Pipe number 1 can fill the tank in 5 h and pipe number 2 in 10 h . Pipe number 3 can drain the water and make empty in $$7\frac{1}{2} \space h$$​ . If all the three pipes are open all together, then the time required for filling the tank completely would be

Correct option is A

Given:

Pipe 1 fills the tank in 5 hours

Pipe 2 fills the tank in 10 hours

Pipe 3 empties the tank in $$7\frac{1}{2} = \frac{15}{2}$$​ hours 

Formula Used: ​

Time = $$\frac{\text{Total work}}{\text{ Net rate}}$$
Solution:

Let total work (capacity of tank) = LCM of 5, 10, and $$\frac{15}{2}$$​ = 30 units

Efficiencies:

Pipe 1 =$$\frac{30}{5}$$​ = 6 units/hour

Pipe 2 = $$\frac{30}{10}$$​ = 3 units/hour

Pipe 3 =$$\frac{30}{15/2} = \frac{30 \times 2}{15}$$​ = 4  units/hour (drain)

Net rate = Fillers - Drainer

= 6 + 3 - 4 = 5 units/hour

Time = $$\frac{30}{5} = 6 \, \text{hours}$$​​