Sets and Relations JEE Main previous year questions with solutions

5 solved JEE Main questions on Sets and Relations, free to read — no sign-in needed. The full chapter has 120 questions; sign in to attempt the remaining 115 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Questions on number of relations and sets
   Let $A = \{1, 4, 7\}$ and $B = \{2, 3, 8\}$. Then the number of elements, in the relation $R = \{((a_1, b_1), (a_2, b_2)) \in ((A \times B) \times (A \times B)) : a_1 + b_2 \text{ divides } a_2 + b_1\}$ is _______.
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   Answer: 18

   Let $A = \{1, 4, 7\}$ and $B = \{2, 3, 8\}$. The relation $R$ consists of pairs $((a_1, b_1), (a_2, b_2))$ from $A \times B$ such that $(a_1 + b_2)$ divides $(a_2 + b_1)$. Let $x = a_1 + b_2$ and $y = a_2 + b_1$. Note that $(a_1, b_2)$ and $(a_2, b_1)$ are both independent elements of $A \times B$. So choosing $((a_1, b_1), (a_2, b_2))$ is equivalent to independently choosing two elements of $A \times B$ with sums $x$ and $y$ such that $x \mid y$. All possible sums $a + b$ with $a \in A$ and $b \in B$: $1 + 2 = 3$, $1 + 3 = 4$, $1 + 8 = 9$ $4 + 2 = 6$, $4 + 3 = 7$, $4 + 8 = 12$ $7 + 2 = 9$, $7 + 3 = 10$, $7 + 8 = 15$ Frequencies of each sum: $f(3) = 1$, $f(4) = 1$, $f(6) = 1$, $f(7) = 1$, $f(9) = 2$, $f(10) = 1$, $f(12) = 1$, $f(15) = 1$ For each possible divisor $x$, count the number of valid pairs $(x, y)$ with $x \mid y$, each contributing $f(x) \cdot f(y)$: $x = 3$: divides $3, 6, 9, 12, 15$ Ways $= 1 \cdot (1 + 1 + 2 + 1 + 1) = 6$ $x = 4$: divides $4, 12$ Ways $= 1 \cdot (1 + 1) = 2$ $x = 6$: divides $6, 12$ Ways $= 1 \cdot (1 + 1) = 2$ $x = 7$: divides $7$ Ways $= 1 \cdot 1 = 1$ $x = 9$: divides $9$ Ways $= 2 \cdot 2 = 4$ $x = 10$: divides $10$ Ways $= 1 \cdot 1 = 1$ $x = 12$: divides $12$ Ways $= 1 \cdot 1 = 1$ $x = 15$: divides $15$ Ways $= 1 \cdot 1 = 1$ Total number of elements in $R$: $6 + 2 + 2 + 1 + 4 + 1 + 1 + 1 = 18$ Hence, the answer is $18$.
2. Q2JEE Main 2025 (08 Apr, Shift 2)Questions on Symmetric Transitive and Reflexive Properties
   Let $A=\{0,1,2,3,4,5\}$. Let $R$ be a relation on A defined by $(x, y) \in R$ if and only if max $\{x, y\} \in\{3,4\}$. Then among the statements $\left(\mathrm{S}_1\right)$ : The number of elements in R is 18 , and $\left(\mathrm{S}_2\right)$ : The relation R is symmetric but neither reflexive nor transitive
   1. A.both are true
   2. B.both are false
   3. C.only $\left(\mathrm{S}_2\right)$ is true
   4. D.only $\left(\mathrm{S}_1\right)$ is true
   Show answer & solution

   Answer: (C)

   $A=\{0,1,2,3,4,5\}$ $\mathrm{R} \equiv\{(0,3),(3,0),(0,4),(4,0),(1,3),(3,1),(1,4)$, $(4,1),(2,3),(3,2),(2,4),(4,2),(3,3),(3,4),(4,3)$, $(4,4)\}$ Total 16 elements Not reflexive as $(0,0), \ldots \ldots,(2,2) \notin \mathrm{R}$ Symmetric $\because \forall$ all a,b $(a, b) \&(b, a) \in R$ Not transitive $\because(0,3),(3,1) \in R$ but $(0,1) \notin \mathrm{R}$ $\Rightarrow$ Only $\mathrm{S}_2$ correct
3. Q3JEE Main 2023 (11 Apr, Shift 1)Questions on Venn Diagram
   An organization awarded $48$ medals in event $A''$, $25$ in event $B''$ and $18$ in event $C''$. If these medals went to total $60$ men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?
   1. A.$15$
   2. B.$21$
   3. C.$10$
   4. D.$9$
   Show answer & solution

   Answer: (B)

