Boats, Streams, Pipes and Cisterns

Overview

Boats, Streams, Pipes and Cisterns appears about 0.5 times per paper across the last four AFCAT solved papers, placing it in the solid add-on band of Numerical Ability.

Boats-and-streams and pipes-and-cisterns are the two lightest topics inside AFCAT Numerical Ability. Each paper carries roughly half a question from this bucket — sometimes one, often none. But the formulas are short, the arithmetic is clean, and the questions almost never hide their structure. If you have a good week of revision, this should be a 30-second-per-question topic that you almost never get wrong.

Both halves share one big idea: there are two speeds (or two rates) that either add or subtract. In boats, the current of the stream either helps you (downstream) or fights you (upstream). In pipes, an inlet adds water and an outlet takes water away. Once you set up the rates with the correct sign, every problem reduces to one or two lines of arithmetic.

This page covers every variant the AFCAT papers have shown so far — basic boat speed, ratio-of-times shortcuts, average speed on a round trip, two pipes and a leak, time-shifted pipe problems where one tap closes partway through, and the classic tank-overflows-after-leak puzzle. By the end you should be able to recognise the variant in five seconds and finish the sum in under a minute.

Why these are reliable add-on marks

The boats-streams-pipes-cisterns cluster carries an average of about 0.5 marks per AFCAT paper. That looks tiny next to percentages or interest, but it has three properties that make it worth a focused 30-minute revision the day before the exam.

• The formula list is short. You only need the b ± s pair for boats and the +1/T, −1/T signed-rate trick for pipes. Everything else is derivation.
• The arithmetic is gentle. Numbers in this topic are almost always whole numbers or simple fractions like 1/2, 1/3, 2/5. Speeds are usually in single or low double digits.
• The question structure is predictable. AFCAT has not asked a true off-pattern boat or pipe question in any recent paper. If you have seen the variant once, you will recognise it.

A typical AFCAT candidate spends roughly two minutes per Numerical Ability question. If you can finish a boat or pipe question in 45 seconds, you bank 75 seconds for the harder topics. That is the real value of this chapter — not the half-mark itself but the time it frees elsewhere.

Boats and streams basics

A boat moves through water at its still-water speed — the speed it would have on a perfectly calm lake with no current. Call this b, usually given in km/h. A river has a current or stream of its own, flowing in one direction with speed s.

When the boat moves downstream, the current pushes it forward. Its effective ground speed is the sum of its own speed and the current's speed: downstream speed = b + s.

When the boat moves upstream, the current pushes against it. Its effective ground speed drops: upstream speed = b − s. The boat must be stronger than the current (b > s) or it cannot make progress against it at all.

The same logic appears in other physical settings. A swimmer in a river, a plane in a tail-wind or head-wind, a person on an escalator that is moving — all are b ± s problems in disguise. AFCAT mostly uses the boat-and-stream wording, but be ready to spot the structure under any cover story.

Boat formulas: extracting b and s from a pair

Many AFCAT problems give you the downstream and upstream speeds and ask for the still-water speed or the stream speed. Set up two simple equations:

• d = b + s
• u = b − s

Adding the two gives 2b, so b = (d + u) / 2. Subtracting gives 2s, so s = (d − u) / 2. The same averaging trick lets you go the other way: if you know b and s, you simply read off d and u.

The really useful observation is that b is always the average of d and u, and s is always half the gap between them. If you ever feel stuck, ask: what is the average of the two speeds, and what is half their difference? Those are b and s respectively.

Notice that distance and time only enter through the speed formula d = distance / time. Once you have a clean d and u, the rest of the problem is averaging.

Ratio-of-times problems

A second AFCAT favourite is the ratio variant. The same distance is covered downstream and upstream; you are told the ratio of the two times taken. You must find b : s.

The trick is that for a fixed distance, time is inversely proportional to speed. So if downstream time : upstream time = m : n, then downstream speed : upstream speed = n : m. From there you can read off b and s by the averaging method above.

