Pipes A and B can fill a tank in 30 hours and 40 hours. respectively. Pipe C is an emptying pipe. When the tank is  empty and all the three pipes are opened together for 20 hours, then $$\frac{19}{30}$$ portion of the tank remains unfilled. Pipe B and C together can empty 30% of the tank in (hours):

Correct option is B

Given:

Pipe A fills the tank in 30 hours.

Pipe B fills the tank in 40 hours.

Pipe C is an emptying pipe.

After 20 hours of all three pipes being opened, $$\frac{19}{30}$$​ of the tank remains unfilled.

Formula Used:

Total work = work rate $$\times$$ time

Solution: 

Let rate of Pipe C = x 

Total capacity = LCM(30,40) = 120 units

Rate of Pipe A = $$\frac{120}{30}$$​ = 4 units per hour

​Rate of Pipe B = $$\frac{120}{40}$$​ = 3 units per hour

Total work done in 20 hours: filled part = 1 - $$\frac{19}{30}$$ = $$\frac{11}{30}$$​​

$$= \frac{11}{30} \times 120 = 44 \text{ units filled}$$​​

Now, total work done:

(4 + 3 − x) × 20 = 44

(7 − x) × 20 = 44

7 - x =$$\frac{44}{20}$$​ = 2.2

x = 7 − 2.2 = 4.8 units/hour

Net rate of B and C = 3 − 4.8  = -1.8 units/hour (emptying).

Time to empty 30% of the tank (36 units):

$$= \frac{36}{1.8} = 20 \text{ hours}$$​​