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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 31-60 of 194 questions
31 Easy
31 of 194
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.
32 Easy
32 of 194
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.
33 Easy
33 of 194
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.
34 Easy
34 of 194
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.
35 Easy
35 of 194
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.
36 Easy
36 of 194
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.
37 Easy
37 of 194
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.
38 Easy
38 of 194
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.
39 Easy
39 of 194
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.
40 Easy
40 of 194
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.
41 Easy
41 of 194
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
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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.
42 Easy
42 of 194
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}.
43 Easy
43 of 194
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.
44 Easy
44 of 194
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.
45 Easy
45 of 194
Application +4 | -1
A class has 175 students. The following data shows the number of students opting one or more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. Hoe many students have offered Mathematics alone?
View answer and explanation
Correct option: C
60
The correct answer is C (60) because it is calculated by using the principle of inclusion-exclusion and subtracting the number of students taking combined subjects from the total number of students taking mathematics.
46 Easy
46 of 194
Analyzing +4 | -1
Let AA and BB be two sets. The (A∪B)′∪(A′∩B)=(A \cup B)^{\prime} \cup\left(A^{\prime} \cap B\right)=
View answer and explanation
Correct option: A
A′A^{\prime}
Option A (A′A^{\prime}) is correct because the expression simplifies to the complement of set A using De Morgan's laws and set identities.
47 Easy
47 of 194
Understanding +4 | -1
Let R{R} be the set of real numbers, If f:R→Rf: {R} \rightarrow {R} is defined by f(x)=exf(x)=e^x, then ff is
View answer and explanation
Correct option: B
Injective but not surjective
Option B (Injective but not surjective) is correct because the exponential function is one-to-one but its range is only positive real numbers.
48 Easy
48 of 194
Understanding +4 | -1
Which of the following is true?
View answer and explanation
Correct option: B
A∩ϕ=ϕA \cap \phi = \phi
Option B (A∩ϕ=ϕA \cap \phi = \phi) is correct because the intersection of any set A with the empty set is always the empty set.
49 Easy
49 of 194
Analyzing +4 | -1
Let A={x:∣x∣<2}A=\{x:|x|<2\} and B={x:∣x−2∣≥3}B=\{x:|x-2| \geq 3\}. Then
View answer and explanation
Correct option: D
B−A=R−(−2,5)B-A=R-(-2,5)
Option D (B−A=R−(−2,5)B-A=R-(-2,5)) is the correct answer because it accurately represents the set of elements in B but not in A.
50 Easy
50 of 194
Analyzing +4 | -1
If A={1,3,5,7,9,11,13,15,17},B={2,4…,18}A = \left\{ {1,3,5,7,9,11,13,15,17} \right\},B = \left\{ {2,4 \ldots ,18} \right\} and NN is the universal set, then A′∪((A∪B)∩B′)A' \cup \left( {\left( {A \cup B} \right) \cap B'} \right) is
View answer and explanation
Correct option: B
NN
Option B (N) is correct because the expression simplifies to A' ∪ A, which by definition equals the universal set N.
51 Easy
51 of 194
Application +4 | -1
In a city, three daily newspapers A,B,CA, B, C are published. 42%42 \% of the people in that city read AA, 51%51 \% read BB and 68%68 \% read C. 30%30 \% read A{A} and B;28%{B} ; 28 \% read B{B} and C{C}; 36%36 \% read A{A} and C;8%{C} ; 8 \% do not read any of the three newspapers. The percentage of persons who read all the three papers is
View answer and explanation
Correct option: A
25%25 \%
Option A (2525 \\%) is correct because it uses the principle of inclusion-exclusion for three sets to find the percentage of people who read all three newspapers.
52 Easy
52 of 194
Understanding +4 | -1
Let A1,A2,A3…,A100{A_1},{A_2},{A_3} \ldots ,{A_{100}} be 100 sets such that n(Ai)=i+1n\left( {{A_i}} \right) = i + 1 and A1⊂A2⊂A3⊂…⊂A100,{A_1} \subset {A_2} \subset {A_3} \subset \ldots \subset {A_{100}}, then Ui=1100AiU_{i = 1}^{100}{A_i} contains… elements
View answer and explanation
Correct option: C
101
Option D (101) is correct because the union of nested sets is the largest set, and n(A_100) = 100 + 1 = 101.
53 Easy
53 of 194
Understanding +4 | -1
If A={x:x=2n+1,n∈Z}A=\{x: x=2 n+1, n \in Z\} \quad and B={x:x=2n,n∈Z}B=\{x: x=2 n, n \in Z\}, then A∪B=A \cup B=
View answer and explanation
Correct option: B
ZZ
Option B (Z) is correct because the union of the set of odd integers and the set of even integers is the set of all integers.
54 Easy
54 of 194
Analyzing +4 | -1
If X={8n−7−1:n∈N):n∈N}\left.X=\left\{8^n-7-1: n \in N\right): n \in N\right\} and Y={49(n−1):n∈N}Y=\{49(n-1): n \in N\} then:
View answer and explanation
Correct option: A
X⊆YX \subseteq Y
Option A (X⊆YX \subseteq Y) is correct because every element in X is also an element in Y, meaning X is a subset of Y.
55 Easy
55 of 194
Application +4 | -1
In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many persons like both coffee and tea?
View answer and explanation
Correct option: A
19
Option A (19) is correct because it is the number of people who like both coffee and tea, calculated using the principle of inclusion-exclusion.
56 Easy
56 of 194
Understanding +4 | -1
Let P{P} be a set of squares, Q{Q} be set of parallelograms, R{R} be a set of quadrilaterals and S{S} be a set of rectangles. Consider the following:

I. P⊂Q{P} \subset {Q} \quad II. R⊂P{R} \subset {P} \quad III. P⊂S{P} \subset {S} \quad IV. S⊂R{S} \subset {R}

Which of the above are correct?
View answer and explanation
Correct option: B
I, III and IV
Option B (I, III and IV) is correct because squares are parallelograms and rectangles, and rectangles are quadrilaterals.
57 Easy
57 of 194
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.
58 Easy
58 of 194
Application +4 | -1
X{X} and Y{Y} are two sets such that X{X} has 40 elements, X∪{X} \cup Y{Y} has 60 elements and X∩Y{X} \cap {Y} has 10 elements, then number of elements Y{Y} has
View answer and explanation
Correct option: C
30
Option C (30) is correct because it uses the principle of inclusion-exclusion to find the number of elements in set Y.
59 Easy
59 of 194
Understanding +4 | -1
If A\mathrm{A} and B\mathrm{B} are two sets, then (A−B)∪(B−A)(\mathrm{A}-\mathrm{B}) \cup(\mathrm{B}-\mathrm{A}) ∪(A∩B)\cup(A \cap B) is equal to
View answer and explanation
Correct option: B
A∪BA \cup B
Option B (A∪BA \cup B) is correct because the expression represents all elements in A, all elements in B, and the intersection of A and B, which is the union of A and B.
60 Easy
60 of 194
Application +4 | -1
Out of 800 boys in a school 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey; 24 played all the three games. The number of boys who did not play any game is
View answer and explanation
Correct option: A
160
Option A (160) is correct because it is calculated by subtracting the number of boys who played at least one game from the total number of boys.
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