U}=\{1,2,3\} and A={1,2}{A}=\{1,2\} then [P(A)]′=[{P}({A})]^{\prime}=
View answer and explanation
Correct option: B
{{3},{2,3},{1,3},{1,2,3}}\{\{3\},\{2,3\},\{1,3\},\{1,2,3\}\}
Option B is correct because [P(A)]' is the complement of P(A) with respect to P(U), where U={1,2,3} and A={1,2}.
163 Medium
163 of 273
Application +4 | -1
If S{S} and T{T} are two sets such that S{S} has 21 elements, T{T} has 32 elements, and S∩T{S} \cap {T} has 11 elements, then number of elements S∪T{S} \cup {T} has
View answer and explanation
Correct option: A
42
Option A (42) is correct because it correctly uses the inclusion-exclusion principle to find the number of elements in the union of sets S and T.
164 Easy
164 of 273
Analyzing +4 | -1
Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If KK% of them are suffering from both ailments, then KK cannot belong to the set
View answer and explanation
Correct option: C
{79, 81, 83, 85}
The correct answer is C ({79, 81, 83, 85}) because the percentage of patients suffering from both ailments must lie within a specific range, and this set falls outside that range.
165 Easy
165 of 273
Understanding +4 | -1
Let AA and BB have 3 and 6 elements respectively. What can be the minimum number of elements in A∪B?A \cup B?
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Correct option: B
6
Option B (6) is correct because the minimum number of elements in A∪BA \cup B occurs when A⊂BA \subset B, in which case A∪B=BA \cup B = B, and n(B)=6n(B) = 6.
166 Easy
166 of 273
Analyzing +4 | -1
Let P={1,2,3,…,100}P=\{1,2,3, \ldots, 100\} be the number of non-empty subsets AA of SS such that the product of elements in AA is even is
View answer and explanation
Correct option: A
250(250−1)2^{50}\left(2^{50}-1\right)
Option A (250(250−1)2^{50}(2^{50}-1)) is correct because it calculates the total number of non-empty subsets minus the number of subsets containing only odd numbers.
167 Medium
167 of 273
Application +4 | -1
20 teachers of a school either teach mathematics or physics. 12 of them teach mathematics while 4 teach both the subjects. Then the number of teachers teaching physics only is
View answer and explanation
Correct option: B
8
Option B (8) is correct because it is calculated by subtracting the number of teachers teaching mathematics and both subjects from the total number of teachers.
168 Hard
168 of 273
Analyzing +4 | -1
Suppose A1,  A2,…,A30{A_1},\;{A_2}, \ldots ,{A_{30}} are thirty sets, each having 5 elements and B1,  B2,  …,  Bn{B_1},\;{B_2},\; \ldots ,\;{B_n} are nn sets each with 3 elements, let
∪i=130Ai=∪j=1nBj=S\mathop \cup \nolimits_{i = 1}^{30} {A_i} = \mathop \cup \nolimits_{j = 1}^n {B_j} = S and each element of S belongs to exactly 10 of the Ai ′s{A_i}{\,^\prime }s and exactly 9 of the Bj{B_j} 's. Then, nn is equal to
View answer and explanation
Correct option: C
45
Option C (45) is correct because by equating the total number of elements counted in sets A_i and B_j, we can solve for n, the number of sets B_j.
169 Medium
169 of 273
Application +4 | -1
There are 600 student in a school. If 400 of them can speak Telugu, 300 can speak Hindi, then the number of students who can speak both Telugu and Hindi is:
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Correct option: A
100
Option A (100) is correct because it is calculated using the formula for the number of elements in the union of two sets: n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
170 Easy
170 of 273
Understanding +4 | -1
A set S\mathrm{S} contains 3 elements, the number of subsets of which of following sets is 256 .
View answer and explanation
Correct option: C
P(P(S))\mathrm{P}(\mathrm{P}(\mathrm{S}))
Option C (P(P(S))\mathrm{P}(\mathrm{P}(\mathrm{S}))) is correct because the power set of the power set of S has 256 subsets.
171 Easy
171 of 273
Analyzing +4 | -1
Let RR be a relation on N\mathbf{N} defined by R={(m,n):m,n∈N and m=n2}.R=\left\{(m, n): m, n \in N \text { and } m=n^2\right\}.
Which of the following is true.
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Correct option: D
None of these
Option D (None of these) is correct because none of the given properties (reflexivity, symmetry, transitivity) hold true for the relation R on the set of natural numbers.
