7. For the first $120 \mathrm{~km}$ of a certain distance, a car travels at a speed of $90 \mathrm{~km} / \mathrm{h}$. Then the speed decreases to $64 \mathrm{~km} / \mathrm{h}$ and the car travels at this speed for 1 hour and 15 minutes. At what speed must the car travel the remaining one-sixth of the distance so that the average speed for the entire journey is $80 \mathrm{~km} / \mathrm{h}$?

No use of a pocket calculator or any manuals is allowed.
First method: In the first stage, the car drove $120 \mathrm{~km}$ at a speed of $90 \mathrm{~km} / \mathrm{h}$.

This took $120: 90=\frac{4}{3}$ hours.

1 POINT

In the second stage, the car drove for 1 hour and 15 minutes $=\frac{5}{4}$ hours at a speed of $64 \mathrm{~km} / \mathrm{h}$.

It covered $\frac{5}{4} \cdot 64=80 \mathrm{~km}$.

1 POINT

In both stages combined, it covered $200 \mathrm{~km}$, which is $\frac{5}{6}$ of the entire journey.

The entire journey is 200: $\frac{5}{6}=240 \mathrm{~km}$,

The average speed will be $80 \mathrm{~km} / \mathrm{h}$ if the total travel time is $240: 80=3$ hours. 2 POINTS

For the first two stages, it took $\frac{4}{3}+\frac{5}{4}=\frac{31}{12}=2 \frac{7}{12}$ hours.

1 POINT

The third stage needs to be completed in $3-2 \frac{7}{12}=\frac{5}{12}$ hours.

1 POINT

The speed in this stage needs to be $40: \frac{5}{12}=96 \mathrm{~km} / \mathrm{h}$.

2 POINTS

TOTAL 10 POINTS