G3 SEC Mathematics Filter Ratios & Data

Two light filters have different detector readings. Filter B gives the larger output, yet filter A gives the larger output-to-input percentage. Which filter performed better? The correct Mathematics answer begins by asking what ‘better’ means. An absolute reading, a proportion, and a claim about a real physical filter are not interchangeable.

This guide turns the fictional planetarium experiment from the previous Physics volume into a sustained G3 SEC Mathematics workshop. You will correct a fixed background reading, form ratios, compare percentage points, read repeated trials, choose appropriate graphs and defend a conclusion. The data are invented teaching figures. They are deliberately designed to show how a plausible answer can be mathematically wrong when its denominator or comparison condition is ignored.

Exam year: 2027 Singapore–Cambridge SEC G3 Mathematics, code K310 (earlier reference 4052).
Two written papers, 90 marks and 2 hours 15 minutes each, weighted equally; Paper 2 ends with a real-world application task.
Practice fractions, percentages, central tendency, graphical representation, units and interpretation.
Never compare unlike input conditions merely because output values are easy to see.
Use every worked answer to build an independent habit: define, calculate, check, interpret.

1. Where this fits in the official 2027 Mathematics syllabus

The official 2027 SEC G3 Mathematics K310 syllabus covers number, algebra, geometry and measurement, and statistics and probability. Students analyse tables and graphs, calculate and interpret mean and measures of spread, use percentages and ratios, and explain misleading statistical representations. The specification gives two papers of 90 marks each, lasting 2 hours 15 minutes each; each carries 50% of the assessment. The last question of Paper 2 focuses on applying Mathematics in a real-world scenario. This invented filter audit uses that style of thinking but is not a past paper and does not forecast a 2027 question.

The Mathematics syllabus also expects relevant working. Non-exact numerical answers are normally given to three significant figures and angles to one decimal place, unless otherwise instructed. Here, exact fractions and suitably rounded percentages are useful. A calculator can perform divisions, but it cannot decide which denominator represents the question. That responsibility remains with the student.

2. Begin by defining the question

A detector output of 92 arbitrary units is greater than one of 83. This is a true numerical comparison. It does not answer which filter passes a larger proportion because the two inputs were different. ‘Which is brighter after filtering?’ and ‘Which has a greater transmitted-to-input ratio?’ are two distinct questions. Write the target quantity in words before calculating. If the final sentence changes the quantity being discussed, the conclusion is invalid even if the arithmetic is flawless.

Your first written line should be ‘Compare corrected output divided by corrected input.’ That phrase fixes a relationship, not merely an operation. It also reminds you that the detector readings are not laboratory-grade claims about physical energy.

3. Keep the data pairs intact

The fictional raw trial table is A: input 128, output 83; B: input 148, output 92; and background 8 in every reading. Keep each output paired with its own input. A common error is to compare A’s output with B’s input or to use one shared denominator without justification. This matters in rates, percentages, probability and performance tables throughout the Mathematics syllabus.

Label columns with meaningful headings: filter, input reading, output reading, background, corrected input, corrected output, ratio. A compact ledger prevents more mistakes than a page of unlabelled calculations.

4. Subtract the background before dividing

For A, corrected input is 128 − 8 = 120 and corrected output is 83 − 8 = 75. For B, corrected input is 148 − 8 = 140 and corrected output is 92 − 8 = 84. This subtraction is an explicit rule of our simplified exercise. The number 8 is not a percentage and must not be subtracted from the final percentage directly. Sequence matters: correct the component measurements, then construct the ratio.

Write brackets: (83 − 8)/(128 − 8), not 83/128 − 8. Brackets communicate the structure of a calculation and prevent a particularly damaging order-of-operations error.

5. Find filter A’s proportion

A’s corrected output-to-input ratio is 75/120 = 5/8 = 0.625. Multiplying by 100 gives 62.5%. The fraction 5/8 is exact for the figures provided. Students who memorise ‘percentage = part over whole times 100’ still need to name the part and whole correctly. In this calculation the corrected output is the numerator and corrected input is the denominator.

Check that the proportion is between zero and one for this model. If your calculation gives 160%, inspect whether you inverted 75 and 120. An answer that violates the expected range is a useful clue, not a reason to erase the working.

6. Find filter B’s proportion

B’s ratio is 84/140 = 3/5 = 0.600, so its percentage is 60.0%. The arithmetic is straightforward; the interpretation is the real test. A has the higher corrected ratio by 2.5 percentage points, while B has the higher corrected absolute output by 9 arbitrary units. Both statements can be true because B also had the higher input reading.

Write two separate comparison sentences and label each. Do not pretend that one statistic answers both questions. In an examination, such precision demonstrates command of the mathematical model.

