a) 4 : 3
b) 5 : 3
c) 6 : 5
d) 7 : 3

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