The sum of digits of two digit number is 12. If the new number formed by reversing the digits is less than the original number by 54, then the original number is 

Let unit’s digit $=x$

Ten’s digit $=y$

As per the problem

Sum of the digits $=12$

$x+y=12\cdots \cdots \left(1\right)$

Original number $=1\left(x\right)+10\left(y\right)=x+10y$

New number formed by reversing $=10x+y$

$\begin{array}{l}10x+y=x+10y-54\\ 9x-9y=-54\\ x-y=-6\cdots \cdots \left(2\right)\end{array}$

By solving  $\left(1\right)\&\left(2\right)$

$\begin{array}{l}x+y=12\\ \underset{\_}{x-y=-6}\\ 2x=6\Rightarrow x=3\\ x+y=12\\ y=12-x=12-3=9\end{array}$

Original number $=3+10\left(9\right)=3+90=93$