The sum of the parents' ages is twice the sum of their children’s ages. Five years ago, the sum of the parents' ages was four times the sum of their children’s ages. In fifteen years, the sum of the parents' ages will be equal to the sum of their children’s ages. How many children were in the family?
Alfred is four times as old as his nephew Franco. 5 years ago, the sum of their ages is equal to the present age of Alfred. How old is each?
If x = present age of a person
x – 3 = age of the person 3 years ago
x + 5 = age of the person 5 years from now or 5 years hence
The difference of the ages of two persons is constant at any time.
Number Problem, digit problem, money problem, lever problem, geometry related, Clock Problem, Age Problem, mixture problem, Work Problem, Motion Problem
In an organization there are CE’s, EE’s and ME’s. The sum of their ages is 2160; the average age is 36; the average age of CE’s and EE’s is 39; the average age of EE’s and ME’s is 32 and 8/11; the average age of the CE’s and ME’s is 36 and 2/3. If each CE had been 1 year older, each EE 6 years and each ME 7 years older, their average age would have been greater by 5 years. Find the number of CE, EE, and ME in the group and their average ages.