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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:

6 hrs
20/3 hrs
7 hrs
15/2 hrs

Correct Answer: 7 hrs

Explanation:

$A + B' s 1 h o u r w o r k = \frac{1}{12} + \frac{1}{15} = \frac{9}{60} = \frac{3}{20}$

$A + C' s 1 h o u r w o r k = \frac{1}{12} + \frac{1}{20} = \frac{8}{60} = \frac{2}{15}$

$p a r t f i l l e d i n 2 h r s = \frac{3}{20} + \frac{2}{15} = \frac{17}{60}$

$p a r t f i l l e d i n 6 h r s = 3 * \frac{17}{60} = \frac{17}{20}$

$r e m a i n i n g p a r t = 1 - \frac{17}{20} = \frac{3}{20}$

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

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