Let's find the rates at which each tap can fill or empty the tank:

Tap-1 can fill the tank in 8 hours, so its rate is 1 tank per 8 hours, which can be represented as 1/8 tank per hour.

Tap-2 can empty the tank in 16 hours, so its rate is -1 tank per 16 hours, which can be represented as -1/16 tank per hour since it's emptying the tank.

Now, when both taps are opened simultaneously, their rates are combined. So, the combined rate of filling the tank when both taps are open is:

Rate = (Rate of Tap-1) + (Rate of Tap-2) Rate = (1/8) + (-1/16)

Now, let's find a common denominator to add the rates:

Rate = (2/16) - (1/16) Rate = 1/16 tank per hour

So, when both taps are open, they fill the tank at a rate of 1/16 tank per hour.

Now, we want to know how long it takes to fill the tank completely. Let's denote the time it takes to fill the tank as "t" hours.

During the first 8 hours, both taps are open, so they fill the tank at a rate of 1/16 tank per hour for 8 hours:

(1/16) * 8 = 1/2 tank (filled in the first 8 hours)

Now, for the remaining time "t - 8" hours, only Tap-1 is open, and it fills the tank at a rate of 1/8 tank per hour:

(1/8) * (t - 8) = (t - 8)/8 tank (filled in the remaining time)

The total tank capacity is 1 (since we want to fill the tank completely).

So, we can write the equation:

1/2 (filled in the first 8 hours) + (t - 8)/8 (filled in the remaining time) = 1 (total capacity)

Now, solve for "t":

1/2 + (t - 8)/8 = 1

Multiply both sides by 8 to get rid of the fractions:

4 + t - 8 = 8

Now, isolate "t":

t - 4 = 8

t = 8 + 4 t = 12

So, it will take 12 hours to fill the tank completely if both taps are opened simultaneously, but Tap-2 is closed after 8 hours.