A man and a boy can do a piece of work in $36$ days. If the man works alone for the last $10$ days, the work is completed in $40$ days. How long would the boy take to do it alone?

Let the one-day work of the man be $M$ and the boy be $B$.
Together, they can complete the work in $36$ days, so:

$$\text{Total Work}=36(M+B)$$

In the second scenario, the total time taken is $40$ days, and the man works alone for the last $10$ days. This implies that they worked together for the first $40-10=30$ days.

$$\text{Work Done}=30(M+B)+10M$$

Equating the total work expressions:

$$36(M+B)=30(M+B)+10M$$

$$36M+36B=30M+30B+10M$$

$$36M+36B=40M+30B$$

$$6B=4M⟹3B=2M⟹M=\frac{3}{2}B$$

Substitute $M$ in terms of $B$ into the total work formula:

$$\text{Total Work}=36\left(\frac{3}{2}B+B\right)=36\times \frac{5}{2}B=90B$$

Since the boy does $B$ work per day, the total time taken by the boy alone is:

$$\text{Time}=\frac{90B}{B}=90\text{ days}$$