The sum of digits of two digit number is $12$. When the digits are reversed, the new number is $36$ less than the original number. Find the original number.

The correct option is D ($84$)

Let $2$-digit number be $xy$
the unit place digit $=y$
Ten's place digit $=x$
As per the condition we write the equation as,
New number $=$ original number $-$ $36$
$10y+x=10x+y-36$
Grouping the like terms,
$\Rightarrow 10y-y=10x-x-36$
$\Rightarrow 9y=9x-36$
We can divide both the sides by $9$,
$\Rightarrow y=x-4$ ---(1)
We know that the sum of the digits, $x+y=12$ ---(2)
Substitute the value of $y$ in equation (2)
$\Rightarrow x+x-4=12$
$\Rightarrow 2x=16$
$\Rightarrow x=8$
Put $x=8$ in equation (1)
$\Rightarrow y=8-4$
$\Rightarrow y=4$

Hence, the required number is, $xy=84$

So, option D is correct.