A is twice as good a workman as B. A and B together complete a piece of work in $28$ days. In how many days will A alone do the same piece of work?

Let the work efficiency of B be $1\text{unit/day}$.
Since A is twice as good as B, the efficiency of A $=2\text{units/day}$.

Combined efficiency of A and B $=1+2=3\text{units/day}$.
They complete the work together in $28$ days.

$$\text{Total Work}=\text{Combined Efficiency}\times \text{Number of Days}$$

$$\text{Total Work}=3\times 28=84\text{units}$$

Time taken by A alone to complete the work is:

$$\text{Days}=\frac{\text{Total Work}}{\text{Efficiency of A}}=\frac{84}{2}=42\text{ days}$$