Tank - Pipes

Let A fill the tank at x liters/hour and B drain it at y liters/hour 
Now as per Condition 1 : 
We get Volume filled till 10pm = 8x-7y       (1) .
Here A operates for 8 hours and B operates for 7 hours .
As per condition 2
We get Volume filled till 6pm = 4x-2y        (2)
Here A operates for 4 hours and B operates for 2 hours .
Now equating (1) and (2)
we get 8x-7y =4x-2y
so we get 4x =5y 
y =4x/5
So volume of tank = $$8x-7\times\ \frac{4x}{5}=\frac{12x}{5}$$
So time taken by A alone to fill the tank = $$\frac{\frac{12x}{5}}{x}=\frac{12}{5}hrs\ $$
= 144 minutes

Let the time taken by A to fill the tank alone be x hours, which implies the time taken by B to empty the tank alone is (x-1) hours (B is the drainage pipe), and the time taken by C to fill the tank is y hours.

It is given that when pipes A, B, and C are turned on together, the empty tank is filled in two hours. 

Hence, $$\frac{1}{x}-\frac{1}{x-1}+\frac{1}{y}=\frac{1}{2}$$ .... Eq(1)

It is given that if pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank.

Hence, B worked for 1 hour, and C worked for 2 hours 15 minutes, which is equal to $$\frac{9}{4}$$ hours.

In 1 hour, B worked $$-\frac{1}{x-1}$$ units, and in $$\frac{9}{4}$$ hours, C worked $$\frac{9}{4y}$$ units.

Hence, $$\frac{9}{4y}-\frac{1}{x-1}=1$$ .... Eq(2)

Solving both equations, we get $$y=\frac{3}{2}$$, and $$x=3$$

Hence, the time taken by C is $$\frac{3}{2}$$ hours, which is equal to $$90$$ minutes.

The correct option is A

Let the rate of flow of water from pipe A be $$4x$$ units per hour. From the ratio of the rates of flow of water (4:9:36), we have that the rate of flow of water from pipe B should be $$9x$$ units per hour, and from pipe C should be $$36x$$ units per hour.

Since pipe A fills the empty tank in 15 hours, the capacity of the tank must be $$15\times 4x = 60x$$ units.

When all three pipes work together, their combined capacity would be $$4x+9x+36x = 49x$$ units per hour.

Therefore, the time it would take for the three pipes to fill the empty tank working together is:  $$\dfrac{60x}{49x}$$ hours

In minutes, this time is equivalent to $$\dfrac{60x}{49x}\times 60 \approx 73.47$$ minutes.

Option A is the closest, and is the correct answer.