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:

Answer: Option C

$$\begin{array}{rl}& \left(\text{A + B}\right)\text{’s 1 hour work}\\ & \text{ = }\frac{1}{12}+\frac{1}{15}=\frac{9}{60}=\frac{3}{20}\\ & \left(\text{A + C}\right)\text{’s 1 hour work}\\ & \text{ = }\frac{1}{12}+\frac{1}{20}=\frac{8}{60}=\frac{2}{15}\\ & \text{Part filled in 2 hrs}\\ & \text{ = }\frac{3}{20}+\frac{2}{15}=\frac{17}{60}\\ & \text{Part filled in 6 hrs}\\ & \text{ = }3\times \frac{17}{60}=\frac{17}{20}\\ & \text{Remaining part}\\ & \text{ = }1-\frac{17}{20}=\frac{3}{20}\end{array}$$
Now it is the turn of A and B and
$$\frac{3}{20}$$ part is filled by A and B in 1 hour
∴ Total time taken to fill tank
= (6 + 1) hrs
= 7 hrs