• Assume the chosen fill level holds 150 litres of water.
• Assume each mug holds 300 millilitres and no water is spilled.

1. Convert the bath volume: 150 litres = 150,000 millilitres.
2. Divide by the mug volume: 150,000 ÷ 300 = 500 mugs.

Keep the mug size fixed. If you instead assume a bath volume of 120–180 litres, the estimate becomes 400–600 mugs. A larger bath must need more mugs; a larger mug must mean fewer refills.

Convert both volumes to the same unit before dividing. The result describes the assumed fill level, not the capacity of every bath.

• Assume the route is 4 kilometres long, including its turns.
• Assume an average walking speed of 4 kilometres per hour, with no extra stops.

1. Use time = distance ÷ speed.
2. 4 kilometres ÷ 4 kilometres per hour = 1 hour, or 60 minutes.

With the same assumed route, a speed of 3 kilometres per hour gives 80 minutes; 5 kilometres per hour gives 48 minutes. Under those alternative speed assumptions, a range of roughly 50–80 minutes is more informative than a precise arrival time.

Route distance can differ from straight-line distance. If your route includes waiting at crossings, include that time separately or account for it in the average speed.

• Assume 120 centimetres of usable shelf length.
• Assume books average 3 centimetres thick and contain 250 pages each.
• Assume the books stand upright in one row with no gaps or bookends.

1. Estimate the book count: 120 ÷ 3 = 40 books.
2. Multiply by pages per book: 40 × 250 = 10,000 pages.

Holding the assumed page count fixed, an average thickness of 2–4 centimetres gives 60–30 books, or 15,000–7,500 pages. This is a sensitivity check, not a measured range for real books.

Related assumptions can move together. Thicker books may also have more pages, so changing thickness while fixing page count is useful for isolating an input but may not describe a realistic collection.

• Assume the box measures 20 centimetres along each inside edge.
• Assume spherical marbles with a diameter of 2 centimetres, so their radius is 1 centimetre.
• Assume the marbles themselves occupy 60% of the box volume; the rest is empty space.

1. Box volume = 20 × 20 × 20 = 8,000 cubic centimetres.
2. Using sphere volume = 4πr³ ÷ 3, one marble occupies approximately 4.2 cubic centimetres.
3. Volume occupied by marbles = 8,000 × 0.60 = 4,800 cubic centimetres.
4. Number of marbles ≈ 4,800 ÷ 4.2 ≈ 1,140.

If the assumed occupied fraction changes to 50–70%, the calculation gives approximately 950–1,330 marbles. Ignoring all gaps would give about 1,900, which should be an upper bound for this volume model.

Dividing container volume by object volume is only a starting point. Packing, object shape and the container edges affect the usable space.

• Assume 600 students attend lunch.
• Assume 75% choose water and each of those students drinks 0.4 litres.
• Assume no additional water is needed for staff, spills or unused servings.

1. Water drinkers = 600 × 0.75 = 450 students.
2. Water served = 450 × 0.4 = 180 litres.

If the assumed share choosing water were 50–100%, with everything else fixed, demand would be 120–240 litres. Check that the number of water drinkers never exceeds the number of students.

A population is not the same as a customer count. Estimate the fraction participating before multiplying by consumption per person.

• Assume a town has 100,000 residents.
• Assume 60% of residents travel to work or school on the chosen weekday.
• Assume 10% of those travellers cycle, each making two one-way trips.

1. Travellers = 100,000 × 0.60 = 60,000 people.
2. Cyclists = 60,000 × 0.10 = 6,000 people.
3. One-way bicycle trips = 6,000 × 2 = 12,000 trips per weekday.

If the assumed cycling share is 5–15%, the estimate becomes 6,000–18,000 trips. An answer of 6,000 under the original assumptions counts cyclists, not their outbound and return journeys.

State exactly what you count. People, return journeys and one-way trips are different units, even when they describe the same activity.

• Assume the roof has a horizontal footprint of 10 metres by 8 metres.
• Assume 10 millimetres of rain falls uniformly across that footprint.
• Assume 75% of that water reaches the tank and the tank has enough spare capacity.

1. Catchment area = 10 × 8 = 80 square metres.
2. Rain depth = 10 millimetres = 0.01 metres.
3. Water falling on the footprint = 80 × 0.01 = 0.8 cubic metres = 800 litres.
4. Water collected = 800 × 0.75 = 600 litres.

With assumed rainfall of 5–20 millimetres and the same roof and collection fraction, the result is 300–1,200 litres. Doubling rain depth should double the estimate.

Use the horizontal catchment footprint for this model, rather than the sloping surface area. Area multiplied by depth becomes volume only when the length units agree.

• Assume an audience of 4,000 people, with 30% placing one order each.
• Assume the interval lasts 20 minutes and each counter serves one order every 20 seconds.
• Assume a planning target of 75% of theoretical counter capacity to leave room for delays.

1. Demand = 4,000 × 0.30 = 1,200 orders.
2. The interval lasts 20 × 60 = 1,200 seconds.
3. Theoretical capacity per counter = 1,200 ÷ 20 = 60 orders.
4. Planned capacity per counter = 60 × 0.75 = 45 orders.
5. Counters needed = 1,200 ÷ 45 ≈ 26.7; round up to 27 staffed counters.

Changing the assumed buying share to 20–40% gives 800–1,600 orders. At the same planned capacity, round 800 ÷ 45 and 1,600 ÷ 45 upward to get 18–36 counters.

Capacity estimates need both a time window and a service rate. Round up when a fraction of a counter cannot operate, and remember that sufficient total capacity does not guarantee short queues.