John is now four times as old as his daughter and six times as old as his son.Twelve years from now,the sum of his daughter and son will differ from his age by 9 years.Determine their present ages

Question

Linear Equations

Answer 1

Let J=John's current age, and D and S be his children's ages.

J=4D=6S, D=J/4, S=J/6.

Their ages in 12 years' time will be J+12, D+12, S+12. The sum of the children's ages wil be D+S+24.

So |D+S+24-(J+12)|=9, |J/4+J/6-J+12|=9,

|12-7J/12|=9⇒12-7J/12=9 or 7J/12-12=9, and we know J>0 and must be an integer.

7J/12=3, 7J=36, J=36/7; or 7J/12=21, J=36, which seems to be the right answer.

John is 36, son is 6 and daughter is 9.

In 12 years' time, John will be 48, his son will be 18 and his daughter will be 21. The children's combined ages will be 39 so John will be 9 years older than their combined ages.