$\begin{array}{lcc} & \text{Inlet pipe (I)} & & \text{Outlet pipe + Inlet pipe (O+I)} \\ \text{Time:} & 8\;\text{hours} & & 10\;\text{hours} \\ \text{Total work:} & \text{LCM}(8,10) & = & 80\;\text{liters} \\ \text{(Capacity of tank)} & & & \\ \text{Efficiency:} & 10\;\text{liter/hour} & & 8\;\text{liter/hour} \end{array}$
Now, $\text{I}_{\text{efficiency}} + \text{O}_{\text{efficiency}} = 8$
$\Rightarrow 10 + \text{O}_{\text{efficiency}} = 8$
$\Rightarrow \boxed{\text{O}_{\text{efficiency}} = -2}$
So, $\text{Time} = \dfrac{\frac{\text{Capacity of tank}}{2}}{2}$
$\Rightarrow \text{Time} = \dfrac{80}{4} = 20$ hours.
$\therefore$ If only the outlet pipe is open, then in $20$ hours the full tank becomes half-full.
Correct Answer$:\text{A}$