The sum of the digits of a two-digit number is 12.

Solution:

Let the digit in the units place be $\mathrm{x}$.

Digit in the tens place $=(12-\mathrm{x})$

$\therefore$ Original number $=10(12-\mathrm{x})+\mathrm{x}=120-9 \mathrm{x}$

On reversing the digits, we have $\mathrm{x}$ at the tens place and $(12-\mathrm{x})$ at the units place.

$\therefore$ New number $=10 \mathrm{x}+12-\mathrm{x}=9 \mathrm{x}+12$

New number $-$ Original number $=54$

$\Rightarrow 9 x+12-(120-9 x)=54$

$\Rightarrow 9 x+12-120+9 x=54$

$\Rightarrow 18 x-108=54$

$\Rightarrow 18 x=54+108$

$\Rightarrow 18 x=162$

$\Rightarrow x=\frac{162}{18}=9$

Therefore, the digit in the units place is $9 .$

$D$ igit in tens place $=(12-\mathrm{x})=(12-9)=3$

Therefore, the original number is $39 .$

Check :

The original number is $39 .$

$S$ um of the digits in the original number $=(3+9)=12$

$N$ ew number obtained on reversing the digit $s=93$

New number - Original number = ( $93-39)=54$

Thus, both the given conditions are satisfied by $39 .$

Hence, the original number $i s 39 .$