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JEE • MATHEMATICS • SET THEORY

Set theory practice · Mixed difficulty

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Practice format and chapter quiz

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 31-60 of 273 questions

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31 Easy
31 of 273
Understanding +4 | -1
If RR and R′R^{\prime} are symmetric relations (not disjoint) on a set AA, then the relation R∩R′R \cap R^{\prime} is
View answer and explanation
Correct option: B
Symmetric
Option B (Symmetric) is correct because the intersection of two symmetric relations is also symmetric.
32 Easy
32 of 273
Analyzing +4 | -1
If A,BA,B and CC are three non-empty sets such that AA and BB are disjoint and the number of elements contained in AA is equal to those contained in the set of elements common to the sets AA and CC, then n(A∪B∪C)n\left( {A \cup B \cup C} \right) is necessarily equal to
View answer and explanation
Correct option: A
n(B∪C)n\left( {B \cup C} \right)
Option A is correct because, given the conditions, the number of elements in the union of A, B, and C is equal to the number of elements in the union of B and C.
33 Easy
33 of 273
Application +4 | -1
If n(u)=100,n(A)=50,  n(B)=20n\left( u \right) = 100,n\left( A \right) = 50,\;n\left( B \right) = 20 and n(A∩B)=10,n\left( {A \cap B} \right) = 10, then n{(A∪B)c}n\left\{ {{{\left( {A \cup B} \right)}^c}} \right\}
View answer and explanation
Correct option: C
40
The correct answer is C (40) because it is calculated by subtracting the number of elements in the union of A and B from the total number of elements in the universal set.
34 Easy
34 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: C
8
Option C (8) is correct because a set with n elements has 2n2^n subsets, and in this case, n=3, so 23=82^3 = 8.
35 Easy
35 of 273
Analyzing +4 | -1
Let AA and BB be two sets that A∩X=B∩X=ϕA \cap X=B \cap X=\phi and A∪X=B∪XA \cup X=B \cup X for some set XX. Then,
View answer and explanation
Correct option: A
A=BA=B
Option A (A=BA=B) is correct because with the given conditions A∩X=B∩X=ϕA \cap X = B \cap X = \phi and A∪X=B∪XA \cup X = B \cup X, it can be shown that set A is equal to set B.
36 Easy
36 of 273
Understanding +4 | -1
If AA and BB are two sets, then A∩(A∪B)A \cap \left( {A \cup B} \right) equals
View answer and explanation
Correct option: A
AA
Option A (AA) is correct because A∩(A∪B)A \cap (A \cup B) simplifies to A, as A is a subset of A∪BA \cup B.
37 Easy
37 of 273
Analyzing +4 | -1
In a certain town 25% families own a phone and 15% own a car, 65% families own neither a phone nor a car. 2000 families own both a car and a phone.

Consider the following statements in this regard:

