If twice the son's age be added to the father's age,the sum is $70 \text{ years}$ and if twice the father's age is added to the son's age,the sum is $95 \text{ years}$. Then the father's age is (in $\text{years}$):

$A$ man's age is $125 \%$ of what it was $10 \, \text{years}$ ago, but $83 \frac{1}{3} \%$ of what it will be after $10 \, \text{years}$. What is his present age? (in $\text{years}$)

The average age of women and child workers in a factory was $15 \text{ years}$. The average age of all the $16$ children was $8 \text{ years}$ and the average age of women workers was $22 \text{ years}$. If $10$ women workers were married,then the number of unmarried women workers is:

The ratio of the present ages of $A$ and $B$ is $7:9$. $6$ years ago,the ratio of $\frac{1}{3}$ of $A$'s age at that time and $\frac{1}{3}$ of $B$'s age at that time was $1:2$. What will be the ratio of $A$'s to $B$'s age $6$ years from now?

The age of a father is twice that of his elder son. Ten years hence,the age of the father will be $3$ times that of the younger son. If the difference of ages of two sons is $15 \text{ years}$,the age of the father is (in $\text{years}$)

Three times the present age of a father is equal to eight times the present age of his son. $8$ years hence,the father will be twice as old as his son at that time. What are their present ages?