A man can row three-quarters of a kilometer against the stream in $$11\frac{1}{4}$$ minutes and down the stream in $$7\frac{1}{2}$$ minutes. The speed (in km/hr) of the man in still water is:

Solution(By Examveda Team)

We can write three - quarters of a kilometer as 750 meters and $$11\frac{1}{4}$$ minutes as 675 seconds
$$\eqalign{ & {\text{Rate}}\,{\text{upstream}} \cr & = {\frac{{750}}{{675}}} m/\sec = \frac{{10}}{9}m/\sec \cr & {\text{Rate}}\,{\text{downstream}} \cr & = {\frac{{750}}{{450}}} m/\sec = \frac{5}{3}m/\sec \cr & \therefore {\text{Rate}}\,{\text{in}}\,{\text{still}}\,{\text{water}} \cr & = \frac{1}{2}\left( {\frac{{10}}{9} + \frac{5}{3}} \right)m/\sec \cr & = \frac{{25}}{{18}}\,m/\sec \cr & = \left( {\frac{{25}}{{18}} \times \frac{{18}}{5}} \right)km/hr \cr & = 5\,km/hr \cr} $$