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:

$$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s 1 hour work}} \cr & {\text{ = }} {\frac{1}{{12}} + \frac{1}{{15}}} = \frac{9}{{60}} = \frac{3}{{20}} \cr & \left( {{\text{A + C}}} \right){\text{'s 1 hour work}} \cr & {\text{ = }} {\frac{1}{{12}} + \frac{1}{{20}}} = \frac{8}{{60}} = \frac{2}{{15}} \cr & {\text{Part filled in 2 hrs}} \cr & {\text{ = }} {\frac{3}{{20}} + \frac{2}{{15}}} = \frac{{17}}{{60}} \cr & {\text{Part filled in 6 hrs}} \cr & {\text{ = }} {3 \times \frac{{17}}{{60}}} = \frac{{17}}{{20}} \cr & {\text{Remaining part}} \cr & {\text{ = }} {1 - \frac{{17}}{{20}}} = \frac{3}{{20}} \cr} $$
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