This collapses to a single ready-made formula. If t_down : t_up = m : n, then:

• d : u = n : m
• b : s = (n + m) : (n − m)

Remember the second one. It is one of the few formulas worth memorising in this chapter because AFCAT recycles it.

The ratio b : s is the only thing you can extract from a time ratio alone. To get actual speeds in km/h, you need at least one absolute number — a distance, a time, or one of the speeds.

Round trip and average speed

A round trip means the same distance is covered once downstream and once upstream. Total time = distance / d + distance / u. Average speed for the whole journey is total distance over total time, which works out to the harmonic mean of d and u:

average speed = 2 d u / (d + u)

Do not average d and u arithmetically — that gives b, not the round-trip average speed. The two are different unless s = 0 (no current at all). The harmonic mean is always smaller than the arithmetic mean because the slow upstream leg dominates the total time.

For a quick sanity check: if d = 10 km/h and u = 5 km/h, the round-trip average is 2 × 10 × 5 / 15 = 100/15 ≈ 6.67 km/h, which is below b = 7.5 km/h. The slow leg drags the average down.

Pipes and cisterns basics

A cistern is a tank. Pipes can either fill it (inlets) or empty it (outlets, leaks, drains). The trick that makes every pipe problem simple is the per-hour rate view.

• If pipe A fills a tank in T hours, then in one hour it fills 1/T of the tank. Its rate is +1/T tank per hour.
• If pipe B empties a full tank in T hours, then in one hour it removes 1/T of the tank. Its rate is −1/T tank per hour.
• When several pipes operate together, the net rate is the algebraic sum of the individual rates.
• Time to fill (or empty) the whole tank is 1 / net rate.

This is the only model you need. Every pipe problem in AFCAT — leaks, multiple inlets, time-shifted operations — is a signed-rate addition followed by one division.

For arithmetic, it is often easier to work in LCM units rather than fractions. If the times are 6, 8 and 12 hours, take the tank to be 24 units. The pipes then deliver 4, 3 and 2 units per hour, which is much friendlier than 1/6, 1/8 and 1/12.

Pipe formulas and combined times

Three patterns cover almost every AFCAT pipe question.

Two inlets together. Pipe A fills in T₁ hours, pipe B in T₂ hours. Combined rate = 1/T₁ + 1/T₂. Combined time = T₁ T₂ / (T₁ + T₂). This is the product-over-sum formula, identical to time-and-work for two workers.

Inlet plus outlet (leak). Inlet fills in T₁, outlet drains in T₂. Net rate = 1/T₁ − 1/T₂. Net time = T₁ T₂ / (T₂ − T₁), assuming T₂ > T₁ so the tank does fill. If T₂ ≤ T₁, the outlet is at least as fast as the inlet and the tank never fills.

Multiple inlets and outlets. Add up all the positive and negative rates. The signed sum is the net rate; divide 1 by it to get the time.

Cistern with multiple inlets and outlets

When three or more pipes are involved, the LCM trick really pays off. Pick the LCM of the individual times as the tank's capacity in arbitrary units, then write each pipe's rate as an integer.

Worked example: three pipes can fill a tank in 6, 8 and 12 hours; a fourth pipe empties it in 24 hours. Take the tank to be LCM(6, 8, 12, 24) = 24 units. The four rates are +4, +3, +2 and −1 units per hour, so the net rate is +8 units per hour. Time to fill = 24 / 8 = 3 hours.

This LCM-of-times method is the single most useful pipe-and-cistern habit. It eliminates fractions and makes mental arithmetic possible. Use it every time you have three or more pipes.

Time-shifted pipe problems

The trickiest pipe variant — and the one that catches careless candidates — is the time-shifted problem. One pipe opens first; another joins later; a third closes after some time. Build the solution in stages.

1. Figure out the rate of each pipe (in tank-per-hour or in LCM units per hour).
2. Identify each stage of the timeline — a stage is a span during which the set of open pipes does not change.
3. For each stage, compute the volume filled = (rate during that stage) × (duration of that stage).
4. Set the sum of stage-volumes equal to the total tank capacity and solve for the unknown.

