A boat takes half time in moving a certain distance downstream than upstream. The ratio of the speed of the boat in still water and that of the current is?

Explanation:-

$$\eqalign{ & {\text{According to question}} \cr & {\text{Downstream speed = }}x + y \cr & {\text{Upstream speed = }}x - y \cr & {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & \therefore x + y = \frac{{\text{D}}}{{\text{T}}}................(i) \cr & \,\,\,x + y = \frac{{\text{D}}}{{{\text{2T}}}}................(ii) \cr & {\text{Solve equation (i) and (ii) }} \cr & x{\text{ = }}\frac{{{\text{3D}}}}{{{\text{4T}}}},\,\,\,\,y = \frac{{\text{D}}}{{{\text{4T}}}} \cr & \therefore \frac{x}{y}\,\, = \frac{{{\text{3D}}}}{{{\text{4T}}}} \times \frac{{4{\text{T}}}}{{\text{D}}} = \frac{3}{1} = 3:1 \cr & {\text{Downstream (D)}}\,\,\,\,\,\,\,\,\,{\text{Upstream (U)}} \cr & {\text{T}} \to \,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr & {\text{S}} \to \,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cr & {{\text{S}}_{\text{B}}} = \frac{{{{\text{S}}_{\text{D}}}{\text{ + }}{{\text{S}}_{\text{U}}}}}{2} = \frac{{2 + 1}}{2} = \frac{3}{2} \cr & {{\text{S}}_{\text{C}}} = \frac{{{{\text{S}}_{\text{D}}} - {{\text{S}}_{\text{U}}}}}{2} = \frac{{2 - 1}}{2} = \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{S}}_{\text{B}}}\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,{{\text{S}}_{\text{C}}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \,\,\,\frac{3}{2}\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,3\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,1 \cr}$$