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

Given: R is defined on 1,2,3,4 1,2,3,4 by R= ((a,b),(c,d)): 2a+3b=3c+4d Find: The number of elements in R. The solution counts ordered pairs by comparing the values of 2a+3b and 3c+4d for all a,b,c,d 1,2,3,4. Step 1: List the possible values of 2a+3b for a,b 1,2,3,4. Step 2: List the possible values of 3c+4d for c,d 1,2,3,4. Step 3: Match equal values from both expressions. For every common value, count the ordered pairs ((a,b),(c,d)) that satisfy the equation 2a+3b=3c+4d From the extracted solution, the total number of such matching ordered pairs is 18 Therefore, the number of elements in R is 18, so the correct option is A.

Let R be a relation defined on the set 1,2,3,4 1,2,3,4 by R= ((a,… — Solution | JEE Mathematics

MCQMediumJEE 2026Relations

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

A
$18$
B
$12$
C
$6$
D
$15$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $R$ is defined on $\{1,2,3,4\}\times\{1,2,3,4\}$ by

R=\{((a,b),(c,d)) : 2a+3b=3c+4d\}

Find: The number of elements in $R$ .

The solution counts ordered pairs by comparing the values of $2a+3b$ and $3c+4d$ for all $a,b,c,d \in \{1,2,3,4\}$ .

Step 1: List the possible values of $2a+3b$ for $a,b \in \{1,2,3,4\}$ .

Step 2: List the possible values of $3c+4d$ for $c,d \in \{1,2,3,4\}$ .

Step 3: Match equal values from both expressions. For every common value, count the ordered pairs $((a,b),(c,d))$ that satisfy the equation

2a+3b=3c+4d

From the extracted solution, the total number of such matching ordered pairs is

18

Therefore, the number of elements in $R$ is $18$ , so the correct option is A.

Common mistakes

A common mistake is to count only the common numerical values of $2a+3b$ and $3c+4d$ . That is wrong because the relation contains ordered pairs $((a,b),(c,d))$ , not just the shared values. Instead, count all combinations of pairs producing each common value.
Another mistake is to treat $\{1,2,3,4\}\times\{1,2,3,4\}$ as having only four elements. That is incorrect because it contains ordered pairs, so there are $16$ possible choices for $(a,b)$ and $16$ possible choices for $(c,d)$ . Work with ordered pairs throughout.
Students may also match values without systematic listing and miss repetitions. This is wrong because different ordered pairs can give the same value of $2a+3b$ or $3c+4d$ . Instead, tabulate all values carefully and count multiplicities.

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Answer

Step-by-step solution

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