1. 10 % families own both a car and a phone

2. 35 % families own either a car or a phone

3. 40,000 families live in the town

Which of the following statements are correct?
View answer and explanation
Correct option: C
2 and 3
Option C (2 and 3) is correct because after calculating the percentages and total families, statements 2 and 3 are true, while statement 1 is false.
38 Easy
38 of 273
Understanding +4 | -1
Let AA be a set represented by the squares of natural number and x,yx,y are any two elements of AA. Then,
View answer and explanation
Correct option: B
xy∈Axy \in A
Option B (xy∈Axy \in A) is correct because if x and y are squares of natural numbers, their product is also a square of a natural number.
39 Easy
39 of 273
Understanding +4 | -1
If A={(x,y):x2+y2≤1,x,y∈R}A=\left\{(x, y): x^2+y^2 \leq 1, x, y \in R\right\} and B={(x,y):x2+y2≤4;x,y∈R}B=\left\{(x, y): x^2+y^2 \leq 4 ; x, y \in R\right\}, then
View answer and explanation
Correct option: C
A−B=ϕA-B=\phi
Option C (A−B=ϕA-B=\phi) is correct because A is a subset of B, meaning all elements of A are also in B, so removing B from A results in an empty set.
40 Easy
40 of 273
Analyzing +4 | -1
Let P={θ:sin⁡θ−cos⁡θ=2cos⁡θ}P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\} and, Q={θ:sin⁡θ+cos⁡θ=2sin⁡θ}Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\} be two sets. Then,
View answer and explanation
Correct option: D
P=QP=Q
Option D (P=QP=Q) is the correct answer because the two sets are equivalent, meaning they contain the same elements.
41 Easy
41 of 273
Application +4 | -1
If A={1,  2,  3,  4,  5},  B={2,  4,  6},  C={3,  4,  6}A = \left\{ {1,\;2,\;3,\;4,\;5} \right\},\;B = \left\{ {2,\;4,\;6} \right\},\;C = \left\{ {3,\;4,\;6} \right\}, then (A∪B)∩C\left( {A \cup B} \right) \cap C is
View answer and explanation
Correct option: A
{3, 4, 6}\left\{ {3,{\text{ }}4,{\text{ }}6} \right\}
The correct answer is A ({3, 4, 6}) because it is the intersection of the union of A and B with C.
42 Easy
42 of 273
Application +4 | -1
Let S = {1, 2, 3, … , 100}. The number of non – empty subsets A of S such that the product of elements in A 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}}\left( {{2^{50}} - 1} \right)) is correct because it calculates the total subsets minus subsets with only odd numbers.
43 Easy
43 of 273
Analyzing +4 | -1
Let S={x∈[−6,3]−{−2,2}:∣x+3∣−1∣x∣−2≥0}S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} and T={x∈Z:x2−7∣x∣+9≤0}T=\left\{x \in Z: x^2-7|x|+9 \leq 0\right\}. Then the number of elements in S∩TS \cap T is
View answer and explanation
Correct option: D
3
Option D (3) is correct because the intersection of sets S and T contains 3 elements: -5, -4, and 3.
44 Medium
44 of 273
Application +4 | -1
If n(A)=115,n(B)=326,n(A−B)=47n(A)=115, n(B)=326, n(A-B)=47, then what is n(A∪B)n(A \cup B) equal to?
View answer and explanation
Correct option: A
373
Option A (373) is correct because it is calculated using the formula for the number of elements in the union of two sets, given the number of elements in each set and their difference.
45 Medium
45 of 273
Analyzing +4 | -1
Which of the following is true?
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because none of the given set relations are true based on the definitions of set membership and subsets.
46 Easy
46 of 273
Understanding +4 | -1
Let AA be a set containing 10 distinct elements, the total number of distinct functions from AA to AA is
View answer and explanation
Correct option: B
101010^{10}
Option B (101010^{{10}}) is correct because each of the 10 elements in A can be mapped to any of the 10 elements in A, resulting in 101010^{{10}} possible functions.
47 Easy
47 of 273
Application +4 | -1
Given n(U)=20,  n(A)=12,  n(B)=9,n(A∩B)=4n\left( U \right) = 20,\;n\left( A \right) = 12,\;n\left( B \right) = 9,n\left( {A \cap B} \right) = 4, where UU is the universal set, AA and BB are subsets of UU, then n[(A∪B)c]n\left[ {{{\left( {A \cup B} \right)}^c}} \right] equals to
View answer and explanation
Correct option: D
3
The correct answer is D (3) because it is calculated by subtracting the number of elements in the union of A and B from the total number of elements in the universal set.
48 Medium
48 of 273
Application +4 | -1
In a B School there are 15 teachers who teach marketing or finance. Of these, 8 teach finance and 4 teach both marketing and finance. How many teach marketing but not finance?
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Correct option: C
11
Option C (11) is correct because it is calculated by subtracting the number of teachers teaching both marketing and finance from the total number of teachers teaching marketing or finance.
49 Easy
49 of 273
Application +4 | -1
Let n(U)=700,  n(A)=200,  n(B)=300n\left( \mathcal{U} \right) = 700,\;n\left( A \right) = 200,\;n\left( B \right) = 300 and n(A∩B)=100.n\left( {A \cap B} \right) = 100. Then, n(Ac∩Bc)=n\left( {{A^c} \cap {B^c}} \right) =
View answer and explanation
Correct option: C
300
Option C (300) is correct because it uses De Morgan's Law and the principle of inclusion-exclusion to find the number of elements in the complement of the union of A and B.
50 Medium
50 of 273
Understanding +4 | -1
Consider the following statements.
I. If AnA_{n} is the set of first nn prime numbers, then ⋃n=210An\bigcup\limits_{{n=2}}^{10} A_{n} is equal to {2,3,5,7,11,13,17,19,23,29}\{2,3,5,7,11,13,17,19,23,29\}
II. If A{A} and B{B} are two sets such that n(A∪B)=50{n}({A} \cup {B})=50, n(A)=28,n(B)=32{n}({A})=28, {n}({B})=32, then n(A∩B)=10{n}({A} \cap {B})=10.
Which of these is correct?
View answer and explanation
Correct option: C
Both are true
Option C (Both are true) is correct because the union of the first n prime numbers up to n=10 is the set of the first 10 prime numbers, and the cardinality formula for the union of two sets is correctly applied.
51 Medium
51 of 273
Application +4 | -1
In a group of 500 students, there are 475 students who can speak Hindi and 200 can speak Bengali. What is the number of students who can speak Hindi only?
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Correct option: B
300
Option B (300) is correct because it is calculated by subtracting the number of students who speak Hindi or Bengali from the total number of students.
52 Easy
52 of 273
Understanding +4 | -1
Which of the following collection is a set?
View answer and explanation
Correct option: A
The collection of all girls in your class.
Option A (The collection of all girls in your class) is correct because it is a well-defined collection, meaning there is no ambiguity about who belongs to the set.
53 Easy
53 of 273
Understanding +4 | -1
Consider the following statements.

