SOLUTION: the ages of a father and son add up to 55 and the digits are reverse of each other. the father is older than 11 when the son is born and the son is currently older than 5. How old

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.

(10t+u)+(10u+t) = 55
    10t+u+10u+t = 55
        11t+11u = 55
Divide through by 11
            t+u = 5

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  ≧ 2

From above,

   t+u = 5

Solve for t
     t = 5-u

Substitute in 

  t-u ≧ 2
5-u-u ≧ 2
 5-2u ≧ 2
  -2u ≧ -3
Divide both sides by -2, which reverses the inequality
    u ≦ 1.5
    
u can  DISABLED_event_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

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