Three taps A,B and C can fill a tank in 12,15 and 20 hours respectively. If A is open all the time and B ,C are open for one hour each alternatively, the tank will be full in:

$A+B\text{'}s 1 hour work=\frac{1}{12}+\frac{1}{15}=\frac{9}{60}=\frac{3}{20}$

$A+C\text{'}s 1 hour work=\frac{1}{12}+\frac{1}{20}=\frac{8}{60}=\frac{2}{15}$  

$part filled in 2 hrs=\frac{3}{20}+\frac{2}{15}=\frac{17}{60}$  

$part filled in 6 hrs=3*\frac{17}{60}=\frac{17}{20}$  

  $remaining part =1-\frac{17}{20}=\frac{3}{20}$  

Now, it is  the turn of A and B (3/20) part is filled by A and B in 1 hour.

Therefore, Total  time taken to fill the tank =(6+1)hrs= 7 hrs