A tap fills a cistern in 18 hours. Another tap empties the full tank in 24 hours. How long (in hours) will it take to fill one-fourth of the tank, if the tank is empty initially and both the taps are open together?​

Correct option is D

Given:

Filling tap fills the tank in 18 hours

Emptying tap empties the tank in 24 hours

Tank is initially empty and both taps are open together

Required: Time to fill one-fourth of the tank.

Concept Used:

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

Solution:

Net rate  = $$\frac{1}{18} - \frac{1}{24} = \frac{4 - 3}{72} = \frac{1}{72}$$​​

Time to fill $$\frac{1}{4}$$​ tank $$= \frac{1/4}{1/72} = \frac{72}{4} = 18 \text{ hours}$$ 

Alternate Solution:

Total Capacity = lcm of 18 and 24 = 72 units

Fill rate = 72 ÷ 18 = 4 units/hour

Empty rate = 72 ÷ 24 = 3 units/hour

Net work rate = 4 − 3 = 1 unit/hour

One-fourth of the tank = $$\frac{1}{4} \times 72$$​ = 18 units

Time = $$\frac{18}{1}$$​ = 18 hours