P, Q and R are three towns on a river which flows uniformly. Q is equidistant from P and R. I row from P to Q and back in 10 hours and I can row from P to R in 4 hours. Compare the speed of my boat in still water with that of the river.

Solution(By Examveda Team)

$$\eqalign{ & {\text{Let PQ = QR = }}x{\text{ }}km \cr & {\text{let speed downstream }} \cr & {\text{ = }}a{\text{ }}km/hr \cr & \,\,\,\,\,\,\,\,\,\, \to \,\,{\text{downstream}} \to \cr & {\text{P}}\overline {\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Q}}\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,} \,{\text{R}}\,\,\,\, \cr & {\text{and speed upstream }} \cr & {\text{ = }}b{\text{ }}km/hr{\text{ }} \cr & {\text{then, }}\frac{x}{a} + \frac{x}{b} = 10 \cr & \Rightarrow x = \frac{{10ab}}{{a + b}} \cr & {\text{and }}\frac{{2x}}{a} = 4 \cr & \Rightarrow x = \frac{{4a}}{2} = 2a \cr & {\text{from (i) and (ii) we have:}} \cr & 2a = \frac{{10ab}}{{a + b}} \cr & \Rightarrow 5b = a + b \cr & \Rightarrow a = 4b \cr & {\text{Required ratio }} \cr & {\text{ = }}\frac{{{\text{Speed in still water}}}}{{{\text{Speed of river}}}} \cr & = \frac{{\frac{1}{2}\left( {a + b} \right)}}{{\frac{1}{2}\left( {a - b} \right)}} \cr & = \frac{{\left( {a + b} \right)}}{{\left( {a - b} \right)}} \cr & = \frac{{4b + b}}{{4b - b}} \cr & = \frac{5}{3} \cr} $$