Three pipes, A, B, and C are capable of filling a tank in 6, 8, and 12 hours, respectively. When all three pipes are opened together, they operate for 2 hours before pipe C is closed. How much additional time will it take to completely fill the tank after that?
\(\dfrac{6}{7}\) hours
Step 1 – Hourly rates:
\(A = \dfrac{1}{6}, \quad B = \dfrac{1}{8}, \quad C = \dfrac{1}{12}\)
Step 2 – Combined rate of A + B + C:
\(\dfrac{1}{6} + \dfrac{1}{8} + \dfrac{1}{12} = \dfrac{4 + 3 + 2}{24} = \dfrac{9}{24} = \dfrac{3}{8}\)
Step 3 – Work done in 2 hours by all three:
\(2 \times \dfrac{3}{8} = \dfrac{3}{4}\)
So work remaining = \(1 - \dfrac{3}{4} = \dfrac{1}{4}\).
Step 4 – After C closes, A + B operate together:
\(\dfrac{1}{6} + \dfrac{1}{8} = \dfrac{4 + 3}{24} = \dfrac{7}{24}\)
Step 5 – Additional time required:
\(\dfrac{1/4}{7/24} = \dfrac{1}{4} \times \dfrac{24}{7} = \dfrac{6}{7} \text{ hours}\)
Hence, the correct answer is \(\dfrac{6}{7}\) hours.
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