Question answer bhej dio tha na ki

Solution 
Let the speed of rowing in still water = x km/hr
Let the speed of current = y km/hr
Speed in downstream = (x + y) km/h
Speed in upstream = (x - y) km/h
Therefore, ATQ,
$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{to} \mathrm{cover} 20 \mathrm{km} \mathrm{downstream} = 2 \mathrm{hrs} \\ \Rightarrow \frac{\mathrm{distance}}{\mathrm{speed}} = 2 \\ \Rightarrow \frac{20}{\mathrm{x} + \mathrm{y}} = 2 \\ \Rightarrow \mathrm{x} + \mathrm{y} = 10 .....\left(1\right) \\ \mathrm{Time} \mathrm{taken} \mathrm{to} \mathrm{cover} 4 \mathrm{km} \mathrm{downstream} = 2 \mathrm{hrs} \\ \Rightarrow \frac{\mathrm{distance}}{\mathrm{speed}} = 2 \\ \Rightarrow \frac{4}{\mathrm{x} - \mathrm{y}} = 2 \\ \Rightarrow \mathrm{x} - \mathrm{y} = 2 .....\left(2\right) \\ \mathrm{adding} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ 2\mathrm{x} = 12 \\ \Rightarrow \mathrm{x} = 6 \\ \mathrm{Put} \mathrm{x} = 6 \mathrm{in} \left(1\right), \mathrm{we} \mathrm{get} \\ 6 + \mathrm{y} = 10 \\ \Rightarrow \mathrm{y} = 4 \\ \mathrm{So}, \mathrm{speed} \mathrm{of} \mathrm{rowing} \mathrm{in} \mathrm{still} \mathrm{water} = 6 \mathrm{km}/\mathrm{h} \\ \mathrm{speed} \mathrm{of} \mathrm{current} = 4 \mathrm{km}/\mathrm{h} \end{gathered}$$
Option (a) is correct.