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?

Solution(By Examveda Team)

Let the speed of boat in still water = x km/hr,
and Speed of current = y km/hr
Rate downstream = (x + y) km/hr, and Rate upstream = (x – y) km/hr
Distance = Speed × Time
$$\eqalign{ & \therefore \left( {x - y} \right) \times 2t = \left( {x + y} \right) \times t \cr & \Rightarrow 2x - 2y = x + y \cr & \Rightarrow 2x - x = 2y + y \cr & \Rightarrow x = 3y \cr & \Rightarrow \frac{x}{y} = \frac{3}{1} = 3:1 \cr} $$

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Alternate Solution :
$$\eqalign{ & {\text{According to Question}} \cr & {\text{Downstream Speed}} = x + y \cr & {\text{Upstream Speed}} = x - y \cr & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & \therefore x + y = \frac{D}{T}\, . . . . .\,\left( {\text{i}} \right) \cr & \,\,\,\,x - y = \frac{D}{{2T}}\,. . . . .\,\left( {{\text{ii}}} \right) \cr & {\text{Solve equation}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & x = \frac{{3D}}{{4T}},\,\,\,\,y = \frac{D}{{4T}} \cr & \therefore \frac{x}{y} = \frac{{3D}}{{4T}} \times \frac{{4T}}{D} \cr & \,\,\,\,\frac{x}{y} = \frac{3}{1} \cr & \,\,\,\,x:y = 3:1 \cr} $$