A cistern has a hole in the bottom through which the water is leaking. A tap can fill the cistern in 6 hours and the hole in the bottom can empty the fully filled cistern in 15 hours. If both the tap and the hole are open, then what will be the time taken to completely fill the empty cistern?​

Correct option is C

Given:

Tap fills the cistern in 6 hours

Hole empties the full cistern in 15 hours

Both are open, cistern starts empty

Formula Used:

Net work rate = Inlet rate – Outlet rate

Total work = 1 cistern (100%)

Time = $$\frac{\text{Total work}} {\text{ Net rate}}$$​​

Solution:

Tap rate = $$\frac{1}{6}$$​  , Hole rate = $$\frac{1}{15}$$​​

Net rate = $$\frac{1}{6} - \frac{1}{15}$$​​

​$$= \frac{5 - 2}{30}$$​​

​$$= \frac{3}{30} = \frac{1}{10}$$​​

Time to fill the cistern = $$\frac{1}{\frac{1}{10}} = 10 \text{ hours}$$​​

Alternate Solution:

Let total capacity = LCM of 6 and 15 = 30 units

Tap fills = 30 ÷ 6 = 5 units/hour 

Hole empties = 30 ÷ 15 = 2 units/hour 

Net fill rate  = 5 - 2 = 3 units/hour

Time to fill = $$\frac{30}{3}$$​ = 10 hours