Tap filling Tank .. Time taken?

A tap can fill a tank in 16 hours whereas another tap can empty the tank it in 8 hours. If in a three fourth filled tank both the taps are opened, then how long will it take to empty the tank in this scenario?

I know that time to fill+empty = (1/16)+(1/8)

How to incorporate three fourth filled tank?

If the tank is full and you open both taps, then it will take $$16$$ hours to empty it: indeed in first $$8$$ hours the second tap empties the tank and in the second $$8$$ hours it empties one more tank that filled the first tap in these $$16$$ hours. If the tank was $$3/4$$ in the beginning, it would take $$(3/4)\times16=12$$ hours.
Or you can write the equations, if $$V$$ is the volume of the tank, the “speeds of emptying” of two taps are: $$u_1=-V/16,\qquad u_2=V/8$$
To find the time, we should divide the volume of the water by the “speed of emptying”: $$t = \frac{(3/4)V}{u_1+u_2} = \frac34\frac{V}{\frac V8-\frac V{16}} = \frac34\frac{V}{\frac V{16}} = \frac34\times 16 = 12~\text{hours}$$