A tank is filled in eight hours by three pipes A,B and C. Pipe A is twice as fast as pipe B, and pipe B is twice as fast as pipe C. How much time will pipe B alone take to fill the tank ?

Correct option is A

Given:

1. Three pipes A, B, and C together can fill a tank in 8 hours.

2. Pipe A is twice as fast as pipe B , and pipe B is twice as fast as pipe C .

Formula Used:

1. Work done (tank filled) in 1 hour by a pipe is the reciprocal of the time it takes to fill the tank:

Work in 1 hour by Pipe A = $$\frac{1}{\text{Time taken by A}}$$​

2. Assume:

- Time taken by Pipe C to fill the tank = x hours.

- Time taken by Pipe B = $$\frac{x}{2} hours$$​

- Time taken by Pipe A = $$\frac{x}{4} hours$$​

3. Work done in 1 hour by all three pipes:

Work in 1 hour = $$\frac{1}{\frac{x}{4}} + \frac{1}{\frac{x}{2}} + \frac{1}{x}$$​

Simplify the equation and solve for x 

Solution:

1. Work done in 1 hour by all three pipes:

Work in 1 hour = $$\frac{4}{x} + \frac{2}{x} + \frac{1}{x} = \frac{7}{x}$$​

2. Since all three pipes together fill the tank in 8 hours:

Work in 1 hour by all pipes = $$\frac{1}{8}$$​

Equating the two expressions for work:

$$\frac{7}{x} = \frac{1}{8}$$​

3. Solve for x :

x = $$7 \cdot 8$$​ = 56

4. Time taken by Pipe B :

Time taken by B = $$\frac{x}{2} = \frac{56}{2} = 28 \, \text{hours}$$​