Question 637479
Jack is married to Jill.  Their son, junior, asked each of them to reveal their ages.  Junior's parents decided to tell him but in the form of a puzzle.  Jack told Junior, "If you reverse the digits in my age, you'll get your mother's age."
<pre>
Let a = the tens digit of Jack's age
Let b = the ones digit of Jack's age
Then Jack's age = 10a+b
Jill's age = 10b+a 
</pre>
Jill told her son, "the sum of my age and your dad's age is equal to 11 times the difference in our ages...said Jack "Remember that I am older than your mother." 
<pre>
(10b+a) + (10a+b) = 11[(10a+b)-(10b+a)]

10b + a + 10a + b = 11[10a+b-10b-a]

        11b + 11a = 11(9a-9b)

        11b + 11a = 99a - 99b

             110b = 88a

Divide both sides by 22

               5b = 4a

Divide both sides by a

             {{{(5b)/a}}} = {{{(4a)/a}}}

             {{{(5b)/a}}} = {{{4}}}

Divide both sides by 5

             {{{b/a}}} = {{{4/5}}}

The only possibility for a and b both being digits is

a = 5 and b = 4

So Jack is 54 and Jill is 45.

Checking: The sum of their ages is 99 and the
difference of their ages is 9, and 99 is 11 times 9.


Edwin</pre>