A cistern can be filled with water by a pipe in 10 hours, and it can be emptied by a second pipe in 8 hours. If both the pipes are opened when the cistern is full, the time in which it will be emptied is:

Correct option is D

Given:

Tap fills the cistern in 10 hours

Hole empties the full cistern in 8 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 = $$\frac{1}{10}$$​  , Hole rate = $$\frac{1}{8}$$​​

Net rate = $$\frac{1}{8} - \frac{1}{10}$$​​

​$$= \frac{5 - 4}{40}$$​​

​$$= \frac{1}{40}$$​​

Time to empty the cistern= $$\frac{1}{\frac{1}{40}} = 40 \text{ hours}$$​​

Alternate Solution:

Let total capacity = LCM of 10 and 8 = 40 units

Tap fills = 40 ÷ 10 = 4 units/hour 

Hole empties = 40 ÷ 8 = 5 units/hour 

Net  rate  = 5 - 4 = 1 units/hour

Time to empty the cistern = $$\frac{40}{1}$$ = 40 hours