## Task 1 - 280821

A cargo ship needs exactly 12 days for a shipping route from port $A$ to port $B$. A tanker travels this route in the opposite direction and takes 15 days for it.

The cargo ship departs from port $A$ 6 days later than the tanker from port $B$.

a) How many days after the departure of the cargo ship do the two ships meet, if they travel at a constant speed?

b) What part of the route has each ship covered by then?
a) The tanker covers $\frac{1}{15}$ of the route each day. On the day the freighter sets sail, the tanker has already covered $\frac{6}{15}=\frac{2}{5}$ of the distance, so the distance between the two ships at this point is $\frac{3}{5}$ of the route.

Since $\frac{1}{12}+\frac{1}{15}=\frac{3}{20}$, the two ships approach each other by $\frac{3}{20}$ of the route each day. As $\frac{3}{5} \div \frac{3}{20} = 4$, the two ships will meet 4 days after the freighter sets sail.

b) The freighter covers $\frac{4}{12}=\frac{1}{3}$ of the route in 4 days, and the tanker, due to $1-\frac{1}{3}=\frac{2}{3}$ (or because of its travel time of 10 days and $\frac{10}{15}=\frac{2}{3}$), covers the remaining $\frac{2}{3}$ of the route.