Time & Work, Pipes & Cisterns Notes

Last reviewed 22 Sept 2026 · 6 min read

Work as a rate

If a person finishes a job in $n$ days, then in one day they do $\frac{1}{n}$ of it. Rates add:

\frac{1}{T} = \frac{1}{a} + \frac{1}{b} + \dots ⟹ T = \frac{1}{\frac{1}{a} + \frac{1}{b} + \dots}

For two people, $T = \frac{ab}{a + b}$ .

The LCM method

Instead of fractions, call the whole job the LCM of the given days. Each person's rate is then a whole number of units a day, and the arithmetic stays clean.

For A in 12 days and B in 18 days: total work $= LCM (12, 18) = 36$ units; A does 3 units/day, B does 2, together 5 → $36/5 = 7.2$ days. Most exam problems are fastest this way.

Efficiency

FormulaEfficiency and time

Efficiency $\propto \frac{1}{time}$ : if A is twice as efficient as B, A takes half the time.
If A is $k$ times as efficient as B and together they take $T$ days, then A alone takes $T (\frac{k + 1}{k})$ and B alone $T (k + 1)$ .
Percent more efficient: "A is 25% more efficient than B" means rates are $5 : 4$ , so times are $4 : 5$ .

The men–days–hours relation

FormulaWork equation

\frac{M _{1} D _{1} H _{1}}{W _{1}} = \frac{M _{2} D _{2} H _{2}}{W _{2}}

$M$ men working $H$ hours a day for $D$ days complete work $W$ . Drop $H$ when hours are not mentioned and drop $W$ when the job is the same in both cases. Men and days are inversely proportional; men and work are directly proportional.

Working together in the usual patterns

Someone leaves partway: compute the work done together up to that day, then divide the remainder by the rate of whoever is left.
Someone joins partway: the same, in reverse.
Alternate days: add the two rates as a repeating two-day block and count how many whole blocks fit; finish the remainder with whoever's turn it is.
Wages: the money is shared in the ratio of work done, which for the same period is the ratio of efficiencies.

Pipes and cisterns

The same arithmetic with a sign: a filling pipe is a positive rate, an emptying pipe or a leak is negative.

FormulaCistern rules

Filling pipes $a$ and $b$ hours, waste pipe $c$ hours: $\frac{1}{T} = \frac{1}{a} + \frac{1}{b} - \frac{1}{c}$
Full in $a$ hours without a leak, $b$ hours with it ( $b > a$ ): the leak alone empties a full tank in $\frac{ab}{b - a}$ hours
A pipe closed early: let it run $x$ hours, write (work by pipe 1) + (work by pipe 2) $= 1$ , and solve for $x$
A negative net rate means the tank never fills

Worked examples

Worked ExampleExample 1 — working together

A can do a job in 12 days and B in 18 days. How long together?

Solution. $T = \frac{12 \times 18}{12 + 18} = \frac{216}{30} =$ 7.2 days (LCM method: 36 units, 3 + 2 = 5 a day).

Worked ExampleExample 2 — one rate from the pair

A and B together finish a job in 8 days; A alone takes 12 days. How long does B take alone?

Solution. $\frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$ → 24 days.

Worked ExampleExample 3 — efficiency

A is twice as efficient as B, and together they finish in 12 days. Find their separate times.

Solution. Let B's rate be $r$ , A's $2 r$ ; $3 r = \frac{1}{12} \Rightarrow r = \frac{1}{36}$ . B alone 36 days, A alone 18 days.

Worked ExampleExample 4 — men and days

Ten men finish a job in 15 days. How long will 6 men take?

Solution. $M_{1} D_{1} = M_{2} D_{2} \Rightarrow 10 \times 15 = 6 \times D \Rightarrow D =$ 25 days.

Worked ExampleExample 5 — men, days and hours

Eight men working 6 hours a day finish a job in 10 days. How many days will 12 men working 5 hours a day take?

Solution. $D = \frac{8 \times 6 \times 10}{12 \times 5} = \frac{480}{60} =$ 8 days.

Worked ExampleExample 6 — leaving partway

A alone takes 20 days and B alone 30 days. They work together for 5 days, then A leaves. How long does B take to finish?

Solution. In 5 days together: $5 (\frac{1}{20} + \frac{1}{30}) = 5 \times \frac{5}{60} = \frac{5}{12}$ . Remaining $\frac{7}{12}$ at B's rate $\frac{1}{30}$ → $\frac{7}{12} \times 30 =$ 17.5 days.

Worked ExampleExample 7 — alternate days

A can do a job in 10 days, B in 15. They work on alternate days, A starting. When is it finished?

Solution. A two-day block does $\frac{1}{10} + \frac{1}{15} = \frac{1}{6}$ . Six blocks complete the job, so it takes 12 days.

Worked ExampleExample 8 — wages

A and B finish a job together in 6 days; A alone would take 10. They are paid ₹ 3,000 in all. Find each share.

Solution. B's rate $= \frac{1}{6} - \frac{1}{10} = \frac{1}{15}$ , so the rates are $\frac{1}{10} : \frac{1}{15} = 3 : 2$ . A gets ₹ 1,800, B ₹ 1,200.

Worked ExampleExample 9 — pipes with a waste pipe

Pipe A fills a tank in 6 hours, pipe B in 8 hours, and pipe C empties it in 12 hours. All three are opened together.

Solution. $\frac{1}{6} + \frac{1}{8} - \frac{1}{12} = \frac{4 + 3 - 2}{24} = \frac{5}{24}$ → $\frac{24}{5} =$ 4.8 hours.

Worked ExampleExample 10 — a leak

A cistern fills in 4 hours, but with a leak it takes 5 hours. How long would the leak alone take to empty a full cistern?

Solution. $\frac{1}{4} - \frac{1}{5} = \frac{1}{20}$ → 20 hours (formula: $\frac{ab}{b - a} = \frac{4 \times 5}{1}$ ).

Worked ExampleExample 11 — a pipe closed early

Two pipes fill a tank in 20 and 30 minutes. Both are opened, but the second is closed after some time and the tank fills in 18 minutes. When was it closed?

Solution. $\frac{18}{20} + \frac{x}{30} = 1 \Rightarrow \frac{x}{30} = 0.1 \Rightarrow x =$ 3 minutes.

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Time & Work, Pipes & Cisterns