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