Notes 23 Sep 2026

Survival laboratory

Revising 2n ESO: rebuilding an Arctic base with base areas, volumes, litres, percentages and an equation. Each task has its solution hidden underneath.

Download the problem sheet (PDF, in Catalan) Download the solutions (PDF, in Catalan)

The situation

Headquarters has been told that a huge storm has damaged the research base in the Arctic. Luckily, the rescue team from our bases in the area managed to rescue the scientists who were working there, and they are expected to recover. You are the new team who must rebuild the laboratory, carry on with their research and keep yourselves safe.

You have limited resources, extreme conditions and very little time to make critical decisions. You will have to calculate, plan, optimise and justify every step with scientific and mathematical rigour.

Note. All calculations are rounded to two decimal places.

Prior knowledge
  1. Geometry and measurement: areas of plane figures (square, rectangle, triangle, circle); base area of solids; volumes: prism $V = A_{\text{base}} \cdot h$, cylinder $V = \pi r^2 h$, pyramid $V = \frac{1}{3} A_{\text{base}} \cdot h$.
  2. Units: length, area and volume units, and how they relate to litres: $1\ \text{dm}^3 = 1$ L and $1\ \text{m}^3 = 1000$ L.
  3. Arithmetic and proportion: rule of three, percentages and operations with decimals and with $\pi$.
  4. Equations: turning a statement into an equation, solving for the unknown (for example $a^2 = 12$) and square roots.
  5. Context: interpreting the results (litres, days, free space) and checking that they make sense.

Challenge 1: Designing the research camp

We realise that every module is unusable, although most of the furniture and materials inside can be reused. We have to redesign the camp and, given the urgency, these are the only modules we have:

Square-based prism

2,5 × 2,5 × 6 m

Buildings

Large cylinder

radius 1.2 m · height 2.5 m

Drinking water

Square-based pyramid

3,5 × 3,5 × 2,5 m

Fuel · recyclable waste

Small cylinder

radius 1 m · height 2 m

Organic waste · general waste

For now there will be 12 people at the camp, and the total area available is 300 m². We need:

Modules for the camp buildings

  • 1 sleeping module for every 3 people.
  • 1 kitchen module.
  • 1 laboratory module for every 4 people.
  • 1 storage module for every 6 people.

Extra modules in total

  • 1 drinking water module.
  • 1 fuel module.
  • 1 organic waste module.
  • 1 general waste module.
  • 1 recyclable waste module.

Task 1. Do all the modules fit?

Before going on, check whether all these modules can be placed in the camp; if not, some things will have to go. Step 1: work out how many modules of each type you need. Step 2: work out the base area of each one. Step 3: find the total area and whether they all fit.

See the solution
Step 1

Building modules (prisms), for 12 people:

$$\text{Sleeping: } 12 : 3 = 4 \quad \text{Kitchen: } 1 \quad \text{Lab: } 12 : 4 = 3 \quad \text{Storage: } 12 : 6 = 2$$
$$4 + 1 + 3 + 2 = 10 \text{ prisms}$$

Extras: 1 large cylinder (water), 2 pyramids (fuel and recyclables) and 2 small cylinders (organic and general waste).

10 prisms + 1 large cylinder + 2 pyramids + 2 small cylinders = 15 modules

Step 2

The base of the prisms and pyramids is a square ($\ell^2$) and that of the cylinders is a circle ($\pi r^2$):

NumberShapeBase area (m²)Subtotal (m²)
10Square-based prism$2{,}5^2 = 6{,}25$$62{,}50$
1Large cylinder$\pi \cdot 1{,}2^2 \approx 4{,}52$$4{,}52$
2Square-based pyramid$3{,}5^2 = 12{,}25$$24{,}50$
2Small cylinder$\pi \cdot 1^2 \approx 3{,}14$$6{,}28$
Step 3
$$62{,}50 + 4{,}52 + 24{,}50 + 6{,}28 = 97{,}80\text{ m}^2$$

That is far less than the $300\ \text{m}^2$ available.

If the cylinder bases are only rounded at the end, the total is $97{,}81\ \text{m}^2$.

The modules take up $97{,}80\ \text{m}^2$: they all fit.

Task 2. Free space

Work out what percentage of the camp is left free of modules. For safety it must be more than 30 %. Is it?

See the solution
$$300 - 97{,}80 = 202{,}20\text{ m}^2$$
$$\dfrac{202{,}20}{300} \cdot 100 = 67{,}40\ \%$$

With $97{,}81\ \text{m}^2$ there are $202{,}19\ \text{m}^2$ free, and the percentage is also $67{,}40\ \%$.

$67{,}40\ \%$ is left free, more than $30\ \%$: the camp is safe.

Challenge 2: Checking capacities

In the middle of the Arctic we must be sure we have enough oxygen if we have to seal ourselves in, enough drinking water, and enough room for waste, which can only be emptied when it is collected. To know the capacity we need the volume of each module, remembering that $1\ \text{dm}^3 = 1$ L.

