A tank is attached with 30 pipes, some of these are filling pipes and the rest are emptying pipes. Each filling pipe can fill the tank completely in 24 hours and each emptying pipe can empty the tank completely in 18 hours. When all the pipes are opened together, it takes 1$$\frac 12$$​ hours to fill the tank completely. How many of the given pipes are emptying pipes?

Correct option is B

Given:

Total number of pipes = 30

Some pipes are filling pipes, and the rest are emptying pipes

Each filling pipe can fill the tank completely in 24 hours

Each emptying pipe can empty the tank completely in 18 hours

All pipes together fill the tank in $$1 \frac{1}{2}$$​ hours

Concept Used:

Rate of work (filling or emptying) is inversely proportional to the time taken to complete the task.

Net rate of work when multiple pipes are open together is the sum of the individual rates.

Solution:

Let the number of filling pipes be x, and the number of emptying pipes be 30 - x.

Rate of x filling pipes = $$\frac{x}{24}$$​ tanks per hour

Rate of 30−x emptying pipes = $$\frac{30 - x}{18}$$​ tanks per hour

When all pipes are open, it takes $$1 \frac{1}{2}$$​​ hours (which is $$\frac{3}{2}​$$​ hours) to fill the tank.

Therefore, the net rate of work is $$\frac{2}{3}$$​ of a tank per hour (because it fills the tank in $$\frac{3}{2}$$​​ hours):

​$$\frac{x}{24} - \frac{30 - x}{18} = \frac{2}{3}$$​​

​$$\frac{3x –(30 - x)(4)}{72} = \frac{2}{3}$$​​

​$$\frac{3x – 120 + 4x}{72} = \frac{2}{3}$$​​

​$$\frac{-120 + 7x}{72} = \frac{2}{3}$$​​

7x – 120 = 48

​$$x = \frac{168}{7}$$​​

x = 24

Thus, the number of filling pipes is x = 24,

and the number of emptying pipes is 30 – 24 = 6