The velocity of a motorboat with respect to still water is $7 ms^{-1}$ and speed of the stream is $3 ms^{-1}$. When the boat starts moving upstream a float was dropped from it. The boat travels $4.2$ km up stream turned about and caught up with the float. The time is taken by the boat to reach the float is:

When a float is dropped, it moves with the stream velocity of 3 m/s, so the relative velocity between the returning boat and the float is 7 m/s downstream plus 3 m/s, which is 10 m/s. The boat travels 4.2 km upstream and 4.2 km back to the starting bank, while the float travels downstream during the total time, meaning the boat must cover the float's downstream drift plus the 8.4 km round trip. Calculating the total time gives 4200 m divided by 4 m/s upstream plus 4200 m plus 3 m/s times the first leg's time divided by 10 m/s, yielding 1050 seconds plus 1845 seconds, which is 2895 seconds or 48.25 minutes.

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