Taps A and C can fill a tank in 5 and 20 hrs respectively. Taps B and

Let the total capacity of the tank be 40 units(LCM of all 4 hours)
The rate of filling of tap A is 8 units per hour.
The rate of filling of tap C is 2 units per hour.

The rate of draining of tap B is 4 units per hour.
The rate of draining of tap D is 1 unit per hour.

For the first two hours, A will fill 16 units.
For the next two hours, A and B will fill (16 - 8) = 8 units.
For the next two hours, A, B, and C will fill (8 + 4) = 12 units.
Units left = 4 units.
Rate of filling when all four pipes are opened = 5 units.
Time is taken to fill the tank when all four pipes are opened = 4/5 hours
Thus, the total time is taken in the process = 6 + 4/5 hours = 34/5 hours.

Thus, the correct option is B.

Given: Taps A and C can fill a tank in 5 and 20 hrs respectively. Taps B and D can drain a full tank in 10 and 40 hrs. respectively.

Asked: If A, B, C and D are opened in a gap of 2 hrs. in that order the tank gets full in :

Let the tank be full in x hours

x/5 - (x-2)/10 + (x-4)/20 - (x-6)/40 >= 1
{8x - 4(x-2) + 2(x-4) - (x-6)} /40 >=1
8x - 4x + 8 + 2x - 8 - x + 6 >=40
5x + 6 >=40
x >= 34/5

The tank gets full in 34/5 hours.

IMO B