Three inlets A, B, and C can fill a cistern in 10 hours, 12 hours, and 15 hours, respectively. First, A alone is opened. After 1 hour from the start, B is also opened. After 2 hours from the start (i.e., 1 more hour later), C is also opened. From the start of A, in what total time will the cistern be just full?

Introduction / Context:
This is a staged opening problem. Compute the work done in each stage and then determine how long the final stage with all three running must continue to complete the job.

Given Data / Assumptions:

• A = 1/10 job/hour.
• B = 1/12 job/hour.
• C = 1/15 job/hour.
• Stage 1 (0–1 h): A only. Stage 2 (1–2 h): A + B. Stage 3 (after 2 h): A + B + C until full.

Concept / Approach:
Sum fractional work stage-wise. Let t be the duration of Stage 3. Total = 1/10 + (1/10 + 1/12) + t*(1/10 + 1/12 + 1/15) = 1. Solve for t and convert to hours/minutes.

Step-by-Step Solution:

Verification / Alternative check:
Convert 52 minutes to hours (0.8666 h) and check: 2 + 2.8666… hours total work equals 1 job exactly.

Why Other Options Are Wrong:
2 or 4 hours are too short/long; 4 h 52 min overshoots; only 2 h 52 min fits the exact arithmetic.

Common Pitfalls:
Mistiming the openings; ensure C opens at the 2-hour mark, not after 2 additional hours.

Final Answer:
2 hrs 52 min