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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Sets and Relations

Let the relation R on the set M=\ 1,2,3, , 16\ be given by R=\ (x,...

Let the relation R on the set M={1,2,3,…,16}\mathrm{M}=\{1,2,3, \ldots, 16\} be given by R={(x,y):4y=5x−3,x,y∈M}\mathrm{R}=\{(x, y): 4 y=5 x-3, x, 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

Options

  1. A4
  2. B1
  3. C2
  4. D3

Correct answer

C. 2

Step-by-step solution

Given R={(x,y):4y=5x−3, x,y∈M}R = \{(x, y) : 4y = 5x - 3,\ x, y \in M\} where M={1,2,…,16}M = \{1, 2, \ldots, 16\}.
From 4y=5x−34y = 5x - 3, we get y=5x−34y = \frac{5x-3}{4}.
For yy to be a positive integer, 5x−3≡0(mod4)5x - 3 \equiv 0 \pmod{4}, i.e., x≡3(mod4)x \equiv 3 \pmod{4}.
So x=3,7,11,15x = 3, 7, 11, 15.
x=3⇒y=3x=3 \Rightarrow y=3,
x=7⇒y=8x=7 \Rightarrow y=8,
x=11⇒y=13x=11 \Rightarrow y=13,
x=15⇒y=18∉Mx=15 \Rightarrow y=18 \notin M.
R={(3,3), (7,8), (11,13)}R = \{(3,3),\ (7,8),\ (11,13)\}.
For symmetry, (7,8)∈R(7,8) \in R requires (8,7)∈R(8,7) \in R and (11,13)∈R(11,13) \in R requires (13,11)∈R(13,11) \in R.
(3,3)(3,3) is already symmetric.
Minimum elements to be added =2= 2.
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