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 their present age.

• A. Present age of father $=10$ years
  Present age of daughter $=5$ years
• B. Present age of father $=13$ years
  Present age of daughter $=26$ years
• C. Present age of father $=20$ years
  Present age of daughter $=18$ years
• D. present age of father $=40$ years
  Present age of daughter $=15$ years

Answer 1

The correct option is D present age of father $=40$ years

Present age of daughter $=15$ years
Suppose ten years ago the age of the daughter was $x$ years.

Then, ten years ago the age of father was $6x$ years.
$\therefore$ present age of father $=(6x+10)$ years and
present age of daughter $=(x+10)$ years
After ten years, we have
Father's age $=(6x+10+10)$ years $=(6x+20)$ years
Daughter;s age $(x+10+10)$ years $=(x+20)$ years
$6x+20=2(x+20)$
$\Rightarrow 6x+20=2x+40$
$\Rightarrow 6x-2x=40-20$
$\Rightarrow 4x=20$

$\Rightarrow x=\frac{20}{4}=5$
$\therefore$ Present age of father $=(6\times 5+10)=40$ years
and present age of daughter $=(5+10)=15$ years.