Question 480647
Lets say the 2 digit number is xy
That is to say the "tens" digit is x and the "units" digit is y   (ie. by xy we don't mean x times y)

So the value of the original number is 10x + y (because the x is in the "tens" column, just like 35 is 3 times 10 plus 5 for example)

If you reverse the digits then the value of the new number is 10y + x

.. and we know that this number is 27 more than the original .. so ...

(10y + x) = (10x + y) + 27

Take an x away from each side

10y = 9x + y + 27

Take a y away from each side

9y = 9x + 27

.. and we can divide both sides by 9 to make it a bit simpler

y = x + 3

So we are looking for a 2 digit number in which one digit is 3 more than the other.

We are also told that the sum of the digits is 13

x + y = 13

Since we know that y = x + 3 (from above) we can substitute for y

x + (x + 3) = 13

So x = 5 which means that y = 8

Therefore the original number was 58, which when reversed becomes 85