Let digits of the number are $a$ and $b$ at tenth and ones place respectively.
Such that required number is $10a +b$
Given that $a+b = 9 \qquad \qquad \rightarrow(i)$
and if we replace the digits then the new number is 9 less than the original number,
$\begin{align} 10b + a +9 &= 10a+b \\ \implies 9a – 9b &= 9 \\ \implies ~~ a- b &=1 \qquad \qquad \rightarrow(ii)\end{align} $
$equation~(i)+(ii)~gives, \\ \implies 2a = 10 \implies a=5 \\ \> \\ equation~(i)-(ii)~gives, \\ \implies 2b = 8 \implies b=4 $
Hence, required number is $10\times 5 + 4 = 54$
Option (C) is correct.
