Question 816517
the digits are reverse of each other.
<pre>
Let the father's age = 10t+u
Let the son's age = 10u+t

Neither t nor u can be 0, since the first digit of neither age can be 0.
</pre>
the ages of a father and son add up to 55
<pre>
(10t+u)+(10u+t) = 55
    10t+u+10u+t = 55
        11t+11u = 55
Divide through by 11
            t+u = 5
</pre>
the father is older than 11 when the son is born 
<pre>
So the father is more than 11 years older than his son
(I would certainly hope so!  There aren't many 11 year-old 
fathers.)

 10t+u > 10u+t + 11
 9t-9u > 11
9(t-u) > 11
  t-u  > 11/9
  t-u  > 1&2/9
  t-u  &#8807; 2

From above,

   t+u = 5

Solve for t
     t = 5-u

Substitute in 

  t-u &#8807; 2
5-u-u &#8807; 2
 5-2u &#8807; 2
  -2u &#8807; -3
Divide both sides by -2, which reverses the inequality
    u &#8806; 1.5
    
u can only = 1, since it can't be 0. 

   t+1 = 5
     t = 4

So the father is 10t+u = 10(4)+1 = 40+1 = 41
And the son is 10u+t = 10(1)+4 = 10+4 = 14
</pre>  
the son is currently older than 5. 
<pre>
We knew the son's age was a two-digit number; otherwise
his father's age couldn't be the reverse of the digits 
of his age. So we did not need this information.

Edwin</pre>