Question

Five years hence, father's age will be three times the age of his son. Five years ago, father was seven times as old as his son. Find their present ages

• A. Father's age $=80$ years, Son's age $=25$ years
• B. Father's age $=60$ years, Son's age $=30$ years
• C. Father's age $=75$ years, Son's age $=42$ years
• D. Father's age $=40$ years, Son's age $=10$ years

Answer 1

Answer: (D)

Father's age $=40$ years, Son's age $=10$ years

Let the father present age $=x$ years and son present age $=y$ years

According to the first condition
$5$ years later father will be $(x+5)$ years old and son will be $(y+5)$ years old.
$\Rightarrow x+5=3(y+5)$
$\Rightarrow x+5=3y+15\Rightarrow x=3y+\mathrm{10....}eq1$

According to the second condition
$5$ years ago father was $(x-5)$ years old and son was $(y-5)$ years old.
$\Rightarrow x-5=7(y-5)$
$\Rightarrow x-5=7y-35$
$\Rightarrow x-7y=-\mathrm{30....}eq2$

Put the value of $x$ from $eq1$
$\Rightarrow 3y+10-7y=-30\Rightarrow -4y=-40\Rightarrow y=10$

Put $y=10$ in $eq1$
$\Rightarrow x=3\times 10+10=40$

Hence, Father present age $=40$ years
Son present age $=10$ years