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JEE • MATHEMATICS • RELATIONS

Relations practice · Easy

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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 180 questions
151 Easy
151 of 180
Understanding +4 | -1
Let Y={1,  2,  3,  4,  5},  A{1,  2},  B={3,  4,  5}Y = \left\{ {1,\;2,\;3,\;4,\;5} \right\},\;A\left\{ {1,\;2} \right\},\;B = \left\{ {3,\;4,\;5} \right\} and ϕ\phi denotes null set. If (A×B)\left( {A \times B} \right) denotes cartesian product of the sets AA and BB; then (Y×A)∩(Y×B)\left( {Y \times A} \right) \cap \left( {Y \times B} \right) is
View answer and explanation
Correct option: D
ϕ\phi
Option D (ϕ\phi) is correct because the intersection of the two Cartesian products is an empty set since A and B have no common elements.
152 Easy
152 of 180
Understanding +4 | -1
Let nn be a fixed positive integer. Let a relation RR be defined in ZZ (the set of all integers) as follows: aRba R b iff n/(a−b)n /(a-b), that is, iff a−ba-b is divisible by nn. Then, the relation RR is
View answer and explanation
Correct option: D
An equivalence relation
Option D (An equivalence relation) is correct because the relation R is reflexive, symmetric, and transitive, satisfying the conditions for an equivalence relation.
153 Easy
153 of 180
Analyzing +4 | -1
Let RR be a relation on the set NN be defined by {(x,y)|x,y∈N,2  x+y=41}.\left\{ {\left( {x,y} \right){\text{|}}x,y \in N,2\;x + y = 41} \right\}. Then, RR is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the relation R is neither reflexive, symmetric, nor transitive, as demonstrated by the counterexamples provided.
154 Easy
154 of 180
Analyzing +4 | -1
For any two real numbers xx and yy, we define xRyx R y if and only if sin⁡2x+cos⁡2y=1\sin ^2 x+\cos ^2 y=1. The relation RR is
View answer and explanation
Correct option: D
An equivalence relation
Option D (An equivalence relation) is correct because the relation R is reflexive, symmetric, and transitive.
155 Easy
155 of 180
Analyzing +4 | -1
Let RR be a relation on R\mathbb{R}, given by R={(a,b):3a−3b+7R=\{(a, b): 3 a-3 b+\sqrt{7} is an irrational number}\}. Then RR is
View answer and explanation
Correct option: A
Reflexive but neither symmetric nor transitive
Option A (Reflexive but neither symmetric nor transitive) is correct because the relation satisfies the reflexive property but fails to satisfy symmetric and transitive properties.
156 Easy
156 of 180
Understanding +4 | -1
In a set of ants in a locality, two ants are said to be related iff they walk on a same straight line, then the relation is
View answer and explanation
Correct option: D
Equivalence
Option D (Equivalence) is correct because the relation is reflexive (an ant walks on the same line as itself), symmetric (if ant A walks on the same line as ant B, then ant B walks on the same line as ant A), and transitive (if ant A walks on the same line as ant B, and ant B walks on the same line as ant C, then ant A walks on the same line as ant C).
157 Easy
157 of 180
Analyzing +4 | -1
If A={n:n3+5n2+2n  is  an  integer  and  itself  is  an  integer},A = \left\{ {n:\frac{{{n^3} + 5{n^2} + 2}}{n}\;\mathrm{is\;an\;integer\;and\;itself\;is\;an\;integer}} \right\}, then the number of elements in the set AA, is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the expression is an integer when n is a factor of 2, including both positive and negative factors.
158 Easy
158 of 180
Analyzing +4 | -1
Let R1R_1 and R2R_2 be two relations defined on R\mathbb{R} by a R1b⇔ab≥0R_1 b \Leftrightarrow a b \geq 0 and aR2b⇔a≥ba R_2 b \Leftrightarrow a \geq b. Then,
View answer and explanation
Correct option: D
Neither R1R_1 nor R2R_2 is an equivalence relation
Option D (Neither R1R_1 nor R2R_2 is an equivalence relation) is correct because neither relation satisfies all three properties of an equivalence relation: reflexivity, symmetry, and transitivity.
159 Easy
159 of 180
Application +4 | -1
The relation RR defined on the set A={1,2,3,4,5}A=\{1,2,3,4,5\} by R={(x,y):∣x2−y2∣<16}R=\left\{(x, y):\left|x^2-y^2\right| < 16\right\} is given by
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because none of the given relations satisfy the condition ∣x2−y2∣<16|x^2 - y^2| < 16 for all pairs.
160 Easy
160 of 180
Understanding +4 | -1
Let RR be a reflexive relation on a finite set A having nn elements, and let there be mm ordered pairs in RR. Then,
View answer and explanation
Correct option: A
m⩾nm \geqslant n
Option A (m⩾nm \geqslant n) is correct because a reflexive relation on a set with n elements must have at least n ordered pairs.
161 Easy
161 of 180
Remembering +4 | -1
If a set is denoted as B=ϕ{B}=\phi, then the number of element in B{B} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the number of elements in an empty set, denoted by ϕ\phi, is zero.
162 Easy
162 of 180
Understanding +4 | -1
Let A={2,3,4,5}A=\{2,3,4,5\} and let R={(2,2),(3,3),(4,4),(5,5),(2,3),(3,2),(3,5),(5,3)}R=\{(2,2),(3,3),(4,4),(5,5),(2,3),(3,2),(3,5),(5,3)\} be a relation in AA. Then RR is
View answer and explanation
Correct option: B
Reflexive and symmetry
Option B is correct because the relation R is reflexive (contains (2,2), (3,3), (4,4), (5,5)) and symmetric (if (a,b) is in R, then (b,a) is also in R).
163 Easy
163 of 180
Understanding +4 | -1
Consider the following statements

