The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your 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 $i s 39 .$
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