$A$ and $B$ can do a piece of work in $30$ days while $B$ and $C$ can do the same work in $24$ days, and $C$ and $A$ can do it in $20$ days. They all work together for $10$ days before $B$ and $C$ leave. How many more days will $A$ take to finish the work?

The correct option is B $18$ days
$(A+B{)}^{\prime }s$ one day work $=\frac{1}{30}$
$(B+C{)}^{\prime }s$ one day work $=\frac{1}{24}$
$(C+A{)}^{\prime }s$ one day work $=\frac{1}{20}$
$(A+B)+(B+C)+(C+A)=2(A+B+C)$
$\Rightarrow 2(A+B+C{)}^{\prime }s$ one day work $=\frac{1}{30}+\frac{1}{24}+\frac{1}{20}=\frac{1}{8}$
$\Rightarrow (A+B+C{)}^{\prime }s$ one day work $=\frac{1}{16}$
Work done by $A,B$ and $C$ in $10$ days $=\frac{10}{16}=\frac{5}{8}$
Remaining work $=1-\frac{5}{8}=\frac{3}{8}$

$A^{\prime }s$ one day work $=(A+B+C{)}^{\prime }s$ one day work $-(B+C{)}^{\prime }s$ one day work $=\frac{1}{16}-\frac{1}{24}=\frac{1}{48}$
Now, $\frac{1}{48}$ work is done by $A$ in $1$ day.
Therefore $1$ work will be done by $A$ in $48$ days.
So, $\frac{3}{8}$ work will be done by $A$ in $48\times \frac{3}{8}=18$ days.