$A$ can do a piece of work in $14$ days which $B$ can do in $21$ days. They begin together but $3$ days before the completion of the work, $A$ leaves off. The total number of days to complete the work is:

The correct option is C. $10\frac{1}{5}$

Amount of work done by $A$ in $14$ days $=\frac{1}{14}$ units of work

Amount of work done by $B$ in $21$ days $=\frac{1}{21}$ units of work

$(A+B)$’s $1$ day’s work $=(\frac{1}{14}+\frac{1}{21})=\frac{5}{42}$ units of work

Now, $B$’s $3$ days’ work $=(\frac{1}{21}\times 3)=\frac{1}{7}$

Remaining work $=(1-\frac{1}{7})=\frac{6}{7}$

Now, $\frac{5}{42}$ work is done by $A$ and $B$ in $1$ day

∴ $\frac{6}{7}$ work is done by $A$ and $B$ in $(\frac{42}{5}\times \frac{6}{7})=\frac{36}{5}$days

Hence, total time taken to complete the work $=(3+\frac{36}{5})$ days = $10\frac{1}{5}$ days