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

Options:
A .  6 hours
B .  \(6\frac{2}{3}hours\)
C .  7 hours
D .  \(7\frac{1}{2}hours\)
Answer: Option C

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


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


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


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


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


Now, it is the turn of A and B and \(\frac{3}{20}\)  part is filled by A and B in 1 hour


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