Time & Work

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Pipes & Cisterns

⏱ 7 min read🧩 8 question types🎯 12 practice Q

The idea in one minute

A tank is filled by some pipes and emptied by others. A pipe that fills is an inlet. A pipe or a leak that empties is an outlet.

Treat it exactly like a time-and-work question. Inlets do positive work, outlets do negative work. The quickest method: take the tank's capacity as the LCM of all the times, so every pipe's speed becomes a whole number of units per hour.

The basic idea

Every question gives you pipes and the time each one takes on its own.

An inlet fills the tank. Count its work as plus (+).
An outlet or a leak empties the tank. Count its work as minus (−).
"A fills the tank in 6 hours" means A fills $\frac{1}{6}$ of the tank every hour.

Rule: Fill = plus, empty = minus. Write the sign next to every pipe before you calculate anything.

The LCM method (use it every time)

Fractions like $\frac{1}{12} + \frac{1}{18}$ are slow and easy to get wrong. Give the tank a number of units instead.

Tank capacity = LCM of all the given times.
Speed of each pipe = capacity ÷ its time (units per hour).
Net speed = inlets added, outlets subtracted.
Time = capacity ÷ net speed.

Example: A fills a tank in 12 hours, B in 18 hours. Tank = LCM(12, 18) = 36 units. A fills 3 units/hour, B fills 2 units/hour. Together: 5 units/hour, so $36 \div 5 = 7.2$ hours = 7 hours 12 minutes.

When one pipe empties the tank

Use the same method. Just subtract the emptying pipe.

A fills in 6 hours, B empties in 9 hours. Tank = 18 units. A = +3, B = −2, so net = +1 unit/hour. The tank fills in 18 hours.

For exactly one filling pipe (a hours) and one emptying pipe (b hours):

T = \frac{a \times b}{b - a}

Check: $\frac{6 \times 9}{9 - 6} = 18$ hours.

Watch: If the emptying pipe is faster, the net speed is negative. A full tank will empty; an empty tank will never fill.

Pipes opened or closed in between

Split the question into stages. The tank only cares how much water is already in it.

Stage 1: water filled = speed × time.
Water still needed = capacity − water filled.
Stage 2: water still needed ÷ new net speed.

A (10 h) and B (15 h) run for 2 hours. Tank = 30 units, speed 5, so 10 units are filled and 20 are left. Now a drain C (30 h) opens: net speed = 5 − 1 = 4. The rest takes 5 hours, so the total is 7 hours.

Watch: Read the last line carefully. Some questions ask for the extra time (5 hours), others for the total time (7 hours).

Finding the leak

"A tank normally fills in 8 hours. Because of a leak, it now takes 10 hours. How long will the leak take to empty the full tank?"

Tank = LCM(8, 10) = 40 units. Normal speed = 5, speed with the leak = 4. The difference, 1 unit/hour, is the leak. So the leak empties the tank in 40 hours.

Tip: Leak time = $\frac{t_1 \times t_2}{t_2 - t_1}$ where $t_1$ is the normal time and $t_2$ the slower time: $\frac{8 \times 10}{2} = 40$ .

Pipes opened one after the other

Some questions open the pipes turn by turn, one hour each. Treat one full round as a single unit of time.

A fills in 4 hours, B in 6 hours. They are opened alternately for one hour each, starting with A. Tank = 12 units, A = 3, B = 2, so one round (2 hours) fills 5 units. After 2 rounds (4 hours) 10 units are filled and 2 are left. It is A's turn: 2 units at 3 per hour take $\frac{2}{3}$ hour = 40 minutes. Total = 4 hours 40 minutes.

Watch: Before counting the last round, check whether the tank gets full during the round. Stop there.

When the answer is in litres

If one pipe's speed is given in litres, the capacity is a real number. Write every pipe's speed in litres per hour and solve for the capacity.

A leak empties a full tank in 6 hours. An inlet that fills 4 litres a minute (240 litres an hour) is opened, and now the full tank empties in 8 hours. Leak − inlet = capacity ÷ 8:

\frac{C}{6} - 240 = \frac{C}{8} \Rightarrow \frac{C}{24} = 240 \Rightarrow C = 5760 \text{ litres}

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Two or three filling pipes together

How to spot it:

All the pipes fill the same tank. Each pipe's time is given, and the time with all of them open is asked.

\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3}

Method

Take tank = LCM of the times.
Find each pipe's speed = tank ÷ its time.
Add the speeds.
Time = tank ÷ total speed.

Why it works:

Each pipe adds its own share of water every hour, so the shares simply add up.

Try this

Three pipes can fill a tank in 20, 30 and 60 minutes. If all three are opened together, the tank fills in:

Show solution

Tank = LCM(20, 30, 60) = 60 units.
Speeds: 3, 2 and 1 units per minute.
Total speed = 6 units per minute.
Time = $60 \div 6 = 10$ minutes.