Typical AFCAT framing: pipe A is opened alone for the first two hours, then pipe B is also opened; the tank fills in T more hours; find T. Another framing: pipes A and B together are opened, but B is closed after t hours and A finishes the job; find t.

The arithmetic is always the same — sum of (rate × duration) over stages equals the capacity. Lay out a small table with one row per stage and you cannot go wrong.

Parallel with time-and-work

Pipes-and-cisterns is mathematically identical to time-and-work with one extra feature — some workers are allowed to undo work. A pipe that empties a tank is a worker with negative output.

If you are comfortable with one topic, you are already comfortable with the other. The only mental adjustment is the sign convention. Pipes inherit every shortcut from time-and-work — LCM trick, efficiency ratios, fractional-work-done-per-day.

Tank-full-then-leak-detected problems

This pattern shows up often enough to deserve its own treatment. A pipe takes T₁ hours to fill a tank. Today, because of a leak, the tank actually takes T₂ > T₁ hours to fill. How long would the leak alone take to empty a full tank?

Set up the rates. Inlet rate = 1/T₁. With the leak, the net rate dropped to 1/T₂. The leak's rate (a negative number) is therefore 1/T₂ − 1/T₁, and the leak alone would empty the tank in T₁ T₂ / (T₂ − T₁) hours.

A second variant: the inlet fills the tank, and once full, a small outlet at the top starts overflowing water. Such problems just ask for either the steady-state inflow rate or how long the tank stays full — both follow from the signed-rate sum.

Common AFCAT trap patterns

Four mistakes account for almost every wrong answer in this chapter.

• Sign flip. Writing b − s for downstream by reflex when the boat moves against the description in the question. Always re-read the direction word.
• Forgetting to subtract the leak. A pipe fills in 8 hours; with a leak, it takes 10. Candidates sometimes say the leak takes 10 − 8 = 2 hours. Wrong — leak time is T₁ T₂ / (T₂ − T₁) = 8 × 10 / 2 = 40 hours. Always work in rates, not in time differences.
• Arithmetic-mean confusion on round trips. Average speed on a round trip is the harmonic mean of d and u, not their arithmetic mean. The arithmetic mean gives b, which is a different quantity.
• Units. Speeds in km/h, times in hours, distances in km. If a problem mixes m and km or hours and minutes, convert before computing. AFCAT sometimes deliberately mixes units to catch this.

Time budget

Target finishing time per question in this chapter: 45 to 75 seconds. The numbers are friendly and the formula list is short, so a careful candidate clears these well below the 100-question / 120-minute average.

• 0–10 seconds. Read the question. Identify the variant: pure boat, ratio-of-times, round trip, simple pipe, leak, or time-shifted.
• 10–25 seconds. Write the speeds or rates. For pipes with three or more taps, switch to LCM units.
• 25–60 seconds. Do the arithmetic. For ratio-of-times problems, jump straight to b : s = (m + n) : (n − m).
• 60–75 seconds. Verify the units of the answer match the question. Confirm whether the asked quantity is b, s, d, u, time, distance or volume.

If you find yourself past 90 seconds on a boat-or-pipe question, mark it and move on. With only half a mark per paper on offer here, you cannot afford to bleed time at the cost of a percentages or interest question worth 3 marks.

Worked AFCAT-style examples

Exam-day strategy

1. Memorise just three formulas: b ± s for boats, b : s = (n + m) : (n − m) for time-ratio problems, and 2du/(d+u) for round-trip average speed.
2. For pipes with three or more taps, switch to LCM units immediately — fractions are pointless when the numbers are this small.
3. For time-shifted pipe problems, draw a two-row timeline: rate per stage and duration per stage. Sum the products and equate to the capacity.
4. Read the direction word twice. Most wrong answers in boats are sign-flips, not arithmetic errors.
5. Average speed on a round trip is the harmonic mean, never the arithmetic mean. The arithmetic mean of d and u is b, which is a different quantity.
6. Aim for 45 to 75 seconds per question. If you cross 90, mark and move on — this is a half-mark topic and time is more valuable on percentages or interest.

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