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 |