Answer

10 minutes

Type 2very common3 practice Q

Filling and emptying pipes open together

How to spot it:

At least one pipe fills and at least one pipe (or a leak) empties, all at the same time.

\text{net} = \sum \text{inlets} - \sum \text{outlets}

Method

Mark each pipe + (fills) or − (empties).
Take tank = LCM of the times.
Net speed = inlets − outlets.
Time = tank ÷ net speed.

Why it works:

The emptying pipe removes water while the others add it, so its speed cancels part of theirs.

Try this

Pipes A and B fill a tank in 12 hours and 15 hours. Pipe C empties it in 20 hours. If all three are opened together, the tank fills in:

Show solution

Tank = LCM(12, 15, 20) = 60 units.
A = +5, B = +4, C = −3.
Net speed = 6 units per hour.
Time = $60 \div 6 = 10$ hours.

Answer

10 hours

Type 3common2 practice Q

Find the leak from two fill times

How to spot it:

"It normally fills in t hours, but because of a leak it takes longer." The leak's own emptying time is asked.

T_{\text{leak}} = \frac{t_1 t_2}{t_2 - t_1}

Method

Take tank = LCM of the normal time and the slower time.
Normal speed − slower speed = speed of the leak.
Leak time = tank ÷ leak speed.

Why it works:

The only difference between the two cases is the leak, so the lost speed is the leak's speed.

Try this

A pipe fills a tank in 6 hours. Because of a leak at the bottom, it takes 8 hours. How long will the leak take to empty the full tank?

Show solution

Tank = LCM(6, 8) = 24 units.
Normal speed = 4, speed with the leak = 3.
Leak speed = 1 unit per hour.
Leak time = $24 \div 1 = 24$ hours.

Answer

24 hours

Type 4common2 practice Q

A pipe is opened or closed in between

How to spot it:

Pipes run for some time, then a pipe is added or removed. The remaining time or the total time is asked.

t_2 = \frac{\text{capacity} - \text{speed}_1 \times t_1}{\text{speed}_2}

Method

Stage 1: water filled = speed × time.
Water still needed = tank − water filled.
Stage 2: divide what is left by the new net speed.
Add the stages if the total time is asked.

Why it works:

The tank does not care how the water got there. Each stage is a fresh, smaller question.

Try this

Pipes A and B fill a tank in 12 hours and 16 hours. Both are opened for 4 hours. Then a drain that can empty the full tank in 24 hours is also opened. In how many more hours will the tank be full?

Show solution

Tank = LCM(12, 16, 24) = 48 units. A = 4, B = 3, drain = −2.
First 4 hours: $7 \times 4 = 28$ units, so 20 units are left.
New net speed = 7 − 2 = 5 units per hour.
More time = $20 \div 5 = 4$ hours.

Answer

4 more hours (8 hours in total)

Type 5occasional

Pipes opened one after the other

How to spot it:

The pipes are opened in turns, one hour (or one minute) each: "alternately" or "A, then B, then A…".

Method

Take tank = LCM of the times.
Water filled in one round = sum of the speeds in that round.
Count the full rounds that fit without filling the tank.
Finish the last part pipe by pipe; stop the moment the tank is full.

Why it works:

The pattern repeats every round, so full rounds can be counted in one step. Only the last round needs care.

Try this

Pipe A fills a tank in 3 hours and pipe B in 4 hours. They are opened alternately for one hour each, starting with A. In how much time will the tank be full?

Show solution

Tank = LCM(3, 4) = 12 units. A = 4, B = 3.
One round (A then B, 2 hours) = 7 units.
After 1 round: 7 units in 2 hours, 5 left.
A's hour adds 4 (3 hours, 1 left). B needs $\frac{1}{3}$ hour = 20 minutes for the last unit.

Answer

3 hours 20 minutes

Type 6common

Tank capacity in litres

How to spot it:

One pipe's speed is given in litres per minute or per hour, and the tank's capacity is asked.

\frac{C}{t_{\text{out}}} - \text{inlet (L/h)} = \frac{C}{t_{\text{net}}}

Method

Change every speed to the same unit (litres per hour).
Write each pipe's speed with C as the capacity, e.g. C/5.
Make an equation from the net speed.
Solve for C.

Why it works:

A speed in litres fixes the size of the tank, so the capacity is a real number, not LCM units.

Try this

A leak can empty a full tank in 5 hours. An inlet pipe that fills 3 litres a minute is opened, and now the full tank empties in 20 hours. What is the capacity of the tank?

Show solution

Inlet = 3 × 60 = 180 litres per hour.
Leak − inlet = net emptying: $\frac{C}{5} - 180 = \frac{C}{20}$ .
$\frac{C}{5} - \frac{C}{20} = \frac{3C}{20} = 180$ .
$C = 180 \times \frac{20}{3} = 1200$ litres.

Answer

1200 litres

Type 7occasional

When was a pipe closed?

How to spot it:

Two pipes start together, one is closed after some time, and the tank still fills in a given total time. The closing time is asked.

