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A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed then the inlet pipe fills the empty tank in $8$ hours. If the outlet pipe is open then the inlet pipe fills the empty tank in $10$ hours. If only the outlet pipe is open then in how many hours the full tank becomes half-full?

- $20$
- $30$
- $40$
- $45$

1 vote

$\begin{array}{lcc} & \text{Inlet pipe (I)} & & \text{Outlet pipe + Inlet pipe (O+I)} \\ \text{Time:} & 8\;\text{hours} & & 10\;\text{hours} \\ \text{Total work:} & \text{LCM}(8,10) & = & 80\;\text{liters} \\ \text{(Capacity of tank)} & & & \\ \text{Efficiency:} & 10\;\text{liter/hour} & & 8\;\text{liter/hour} \end{array}$

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$:\text{A}$

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$:\text{A}$