Pipes and Cisterns

Link with Time and Work

Pipes and cisterns are time-and-work problems where:

• An inlet pipe fills the tank → positive work.
• An outlet pipe / leak empties the tank → negative work.

Method (LCM / capacity)

Take tank capacity = LCM of the times.

Example: Pipe A fills in 12 h, pipe B empties in 18 h. Capacity = 36 units. A = +3/h, B = −2/h. Both open → +1/h → 36 hours to fill.

Formulas

• Two inlets (a, b hours): time = ab/(a + b).
• Inlet a, outlet b (b > a): time to fill = ab/(b − a).

Leak problems

A pipe normally fills a tank in 6 h but takes 8 h because of a leak. Capacity = 24. Pipe = 4/h, effective = 3/h → leak = 1/h → leak alone empties the full tank in 24 h.

Exam tips (RRB NTPC)

1. Always mark emptying pipes with a minus sign.
2. If pipes are opened at different times, compute work done in each phase separately.

Analogy

A tank with an inlet and a leak is like a wallet with salary coming in and expenses going out. If ₹3,000 comes in each day and ₹2,000 goes out, you save only ₹1,000 a day — so reaching ₹36,000 takes 36 days, not 12.