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 and C are open for one hour each alternately, the tank will be full in:

The correct option is C

7 hrs

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

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

Part filled in 2 hours $=(\frac{3}{20}+\frac{2}{15})=\left(\frac{17}{60}\right);$

Part filled in 6 hrs $=(3\times \frac{17}{60})=\frac{17}{20}$

Remaining part $=(1-\frac{17}{20})=\frac{3}{20}$

Now, it is the turn of A and B and part is filled by A and B in 1 hour.

$\therefore$ Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.