Let the ten's-digit = x and unit's digit = Y

Therefore, The original number = 10x + y

After interchanging the digits

New number = 10y + x

Now, according to the question,

\( \Large \left(10y + x\right) - \left(10x + y\right) = 27 \)

=> 9y - 9 x = 27

=> y - x = 3 ...(i)

and y + x = 13 ...(ii)

On solving Eqs. (i) and (ii), we get

7 = 8 and x = 5

Therefore, Required number

\( \Large = 10 \times 5 + 8 = 58 \)