June 2011
Problem of the Month

Mother & Daughter Problem
The MENSA Puzzle Calendar

The ages of a mother and a daughter are the same but with the digits reversed.
Twelve years ago, the mother was twice as old as the daughter.

How old are they now?

Solution to the Problem:

The mother is 84 and the daughter 48.

Let x = tens digit of the Mother's age.
Let y = ones digit of the Mother's age.

Then the Mother's age is given by 10x + y and the daughter's age is given by 10y + x.

Twelve years ago:
10x + y -12 = 2 (10y + x -12).

Since x can only be a single digit 1 through 9, we can substitute each value to see which will give a single digit 1 through 9 for y.

Only x = 8 gives a proper value of y = 4, so the mother is 84 and the daughter is 48.

Correctly solved by:

1. James Alarie	Flint, Michigan
2. Chad Fore	Gate City High School, Gate City, Virginia
3. John Funk	Ventura, California
4. Alexandre Roux	France
5. David & Judy Dixon	Bennettsville, South Carolina

Send any comments or questions to: David Pleacher