Let R be a relation defined on the set \ 1,2,3,4\ ×\ 1,2,3,4\ by R=\ ((a, b),(c, d)): 2 a+3 b=3 c+4 d\ . Then the number of elements in R is

Mathematics · Sets and Relations

JEE Main 2026 — 24 January, Morning Shift — Question 12

Let R be a relation defined on the set $\{1,2,3,4\} \times\{1,2,3,4\}$ by $R=\{((a, b),(c, d)): 2 a+3 b=3 c+4 d\}$ . Then the number of elements in R is

Option A:
$6$
Option B:
$18$
Option C:
$12$
Correct
Option D:
$15$

Answer: C

Step-by-step solution

The set has 16 ordered pairs $(a,b)$ with $a,b\in\{1,2,3,4\}$ . The condition is $2a+3b = 3c+4d$ . Compute all values of $2a+3b$ : for $a=1$ : $5,8,11,14$ ; $a=2$ : $7,10,13,16$ ; $a=3$ : $9,12,15,18$ ; $a=4$ : $11,14,17,20$ . Compute all values of $3c+4d$ : for $c=1$ : $7,11,15,19$ ; $c=2$ : $10,14,18,22$ ; $c=3$ : $13,17,21,25$ ; $c=4$ : $16,20,24,28$ . Common values $v$ and their frequencies: $v=7$ (1 each), $10$ (1), $11$ (2 in L, 1 in R), $13$ (1), $14$ (2 in L, 1 in R), $15$ (1), $16$ (1), $17$ (1), $18$ (1), $20$ (1). Sum of products: $1+1+2+1+2+1+1+1+1+1 = 12$ . Thus, there are 12 ordered pairs in $R$ . Hence, the answer is option C.

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Sets and Relations
Topic: Number of Relations