73(1007). A master and his apprentice were supposed to complete a job by a certain deadline. However, when half of the work was done, the apprentice fell ill, and the master, left alone, finished the job 2 days late. How many days would it take for each of them to complete the entire job working alone, if the master would need 5 fewer days than the apprentice?
Solution. Let the master be able to complete the entire job in $x$ days, then the apprentice - in $(x+5)$ days. In one day, the master completed $\frac{1}{x}$ of the work, the apprentice completed $\frac{1}{x+5}$ of the work; when working together, the master and the apprentice completed $\frac{2 x+5}{x(x+5)}$ of the work per day $\left(\frac{1}{x}+\frac{1}{x+5}=\frac{2 x+5}{x(x+5)}\right)$. For half the work, the master and the apprentice spent $\frac{x(x+5)}{2(2 x+5)}$ days $\left(\frac{1}{2}: \frac{2 x+5}{x(x+5)}\right)$, which is 2 days less than the time spent by the master alone on half the work.

Considering that the master, working alone, spent $\frac{x}{2}$ days on half the work, we set up the equation

$$
\frac{x}{2}-\frac{x(x+5)}{2(2 x+5)}=2
$$

After simplifications, we get:

$$
\begin{gathered}
x^{2}-8 x-20=0 \\
x_{1}=10, x_{2}=-2
\end{gathered}
$$

Only the positive root satisfies the condition of the problem: the master could complete the work in 10 days; therefore, the apprentice in 15 days.

Answer: 10 days, 15 days.