Method

Take tank = LCM of the times.
The pipe that was never closed works for the whole time: find its water.
The rest of the water came from the closed pipe.
Its working time = its water ÷ its speed.

Why it works:

Working backwards from the finished tank is faster than guessing the closing time.

Try this

Pipes A and B can fill a tank in 20 minutes and 30 minutes. Both are opened together, but A is closed after some time. The tank is full in 18 minutes. After how many minutes was A closed?

Show solution

Tank = LCM(20, 30) = 60 units. A = 3, B = 2 per minute.
B works all 18 minutes: $2 \times 18 = 36$ units.
A filled the other 60 − 36 = 24 units.
A worked $24 \div 3 = 8$ minutes.

Answer

After 8 minutes

Type 8common

One pipe is faster than the other

How to spot it:

The question compares the pipes ("twice as fast", "takes 5 hours more") and gives only the time together.

Method

Turn the comparison into speeds, e.g. twice as fast → speeds 2 and 1.
Tank = speed together × time together.
Each pipe's time = tank ÷ its speed.

Why it works:

Once speeds are in a ratio, the tank size follows from the time together and everything else is division.

Try this

Pipe A is twice as fast as pipe B. Together they fill a tank in 12 hours. How long will B alone take?

Show solution

Speeds: A = 2, B = 1 unit per hour; together 3.
Tank = $3 \times 12 = 36$ units.
B alone = $36 \div 1 = 36$ hours (A alone = 18 hours).

Answer

36 hours

Formula sheet

Net speed

\text{net} = (\text{inlets}) - (\text{outlets})

Speeds in units per hour, with tank = LCM of the times.

Time to fill

T = \frac{\text{capacity}}{\text{net speed}}

Two inlets

T = \frac{a b}{a + b}

Both pipes fill.

One inlet, one outlet

T = \frac{a b}{b - a}

a = filling time, b = emptying time, b > a.

Leak time from two fill times

T_{\text{leak}} = \frac{t_1 t_2}{t_2 - t_1}

t₁ = normal time, t₂ = time with the leak.

Stage method

t_2 = \frac{\text{capacity} - \text{speed}_1 \times t_1}{\text{speed}_2}

Shortcuts that save time

⚡ One inlet, one outlet: multiply over subtract

For one filling pipe and one emptying pipe, time = product ÷ difference. No LCM needed.

Example

A pipe fills a tank in 10 hours. A leak empties the full tank in 15 hours. With both open, how long does the tank take to fill?

Show solution

Product = $10 \times 15 = 150$ .
Difference = $15 - 10 = 5$ .
Time = $150 \div 5 = 30$ hours.

Answer

30 hours

⚡ Leak from two fill times

The leak is the only thing that changed, so the drop in speed belongs to the leak. Use product ÷ difference of the two fill times.

Example

A cistern fills in 8 hours. Because of a leak it takes 8 hours 40 minutes. The leak alone can empty the full cistern in:

Show solution

8 h 40 min = $\frac{26}{3}$ hours.
Leak speed = $\frac{1}{8} - \frac{3}{26} = \frac{13 - 12}{104} = \frac{1}{104}$ tank per hour.
So the leak empties the tank in 104 hours.

Answer

104 hours

⚡ Check the options before finishing

Filling pipes together are always faster than the fastest pipe alone. Adding an outlet always makes it slower than the inlets alone. Use this to cut two options in a few seconds.

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding the emptying pipe's speed as if it fills the tank.

Put a minus sign in front of every outlet or leak before adding.

Mistake 02

Using $\dfrac{ab}{a+b}$ for one filling and one emptying pipe.

That formula is for two inlets. For fill + empty use $\dfrac{ab}{b-a}$ .

Mistake 03

Giving the stage-2 time when the question asks for the total time.

Underline the last line: extra time or total time. Add the stages if it says total.

Mistake 04

Forgetting the tank is already part-full when a new pipe is opened.

Find the water still needed first, then divide by the new speed.

Mistake 05

In alternate-pipe questions, counting full rounds past the point where the tank is full.

Before each hour, check whether the water left is less than that pipe fills in an hour.

Mistake 06

Mixing minutes and hours (8 h 40 min taken as 8.4 hours).

Convert first: 40 minutes = $\dfrac{2}{3}$ hour, so 8 h 40 min = $\dfrac{26}{3}$ hours.

Quick revision

Read this the night before the exam.

Inlet = plus, outlet or leak = minus.
Tank = LCM of the times; speed = tank ÷ time.
Two inlets: $\dfrac{ab}{a+b}$ . One inlet, one outlet: $\dfrac{ab}{b-a}$ .
Stages: water filled so far, water left, then divide by the new speed.
Leak time = $\dfrac{t_1 t_2}{t_2 - t_1}$ .
Alternate pipes: count full rounds, then check the last round hour by hour.
Net speed zero or negative: an empty tank never fills.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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