I. Let n(A)=mn(A)=m and n(B)=nn(B)=n. Then the total number of non-empty relations that can be defined from AA to BB is 2mn−12^{{mn}}-1

II. If A={1,2,3},B={3,8}{A}=\{1,2,3\}, {B}=\{3,8\}, then (A∪B)×(A∩B)({A} \cup {B}) \times({A} \cap {B}) is equal to {(1,3),(2,3),(3,3),(8,3)}.\{(1,3),(2,3),(3,3),(8,3)\}.

III. If (x2−1,y9+1)=(2,1)\left(\frac{x}{2}-1, \frac{y}{9}+1\right)=(2,1), then the values of x{x} and y{y} respectively are 6 and 0.

Choose the correct option.
View answer and explanation
Correct option: D
All the three statements are true
All three statements are true because they correctly describe the number of relations, the Cartesian product of union and intersection, and the equality of ordered pairs.
164 Easy
164 of 180
Analyzing +4 | -1
Let RR be a relation on the set of integers given by aa RR bb ⇒a=2kb\Rightarrow a=2^k b for some integer k.k. Then, RR is
View answer and explanation
Correct option: A
An equivalence relation
Option A is correct because the relation R is reflexive, symmetric, and transitive, thus satisfying the conditions for an equivalence relation.
165 Easy
165 of 180
Application +4 | -1
The number of equivalence relations in the set {1,2,3}\{1,2,3\} containing (1,2)(1,2) and (2,1)(2,1) is
View answer and explanation
Correct option: A
2
Option A (2) is correct because there are only two equivalence relations that satisfy the given conditions: the smallest one containing (1,2) and (2,1), and the universal relation.
166 Easy
166 of 180
Analyzing +4 | -1
Set builder form of the relation R={(−2,−7),(−1,−4),(0,−1),(1,  2),(2,  5)}R = \left\{ {\left( { - 2, - 7} \right),\left( { - 1, - 4} \right),\left( {0, - 1} \right),\left( {1,\;2} \right),\left( {2,\;5} \right)} \right\} is
View answer and explanation
Correct option: D
{(u,υ):υ=3u−1;−2≤u<3  and  u∈Z}\{ \left( {u,\upsilon } \right):\upsilon = 3u - 1; - 2 \le u < 3\;\mathrm{and\;}u \in Z\}
Option D is correct because it accurately represents the given relation in set-builder form, defining the relationship between elements and their domain.
167 Easy
167 of 180
Analyzing +4 | -1
Let A={1,2,3}A=\{1,2,3\}. Then find the number of relations containing (1,2)(1,2) and (1,3)(1,3), which are reflexive and symmetric but not transitive, is
View answer and explanation
Correct option: A
1
Option A (1) is correct because there is only one relation that contains (1,2) and (1,3), which is reflexive and symmetric but not transitive.
168 Easy
168 of 180
Understanding +4 | -1
The relation ‘is not equal to’ is defined on RR, is
View answer and explanation
Correct option: B
Symmetric only
Option B (Symmetric only) is correct because if a is not equal to b, then b is not equal to a, but a is not equal to itself is false, and transitivity doesn't hold.
169 Easy
169 of 180
Understanding +4 | -1
If n(X)=5n(X)=5 and n(Y)=7n(Y)=7, then the number of relations on X×Y{X} \times {Y} is 25m2^{5 {m}}. The value of &#39;m{m}&#39; is
View answer and explanation
Correct option: A
7
7 is the correct answer because the number of relations on X x Y is 2^(n(X) * n(Y)), so 5m = 5 * 7, thus m = 7.
170 Easy
170 of 180
Application +4 | -1
If a set AA has 10 elements, then the ratio of total no. of reflexive relations to total no. of symmetric relations is
View answer and explanation
Correct option: A
2352^{35}
Option A (2352^{35}) is correct because the ratio of reflexive relations to symmetric relations on a set with 10 elements is calculated as 2(n2-n) / 2^((n2+n)/2) = 2^((n2-3n)/2), and substituting n=10 gives 235.
171 Easy
171 of 180
Understanding +4 | -1
Let R\mathrm{R} be a reflexive relation on a set A\mathrm{A} and I\mathrm{I} be the identity relation on A\mathrm{A}. Then
View answer and explanation
Correct option: B
I⊂RI \subset R
Option B (I⊂RI \subset R) is correct because since R is reflexive, it must contain all the pairs in the identity relation I, meaning I is a subset of R.
172 Easy
172 of 180
Application +4 | -1
If AA is the set of even natural numbers less than 10 and BB is the set of prime numbers less than 5 , then the number of relations from AA to BB is
View answer and explanation
Correct option: A
282^8
Option A (282^8) is correct because the number of relations from set A to set B is 2n(A)×n(B)2^{n(A) \times n(B)}, where n(A) and n(B) are the number of elements in sets A and B, respectively.
173 Easy
173 of 180
Understanding +4 | -1
If a relation RR is defined on the set ZZ of integers as follows:

