2. "Miki and I," said Philip, "can finish the assigned work in 20 days, but if I work with Ivan, we would finish the same work 5 days earlier." "I have a better combination," said Ivan, "If I worked with Miki, we would finish the work a fifth of the time earlier than when working with Philip."

How many days would each of them finish the work alone, and how many days would they finish it together?
Solution. Let's denote by $x, y, z$ the number of days it would take for Miki, Philip, Ivan to complete the work individually. Then $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ is the part of the work that Miki, Philip, and Ivan would complete in one day, respectively. Therefore, due to the conditions of the problem, we have:

$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=\frac{1}{20} \\
\frac{1}{y}+\frac{1}{z}=\frac{1}{15} \\
\frac{1}{x}+\frac{1}{z}=\frac{1}{12}
\end{array}\right.
$$

By adding these three equations, after simplification, we get:

$$
2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{20}+\frac{1}{15}+\frac{1}{12} \quad \Rightarrow \quad \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{10}
$$

This means that working together, they would complete one-tenth of the work, i.e., the entire work together would be completed in 10 days. Now:

$$
\begin{aligned}
& \frac{1}{x}=\frac{1}{10}-\left(\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{10}-\frac{1}{15}=\frac{1}{30} \Rightarrow x=30 \\
& \frac{1}{y}=\frac{1}{10}-\left(\frac{1}{x}+\frac{1}{z}\right)=\frac{1}{10}-\frac{1}{12}=\frac{1}{60} \Rightarrow y=60 \\
& \frac{1}{z}=\frac{1}{10}-\left(\frac{1}{y}+\frac{1}{x}\right)=\frac{1}{10}-\frac{1}{20}=\frac{1}{20} \Rightarrow z=20
\end{aligned}
$$

| 5,6,7,8,9 |  | $21,22,23,24,25$ |  |  |
| :---: | :---: | :---: | :---: | :---: |
| $2 \rightarrow$ | 10 | 20 | 26 | 36 |
| $3 \rightarrow$ | 11 | 19 | 27 | 35 |
| $4 \rightarrow \quad 2$ | 12 | 18 | 28 | 34 |
| $\rightarrow 1$ |  | 17 |  |  |

Thus, Miki would complete the work alone in 30 days, Philip in 60 days, and Ivan in 20 days.