Question

Ten years ago a father was six times as old as his daughter. After $10$ years, he will be twice as old as his daughter. Determine the sum of their present ages.

Answer 1

Let the present age of father and his daughter are x and y respectively.

Given that ten years ago a father was six times as old as his daughter.

$(x-10)=6(y-10)$

$\Rightarrow x-10=6y-60$

$\Rightarrow x-6y+50=0⟶\left(1\right)$

Also given that after 10 years, he will be twice as old as his daughter.

$(x+10)=2(y+10)$

$\Rightarrow x+10=2y+20$

$\Rightarrow x-2y-10=0⟶\left(2\right)$

Subtracting ${eq}^n\left(1\right)$ from $\left(2\right)$, we have

$x-2y-10-(x-6y+50)=0$

$\Rightarrow x-2y-10-x+6y-50=0$

$\Rightarrow 4y-60=0$

$\Rightarrow y=15$

Substituing the value of y in ${eq}^n\left(2\right)$, we have

$x-2\times 15-10=0$

$\Rightarrow x-30-10=0$

$\Rightarrow x=40$

Hence, the present ages of father and his daughter is 40 and 15 respectively.

$\therefore$ sum of their present ages $=x+y=40+15=55$

Hence, the correct answer is 55.