   Let $A,B$ and $C$ denote the set of men who received medals in even $A$, event $B$ and event $C$ respectively. Then, $n\left(A\right)=48,n\left(B\right)=25,n\left(C\right)=18$ , $n\left(A\cup B\cup C\right)=60$ and $n\left(A\cap B\cap C\right)=5$, Now we know that, $n\left(A\cup B\cup C\right)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n\left(A\cap B\right)-n\left(A\cap C\right)-n\left(B\cap C\right)+n\left(A\cap B\cap C\right)$ $\Rightarrow 60=48+25+18-n(A\cap B)-n(A\cap C)-n(B\cap C)+5$ $\Rightarrow n(A\cap B)+n(A\cap C)+n(A\cap C)=48+25+18+5-60=36$ Therefore, the number of people who received medals in exactly two of the three sports will be $36-3\left(A\cap B\cap C\right)=36-15=21$.
4. Q4JEE Main 2026 (02 Apr, Shift 2)Questions on number of relations and sets
   Let $A = \{2, 3, 4, 5, 6\}$. Let R be a relation on the set $A \times A$ given by $(x, y) R (z, w)$ if and only if $x$ divides $z$ and $y \leq w$. Then the number of elements in R is _______.
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   Answer: 120

   Given the set $A = \{2, 3, 4, 5, 6\}$. The relation $R$ is defined on $A \times A$ such that $(x, y) R (z, w)$ if and only if $x$ divides $z$ and $y \leq w$. The number of elements in $R$ is the number of ordered pairs $((x, y), (z, w))$ satisfying these conditions. Since the conditions for $(x, z)$ and $(y, w)$ are independent, the total number of elements in $R$ will be the product of the number of pairs $(x, z)$ satisfying $x$ divides $z$ and the number of pairs $(y, w)$ satisfying $y \leq w$. First, we find the number of pairs $(x, z) \in A \times A$ such that $x$ divides $z$: For $x = 2$, $z \in \{2, 4, 6\}$ ($3$ pairs) For $x = 3$, $z \in \{3, 6\}$ ($2$ pairs) For $x = 4$, $z \in \{4\}$ ($1$ pair) For $x = 5$, $z \in \{5\}$ ($1$ pair) For $x = 6$, $z \in \{6\}$ ($1$ pair) Total number of pairs $(x, z) = 3 + 2 + 1 + 1 + 1 = 8$. Next, we find the number of pairs $(y, w) \in A \times A$ such that $y \leq w$: For $y = 2$, $w \in \{2, 3, 4, 5, 6\}$ ($5$ pairs) For $y = 3$, $w \in \{3, 4, 5, 6\}$ ($4$ pairs) For $y = 4$, $w \in \{4, 5, 6\}$ ($3$ pairs) For $y = 5$, $w \in \{5, 6\}$ ($2$ pairs) For $y = 6$, $w \in \{6\}$ ($1$ pair) Total number of pairs $(y, w) = 5 + 4 + 3 + 2 + 1 = 15$. The total number of elements in the relation $R$ is the product of the number of these pairs: Total elements = $8 \times 15 = 120$. Answer: $120$
5. Q5JEE Main 2026 (24 Jan, Shift 1)Questions on number of relations and sets
   Let R be a relation defined on the set $\{1,2,3,4\} \times\{1,2,3,4\}$ by $\mathrm{R}=\{((a, b),(c, d)): 2 a+3 b=3 c+4 d\}$. Then the number of elements in R is
   1. A.18
   2. B.6
   3. C.15
   4. D.12
   Show answer & solution

   Answer: (D)

   We need $2a+3b = 3c+4d$ where $a,b,c,d \in \{1,2,3,4\}$. Possible values of $2a+3b$: range from $5$ to $20$. Possible values of $3c+4d$: range from $7$ to $28$. For each common value $k$, the number of elements in $R$ equals (number of $(a,b)$ with $2a+3b=k$) $\times$ (number of $(c,d)$ with $3c+4d=k$). $k=7$: $(a,b)=(2,1)$, $(c,d)=(1,1)$ $\Rightarrow 1 \times 1=1$ $k=10$: $(2,2)$, $(2,1)$ $\Rightarrow 1$ $k=11$: $(1,3),(4,1)$, $(1,2)$ $\Rightarrow 2$ $k=13$: $(2,3)$, $(3,1)$ $\Rightarrow 1$ $k=14$: $(1,4),(4,2)$, $(2,2)$ $\Rightarrow 2$ $k=15$: $(3,3)$, $(1,3)$ $\Rightarrow 1$ $k=16$: $(2,4)$, $(4,1)$ $\Rightarrow 1$ $k=17$: $(4,3)$, $(3,2)$ $\Rightarrow 1$ $k=18$: $(3,4)$, $(2,3)$ $\Rightarrow 1$ $k=20$: $(4,4)$, $(4,2)$ $\Rightarrow 1$ Total $= 1+1+2+1+2+1+1+1+1+1 = 12$