I. ϕ∈{ϕ}\phi \in\{\phi\} \quad II. {ϕ}⊆ϕ\{\phi\} \subseteq \phi

Which of the statements given above is/are correct?
View answer and explanation
Correct option: D
Neither I nor II
Option D (Neither I nor II) is correct because the empty set is not an element of a set containing the empty set, and the set containing the empty set is not a subset of the empty set.
54 Easy
54 of 273
Application +4 | -1
Suppose A1,A2,A3,………,A30A_1, A_2, A_3, \ldots \ldots \ldots, A_{30} are thirty sets each having 5 elements and B1,B2,…….,BnB_1, B_2, \ldots \ldots ., B_n are nn sets each with 3 elements. Let ⋃i=130Ai=⋃j=1nBj=S\bigcup_{i=1}^{30} A_i=\bigcup_{j=1}^n B_j=S and each elements of SS belongs to exactly 10 of the Ai′sA_i^{\prime} s and exactly 9 of the Bj′sB_j^{\prime} s. Then nn is equal to:
View answer and explanation
Correct option: C
45
Option C (45) is correct because it calculates the number of sets BjB_j needed to satisfy the given conditions regarding the number of elements and their occurrences in the sets.
55 Easy
55 of 273
Application +4 | -1
X{X} and Y{Y} are two sets such that n(X)=17,n(Y)=23n({X})=17, n({Y})=23, n(X∪Y)=38n({X} \cup {Y})=38 then n(X∩Y)n({X} \cap {Y}) is
View answer and explanation
Correct option: B
2
Option B (2) is correct because it accurately applies the inclusion-exclusion principle to find the intersection of two sets.
56 Easy
56 of 273
Understanding +4 | -1
The set (A\B)∪(B\A)(A \backslash B) \cup(B \backslash A) is equal to
View answer and explanation
Correct option: B
(A∪B)\(A∩B)(A \cup B) \backslash(A \cap B)
Option B ((A∪B)\(A∩B)(A \cup B) \backslash (A \cap B)) is correct because it represents the symmetric difference, which is the union of A and B excluding their intersection.
57 Easy
57 of 273
Application +4 | -1
In a class of 60 students, 34 have passed in Mathematics and 38 in physics. If each of the students pass at least in one of the subjects, then the number of students who have passed in physics only is
View answer and explanation
Correct option: B
26
Option B (26) is correct because it is obtained by using the principle of inclusion-exclusion to find the number of students who passed in both subjects and then subtracting that from the number of students who passed in physics.
58 Easy
58 of 273
Analyzing +4 | -1
Universal set, U={x:x5−6x4+11x3−6x2=0}U = \left\{ {x:{x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0} \right\} And A={x:x2−5x+6=0}A = \left\{ {x:{x^2} - 5x + 6 = 0} \right\}
  B={x:x2−3x+2=0}\;B = \left\{ {x:{x^2} - 3x + 2 = 0} \right\}
Then, (A∩B)′{\left( {A \cap B} \right)^\prime } is equal to
View answer and explanation
Correct option: C
{0,1,3}\left\{ {0,1,3} \right\}
The correct answer is C ({0,1,3}) because after finding the sets U, A, and B, the complement of their intersection (A ∩ B)' is {0,1,3}.
59 Easy
59 of 273
Analyzing +4 | -1
If A,B,CA, B, C be three sets such that A∪B=A∪CA \cup B=A \cup C and A∩B=A∩CA \cap B=A \cap C, then
View answer and explanation
Correct option: B
B=CB=C
Option B (B=CB=C) is correct because given the conditions A∪B=A∪CA \cup B = A \cup C and A∩B=A∩CA \cap B = A \cap C, it can be proven that set B is equal to set C.
60 Easy
60 of 273
Understanding +4 | -1
Let AA and BB two non-empty subsets of a set XX such that AA is not a subset of BB then
View answer and explanation
Correct option: D
AA and the complement of B are non-disjoint
Option D (A and the complement of B are non-disjoint) is correct because if A is not a subset of B, then there exists an element in A that is not in B, meaning it's in the complement of B.
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