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?

A fill the tank = 12 hrs

B fill the tank = 15 hrs

C fill the tank = 20 hrs

Total work done = LCM (12,15,20)= 60

A's efficiency = $$\frac{60}{12}$$ = 5

A's efficiency = $$\frac{60}{15}$$ = 4

C's efficiency = $$\frac{60}{20}$$ = 3

Tank will be fiiled= $$\frac{60}{12}$$ = 4hours

Tap A is open all the time and taps B and C are open for one hour each alternately

so

as we know that , 

efficiency multiplied by time = work 

here tap A open for T time , meanwhile tap B And C open alternatively

$$5\times T + 4\times(\frac{T}{2}) +  3\times(\frac{T}{2})$$= 60

T = 7 hours