Question

$A$ can do a piece of work in $40$ days. He works at it for $8$ days and then $B$ finishes the remaining work in $16$ days. How long will they take to complete the work if they do it together?

Answer 1

Given,

$A$ can do a piece of work in $=40$ days

So, $A$’s $1$ day's work $=\frac{1}{40}$

So, $A$’s $8$ day's work $=\frac{8}{40}=\frac{1}{5}$

$\therefore$ Remaining work $=1-\frac{1}{5}=\frac{4}{5}$

Now, $B$ can do $\frac{4}{5}$ piece of work in $=16$ days

So, $B$’s $1$ day's work $=\frac{\left(4/5\right)}{16}=\frac{4}{5\times 16}=\frac{1}{20}$

Therefore, $(A+B)$’s $1$ day's work $=\frac{1}{40}+\frac{1}{20}$

$=\frac{1+2}{40}=\frac{3}{40}$

$\therefore$ Both $(A+B)$ can do a piece of work in $=\frac{1}{\left(3/40\right)}$ days

$=\frac{40}{3}$ days

$=13\frac{1}{3}$ days