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:

6 hours

6$ \dfrac{2}{3} $hours

7 hours

7$ \dfrac{1}{2} $hours

Explanation:

$\left(A + B\right)$s 1 hours work =$ \left(\dfrac{1}{12} +\dfrac{1}{15} \right) $=$ \dfrac{9}{60} $=$ \dfrac{3}{20} $.

$\left(A + C\right)$s hours work =$ \left(\dfrac{1}{12} +\dfrac{1}{20} \right) $=$ \dfrac{8}{60} $=$ \dfrac{2}{15} $.

Part filled in 2 hrs =$ \left(\dfrac{3}{20} +\dfrac{2}{15} \right) $=$ \dfrac{17}{60} $.

Part filled in 6 hrs =$ \left(3 \times\dfrac{17}{60} \right) $=$ \dfrac{17}{20} $.

Remaining part =$ \left(1 -\dfrac{17}{20} \right) $=$ \dfrac{3}{20} $.

Now, it is the turn of A and B and $ \dfrac{3}{20} $ part is filled by A and B in 1 hour.

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