Two ports A and B are 300 km apart. Two ships leave A for B such that the second leaves 8 hours after the first. The ships arrive at B simultaneously. Find the time the slower ship spent on the trip if the speed of one of them is 10 km/h higher than that of the other.

A__________300 km_______B
Let speed of the 1^st ship is s kmph and speed of 2^nd ship is (s + 10) kmph.
Let 2^nd ship takes time t to cover the distance 300 km then 1^st ship takes (t + 8) hours.

As distance is constant, we have,
$$\frac{{\text{s}}}{{{\text{s}} + 10}} = \frac{{\text{t}}}{{{\text{t}} + 8}}$$
Or, st + 8s = st + 10t
Or, 8s = 10t

If the slower ship took 20 hours (option D) the faster ship would take 12 hours and their respective speeds would be 15 and 25 kmph.