## Task 4 - 220834

A helicopter took off at 4:30 AM from a city $A$ and flew with a speed of $250 \frac{\mathrm{km}}{\mathrm{h}}$ to a city $B$. It stayed there for 30 minutes and then flew back to $A$ on the same route with a speed of $200 \frac{\mathrm{km}}{\mathrm{h}}$, arriving at $A$ on the same day at 11:45 AM.

Determine the length of the route from $A$ to $B$!
The sought length is $x$ kilometers, the times for the outbound and return flights are $t_{1}$ hours and $t_{2}$ hours, respectively. Thus,

$$
x=250 t_{1} \quad \text{and} \quad x=200 t_{2},
$$

so

$$
t_{1}=\frac{x}{250} \quad ; \quad t_{2}=\frac{x}{200}
$$

From the start in A until the arrival back in A, 7 $\frac{1}{4}$ hours have passed; after subtracting the waiting time, the remaining flight time is

$$
\left(t_{1}+t_{2}\right) \text { hours }=6 \frac{3}{4} \text { hours }
$$

From (1) and (2) it follows that

$$
\frac{x}{250}+\frac{x}{200}=\frac{27}{4} \quad \Rightarrow \quad x=750
$$

Thus, the sought length is $750 \mathrm{~km}$.