7. Percentage points are not relative percentages

The difference between 62.5% and 60.0% is 2.5 percentage points. If someone asks how much greater A’s percentage is relative to B’s, then calculate (62.5 − 60.0)/60.0 × 100 ≈ 4.17%. These are different measures. A result reported as ‘4.17 percentage points’ would be incorrect. Choosing the denominator is part of the answer, not optional decoration.

Give the examiner the exact comparison language. Percentage-point change subtracts rates; relative percentage change divides that difference by a base rate. Name the base when it is not obvious.

8. Design a valid comparison

To compare filters fairly, the physical investigation should ideally hold the source setting, detector and geometry steady. The Maths task can still analyse nonmatching inputs by normalising, but the ratio alone cannot guarantee a valid physical comparison. A numerical answer need not claim more than its model supports. Distinguish the given mathematical relationship from the experimental question of whether the measured conditions were genuinely comparable.

A defensible conclusion begins ‘For the supplied readings and correction rule …’. A conclusion beginning ‘This proves filter A always …’ leaves the mathematical evidence behind.

9. Use three trials instead of one

Trial A has raw inputs 128, 126 and 130 and raw outputs 83, 81 and 84, each against background 8. The corrected inputs are 120, 118 and 122; outputs are 75, 73 and 76. Keeping trial pairs together yields 75/120, 73/118 and 76/122. These ratios vary slightly. Repeating the calculation shows whether the one-trial answer is representative.

Calculate each ratio using the complete original fraction before rounding. Early rounding can change a close comparison, especially if a question asks which is larger.

10. A ratio of totals is not automatically a mean of ratios

For trial A, total corrected output is 75 + 73 + 76 = 224 and total corrected input is 120 + 118 + 122 = 360. Their ratio is 224/360 = 0.622222…, or about 62.2%. By comparison the unweighted mean of the three individual ratios is approximately 62.2% but not exactly the same. The ratio of totals weights larger input observations more. The mean of ratios gives each trial equal weight.

These measures answer slightly different questions. State which was requested. If all trial inputs are equal, the results coincide; when denominators differ, they need not.

11. Repeat the B calculation

For B, raw inputs are 148, 150 and 146; raw outputs are 92, 93 and 91; background remains 8. Thus corrected inputs are 140, 142 and 138; outputs are 84, 85 and 83. Totals are 420 and 252 respectively, so aggregate ratio = 252/420 = 0.6 = 60.0%. This gives a transparent comparison with A’s aggregate ratio of 62.2%.

The arithmetic need not be complicated to be demanding. Most lost marks come from a wrong pairing, missing correction or unsupported interpretation rather than from the final division.

12. Choose a mean for the right variable

A’s corrected outputs are 75, 73 and 76. Their arithmetic mean is 224/3 = 74.666…, approximately 74.7 arbitrary units. B’s corrected outputs are 84, 85 and 83, with mean 84.0 units. This shows B’s absolute detector response is larger in these trials. It does not overturn A’s larger output-to-input ratio. A mean has meaning only when the variable and collection conditions are declared.

When comparing means from separate experiments, first inspect whether the groups begin from comparable conditions. The average is not an excuse to ignore the design.

13. Describe spread without declaring a winner

A’s corrected output range is 76 − 73 = 3 units; B’s range is 85 − 83 = 2 units. These show sample spread but do not establish measurement accuracy, scientific causation or universal product consistency. The G3 syllabus includes interquartile range and standard deviation, but using a more advanced statistic does not repair an unfair comparison. A calculator can give a standard deviation even when the experimental question is ill-posed.

Use the simplest useful spread measure unless the task directs otherwise. Tell readers what it measures and what it does not.

14. Calculate corrected ratios trial by trial

A’s three individual ratios are 62.5%, about 61.9%, and about 62.3%. B’s are 60.0%, about 59.9%, and about 60.1%. The overlap question matters more than a dramatic title: within the supplied figures, the sets are separated, but these are just six fictional observations. A general population conclusion or a claim about future measurements requires more than arithmetic.

Keep the rounded ratio beside the original fraction. This makes it possible to audit an answer if a later comparison is sensitive to the final decimal place.

15. Separate an error from a limitation

Subtracting the background after division is an arithmetic-method error. Using one trial to announce universal filter performance is a limitation of inference. Reversing the numerator and denominator is a model error. Drawing a graph with a truncated vertical scale may be a communication problem. Good revision diagnoses these differently. Students often repeat mistakes because they revise an entire chapter when only one precise reasoning step is weak.

