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JEE • MATHEMATICS • SET THEORY

Set theory practice · Easy

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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 61-90 of 194 questions
61 Easy
61 of 194
Understanding +4 | -1
Let ZZ denote the set of integers, then
{x∈Z:∣x−3∣<4}n{x∈Z:∣x−4∣<5}=\left\{ {x \in Z:\left| {x - 3} \right| < 4} \right\}n\left\{ {x \in Z:\left| {x - 4} \right| < 5} \right\} =
View answer and explanation
Correct option: C
{0,1,2,3,4,5,6}\left\{ {0,1,2,3,4,5,6} \right\}
Option C ({0,1,2,3,4,5,6}\{ {0,1,2,3,4,5,6} \}) is correct because it represents the intersection of two sets of integers defined by absolute value inequalities.
62 Easy
62 of 194
Application +4 | -1
Suppose A1,A2,A3,…….A30A_1, A_2, A_3, \ldots \ldots . A_{30} are thirty sets each with 5 elements and B1,B2,B3,…….BnB_1, B_2, B_3, \ldots \ldots . B_n are n sets. Each with three elements such that Ui=130Ai=Uj=1nBj=S\mathop U\limits_{i = 1}^{30} A_i=\mathop U\limits_{j = 1}^{n} B_j=S. If each element of S{S} belongs to exactly 10 of the AiA_i 's and exactly 9 of the BjB_j 's, then the value of n{n} is
View answer and explanation
Correct option: C
45
Option C (45) is correct because it is derived from equating the total number of elements in S calculated using the information from sets A_i and B_j.
63 Easy
63 of 194
Understanding +4 | -1
Let A={1,2,3},B={3,4}A=\{1,2,3\}, B=\{3,4\} and C={4,5,6}C=\{4,5,6\}, then A∪(B∩C)A \cup(B \cap C) is
View answer and explanation
Correct option: B
{1,2,3,4}\{1,2,3,4\}
Option B ({1,2,3,4}) is correct because it represents the union of set A with the intersection of sets B and C.
64 Easy
64 of 194
Application +4 | -1
The number of elements in the set {(a,b):2a2+3b2=35,  a,  b∈Z}\left\{ {\left( {a,b} \right):2{a^2} + 3{b^2} = 35,\;a,\;b \in Z} \right\}, where ZZ is the set of all integers, is
View answer and explanation
Correct option: C
8
The correct answer is C (8) because there are 8 integer pairs (a, b) that satisfy the equation 2a2 + 3b2 = 35.
65 Easy
65 of 194
Analyzing +4 | -1
For any two sets AA and BB, if 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: B
A=BA = B
Option B is correct because given the conditions, the sets A and B must be equal.
66 Easy
66 of 194
Analyzing +4 | -1
Three sets A,B,CA, B, C are such that A=B∩CA=B \cap C and B=C∩AB=C \cap A, then
View answer and explanation
Correct option: C
A⊃BA \supset B
Option B (A⊃BA \supset B) is correct because given A=B∩CA = B \cap C and B=C∩AB = C \cap A, it implies that A and B are equal, and therefore A contains B.
67 Easy
67 of 194
Understanding +4 | -1
I. The collection of all months of a year beginning with the letter J.

II. The collection of ten most talented writers of India.

III. A team of eleven best cricket batsmen of the world.

IV. The collection of all boys in your class.

Which of the above are the sets?
View answer and explanation
Correct option: C
I and IV
Option C (I and IV) is correct because a set must be a well-defined collection, and only the collections in I and IV are well-defined.
68 Easy
68 of 194
Application +4 | -1
In a committee 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. The number of persons speaking at least one of these two languages is
View answer and explanation
Correct option: A
60
Option A (60) is correct because it applies the principle of inclusion-exclusion to find the number of people speaking at least one language.
69 Easy
69 of 194
Understanding +4 | -1
Which of the following is the empty set?
View answer and explanation
Correct option: B
{x:x\left\{x: x\right. is a real number and x2+1=0}\left.x^2+1=0\right\}
Option B ({x:x\{x: x is a real number and x2+1=0}\left.x^2+1=0\right\}) is the empty set because the equation x2+1=0x^2 + 1 = 0 has no real solutions.
70 Easy
70 of 194
Understanding +4 | -1
Let A={x:x  A = \{ x:x\; is a multiple of 3} and B = {x:x{x:x} is a multiple of 5}. Then, A∩BA \cap B is given by
View answer and explanation
Correct option: C
{15, 30, 45, ......}\left\{ {15,{\text{ }}30,{\text{ }}45,{\text{ }}......} \right\}
Option C ({15, 30, 45, ......}\left\{ {15,{\text{ }}30,{\text{ }}45,{\text{ }}......} \right\}) is correct because A∩BA \cap B represents the set of elements that are multiples of both 3 and 5, which are multiples of 15.
71 Easy
71 of 194
Application +4 | -1
Let A={x:x∈R,−1<x<1},B={x:x∈R,x≤0\mathrm{A}=\{\mathrm{x}: \mathrm{x} \in \mathrm{R},-1<\mathrm{x}<1\}, \mathrm{B}=\{\mathrm{x}: \mathrm{x} \in \mathrm{R}, \mathrm{x} \leq 0 or x≥2}\mathrm{x} \geq 2\} and A∪B=R−D\mathrm{A} \cup \mathrm{B}=\mathrm{R}-\mathrm{D}, then the set D\mathrm{D} is
View answer and explanation
Correct option: B
{x:1≤x<2}\{\mathrm{x}: 1 \leq \mathrm{x}<2\}
Option B is correct because it represents the set D, which is the complement of the union of sets A and B with respect to the set of real numbers.
72 Easy
72 of 194
Understanding +4 | -1
The empty set is represented by

