Three taps filling a tank with one tap closed later: Taps A, B, and C together can fill a tank in 6 hours. After all three run for 2 hours, tap C is closed and the remaining part is filled by A and B in 8 more hours. In how many hours can C alone fill the tank?

Introduction / Context:
Classic pipes-and-cisterns setup: total rate is given, then one pipe is turned off. We use the initial joint work to find what remains, infer A+B rate from the second phase, and subtract from total to isolate C’s rate.

Given Data / Assumptions:

• A + B + C together fill in 6 h ⇒ combined rate = 1/6 tank/h.
• All three run for 2 h ⇒ filled = 2 * (1/6) = 1/3 of the tank.
• Remaining = 2/3 of the tank, filled in 8 h by A + B only.

Concept / Approach:
From phase two: (A + B) * 8 = 2/3 ⇒ A + B = 1/12. Since A + B + C = 1/6, we get C = (1/6 − 1/12) = 1/12 tank/h. Time of C alone is reciprocal of its rate.

Step-by-Step Solution:
Total rate (A+B+C) = 1/6In 2 h, work done = 2/6 = 1/3Remaining = 2/3 = (A+B)*8 ⇒ A+B = (2/3)/8 = 1/12Thus C = 1/6 − 1/12 = 1/12Time for C alone = 1 / (1/12) = 12 h

Verification / Alternative check:
Reconstruct: Phase 1 adds 1/3; Phase 2 with A+B at 1/12 for 8 h adds 2/3 → total 1 tank, consistent.

Why Other Options Are Wrong:

• 16 h, 14 h, 18 h contradict the exact subtraction of rates and second-phase computation.

Common Pitfalls:

• Mistaking the remaining fraction after 2 h.
• Adding or subtracting times instead of rates.

Final Answer:
12 h