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."
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
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."
(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
=
=
Divide both sides by 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