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IGCSE · Cambridge (CIE) · 0610

Interpret repeated limewater cloudiness timings

Calculate means and keep the carbon-dioxide conclusion within the method’s limits.

Section 11.1 · Gas exchange in humans

Gas exchange in humans pathway · 11.1

Core and Extended; Biology 0610, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked

Before you begin

  • Recall means and the fair-test variables.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

How is a three-value mean calculated?

What you will learn

  • Read the supplied teaching results
  • Keep ratios distinct from calibration

Interpret repeated limewater cloudiness timings

Invented matched-flow trials: inspired 30,31,29 s; expired 6,5,7 s. Means 30 s and 6 s support a comparative conclusion, not an uncalibrated concentration ratio.Inspired / s30 | 31 | 29 → mean 30Expired / s6 | 5 | 7 → mean 6InferenceExpired has more CO₂
Invented matched-flow trials: inspired 30,31,29 s; expired 6,5,7 s. Means 30 s and 6 s support a comparative conclusion, not an uncalibrated concentration ratio. Original teaching model.
Invented teaching data — equal flow and cloudiness endpoint
Air sampleTrial 1 / sTrial 2 / sTrial 3 / sMean / s
Inspired30312930
Expired6576

Read the supplied teaching results

In the invented fixed-flow investigation, inspired-air timings are 30, 31 and 29 seconds; expired-air timings are 6, 5 and 7 seconds. All use fresh equal limewater and the same cloudiness endpoint. Add each group of three values and divide by three: the means are 30 seconds and 6 seconds respectively.

The expired sample reaches the endpoint sooner in all three paired trials. Under the matched method this supports the conclusion that expired air has more carbon dioxide. Repeats show some variation and make a single unusual endpoint less decisive; they do not remove a systematic error such as an unnoticed difference in flow rate.

Keep ratios distinct from calibration

Thirty divided by six gives a time ratio of five. That describes the observed timings, not automatically a fivefold carbon-dioxide concentration. The reagent reaction, endpoint and apparatus response would need calibration before timing could be translated into an exact gas concentration.

These numbers are teaching data, not a measurement from a real learner. Describe their pattern and method limits without diagnosing lung function. If sample delivery is not matched or one tube begins cloudy, the same numerical means cannot establish a fair air-composition comparison. Check the design before treating arithmetic as sufficient evidence.

Pause and explain

What does the matched 30 s versus 6 s result support?

Put the idea to work

Worked example

Calculate the inspired-air mean for 30, 31 and 29 seconds.

Show the worked solution
  1. Add: 30 + 31 + 29 = 90.
  2. Divide by three trials.

Answer 30 seconds.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Calculate both mean times from the supplied three trials per sample.

Give me a hint

Find 90/3 and 18/3.

Compare my reasoning
  1. Inspired total: 90 seconds.
  2. Expired total: 18 seconds.

Inspired: 30 s. Expired: 6 s.

Look for these in your work

  • Use all repeats
  • Include units
2 · Independent

Write the observed pattern and the supported air-composition conclusion.

Give me a hint

Separate observation from inference.

Compare my reasoning
  1. Expired air clouds comparable limewater sooner.
  2. At matched flow this supports a higher carbon-dioxide fraction.

Observation: expired-air mean is shorter. Inference: expired air contains more carbon dioxide under the controlled comparison.

Look for these in your work

  • State both levels
  • Retain the method conditions
3 · Transfer

Evaluate: 30/6 = 5, therefore expired air has exactly five times the carbon-dioxide concentration.

Give me a hint

A timing ratio is not automatically a concentration calibration.

Compare my reasoning
  1. The quotient correctly describes the time ratio.
  2. No calibration linking time and concentration is supplied.

The time ratio is five, but the exact concentration ratio is not established without calibration.

Look for these in your work

  • Accept the correct arithmetic
  • Reject the unsupported conversion

Common mistakes

  • Dividing a sum by the wrong number of trials.
  • Treating an inverse time ratio as calibrated concentration.
Recall without your notes

What can you explain now?

What does the matched 30 s versus 6 s result support?

Compare with the explanation

Expired air reaches the endpoint faster and contains more CO₂

It is a comparative inference under the controlled method.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Recalculate the means and distinguish time ratio from concentration ratio.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Biology

Course scope and assessment

780 sequenced lessons provide 2340 guided, independent and application problems in characteristics, classification, cells, organisation, specimen size, membrane transport, biological molecules, enzymes, plant and human nutrition, plant transport, circulatory systems, the heart, blood vessels, blood, disease transmission, body defences, immunity, gas exchange in humans, respiration energy uses and investigations, aerobic respiration, anaerobic respiration and recovery, excretion in humans, coordination and response, sense organs, hormones, homeostasis, tropic responses, drugs, asexual reproduction, sexual reproduction, sexual reproduction in plants, sexual reproduction in humans, sex hormones in humans, sexually transmitted infections, chromosomes, genes and proteins, mitosis, meiosis, monohybrid inheritance, variation, adaptive features, selection, energy flow, food chains and webs, nutrient cycles, population, community, ecosystem, population-growth phases, food supply, habitat destruction, pollution, conservation, biotechnology and genetic modification. 61 of 61 syllabus sections have linked teaching and staged practice. The earlier inheritance overview remains introductory, and the monohybrid pathway develops the relevant Core and Supplement requirements. This is not full programme completion or guaranteed exam readiness. Written practice is self-checked.

StudyForA · Learn with understanding

Original teaching content and links to official sources. Check your exam year and qualification before using a resource.