Question

Father and son together complete a task in $15$ days. If one days work of father is $3$ times one days work of this son. Find the number of days required by each alone to complete the task.

Answer 1

Let $F$ and $S$ be the number of days required by father and son repectively.

$\frac{1}{F}+\frac{1}{S}=\frac{1}{15}.....\left(1\right)\left[\text{Given}\right]$

Also,

$\frac{1}{F}=\frac{3}{S}.....\left(2\right)$

For equation $\left(1\right)\&\left(2\right)$, we have

$\frac{3}{S}+\frac{1}{S}=\frac{1}{15}$

$\Rightarrow S=4\times 15=60$

Substituting the value of $S$ in equation $\left(2\right)$, we have

$\frac{1}{F}=\frac{3}{60}$

$\Rightarrow F=20$

Hence the number of days required by father and son each alone is $20$ and $60$ respectively.