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”: $$ t = \frac{(3/4)V}{u_1+u_2} = \frac34\frac{V}{\frac V8-\frac V{16}} = \frac34\frac{V}{\frac V{16}} = \frac34\times 16 = 12~\text{hours} $$
