A tap fills a cistern in 9 hours. Another tap empties the full tank in 90 hours. How long (in hours) will it take to fill twice the volume of the tank, if the tank is empty initially and both the taps are open together?​

Correct option is D

Given:

Tap A (inlet) fills the tank in 9 hours

Tap B (outlet) empties the full tank in 90 hours

We need to find the time to fill twice the tank's volume when both taps are open

Solution:

Work done per hour = $$\left(\frac{1}{9} - \frac{1}{90}\right)$$​​

​$$= \frac{10 - 1}{90}$$​​

​$$= \frac{1}{10}$$​​

Time to fill 2 tanks = 2 × 10 = 20 hours

Alternate Solution:

Let LCM of 9 and 90 = 90 units ( total capacity)

Tap A fills = 90 ÷ 9 = 10 units/hour

Tap B empties = 90 ÷ 90 = 1 unit/hour

Net fill rate = 10 - 1 = 9 units/hour

For double the tank: 2 × 90 = 180 units

Time = $$\frac{180}{9}$$​ = 20 hours