Let the numbers be $x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$. If we add all the two-term sums, we get

$$
4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=90
$$

so the sum of the numbers is

$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=\frac{45}{2}
$$

Due to the ordering of the numbers, $x_{1}+x_{2}=6$ is the smallest sum, and

$$
x_{4}+x_{5}=12
$$

is the largest. Comparing with (1), we get

$$
x_{3}=\frac{45}{2}-\left(x_{1}+x_{2}\right)-\left(x_{4}+x_{5}\right)=\frac{9}{2}
$$

Since $x_{3}+x_{4} \leq x_{3}+x_{5} \leq x_{4}+x_{5}$, the second largest sum is $x_{3}+x_{5}=11$, so, since $x_{3}=\frac{9}{2}$, we have $x_{5}=\frac{13}{2}$. From (2), $x_{4}=12-\frac{13}{2}=\frac{11}{2}$.

Finally, $x_{1}+x_{2}=6$ and $x_{1}+x_{3}=7$ is the next smallest sum, so

$$
x_{1}=7-x_{3}=7-\frac{9}{2}=\frac{5}{2} \quad \text { and } \quad x_{2}=6-\frac{5}{2}=\frac{7}{2}
$$

The numbers, in order, are thus: $\frac{5}{2}, \frac{7}{2}, \frac{9}{2}, \frac{11}{2}$ and $\frac{13}{2}$.

It is easy to verify that all the listed two-term sums occur, since

\[

\]

Blaskó Ján (Budapest, Slovak Gymn., 10th grade)
Arnold thought of 5 numbers, and he told Bendeguz all the sums that can be obtained by adding two of these numbers. These are: $6,7,8,8,9,9,10,10,11,12$. What numbers did Arnold think of?