22(1150). How many hours can each of the three workers complete the work if the productivity of the third worker is equal to half the sum of the productivities of the first and second workers? It is known that if the third worker worked for 48 hours, the first would need 10 hours to finish the work, and the second would need 15 hours.
Solution. Let the entire work be denoted by $A$, the labor productivity of the first worker by $x_{1}$, and the second worker by $x_{2}$. Then, according to the condition, the productivity of the third worker is $\frac{x_{1}+x_{2}}{2}$ and $10 x_{1}=$ $=15 x_{2}$, i.e., $x_{1}=\frac{3}{2} x_{2}, x_{2}=\frac{2}{3} x_{1}$.

The first worker can complete the entire work in $\frac{A}{x_{1}}$ hours, or in

$$
\frac{48 \cdot \frac{x_{1}+x_{2}}{2}+10 x_{1}}{x_{1}}=\frac{24\left(x_{1}+\frac{2}{3} x_{1}\right)+10 x_{1}}{x_{1}}=\frac{50 x_{1}}{x_{1}}=50(\text{h})
$$

The second worker can complete the entire work in $\frac{A}{x_{2}}$ hours, or in

$$
\frac{48 \cdot \frac{x_{1}+x_{2}}{2}+15 x_{2}}{x_{2}}=\frac{24\left(\frac{3}{2} x_{2}+x_{2}\right)+15 x_{2}}{x_{2}}=\frac{75 x_{2}}{x_{2}}=75
$$

The third worker can complete the entire work in

$$
\frac{A}{x_{3}}=\frac{A}{\frac{x_{1}+x_{2}}{2}}=\frac{A}{\frac{1}{2}\left(\frac{A}{50}+\frac{A}{75}\right)}=60(\text{h})
$$

Answer: 50 h, 75 h, 60 h.