Time & Work · Topic 5 of 6

🚰 Pipes & Cisterns — Basics

3 exam question types, fully solved

Pipes & cisterns is just time & work wearing a different coat. The tankis the “work”, an inlet pipe is a worker that fills it, and an outlet(or leak) is a worker that undoes work — so its rate is negative.

Net fill = inflow − outflow

Take the tank capacity = LCM of all the pipe times, then give each pipe a rate in units/hour. Fillers count as , drainers as .

• Two inlets together: A fills in 12 h, B in 6 h. Capacity = LCM = 12; rates and ; together units/h, so full in h. (Same product-over-sum shortcut: h.)

• Inlet + outlet: inlet fills in 6 h , outlet empties in 12 h . Net unit/h, so the tank fills in h. If the net came out negative, the tank can never fill — it drains.

• Three inlets: just add all three positive rates and divide the capacity by the sum.

Reading tip: “empties”, “drains”, “leak” ⇒ minus. Decide each pipe's sign first, then add. That one habit prevents most pipe mistakes.

🚰Pipe & Cistern SimulatorToggle pipes · watch the tank fill or drain

Pipe AINLET

Pipe BINLET

Pipe COUTLET

~6 hr to fill

Pipe A rate (inlet)1/6 per hr = 16.7%/hr

Pipe B rate (inlet)OFF

Pipe C rate (outlet)OFF

Net rate+16.67%/hr

Net rate = (1/A + 1/B) − 1/C · Time to fill = 1 ÷ net rate · If net rate ≤ 0, the outlet is as fast (or faster) than the inlets — the tank never fills.

Inlet fills 30-unit tank at /h, leak drains at /h. Net = 3/h, so it fills in hours.

Q1An inlet fills a tank in 20 minutes and an outlet empties the full tank in 30 minutes. With both open, the tank fills in:

Q2Two inlet taps P and Q fill a tank in 10 hours and 12 hours. Opened together, they fill it in:

Q3Three inlet pipes fill a tank in 6, 8 and 12 hours. With all three open, the tank fills in:

📚

Real exam questions — Pipes & Cisterns Basics

3 question types · 12 solved examples from real SSC papers

Pipes and cisterns is just work-and-time with a sign rule: inlets add and outlets subtract, and the net fill rate decides the time.

How to solve this type

Set capacity = LCM of all the pipe times, then give every pipe a SIGNED rate: an inlet/filler is positive, an outlet/drain/leak is negative. Net rate = inflow

-

outflow; time = capacity

\div

net rate. If the net is positive the tank fills, if negative it empties, and if it is zero the level never changes. For exactly one filler

x

and one drain

y

the shortcut time is

\dfrac{x\times y}{y-x}

. When the tank starts part-full, divide only that fraction of the capacity by the net rate.

A pipe fills a tank in 12 hours; an outlet empties the full tank in 36 hours. With both open, the tank fills in:

A20 hoursB18 hoursC24 hoursD15 hours

A pipe can fill a tank in 30 hours, but due to a leak it takes 50 hours to fill. The leak alone can empty the full tank in:

A75 hoursB60 hoursC70 hoursD85 hours

A tank is 3/4 full. Outlet A can empty the full tank in 8 hours and inlet B can fill it in 12 hours. With both open, the tank empties in:

A15 hoursB18 hoursC12 hoursD9 hours

Pipe A fills a tank in 12 minutes, pipe B in 16 minutes, and outlet C empties it in 8 minutes. If all three are open, the tank:

AEmpties in 48 minBFills in 24 minCEmpties in 24 minDFills in 48 min