Write one short diagnosis beside every correction: ‘wrong denominator’, ‘background not removed’, ‘unfair comparison’ or ‘overclaim’. Then create a new question that targets that single issue.

16. Build a graph that cannot quietly mislead

A two-bar chart showing corrected absolute outputs should use a labelled vertical scale and unit, and it should not silently be titled ‘filter efficiency’. A separate chart of output-to-input percentages answers that different question. If the vertical axis begins near 70 rather than zero, the visible difference between bars may seem enormous. G3 students should recognise when axis choice or graphical representation exaggerates a comparison.

A chart is an argument made with visual choices. Read the title, axes, baseline, labels and source before deciding what the chart supports.

17. Do not connect categories as a continuous experiment

Filters A and B are categories. Drawing a line between their bars may imply meaningful intermediate filters, which the problem does not define. A line graph is useful for quantities changing along a meaningful numerical axis such as time, but it is not automatically the best choice for names. The official Maths syllabus expects students to know purposes, strengths and weaknesses of different representations.

The question ‘Which chart should I use?’ cannot be answered by choosing the most attractive option. Choose the chart that preserves the nature of the variable.

18. Explore missing data honestly

Suppose trial B’s third output was lost. The mean of the remaining two corrected outputs would be (84 + 85)/2 = 84.5, but it would be wrong to describe this as the three-trial mean. You should either identify the missing observation or state that the calculation uses the available two. Do not replace missing data with a value you imagine fits the trend unless the task explicitly defines an estimate.

A precise answer distinguishes observed, estimated and assumed numbers. That habit transfers from school maths to scientific reports and everyday financial comparisons.

19. Use inequalities to make a decision rule

Imagine a fictional display policy requiring an output-to-input ratio of at least 61%. Under the aggregate data A meets the threshold at about 62.2%; B does not at 60.0%. Write the mathematical rule r ≥ 0.61 before testing each. The result applies to this decision criterion, not to every purpose. Another policy might seek a minimum absolute output and produce the opposite choice.

State the rule, apply the same rule to every option, then declare the decision. A defensible decision must reveal its criterion.

20. Work backwards from a target

If a corrected source reading is 160 units and the target output-to-input ratio is at least 62.5%, the corrected output must be at least 0.625 × 160 = 100 units. If the background is 8 and the assumed detector model still holds, the raw output threshold is at least 108 units. This reverse calculation is often more valuable than calculating the percentage of one existing reading.

Check the final requirement: a corrected output is not the same as a raw output. Work backwards carefully through each transformation and label both.

21. Rounding should preserve a close decision

Suppose two ratios were 60.44% and 60.46%. Rounding both to one decimal place could conceal the distinction. If the question asks which is larger, compare unrounded values first and round only for display. Official K310 guidance normally calls for three significant figures in non-exact numbers unless otherwise instructed. The degree of precision should fit the decision and any explicit accuracy requirement.

Use the calculator’s full working value while checking inequalities. Then report a suitable final value rather than implying unlimited measurement certainty.

22. The real-world Paper 2 problem is a reasoning chain

An extended Paper 2 scenario may integrate data, percentages, algebra and an interpretation. You must identify the objective, gather relevant numbers, select an operation, compare alternatives and justify a decision. The filter exercise rehearses that chain. It does not suggest that the official last question will involve a planetarium, a light detector or arbitrary signal units.

Show enough working for a reviewer to reconstruct your calculation. Interpret in the units and circumstances of the question, not in generic words such as ‘A is better’.

23. Transfer from PSLE arithmetic to G3 interpretation

At Primary level, a ratio task may use a small number of tidy quantities. The later challenge is often not arithmetic difficulty but competing interpretations. A larger numerator may pair with a larger denominator. A background correction may change both. A policy threshold may be applied to proportions instead of totals. These cases reward reading as much as calculation.

Families moving from PSLE to secondary study can practise this transition by asking children to explain every number’s role before operating on it. The goal is independence, not faster guessing.

24. The final audit sentence

An examiner can judge a conclusion more easily when it names what was compared, how it was calculated and the condition that limits it. A sound statement is: ‘For the three fictional trials, A’s aggregate corrected detector ratio was about 62.2%, above B’s 60.0%, although B had the higher corrected absolute output because its input was higher.’ It is short enough to be read and complete enough to be checked.

This kind of answer works because it refuses to collapse two incompatible measures into one slogan. It is the mathematical equivalent of honest scientific reporting.

Independent examination practice: show the working before reading the solution

Problem 1: A has raw input 108, raw output 68 and background 8. Find the corrected ratio.

Answer and reason: Subtract 8 from each raw reading: 100 input and 60 output. The corrected ratio is 60/100 = 0.60 = 60%. Record the fraction and final percentage; if the question asks for a fraction of the source signal, retain 3/5.

