Question

Six years ago a father was seven times as old as his son.After six years he will be three times as old as his son.Determine their present ages.

Answer 1

Let the present age of the son$=x$ years

Six years ago, the age of the son$=x-6$ years

Six years ago, the age of the father$=7(x-6)$ years

Present age of the father$=7(x-6)+6$ years

Six years after, the age of son$=x+6$ years

Six years after,the age of father$=[7(x-6)+6]+6$ years

Given: Six years after, the father will be three times as old as his son.

$[7(x-6)+6]+6=3(x+6)$

$\Rightarrow 7x-42+12=3x+18$

$\Rightarrow 7x-3x=18+30$

$\Rightarrow 4x=48$ or $x=\frac{48}{4}=12$

Present age of the son$=x=12$years

Present age of the father$=7(x-6)+6=7(12-6)+6=7\times 6+6=42+6=48$ years