Taps A, B, C fill a tank in 12 h, 15 h, and 20 h respectively. A stays open continuously; B and C open alternately for one hour each. In how many hours will the tank be full?

Problem restatement
Compute total time when A runs continuously and B, C alternate hourly.

Given data

• A = 1/12 per hour
• B = 1/15 per hour
• C = 1/20 per hour

Concept/Approach
Evaluate work done in a 2-hour cycle (A+B then A+C). Multiply full cycles until near completion, then add the next hour as needed.

Step-by-step calculation
2-hour work = (A+B) + (A+C) = 2A + B + C= 2/12 + 1/15 + 1/20 = 1/6 + 1/15 + 1/20LCM 60 ⇒ (10 + 4 + 3)/60 = 17/60 per 2 hoursAfter 6 hours (3 cycles): 3 × 17/60 = 51/60Next hour is A+B: 1/12 + 1/15 = 9/60 ⇒ 51/60 + 9/60 = 1Total time = 7 hours

Common pitfalls
Assuming 6 hours (stopping at cycles) without considering the next alternating hour.

Final Answer
7 hours