A number consists of two digits whose sum is 9.

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

$\because$ Sum of two digits $=9$

$\therefore$ Tens digit $=(9-\mathrm{x})$

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

Reversed number $=10 \mathrm{x}+(9-\mathrm{x})$

According to the question,

$10 \times(9-\mathrm{x})+\mathrm{x}-27=10 \mathrm{x}+(9-\mathrm{x})$

or $90-10 \mathrm{x}+\mathrm{x}-27=10 \mathrm{x}+9-\mathrm{x}$

or $9 \mathrm{x}+9 \mathrm{x}=90-27-9$

or $18 \mathrm{x}=54$

or $\mathrm{x}=\frac{54}{18}=3$

$\therefore$ The n umber $=10 \times(9-3)+3=63$