Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. Tap A is open all the time and taps B and C are open for one hour each alternately. In how many hours will the tank be full?

Correct option is A

Given:

Tap A fill tank = 12 hours

Tap B fill tank = 15 hours

Tap C fill tank = 20 hours

Solution:

Let the rates of taps A, B, and C be as follows:

Rate of A = $$\frac{1}{12}$$​ of the tank per hour

Rate of B =$$\frac{1}{15}$$​ of the tank per hour

Rate of C = $$\frac{1}{20}$$​ of the tank per hour

In 2 hours, tap A is always open, and taps B and C work alternately. In the first hour, taps A and B are working together, and in the second hour, taps A and C are working together.

The amount of the tank filled by A and B in the first hour is:

Rate of A + Rate of B = $$\frac{1}{12} + \frac{1}{15}$$​​

​$$= \frac{5}{60} + \frac{4}{60} = \frac{9}{60}$$​​

The amount of the tank filled by A and C in the second hour is:

Rate of A + Rate of C =$$\frac{1}{12} + \frac{1}{20}$$​​

​$$= \frac{5}{60} + \frac{3}{60} = \frac{8}{60}$$​​

Therefore, the amount of the tank filled in 2 hours is:

Amount filled in 2 hours=$$\frac{9}{60} + \frac{8}{60}$$​​​$$= \frac{21}{60} = \frac{17}{60}$$   

Amount filled in 6 hours = $$\frac{51}{60}$$

Remaining work  = 1 - $$\frac{51}{60}$$  = $$\frac{9}{60}$$

A + B complete the remaining work = $$\frac{\frac{9}{60}}{\frac{9}{60}}$$  = 1 hour​

Total time = 1 + 6 = 7 hours 

Thus, the tank will be full in approximately 7 hours.