Let the relation R on the set M =\ 1,2,3, ldots ldots .16\ be given by R =\ ( x , y ): 4 y =5 x -3, x , y in M \ . Then the minimum number of elements required to be added in R , in order to make the relation symmetric, is equal to

Mathematics · Sets and Relations

JEE Main 2026 — 22 January, Morning Shift — Question 2

Let the relation R on the set $\mathrm{M}=\{1,2,3, \ldots \ldots .16\}$ be given by $\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): 4 \mathrm{y}=5 \mathrm{x}-3, \mathrm{x}, \mathrm{y} \in \mathrm{M}\}$ . Then the minimum number of elements required to be added in R , in order to make the relation symmetric, is equal to

Option A:
$1$
Option B:
$2$
Correct
Option C:
$4$
Option D:
$3$

Answer: B

Step-by-step solution

Given: $M = \{1,2,3,\dots,16\}$

Relation: $4y = 5x - 3$

⇒ $y = \frac{5x-3}{4}$

For y to be integer:

5x - 3 \equiv 0 \pmod{4}

Since 5 ≡ 1 (mod 4),

x - 3 \equiv 0 \pmod{4}

x \equiv 3 \pmod{4}

So,

x = 3,7,11,15

Corresponding y:

x=3 ⇒ y=3 x=7 ⇒ y=8 x=11 ⇒ y=13 x=15 ⇒ y=18 (not in M)

So R = $\{(3,3),(7,8),(11,13)\}$

For symmetry, we need:

(8,7),(13,11)

(3,3) already symmetric.

Thus minimum elements to be added = $\boxed{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Sets and Relations
Topic: Number of Relations