Alternate operation of two inlets and one outlet (A, then B, then C) Taps A and B are inlets (20 min and 30 min), tap C is an outlet (15 min). They are operated alternately for one minute each in the repeating order A → B → C → A → …. How long will it take to fill the tank (stop immediately when full)?

Introduction / Context:
Minute-by-minute alternation requires accounting for partial minutes at the end, since the tank may become full during an inlet’s minute before the next outlet minute would start. We therefore compute cycle gains and then finish with a fractional minute if needed.

Given Data / Assumptions:

• A: +1/20 per min; B: +1/30 per min; C: −1/15 per min.
• Order repeats A → B → C, each exactly 1 min.
• Stop as soon as the tank becomes full.

Concept / Approach:
Net gain per 3-minute cycle = 1/20 + 1/30 − 1/15. After many full cycles, a final partial minute by A finishes the remaining fraction.

Step-by-Step Solution:

Verification / Alternative check:
Any longer (e.g., waiting for B or C) would overshoot or reduce level; stopping mid-minute when full is appropriate.

Why Other Options Are Wrong:
180, 167, 185 are rounded or mis-sequenced approximations; 532/3 is exact.

Common Pitfalls:
Forcing whole-minute stops, which is unrealistic for continuous flow; ignoring the outlet’s negative contribution.

Final Answer:
532/3 min