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

Let the work efficiency of $\text{B}$ be $1\text{ unit/day}$. Since $\text{A}$ is twice as efficient as $\text{B}$, $\text{A}$ 's efficiency is $2\text{ units/day}$.
Their combined daily output efficiency is:

$$\text{Combined Efficiency}=1+2=3\text{ units/day}$$

They complete the work project together in $28\text{ days}$. Therefore, total quantity of work is:

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

The time required for $\text{A}$ to complete this total work completely alone is:

$$\text{Time taken by A}=\frac{\text{Total Work}}{\text{A’s efficiency}}=\frac{84\text{ units}}{2\text{ units/day}}=42\text{ days}$$