Pipes A and B together can fill a cistern in 10 hours. Pipes B and C together can fill the same cistern in 20 hours, and pipes C and A together can fill it in 12 hours. Find the time taken by pipe B alone to fill the cistern.
30 hours
Let the filling rates per hour be \(A, B, C\). Then \(A+B=\frac{1}{10}\), \(B+C=\frac{1}{20}\), and \(C+A=\frac{1}{12}\).
Add all three: \(2(A+B+C) = \frac{1}{10}+\frac{1}{20}+\frac{1}{12}\). Using LCD 60: \(\frac{6}{60}+\frac{3}{60}+\frac{5}{60}=\frac{14}{60}=\frac{7}{30}\).
So \(A+B+C = \frac{7}{60}\).
Pipe B alone: \(B = (A+B+C) - (C+A) = \frac{7}{60} - \frac{1}{12} = \frac{7}{60} - \frac{5}{60} = \frac{2}{60} = \frac{1}{30}\).
So B fills the cistern in 30 hours.
Hence, pipe B alone takes 30 hours to fill the cistern.
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