Mathematics - Sets and Relations Question with Solution | TestHub

Given,

Set $A=\left\{1,2,3,4\right\}$

And relation on $A\times A$ as $2a+3b=4c+5d$

Now, Maximum value of $2a+3b=20$ at $\left(4, 4\right)$

And Minimum value of $4c+5d=9$ at $\left(1, 1\right)$

So, $4c+5d$ can be equal to $9, 13, 14, 17, 18, 19$

Now $2a+3b$ can be $9$ if  $\left(a, b\right)=\left(3, 1\right)\Rightarrow \left(c, d\right)=\left(1, 1\right)$

Similarly, $2a+3b$ can be $13$, if $\left(a, b\right)=\left(2, 3\right)\Rightarrow \left(c, d\right)=\left(2, 1\right)$

And $2a+3b$ can be $14$, if $\left(a, b\right)=\left(4, 2 \right) \text{or }\left(1, 4\right)$

$\Rightarrow \left(c, d\right)=\left(1, 2\right)$

Similarly, $2a+3b$ can be $17$, if $\left(a, b\right)=\left(4, 3\right)\Rightarrow$ $\left(c, d\right)=\left(3, 1\right)$

And $2a+3b$ can be $18$ if $\left(a, b\right)=\left(3, 4\right)\Rightarrow$

$\left(c, d\right)=\left(2, 2\right)$

Hence, there are total $6$ elements which satisfy the given relation.