$\mathrm{Az}$

$$
x^{2}-5 x+q=0 \quad \text { and } \quad x^{2}-7 x+2 q=0
$$

in these equations, determine $q$ so that one of the roots of the second equation is twice as large as one of the roots of the first equation.
If the roots of the first equation are $x_{1}$ and $x_{2}$, and the roots of the second equation are $x_{3}$ and $x_{4}$, then

$$
\begin{gathered}
x_{1} x_{2}=q, \quad x_{3} x_{4}=2 q \\
x_{1}+x_{2}=5, \quad x_{3}+x_{4}=7
\end{gathered}
$$

Let

$$
x_{3}=2 x_{1}
$$

then

$$
2 x_{1} x_{4}=2 q=2 x_{1} x_{2}
$$

which means

$$
x_{2}=x_{4}
$$

so

$$
x_{1}+x_{2}=5
$$

and

$$
2 x_{1}+x_{2}=7
$$

thus

$$
x_{1}=2, x_{2}=3 \text { and } q=6
$$

(Ernó Fried, Budapest.)

The problem was also solved by: Bauer E., Bendl K., Csada I., Ehrenfeld N., Erdős V., Felhőssy J., Füstös P., Kirchknopf E., Kiss J., Kürth R., Lusztig M., Neumann F., Paunz A., Rosenthal M., Sárközy P., Spitzer L. Steiner L., Szóke D., Tóth B., Wáhl V., Wellis D.