One of the two inlet pipes works twice as efficiently as the other. The two, working alongside a drain pipe that can empty a cistern all by itself in 8 hours, can fill the empty cistern in 8 hours. How many hours will the less efficient inlet pipe take to fill the empty cistern by itself?

Let the time taken by the less efficient inlet pipe be t hours. Hence, it fills $$\frac{1}{t}$$ of the cistern per hour. 

Now, we are given that the other pipe is twice as efficient. So, the more efficient pipe fills $$\frac{2}{t}$$ of the cistern per hour. 

Now, the draining pipe can drain the whole cistern in 8 hours, so it means it drains $$\frac{1}{8}$$ of the cistern per hour. 

We are given that if all the pipes are opened, the cistern can be filled in 8 hours. It means, when all the pipes are opened, the cistern fills at the rate of $$\frac{1}{8}$$ of the cistern per hour. 

Now, adding the rate of filling / draining per hour for all three pipes should be equal to 1/8.

$$\dfrac{1}{t}+\dfrac{2}{t}-\dfrac{1}{8}=\dfrac{1}{8}$$

$$\dfrac{1}{t}+\dfrac{2}{t}=\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}$$

$$\dfrac{3}{t}=\dfrac{1}{4}$$

With that, $$t=12$$

We assumed that the less efficient inlet pipe takes t hours or we can say, it takes 12 hours.