A tap fills a cistern in 15 hours. Another tap empties the full tank in 17 hours. How long (in hours) will it take to fill two-third of the tank, if the tank is empty initially and both the taps are open together?​

Correct option is B

Given:

Filling tap fills the tank in 15 hours

Emptying tap empties the tank in 17 hours

Tank is initially empty and both taps are opened together

Required: Time to fill two-thirds of the tank

Formula Used:

Time = Required work ÷ Rate

Solution: 

Net work per hour = $$\frac{1}{15} - \frac{1}{17}$$​​

​$$= \frac{17 - 15}{255} = \frac{2}{255}$$​​

To fill two-thirds of the tank:

Time = $$\frac{2}{3} \div \frac{2}{255} = \frac{2}{3} \times \frac{255}{2}$$​ = 85 hours 

Alternate Solution: ​

Let total tank capacity = LCM(15, 17) = 255 units

Tap A fills 255 units in 15 hrs →$$\frac{255}{15}$$​ = 17 units/hr

Tap B empties 255 units in 17 hrs → $$\frac{255}{17}$$​ = 15 units/hr

Net work per hour = 17 - 15 = 2 units/hr

Two-thirds of the tank = $$\frac{2}{3} \times 255$$​ = 170 units

Time = $$\frac{170}{2}$$​ = 85 hours