## Task 1 - 200621

An airplane flying from A to B at a constant speed was still $2100 \mathrm{~km}$ away from B at 10:05 AM, and only $600 \mathrm{~km}$ away at 11:20 AM.

At what time will it arrive in B if it continues to fly at the same speed?
The time from 10:05 AM to 11:20 AM is $1 \mathrm{~h} 15 \mathrm{~min}=75 \mathrm{~min}$.

Due to $2100-600=1500$, the airplane has covered $1500 \mathrm{~km}$ during this time. Since $75: 15=5$, it takes $5 \mathrm{~min}$ for every $100 \mathrm{~km}$. Since $6 \cdot 5=30$, it will take 30 min for the remaining $600 \mathrm{~km}$. Therefore, it will arrive in B 30 min after 11:20 AM, i.e., at 11:50 AM.