JEE 2023 — Mathematics PYQ

Given that $A=\{1,2,3,4\}$

and $R=\{((a,b),(c,d)):2a+3b=4c+5d\}$

$\Rightarrow 2a=\{2,4,6,8\},3b=\{3,6,9,12\},4c=\{4,8,12,16\}$

and $5d=\{5,10,15,20\}$

$\Rightarrow 2a+3b=\left\{\begin{array}{c}5,8,11,14\\ 7,10,13,16\\ 9,12,15,18\\ 11,14,17,20\end{array}\right\}$ and $4c+5d=\left\{\begin{array}{c}9,14,19,24\\ 13,18,23,28\\ 17,22,27,32\\ 21,26,31,36\end{array}\right\}$

Hence, possible values for which $2a+3b=4c+5d$

are $9,13,14,14,17,18$.

So, total number of pairs are $6$.