A can do $\frac{3}{5}\text{th}$ of a work in $12$ days, and B can do $\frac{1}{3}\text{rd}$ of that work in $15$ days. They worked together for $12$ days and then A left the work. B alone will complete the remaining work in how many days?

First, let us find the total time taken by A and B individually to complete the entire work:Time taken by A to complete the whole work = $12\times \frac{5}{3}=20\text{ days}$ Time taken by B to complete the whole work = $15\times \frac{3}{1}=45\text{ days}$ Let the total work be the LCM of $20$ and $45$, which is $180\text{ units}$.Efficiency of A = $\frac{180}{20}=9\text{ units/day}$ Efficiency of B = $\frac{180}{45}=4\text{ units/day}$ Combined efficiency of A and B = $9+4=13\text{ units/day}$ They worked together for $12$ days:
$$\text{Work done in 12 days}=13\times 12=156\text{ units}$$
Remaining work to be completed:
$$\text{Remaining Work}=180-156=24\text{ units}$$
Now, A leaves and B completes the remaining work alone:
$$\text{Time taken by B}=\frac{\text{Remaining Work}}{\text{Efficiency of B}}=\frac{24}{4}=6\text{ days}$$