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.

Correct Answer: 20 hours
A__________300 km_______BLet speed of the 1st ship is s kmph and speed of 2nd ship is (s + 10) kmph. Let 2nd ship takes time t to cover the distance 300 km then 1st 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 + 10tOr, 8s = 10tIf 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.