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