Let ten digits be x and one digit be y.

10x+y- 10y-x=27

9x-9y=27

9*(x-y)=27

x-y=3

Substitute x=2y into x-y=3.

2y-y=3

y=3

x=6

So this two-digit number is 63.

I don't think it's interesting to solve it by equations. I'll teach you another method of logical reasoning:

Because ten digits are twice as many as single digits, then ten digits must be even.

So this number must be even after the exchange.

And the difference between the two numbers is that 27 is singular (odd), so it can be determined that the original single digit must be singular (odd).

Then there are only two numbers, 2 1 and 63.

And 2 1 itself is less than 27, so it can only be 63.