(a,b)∈R⇔a2+b2=25.(a, b) \in R \Leftrightarrow a^2+b^2=25.

Then, Domain (R)=(R)=
View answer and explanation
Correct option: C
{0,±3,±4,±5}\{0, \pm 3, \pm 4, \pm 5\}
Option C ({0,±3,±4,±5}\{0, \pm 3, \pm 4, \pm 5\}) is correct because it includes all possible integer values of 'a' for which there exists an integer 'b' such that a2+b2=25a^2 + b^2 = 25.
174 Easy
174 of 180
Analyzing +4 | -1
In the set A={1,  2,  3,  4,  5},A = \left\{ {1,\;2,\;3,\;4,\;5} \right\}, a relation RR is defined by R={(x,y):x,y∈A  and  x<y}.R = \left\{ {\left( {x,y} \right):x,y \in A\;\mathrm{and\;}x < y} \right\}. Then, RR is
View answer and explanation
Correct option: C
Transitive
Option C is correct because the relation R is transitive, meaning if x < y and y < z, then x < z.
175 Easy
175 of 180
Understanding +4 | -1
The relation R={(1,  1),  (2,  2),  (3,  3)}R = \left\{ {\left( {1,\;1} \right),\;\left( {2,\;2} \right),\;\left( {3,\;3} \right)} \right\} on the set {1,  2,  3}\left\{ {1,\;2,\;3} \right\} is
View answer and explanation
Correct option: C
An equivalence relation
Option C (An equivalence relation) is correct because the relation is reflexive, symmetric, and transitive.
176 Easy
176 of 180
Understanding +4 | -1
Let R={(3,3),(6,6),(9,9),(12,12),(6,12)R=\{(3,3),(6,6),(9,9),(12,12),(6,12), (3,9),(3,12),(3,6)}(3,9),(3,12),(3,6)\} be a relation on the set A={3,6,9,12}.A=\{3,6,9,12\}. Then relation is
View answer and explanation
Correct option: D
Reflexive and transitive only
Option D (Reflexive and transitive only) is correct because the relation R satisfies the reflexive and transitive properties but not the symmetric property.
177 Easy
177 of 180
Understanding +4 | -1
Let P={(x,y):∣x2+y2∣=1,x,y∈R}P=\left\{(x, y):\left|x^2+y^2\right|=1, x, y \in R\right\}. Then P\mathrm{P} is
View answer and explanation
Correct option: B
Symmetric
Option B (Symmetric) is correct because if (x,y)∈P(x, y) \in P, then ∣x2+y2∣=1|x^2 + y^2| = 1, which implies ∣y2+x2∣=1|y^2 + x^2| = 1, so (y,x)∈P(y, x) \in P.
178 Easy
178 of 180
Understanding +4 | -1
If A={x∈R:0<x<1} and B={x∈R:−1<x<1},A = \{ x \in R:0 < x < 1\} \,{\text{and}}\,B = \{ x \in R: - 1 < x < 1\} , then A×BA \times B is the set of all points lying
View answer and explanation
Correct option: A
Inside the rectangle having vertices at (1,1),(0,1),(0,−1)(1,1),(0,1),(0,-1) and (1,−1)(1,-1)
Option A (Inside the rectangle having vertices at (1,1),(0,1),(0,−1)(1,1),(0,1),(0,-1) and (1,−1)(1,-1)) is correct because it represents the region defined by the Cartesian product of the given intervals for A and B.
179 Easy
179 of 180
Application +4 | -1
A survey shows that 73%73 \% of the persons working in an office like coffee, whereas 65%65 \% like tea. If xx denotes the percentage of them who like both coffee and tea, then xx cannot be
View answer and explanation
Correct option: A
36
Option A (36) is correct because the percentage of people who like both coffee and tea, x, must be greater than or equal to 38 and less than or equal to 65, thus 36 is not a possible value.
180 Easy
180 of 180
Analyzing +4 | -1
Let RR be a relation on the set of integers given by aRb⇔a=2k.baRb \Leftrightarrow a = {2^k}.b for some integer kk. Then, RR is
View answer and explanation
Correct option: A
An equivalence relation
Option C (Reflexive and transitive but not symmetric) is correct because the relation satisfies the reflexive and transitive properties but fails the symmetric property.
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