Task 1. The capacity of the modules

Work out the capacity of the modules, in m³ and in litres.

See the solution

Since $1\ \text{m}^3 = 1000\ \text{dm}^3 = 1000$ L, we multiply by $1000$ to get litres. Volume of each module:

NumberShapeCalculationVolume (m³)Litres
10Square-based prism$2{,}5 \cdot 2{,}5 \cdot 6$$37{,}50$$37\,500$
1Large cylinder$\pi \cdot 1{,}2^2 \cdot 2{,}5$$11{,}31$$11\,310$
2Square-based pyramid$\tfrac{1}{3} \cdot 3{,}5^2 \cdot 2{,}5$$10{,}21$$10\,210$
2Small cylinder$\pi \cdot 1^2 \cdot 2$$6{,}28$$6\,280$

A pyramid has a third of the volume of the prism with the same base and height; that is where the $\tfrac{1}{3}$ comes from.

Task 2. The air in the buildings

How much air is there in the building modules?

See the solution

The buildings are the 10 prisms:

$$10 \cdot 37{,}50 = 375\text{ m}^3 = 375\,000 \text{ L}$$

$375\,000$ L of air

Task 3. Water, fuel and waste

How much drinking water, fuel and waste of each type can we store?

See the solution
WhatModuleCapacity
Drinking water1 large cylinder$11\,310$ L
Fuel1 pyramid$10\,210$ L
Recyclable waste1 pyramid$10\,210$ L
Organic waste1 small cylinder$6\,280$ L
General waste1 small cylinder$6\,280$ L

Challenge 3: Survival

Before sending the expedition we must check that we meet the survival requirements.

A. Oxygen during a lockdown

In an extreme storm or a chemical leak, the 12 people will have to shut themselves in the building modules, which are connected to each other. We want to know how long they could last on the oxygen without renewing the air.

Data: a person at rest needs about $0{,}5$ L of oxygen per minute, and the air we breathe is $21\ \%$ oxygen.

Task A1

Using the data from challenge 2, how much air is there in the 10 modules?

See the solution
$$10 \cdot 37{,}50 = 375\text{ m}^3 = 375\,000 \text{ L}$$

$375\,000$ L

Task A2

How much oxygen is there in that air?

See the solution

$21\ \%$ of the air is oxygen:

$$0{,}21 \cdot 375\,000 = 78\,750 \text{ L}$$

$78\,750$ L of oxygen

Task A3

How much oxygen do the 12 people need every minute?

See the solution
$$12 \cdot 0{,}5 = 6 \text{ L/min}$$

$6$ L/min

Task A4

Work out how many minutes we have before the oxygen runs out, and convert it to hours and days.

See the solution
$$\dfrac{78\,750}{6} = 13\,125 \text{ min}$$
$$\dfrac{13\,125}{60} = 218{,}75 \text{ h}$$
$$\dfrac{218{,}75}{24} \approx 9{,}11 \text{ days}$$

$13\,125$ min $= 218{,}75$ h $\approx 9{,}11$ days

Task A5

How many days will they last? Do you think survival depends only on oxygen?

See the solution

Going by oxygen, about 9 days. But it does not depend only on oxygen:

  • The carbon dioxide we breathe out builds up and becomes dangerous long before the oxygen runs out.
  • Furniture and materials take up part of the modules: there is less air than the volume we calculated.
  • Water, food, heating with the fuel and humidity also matter.

About 9 days, but the real limit would come sooner.

B. A smaller fuel pyramid

We would like to make the fuel pyramid smaller, because we will not need that much and it is an extra hazard. Can you find an equation that gives a new side length for the square base so that the fuel is exactly $10\ \text{m}^3$? The height is still $2{,}5$ m.

a = 3,50 m  →  V = ⅓ · a² · 2,5 = 10,21 m³
See the solution

Volume of a square-based pyramid with side $a$ and height $h$:

$$V = \dfrac{1}{3} \cdot a^2 \cdot h$$

With $V = 10$ and $h = 2{,}5$, the equation is:

$$\dfrac{1}{3} \cdot a^2 \cdot 2{,}5 = 10$$

We solve for $a^2$ and take the square root (the side must be positive):

$$a^2 = \dfrac{10 \cdot 3}{2{,}5} = 12 \;\Longrightarrow\; a = \sqrt{12} \approx 3{,}46 \text{ m}$$

Check: with $a = 3{,}46$ we get $V \approx 9{,}98\ \text{m}^3$; the difference comes from rounding. Notice the side only shrinks by 4 cm, from $3{,}5$ to $3{,}46$ m: the original volume, $10{,}21\ \text{m}^3$, was already very close to 10.

The side of the base must be $\sqrt{12} \approx 3{,}46$ m.