JEE Main 202621 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Sets and Relations

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Sets and Relations

Let A=\ 2,3,5,7,9\ . Let R be the relation on A defined by x R y if...

Let

\mathrm{A}=\{2,3,5,7,9\}

. Let R be the relation on A defined by

x \mathrm{R} y

if and only if

2 x \leq 3 y

. Let

l

be the number of elements in R, and m be the minimum number of elements required to be added in R to make it a symmetric relation. Then

l+\mathrm{m}

is equal to :

Options

Correct answer

D. 25

Step-by-step solution

Given

A = \{2, 3, 5, 7, 9\}

and

xRy

iff

2x \le 3y

, i.e.,

y \ge \frac{2x}{3}

.
For

x=2

y \in \{2,3,5,7,9\}

→ 5 pairs
For

x=3

y \in \{2,3,5,7,9\}

→ 5 pairs
For

x=5

y \in \{5,7,9\}

→ 3 pairs
For

x=7

y \in \{5,7,9\}

→ 3 pairs
For

x=9

y \in \{7,9\}

→ 2 pairs
So

l = 18

.
For symmetry, pairs in

R

whose reversal is missing:

(2,5),(2,7),(2,9),(3,5),(3,7),(3,9),(5,9)

require adding

(5,2),(7,2),(9,2),(5,3),(7,3),(9,3),(9,5)

→

m = 7

l + m = 18 + 7 = 25

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