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?

Answer: Option D

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
$$\begin{array}{rl}& \therefore \left(x-y\right)\times 2t=\left(x+y\right)\times t\\ & \Rightarrow 2x-2y=x+y\\ & \Rightarrow 2x-x=2y+y\\ & \Rightarrow x=3y\\ & \Rightarrow \frac{x}{y}=\frac{3}{1}=3:1\end{array}$$

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