I. ϕ\phi \quad II. {ϕ}\{\phi\} \quad III. {}\{\} \quad IV. {{}}\{\{\}\}
View answer and explanation
Correct option: B
I and III
Option B (I and III) is correct because the empty set is represented by either the symbol ϕ\phi or by {}.
73 Easy
73 of 194
Analyzing +4 | -1
If a  N={a  x  :x∈N}a\;N = \left\{ {a\;x\;:x \in N} \right\} and b  N∩c  N=d  N,b\;N \cap c\;N = d\;N, where b,c∈Nb,c \in N then
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because dd is the least common multiple of bb and cc, not necessarily their product when bb and cc are not relatively prime.
74 Easy
74 of 194
Remembering +4 | -1
In the above question, the number of families which buy none of A,BA,B and CC is
View answer and explanation
Correct option: A
4000
Option A (4000) is correct because it is calculated by subtracting the number of families buying at least one of A, B, or C from the total number of families.
75 Easy
75 of 194
Analyzing +4 | -1
Let A=[x:x∈R,∣x∣<1]B=[x:x∈R,∣x−1∣≥1]A=[x: x \in R,|x|<1] B=[x: x \in R,|x-1| \geq 1] and A∪B=R−DA \cup B=R-D, then the set D\mathrm{D} is:
View answer and explanation
Correct option: B
[x:1≤x<2][x: 1 \leq x<2]
Option B ([x:1≤x<2][x: 1 \leq x<2]) is correct because it represents the set D such that A∪B=R−DA \cup B = R - D, where A and B are defined by the given inequalities.
76 Easy
76 of 194
Understanding +4 | -1
Suppose n(A)=3,n(B)=6n(A)=3, n(B)=6 and A⊆BA \subseteq B. Then, the number of elements in A∪BA \cup B is equal to
View answer and explanation
Correct option: C
6
Option C (6) is correct because when A is a subset of B, the union of A and B is simply B, and n(B) = 6.
77 Easy
77 of 194
Understanding +4 | -1
Which one of the following is correct?
View answer and explanation
Correct option: A
A∪P(A)=P(A)\mathrm{A} \cup \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A})
Option A (A∪P(A)=P(A)\mathrm{A} \cup \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A})) is correct because A is a subset of its power set, so the union of A and its power set is just the power set.
78 Easy
78 of 194
Understanding +4 | -1
The set of all letters of the word ‘SCHOOL’ is
represented by

I. {S, C, H, O, O, L}

II. {S, C, H, O, L}

III. {C, H, L, O, S}

IV. {S, C, H, L}

The correct code is
View answer and explanation
Correct option: C
II and III
Option C (II and III) is correct because a set contains distinct elements, and the order of elements in a set does not matter.
79 Easy
79 of 194
Application +4 | -1
Let ⋃i=150Xi=⋃j=1nYj=Z\bigcup\limits_{i=1}^{50} X_i=\bigcup\limits_{j=1}^n Y_j=Z, where each XiX_i contains 10 elements and each YjY_j contains 5 elements. If each element of ZZ is an element of exactly 20 of sets Xi’s X_i^{\text {'s }} and exactly 6 of sets Yj’s Y_j^{\text {'s }}, then nn is equal to
View answer and explanation
Correct option: C
30
Option C (30) is correct because by equating the total number of elements counted in two ways, we can solve for n.
80 Easy
80 of 194
Understanding +4 | -1
If An{A_n} is the set of first nn prime numbers, then ∪n=210An=\cup _{n = 2}^{10}{A_n} =
View answer and explanation
Correct option: B
{2,3,5,7,11,13,17,19,23,29}\left\{ {2,3,5,7,11,13,17,19,23,29} \right\}
Option B ({2,3,5,7,11,13,17,19,23,29}\left\{ {2,3,5,7,11,13,17,19,23,29} \right\}) is correct because the union of the first n prime numbers from n=2 to n=10 is simply the set of the first 10 prime numbers.
81 Easy
81 of 194
Analyzing +4 | -1
Let A,B,CA, B, C be three sets such that A∪B∪C=UA \cup B \cup C=U, where UU is the universal set. Then, {(A−B)∪(B−C)∪(C−A)}′{\{ (A - B) \cup (B - C) \cup (C - A)\} ^\prime }
View answer and explanation
Correct option: C
A∩B∩CA \cap B \cap C
Option C (A∩B∩CA \cap B \cap C) is correct because the complement of the union of differences simplifies to the intersection of all three sets.
82 Easy
82 of 194
Understanding +4 | -1
Consider the following statements.

