A pipe can fill a tank in 9 hours. Another pipe can empty the filled tank in 63 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:

First pipe fills the tank in 9 hours.

Second pipe empties the tank in 63 hours.

Both pipes are opened simultaneously.

Solution: 

The rate of the first pipe (filling) = $$\frac{1}{9}$$​​

The rate of the second pipe (emptying) = $$\frac{1}{63}$$​​

Net rate = $$\frac{1}{9} - \frac{1}{63}$$​​

​$$= \frac{7}{63} - \frac{1}{63}$$​​

​$$= \frac{6}{63}$$​​

​$$= \frac{2}{21}$$​​

Now, 

Time required to fill two-thirds of the tank;

​$$= \frac{\frac{2}{3}}{\frac{2}{21}}$$​​

​$$= \frac{2}{3} \times \frac{21}{2}$$​​

​$$= 7 \, \text{hours}$$

Alternate Solution:

The LCM of 9 and 63 is 63.

The rate of filling = $$\frac{63}{9}$$​ = 7 units per hour.

The rate of emptying = $$\frac{63}{63}$$​ = 1 unit per hour.

The net rate of filling = 7 - 1 = 6 units per hour.

We need to fill two-thirds of the tank, so the amount to be filled is $$\frac{2}{3}$$​ units.

Time required to fill $$\frac{2}{3}$$ of the tank = $$\frac{\frac{2}{3}\times 63}{6}$$​​

​$$= \frac{2}{3} \times \frac{63}{6} = \frac{1}{7} \, \text{hours}$$​​

So, the time taken to fill two-thirds of the tank is 7 hours.

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