A takes $50\%$ more time to complete a work than B. If together they take $27\text{ days}$ to complete the work, how much time shall B take to do it alone?

Let the time taken by B to complete the work be $2t$ days.
Then, time taken by A = $2t+50\%\text{ of }2t=3t$ days.

Ratio of time taken (A : B) = $3:2$.
Since efficiency is inversely proportional to time, the ratio of efficiencies (A : B) = $2:3$.

Let efficiency of A = $2$ units/day and efficiency of B = $3$ units/day.
Total work done together in $27\text{ days}$:

$$\text{Total Work}=(2+3)\times 27=5\times 27=135\text{ units}$$

Time taken by B alone to complete the work:

$$\text{Time}=\frac{\text{Total Work}}{\text{Efficiency of B}}=\frac{135}{3}=45\text{ days}$$