Fill That Tank

Let us assume that the volume of the water tank is $\color{#3D99F6}x$ volumes .

• When pipe 2 is closed, time taken by pipe 1 to fill the tank = 1 hour

$\implies$ Volume filled by pipe 1 in 1 hour = $\color{#3D99F6}x$ volumes

• When pipe 1 is closed, time taken by the pipe 2 to empty the tank = 2 hours

$\implies$ Volume emptied by pipe 2 in 1 hour = $\color{#E81990}\dfrac{x}{2}$ volumes

So, it is clear that in 1 hour pipe 1 fills $\color{#3D99F6}x$ volumes and pipe 2 empties $\color{#E81990}\dfrac{x}{2}$ volumes. When both are open the volume of tank filled in 1 hour :

$\implies {\color{#3D99F6}x} - {\color{#E81990}\dfrac{x}{2}} = {\color{#D61F06}\dfrac{x}{2}} volumes$

Now, $\text{in one hour if half of the tank gets filled then it is obvious that the whole tank gets filled in 2 hours.}$

$\color{#20A900}\text{Therefore, when both the pipes are open the water tank gets filled in 2 hours.}$

Let $t$ be the time to fill the tank if both pipes are open. Then,

$\dfrac{t}{1}-\dfrac{t}{2}=1$

$\dfrac{2t-t}{2}=1$

$2t-t=2$

$t=\boxed{2~hours}$

The answer is $\boxed 2$ hours. In 2 hours, pipe 1 fills up 2 tankful of water and pipe 2 drains off 1 tankful of water, therefore remaining 1 tankful.