$A$ and $B$ together can do a piece of work in $25$ days. If $B$ works alone for the last $10$ days, it is completed in $30$ days. In how many days can $A$ alone do it?

Let $A$ 's daily work be $A$ and $B$ 's daily work be $B$.
Together, they complete the work in $25$ days:

$$\text{Total Work}=25(A+B)$$

In the second scenario, the work is completed in $30$ days with $B$ working alone for the last $10$ days. This means they worked together for the first $30-10=20$ days.

$$\text{Work Done}=20(A+B)+10B$$

Equating the two:

$$25(A+B)=20(A+B)+10B$$

$$25A+25B=20A+20B+10B$$

$$25A+25B=20A+30B$$

$$5A=5B⟹A=B$$

Since $A=B$, substitute $B=A$ into the total work equation:

$$\text{Total Work}=25(A+A)=50A$$

Since $A$ completes $A$ units of work per day, the time taken by $A$ alone is:

$$\text{Time}=\frac{50A}{A}=50\text{ days}$$