Time & Work

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Subtopic 3 of 5·← Efficiency, 'Twice as Good' & Ratio Cases·Men–Days–Hours Chain & Provisions →

Pipes & Cisterns

Inlets fill (+rate), outlets/leaks drain (−rate). Treat the tank as 1 unit (or LCM litres).

$\text{net rate} = \sum_{\text{inlets}} \frac{1}{T_i} - \sum_{\text{outlets}} \frac{1}{T_j}, \qquad \text{time} = \frac{1}{\text{net rate}}$

Two pipes, one filling one emptying: net time $= \frac{ab}{b - a}$ (a = fill time, b > a = empty time).

Stage-based problems: compute the work already done in the first stage, then re-solve the remainder with the new set of pipes.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

Pipes are workers with signs

A pipe that fills a tank is an inlet; its rate is positive. A pipe (or a leak) that empties the tank is an outlet; its rate is negative. Measure the tank in LCM units — litres or just "units" — exactly as in work problems. A fills in 12 hours → $+1$ tank per 12 h. B empties in 24 hours → $-\frac{1}{24}$ per hour. Everything else is bookkeeping: $\text{net rate} = (\text{sum of inlet rates}) - (\text{sum of outlet rates}), \qquad \text{time} = \frac{\text{tank}}{\text{net rate}}$

One inlet, one outlet

A fills in a hours, B empties in b hours ( $b > a$ ). Net rate $= \frac1a - \frac1b$ , so the tank fills in $T = \frac{ab}{b - a}$ 10 and 15 → $\frac{150}{5} = 30$ hours. Note the minus sign in the denominator — the two-inlet formula $\frac{ab}{a+b}$ is a different question.

More than one of each

Add every inlet rate, subtract every outlet rate. 8 h and 12 h filling, 24 h emptying: take the tank as 24 units → $3 + 2 - 1 = 4$ units/h → 6 hours. Watch the sign of the net rate: if the outlets win, the tank never fills (some questions ask exactly this — the answer is "it will never fill").

Staged problems (pipes opened or closed mid-way)

Work stage by stage:

Compute the work done in stage 1 = (stage-1 rate) × (stage-1 time).
Find what fraction of the tank is still empty.
Re-solve the remainder as a fresh, smaller tank with the new set of pipes. A (10 h) and B (15 h) run 2 hours: 30-unit tank, rate 5 → 10 units done, 20 left. A drain C (30 h) now opens: net $5 - 1 = 4$ units/h → 5 hours more, 7 hours total. Always read carefully whether the question wants the extra time or the total time — both are asked.

The reverse leak question

"A tank fills in 8 hours normally, but a leak at the bottom makes it take 8 hours 40 minutes. How long would the leak take to empty a full tank?" Treat the leak as an unknown outlet: its rate = normal fill rate − slow fill rate $= \frac18 - \frac{3}{26} = \frac{1}{104}$ → the leak empties the full tank in 104 hours.

The empty tank with a leak from the start

Tank full, leak alone empties it in x hours, inlet alone fills the empty tank in y hours — with both acting from empty, fill time $= \frac{xy}{x - y}$ (same fill–drain formula; the leak must be slower than the inlet for filling to be possible at all).

Common traps

Adding an outlet's rate instead of subtracting it.
Using $\frac{ab}{a+b}$ (two-inlets formula) for a fill–drain pair.
Reporting the stage-1 time or the stage-2 time when the total was asked.
Forgetting the tank may already be part-full when the new pipe opens.

Quick revision

Inlets +, outlets −; net rate = inlets − outlets.
Fill–drain pair: $T = \frac{ab}{b-a}$ .
Stage it: work done in stage 1, then a fresh smaller tank.
Reverse leak: leak rate = fast rate − slow rate; leak time = reciprocal.
Net rate ≤ 0 → the tank never fills.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Two or three inlets togethervery common2 practice Q

How to spot it:

Two or three pipes all filling the same tank, individual fill times given, combined fill time asked.

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

Tank = LCM of the fill times; rates in units/hour.
Add the rates; time = tank ÷ total rate.
Two inlets only: use $\frac{ab}{a+b}$ directly.

Why: every pipe pours its own share each hour, exactly like workers on a job.

Example: Two pipes can fill a tank in 10 hours and 15 hours respectively. With both open together, the tank is filled in:

Tank $= 30$ units; rates 3 and 2 → 5/h → $30 \div 5 = 6$ hours.

Type 2: Inlets and outlets together (net rate)very common3 practice Q

How to spot it:

Filling pipes and an emptying pipe (or leak) are open at the same time; the fill time is asked.

