Set up a $4\times 4$ table of the possible distances that may have been covered:
$$\qquad\ \ \ 10\qquad \ \ \ 15 \qquad\ \ 20 \qquad\ \ \ 25$$
$$30 \qquad 300\qquad 450 \qquad\ 600 \qquad\ 750$$
$$40 \qquad 400\qquad 600 \qquad\ 800 \qquad 1000$$
$$50 \qquad 500\qquad 750 \qquad 1000 \qquad 1250$$
$$60 \qquad 600\qquad 900 \qquad 1200 \qquad 1500$$
The upper row gives the possible speends, and the left column gives the possible times. The 16 products are the possible distances, and the 4 distances that have actually been covered must lie in different rows and in different columns. Now start to eliminate.
By (2), the longest distance is not in the right column or in the bottom row, so it can be at most $1000$. By (2) and (4) we can eliminate three entries in the last column, as well as the entry $1200=20\cdot 60$ in the bottom row. This leaves only one possible time for the person with speed $25$, namely $30$ hours with covered distance $25\cdot 30=750$.
Eliminate the other three entries in the row of time $30$, as well as the entry $15\cdot 50=750$. We are left with the table:
$$\qquad\ \ \ 10\qquad \ \ \ 15 \qquad\ \ 20 \qquad\ \ \ 25$$
$$30 \qquad - \qquad\ - \qquad\ -\ \qquad\ 750$$
$$40 \qquad 400\qquad 600 \qquad\ 800 \qquad -$$
$$50 \qquad 500\qquad -\qquad 1000 \qquad\ -$$
$$60 \qquad 600\qquad 900 \qquad\ -\ \qquad\ -$$
Now show that it is not possible to find a solution which contains the distance $20\cdot 40=800$.
It follows that the person with speed $20$ must have travelled for time $50$. The rest is now easy, and rule (1) then gives you the names to go with the distances.
