Two taps can fill a cistern in 2 hours and 64 hours, respectively. A third tap can empty it in 2 hours. How long (in hours) will it take to fill half of the empty cistern, if all of them are opened together?​

Correct option is B

Given:

Tap A fills the cistern in 2 hours 

Tap B fills the cistern in 64 hours  

Tap C empties the cistern in 2 hours 

Formula Used: 

Time = $$\frac{\text{Work}}{\text{Total Rate}}$$​​

Solution: 

Rate of tap A = $$\frac{1}{2}$$​​

Rate of tap B = $$\frac{1}{64}$$​​

Rate of  tap C = $$-\frac{1}{2}$$​ (negative since it's an emptying tap)

Total rate = $$\frac{1}{2} + \frac{1}{64} - \frac{1}{2} = \frac{1}{64}$$​​

Time to fill half cistern = $$\frac{1}{2} \div \frac{1}{64} = \frac{1}{2} \times 64$$​ = 32 hours

Alternate Solution: ​

LCM of 2, 64 = 64 units (assume total capacity of cistern = 64 units)

Tap A fills = $$\frac{64}{2}$$​ = 32 units/hour

Tap B fills = $$\frac{64}{64}$$​ = 1 unit/hour

Tap C empties = $$\frac{64}{2}$$​ = 32 units/hour

Net efficiency = 32 + 1 - 32 = 1 unit/hour

Half of the cistern = $$\frac{64}{2}$$​ = 32 units

Time to fill 32 units = $$\frac{32}{1}$$​ = 32 hours