​One pipe can fill a tank three times as fast another one. Both the pipes opened together can fill the tank completely in 36 minutes. In how many hours will the faster pipe alone fill the empty tank completely?

Correct option is B

Given:

Time taken by pipe to fill the tank = 3 times faster than other

Both can fill the pipe in 36minutes

Formula Used:

Let total work to fill the tank be 1.

Then work done by a pipe in 1 hour =$$\frac{1}{Total\ time\ to\ fill}$$​​

Solution:

Let Work done  by one pipe in 1 minute will be $$\frac{Total\ Work}{Total\ time\ } =\frac{1}{x}$$​​

Work done  by another pipe in 1 minute will be $$\frac{Total\ Work}{Total\ time} =\frac{1}{x}\times 3 = \frac{3}{x}$$​​

Work done  by both pipes  together in 1 minute will be =$$\frac{1}{x} + \frac{3}{x} = \frac{1+3}{x}= \frac{4}{x}$$​

Tank will be filled by both pipes working together $$= \frac{total\ work }{work\ done\ in\ 1\ minute}=\frac{1}{\frac{4}{x}}=\frac{x}{4}$$​​

And $$\frac{x}{4} = 36$$​​

x= 144minutes

Time taken by faster pipe to fill the tank =$$\frac{x}{3} =\frac{144}{3} =48minutes$$​​

Time in hour =$$\frac{48}{60} =\frac{4}{5}hour$$​​

Alternative Method:

Total Work = Efficiency × Time

Efficiency of first pipe =1

Efficiency of second pipe =1 $$\times$$​ 3 = 3

Total efficiency = 1+3 = 4

Total time taken by both to fill the tank = 4 $$\times$$​ 36 = 144

Time taken by faster pipe to fill the tank = $$\frac{144}{3}$$​ =48minutes

Time in hour =$$\frac{48}{60} =\frac{4}{5}hour$$​​