172 Easy
172 of 273
Analyzing +4 | -1
If A={θ:2cos⁡2θ+sin⁡θ≤2}A=\left\{\theta: 2 \cos ^2 \theta+\sin \theta \leq 2\right\} and B={θ:π2≤θ≤3π2}B=\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\right\}, then A∩BA \cap B is equal to
View answer and explanation
Correct option: C
{θ:π/2≤θ≤5π/6}∪{θ:π≤θ≤3π/2}\{\theta: \pi / 2 \leq \theta \leq 5 \pi / 6\} \cup\{\theta: \pi \leq \theta \leq 3 \pi / 2\}
Option C ({θ:π/2≤θ≤5π/6}∪{θ:π≤θ≤3π/2}\{\theta: \pi / 2 \leq \theta \leq 5 \pi / 6\} \cup \{\theta: \pi \leq \theta \leq 3 \pi / 2\}) is correct because it represents the intersection of the solution sets for the trigonometric inequality and the given interval.
173 Easy
173 of 273
Understanding +4 | -1
The value of n[P(P(P(ϕ)))]n\left[ {P\left( {P\left( {P\left( \phi \right)} \right)} \right)} \right] is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because it represents the cardinality of the power set of the power set of the power set of the empty set.
174 Easy
174 of 273
Analyzing +4 | -1
Let S={1,  2,  3,  4}S = \left\{ {1,\;2,\;3,\;4} \right\}. The total number of unordered pairs of disjoint subsets of SS is equal to
View answer and explanation
Correct option: D
41
Option D (41) is correct because the total number of unordered pairs of disjoint subsets of S is calculated using the formula (3^n + 1)/2 where n is the number of elements in S.
175 Easy
175 of 273
Application +4 | -1
Let X={1,2,3,4,5}\mathrm{X=}\{1,2,3,4,5\}. The number of different ordered pairs (Y,Z)(\mathrm{Y}, \mathrm{Z}) that can formed such that Y⊆X,Z⊆X\mathrm{Y} \subseteq \mathrm{X}, \mathrm{Z} \subseteq \mathrm{X} and Y∩Z\mathrm{Y} \cap \mathrm{Z} is empty is:
View answer and explanation
Correct option: B
353^5
Option B (353^5) is correct because each element in X has three possibilities: belonging to Y, belonging to Z, or belonging to neither, given that Y and Z are disjoint subsets of X.
176 Easy
176 of 273
Application +4 | -1
If A={1,  2,  3,  4},A = \left\{ {1,\;2,\;3,\;4} \right\}, then the number of subsets of AA that contain the element 2 but not 3, is
View answer and explanation
Correct option: B
4
Option C (4) is correct because there are 22 = 4 subsets that contain the element 2 but not 3, formed from the remaining elements 1 and 4.
177 Easy
177 of 273
Understanding +4 | -1
If A={x:x=3n,n∈Z}A=\{x: x=3 n, n \in Z\} and B={x:x=4n,n∈Z}B=\{x: x=4 n, n \in Z\} then A∩BA \cap B is
View answer and explanation
Correct option: D
None of these
Option D ({x:x=12n,n∈Z}\{x: x=12 n, n \in Z\}) is correct because the intersection of multiples of 3 and 4 are multiples of 12.
178 Easy
178 of 273
Application +4 | -1
If A={2,4}A=\{2,4\} and B={3,4,5}B=\{3,4,5\}, then (A∩B)×(A∪B)(A \cap B) \times(A \cup B) is
View answer and explanation
Correct option: D
{(4,2),(4,3),(4,4),(4,5)}\{(4,2),(4,3),(4,4),(4,5)\}
Option D ({ (4,2),(4,3),(4,4),(4,5) }) is correct because it accurately represents the Cartesian product of the intersection of A and B with the union of A and B.
179 Medium
179 of 273
Understanding +4 | -1
The cardinality of the set P{P[P(ϕ)]}\mathrm{P}\{\mathrm{P}[\mathrm{P}(\phi)]\} is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the power set of the power set of the power set of the empty set has 4 elements.
180 Easy
180 of 273
Application +4 | -1
If AA and BB are two sets such that n(A)=7,n(B)=6n\left( A \right) = 7,n\left( B \right) = 6 and (A∩B)≠ϕ.\left( {A \cap B} \right) \ne \phi . Then the greatest possible value of n(A  Δ  B),n\left( {A\;\mathrm{\Delta \;}B} \right), is
View answer and explanation
Correct option: A
11
Option A (11) is correct because the greatest possible value of n(AΔB)n(A \Delta B) occurs when n(A∩B)n(A \cap B) is minimized, which is 1 in this case.
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