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Let's assume the unit digit of the number is $y$ & $x$ is the digit of ten's place.

∴ The number is $10x+y$

Now, given $x+y =12$

if the number is $yx$, then $\{(10y+x)-(10x+y)\}=54$

Now, we know, $x+y =12$

∴ $y=12-x$

Now, $\{(10y+x)-(10x+y)\}=54$

$10y+x-10x-y=54$

$10(12-x)+x-10x-(12-x)=54$

Or,$120-10x+x-10x-12+x=54$

Or, $120-20x+2x-12=54$

Or,$108-18x=54$

Or,$108-54=18x$

Or,$18x=54$

Or,$x=3$

∴$ y = 12-x = 12-3 = 9$

The original number is $39$ which is option A)
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