Pipes A, B, and C together can fill a cistern in 6 hours. After they work together for exactly 2 hours, Pipe C is closed and A and B continue to fill the cistern. With only A and B running, they take 7 more hours to finish filling. How long would Pipe C alone take to fill the cistern?

Introduction / Context:Let a, b, c be the hourly fill rates of A, B, and C. The joint rate and the two-phase timeline give two equations, from which we isolate c and invert to get C’s solo time.

Given Data / Assumptions:

• a + b + c = 1/6 job/hour.
• After 2 hours together, remaining work is filled by A+B in 7 hours ⇒ (a + b) * 7 = remaining.

Concept / Approach:Work done in first 2 hours = 2 * (1/6) = 1/3. Remaining = 2/3. Thus (a + b) * 7 = 2/3 ⇒ a + b = 2/21. Then c = 1/6 − 2/21.

Step-by-Step Solution:

Verification / Alternative check:Check totals: First 2 h give 1/3; then 7 h at 2/21 give 14/21 = 2/3; total 1 job.

Why Other Options Are Wrong:10/12/16 hours do not match the derived c = 1/14.

Common Pitfalls:Confusing “7 hours to finish” as total time rather than the second phase duration.

Final Answer:14 hours