A pipe can fill a tank in 9 hours. Another pipe can empty the filled tank in 27 hours. If both the pipes are opened simultaneously, then the time (in hours) in which the tank will be two-third filled, is:

Correct option is D

Given:

Pipe 1 fills the tank in 9 hours

Pipe 2 empties the tank in 27 hours

Both pipes open simultaneously

Find time to fill $$\frac{2}{3}$$​ of the tank

Formula Used:

Combined rate = $$\frac{1}{\text{Time to fill}} - \frac{1}{\text{Time to empty}}$$​​

Time = $$\frac{\text{Fraction to fill}}{\text{Combined rate}}$$​​

Solution:

Filling rate = $$\frac{1}{9}$$​ tank/hour

Emptying rate = $$\frac{1}{27}$$​ tank/hour

Combined rate =$$\frac{1}{9} - \frac{1}{27}$$​​

​$$= \frac{3 - 1}{27}$$​​

$$= \frac{2}{27}$$​ tank/hour

Time to fill $$\frac{2}{3}$$​ of tank:

t = $$\frac{\frac{2}{3}}{\frac{2}{27}} = \frac{2}{3} \times \frac{27}{2} = 9 \text{ hours}$$​​

Time taken to fill two-third of the tank = 9 hours

Alternate Solution: ​

LCM of 9 and 27 = 27 hours

In 27 hours, pipe 1 fills = $$\frac{27}{9}$$​ = 3 tanks

In 27 hours, pipe 2 empties = $$\frac{27}{27}$$​ = 1 tank

Net filling in 27 hours = 3 - 1 = 2 tanks

Time to fill 1 tank =$$\frac{27}{2}$$​ = 13.5 hours

Time to fill $$\frac{2}{3}$$​ tank =$$\frac{2}{3} \times 13.5$$​ = 9 hours