Two families lived next door to each other —the Abramowitzs and the Bernhardts. The total ages of the four members of the Abramowitz family amounted to one hundred years, and the total ages of the four members of the Bernhardt family also amounted to the same.

It was found in the case of each family that the sum obtained by adding the squares of each of the children's ages to the square of the mother's age equaled the square of the father's age. In the case of the Abramowitz family, however, Emmylou was one year older than her brother Eric, whereas Francine Bernhardt was two years older than her brother Frank.

What was the age of each of the eight individuals?

(In reply to Computer Assisted Solution (spoiler) by Charlie)

Solved independantly using a spreadsheet, with some trial and error, I arrived at the same ages.

39 + 34 + 14 + 13 = 100
39²=1521,  34²+14²+13²=1156+196+169=1521

42 + 40 + 10 + 8 = 100
42²=1764,  40²+10²+8²=1600+100+64=1764 

Mr.Abramowitz 39
Mrs.Abramowitz 34
Emmylou 14
Eric 13

Mr. Bernhardt 42
Mrs. Bernhardt 40
Francine 10
Frank 8

Edited on July 6, 2006, 4:15 pm

Posted by Dej Mar on 2006-06-21 12:10:03