I. The vowels in the English alphabet.

II. The collection of books.

III. The rivers of India.

IV. The collection of most talented batsmen of India.

Which of the following is/are well-defined collections?
View answer and explanation
Correct option: C
I and III
Option C (I and III) is correct because the vowels in the English alphabet and the rivers of India are well-defined collections.
83 Easy
83 of 194
Understanding +4 | -1
A,BA,B and CC are three non-empty sets. If A⊂BA \subset B and B⊂C,B \subset C, then which of the following is true?
View answer and explanation
Correct option: C
A∪B=B∩CA \cup B = B \cap C
The correct answer is A∪B=B∩CA \cup B = B \cap C because if A⊂BA \subset B and B⊂CB \subset C, then A∪B=BA \cup B = B and B∩C=BB \cap C = B, thus A∪B=B∩CA \cup B = B \cap C.
84 Easy
84 of 194
Understanding +4 | -1
Let XX be a family of sets and RR be a relation on XX defined by 'AA is disjoint from B′B'. Then, RR is
View answer and explanation
Correct option: B
Symmetric
Option B (Symmetric) is correct because if A is disjoint from B, then B is also disjoint from A, satisfying the definition of a symmetric relation.
85 Easy
85 of 194
Understanding +4 | -1
If A={1,3,5,7,9,11,13,15,17},B={2,4,…18}{A}=\{1,3,5,7,9,11,13,15,17\}, {B}=\{2,4, \ldots 18\} and N{N} is the universal set then A′∪(A∪(B∩B′))A^{\prime} \cup\left(A \cup\left(B \cap B^{\prime}\right)\right) is
View answer and explanation
Correct option: D
N{N}
Option D is correct because A' ∪ (A ∪ (B ∩ B')) simplifies to the universal set N, since B ∩ B' is the empty set and A ∪ ∅ = A, then A' ∪ A = N.
86 Easy
86 of 194
Application +4 | -1
Two finite sets have m and n elements. The number of elements in the power set of first set is 48 more than the total number of elements in the power set of the second set. Then, the value of MM and NN are
View answer and explanation
Correct option: C
6,4
Option C (6,4) is correct because it satisfies the equation derived from the problem statement, where the difference in the number of elements in the power sets is 48.
87 Easy
87 of 194
Application +4 | -1
In a class, 70 students wrote two tests viz; test-I and test-II. 50% of the students failed in test-I and 40% of the students in test-II. How many students passed in both tests?
View answer and explanation
Correct option: B
7
The correct answer is B (7) because it is calculated by using the principle of inclusion-exclusion.
88 Easy
88 of 194
Analyzing +4 | -1
If A={p:p=(n+2)(2n5+3n4+4n3+5n2+6)n2+2n,n,p∈Z+}A = \left\{ {p:p = \frac{{\left( {n + 2} \right)\left( {2{n^5} + 3{n^4} + 4{n^3} + 5{n^2} + 6} \right)}}{{{n^2} + 2n}},n,p \in {{\text{Z}}^ + }} \right\} then the number of elements in the set AA, is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the expression simplifies to p=2n4+3n3+4n2+5n+6np = 2n^4 + 3n^3 + 4n^2 + 5n + \frac{6}{n}, and pp is an integer only when nn is a factor of 6, giving 4 possible values.
89 Easy
89 of 194
Understanding +4 | -1
If SS is the set of squares and RR is the set of rectangles, then (S∪R)−(S∩S)\left( {S \cup R} \right) - \left( {S \cap S} \right) is
View answer and explanation
Correct option: D
Set of rectangles but not squares
Option D (Set of rectangles but not squares) is correct because the set difference (S ∪ R) - (S ∩ R) results in elements that are in R but not in S, which are rectangles that are not squares.
90 Easy
90 of 194
Analyzing +4 | -1
If a set AA contains n elements, then which of the following cannot be the number of reflexive relations on the set AA?
View answer and explanation
Correct option: D
2n+1{2^{n + 1}}
The correct answer is 2n+1{2^{n + 1}} because the number of reflexive relations on a set with n elements is 2n2−n{2^{{n^2} - n}}, and n2−n=n+1{n^2} - n = n + 1 has no solution in N.
Revision notes and key takeaways

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