24. Let $t_{1}$ be the time that has passed since the whales parted until the moment the faster whale turns around, and $t_{2}$ be the time that has passed from the moment of the turn until the whales meet again (if we take one hour as the unit of time, then $\left.t_{1}+t_{2}=3 / 4\right)$.

The distance between the place where the whales parted and the place where they meet again is

$$
6\left(t_{1}+t_{2}\right)=10 t_{1}-10 t_{2}
$$

Thus,

$$
4 t_{1}+4 t_{2}=3, \quad 4 t_{1}-16 t_{2}=0
$$

Therefore, $20 t_{2}=3$ and $t_{2}=3 / 20=9$ min.

Hence, the whale turned around at 9:51.
## 24. Two Whales

Two whales are swimming in Antarctic waters in a straight line at a speed of 6 km/h. Suddenly, one of them, without changing direction, started swimming faster, at a speed of 10 km/h. Then he suddenly turned around and swam back towards the second whale, which did not change its speed or direction. Suppose the whales parted at a quarter past nine and met again at 10 o'clock.

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-050.jpg?height=631&width=1338&top_left_y=427&top_left_x=376)

At what time did the faster whale turn around and head back in the opposite direction?