\text{net rate} = \sum \frac{1}{T_{\text{in}}} - \sum \frac{1}{T_{\text{out}}}

Mark each pipe + (fills) or − (empties) before any arithmetic.
Tank in LCM units; add inlets, subtract outlets.
Time = tank ÷ net rate. A fill–drain pair alone is $\frac{ab}{b-a}$ .

Why: an outlet undoes work, so it must cancel part of the inlets' contribution.

Example: Two pipes fill a tank in 8 hours and 12 hours, while a third pipe empties it in 24 hours. With all three open, the tank fills in:

Tank = 24 units → $3 + 2 - 1 = 4$ units/h → 6 hours.

Type 3: Reverse: find the leak / outlet from two fill timescommon2 practice Q

How to spot it:

'The tank fills in t hours normally but takes t′ hours because of a leak' — the leak's solo emptying time is asked.

\frac{1}{T_{\text{leak}}} = \frac{1}{t} - \frac{1}{t'}, \quad T_{\text{leak}} = \frac{t \cdot t'}{t' - t}

Fast fill rate $= \frac1t$ , slow (with leak) $= \frac{1}{t'}$ .
Leak rate $=$ fast − slow; invert for the leak's own emptying time.
Put t′ in the same units as t (40 minutes = $\frac23$ hour).

Why: the only difference between the two runs is the leak, so the rate gap belongs to the leak alone.

Example: A cistern normally fills in 8 hours, but a leak at the bottom makes it take 8 hours 40 minutes. The leak alone can empty the full cistern in:

Slow time $= \frac{26}{3}$ h. Leak rate $= \frac18 - \frac{3}{26} = \frac{13-12}{104} = \frac{1}{104}$ → 104 hours.

Type 4: Staged pipes — opened or closed mid-waycommon2 practice Q

How to spot it:

Pipes run for a few hours, then another pipe opens (or one closes); the remaining or total time is asked.

W_1 = \text{rate}_1 \times t_1, \qquad t_2 = \frac{1 - W_1}{\text{rate}_2}

Stage 1: work done = rate × time; find the empty part left.
Stage 2: recompute the net rate with the new set of pipes.
Divide the remainder by the stage-2 rate; add stage times if the total is asked.

Why: the tank remembers only how much water is in it — stages are independent rate problems.

Example: Pipes A and B fill a tank in 10 hours and 15 hours. Both run for 2 hours; then a drain pipe C, which can empty the full tank in 30 hours, is also opened. The total time to fill the tank is:

Tank = 30 units; stage 1: $5 \times 2 = 10$ done, 20 left. Stage 2: $5 - 1 = 4$ units/h → 5 h more → total 7 hours.

Formulas

Net filling rate

\frac{1}{T} = \sum \frac{1}{T_i^{\,+}} - \sum \frac{1}{T_j^{\,-}}

Fill + drain pair

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

Work done in stage 1

W_1 = \text{rate}_1 \times t_1

Tank as LCM units

\text{tank} = \text{LCM of times; rates in units/hour}

Shortcut tricks

⚡ Sign discipline

Empty pipes subtract. Write + and − before adding anything.

Example: Two pipes fill a tank in 8 hours and 12 hours; a third empties it in 6 hours. With all three open, the tank fills in:

$\frac{1}{8} + \frac{1}{12} - \frac{1}{6} = \frac{3 + 2 - 4}{24} = \frac{1}{24}$ ⇒ 24 hours.

⚡ Stage the problem

Finish stage 1 first, then treat the remainder as a fresh tank.

Example: Pipes A (12 h) and B (16 h) fill a tank. Both run for 4 hours, then a drain (24 h) is also opened. The tank is full after how many more hours?

In 4 h: $\frac{7}{48} \times 4 = \frac{7}{12}$ done. New rate $= \frac{7}{48} - \frac{1}{24} = \frac{5}{48}$ , so the remainder takes $\frac{5}{12} \div \frac{5}{48} = 4$ h ⇒ total 8 hours.

⚡ Fill-and-drain pair formula

One inlet, one outlet: net time = ab/(b − a).

Example: A pipe fills a tank in 10 hours; a leak empties the full tank in 15 hours. With both, the tank fills in:

$\frac{10 \times 15}{15 - 10} = 30$ hours.

Where students lose marks

Adding an outlet's rate instead of subtracting it.
Forgetting the tank may already be part-full when a new pipe opens.
Using $\frac{ab}{a+b}$ for a fill-drain pair (that formula is for two inlets).
Reporting the stage-1 time instead of the total time.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

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