13.391 Two cars start simultaneously from points A and B and meet at 12 o'clock noon. If the speed of the first car is doubled while the speed of the second car remains the same, the meeting will occur 56 minutes earlier. If, however, the speed of the second car is doubled while the speed of the first car remains the same, they will meet 65 minutes earlier. Determine the time of the meeting in the case where the speed of both was doubled.
Solution. Let $x$ and $y$ be the initial speeds (km/min) of the cars, and $2z$ be the time (in minutes) after which they met after leaving from A and B. Then $2z(x+y)$ is the distance between A and B, hence

\[
\left\{
\begin{array}{l}
2z(x+y) = (2z-56)(2x+y) \\
2z(x+y) = (2z-65)(x+2y)
\end{array}
\right.
\]
\[
\left\{
\begin{array}{l}
z x = 28(2x + y) \\
2 z y = 65(x + 2y)
\end{array}
\right.
\]

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-340.jpg?height=157&width=665&top_left_y=322&top_left_x=100)

By the problem's context, $z$ is the time it would take for the cars to meet if their speed were twice as fast. Hence, $z = 91$ minutes.

Answer: The meeting time is 10 hours and 29 minutes.