Question

# Ten years ago, father was the twelve times as old as his son and ten years later will be twice as old as his son will be. Find their present ages.

Solution

## Let the present ages of father and son be x years and y years respectively.Ten years ago, Father's age $$=(x-10)$$ yearsSon's age $$=(y-10)$$ years$$\therefore x-10=12(y-10)\Rightarrow x-12y+110=0$$ ...(i)Ten years later, Father's age $$=(x+10)$$ yearsSon's age $$=(y+10)$$$$\therefore x+10=2(y+10)\Rightarrow x-2y-10=0$$ ...(ii)Subtracting (ii) from (i), we get$$-10y+120=0\Rightarrow 10y=120\Rightarrow y=12$$Putting $$y=12$$ in (i), we get$$x-144+110=0\Rightarrow x=34$$Thus, the present age of the father is $$34$$ years and the present age of the son is $$12$$ years.Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More