J P Mohindru and Bharat Mohindru Solutions for Chapter: Pair of Linear Equations in Two Variables, Exercise 6: EXERCISE

A father is three times as old as his son. In $12$ years time, he will be twice as old as his son. Find the present ages of father and the son.

Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Two years ago, a man was five times as old as his son. Two years later, his age will be $8$ more than three times the age of his son. Find their present ages.

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

The present age of a woman is $3$ years more than three times the age of her daughter. Three years later, the woman's age will be 10 years more then twice the age of her daughter. Find their present ages.
 

$\mathrm{A}$ is elder to $\mathrm{B}$ by $2$ years. $\mathrm{A}\text{'}\mathrm{s}$ father $\mathrm{F}$ is twice as old as $\mathrm{A}$ and $\mathrm{B}$ is twice as old as his sister $\mathrm{S}$. If the ages of the father and sister differ by $40$ years, find the age of $\mathrm{A}$.
 

Father's age is three times the sum of ages of his two children. After $5$ years his age will be twice the sum of ages of two children. Find the age of the father. 
 

If twice the son's age in years is added to the father's age, the sum is $70.$ But, if twice the father's age is added to the son's age, the sum is $95.$ Find the ages of father and son.