Try the same skill again with one changed condition before moving on. For this problem, the most helpful self-check is to name every input, identify the intended denominator and explain the final quantity in a sentence. If your answer differs, do not only copy the supplied result. Track the first incorrect line, test the corrected step on another example, and keep the verified method. Mathematics becomes reliable when the reasoning survives new numbers and wording.

Problem 2: A revised trial uses input 140, output 88, background 8. Find the proportion.

Answer and reason: Corrected input = 132; corrected output = 80. The ratio is 80/132 = 20/33 ≈ 0.60606, or 60.6% to three significant figures. Do not round the numerator or denominator before the division.

Problem 3: What is the difference between 62.2% and 60.0%?

Answer and reason: It is 2.2 percentage points when quoted to the displayed decimal place. A relative percentage difference would be about (62.2 − 60.0)/60.0 × 100 = 3.67%, and must be described with its reference base.

Problem 4: Why is 92 larger than 83 but the B ratio smaller?

Answer and reason: The original inputs are different. B’s corrected input is 140 against A’s 120. Dividing corrected outputs by their respective corrected inputs shows the different proportions. A larger part does not imply a larger part-to-whole ratio.

Problem 5: Find the mean corrected output for A’s three repeated readings.

Answer and reason: The corrected outputs are 75, 73 and 76. Mean = (75 + 73 + 76)/3 = 224/3 ≈ 74.7 arbitrary units. This describes output level, not percentage transmission.

Problem 6: Find the range of B’s corrected outputs.

Answer and reason: Values are 84, 85 and 83. Range = highest − lowest = 85 − 83 = 2 units. State the unit and avoid asserting that small range alone proves unbiased measurement.

Problem 7: Find A’s ratio of corrected totals.

Answer and reason: The totals are 224 output and 360 input. Their ratio is 224/360 = 28/45 ≈ 0.6222 = 62.2%. If instructed to find the unweighted average of trial percentages, calculate that separately.

Problem 8: Find B’s ratio of corrected totals.

Answer and reason: The totals are 252 output and 420 input. Ratio = 252/420 = 3/5 = 60%. Do not confuse total output 252 with the input 420 or with the raw detector readings.

Problem 9: A school requires at least 61% corrected ratio. Which filter meets the sample-based rule?

Answer and reason: A is approximately 62.2%, so it meets the stated rule. B is 60.0%, so it fails it. This verdict is only for the invented observations and threshold, not an unconditional recommendation of real equipment.

Problem 10: If corrected input is 200 and minimum ratio is 0.64, what minimum corrected output is required?

Answer and reason: Multiply 200 × 0.64 = 128 units. If the raw reading has a background of 8 units under the model, the raw detector reading would have to be at least 136. Distinguish the corrected target from the instrument display.

Problem 11: Which graph correctly communicates output-to-input ratios?

Answer and reason: A labelled bar chart with categories A and B on the horizontal axis, percentages on a vertical axis and a nonmisleading scale is appropriate. A chart of raw outputs answers a different question. Do not claim a wavelength transmission curve from two category names.

Problem 12: How would you explain an average based on only two of three trials?

Answer and reason: Name the missing observation and say the mean is for the two available trials only. Do not invent the missing reading or report a three-trial mean. In data handling, a correct disclosure may matter more than a tidy-looking number.

Problem 13: A ratio is 0.59859. Round to three significant figures.

Answer and reason: It rounds to 0.599 or 59.9% when expressed as a percentage to three significant figures. Keep the unrounded value for deciding which of two close options is larger.

Problem 14: A chart truncates the vertical axis from 70 to 90. What could go wrong?

Answer and reason: A difference of a few units may appear much larger visually than it is relative to the total size of the readings. Check labels, baseline, scale and whether the chart shows outputs or proportions before accepting its caption.

Problem 15: How would you repair ‘the more expensive filter has higher transmission’ when no price is given?

Answer and reason: Remove the assertion about price because the dossier contains no cost data. A numerical model cannot supply a missing commercial variable. Say what the provided corrected ratios actually show and note that price comparison would require new evidence.

Problem 16: Why should a learner show brackets in (83 − 8)/(128 − 8)?

Answer and reason: They establish that both raw readings are corrected before division. Without brackets a reader might perform a different calculation. Clean notation makes the reasoning visible and protects the answer under examination pressure.

Questions learners and parents often ask

Does a calculator make this easy?

It makes division quick, but most errors here arise from definitions. A calculator will not tell you whether the task requires a raw output, corrected output, ratio of totals, mean of ratios or a fair comparison.

Why study this after the planetarium Physics guide?

The Physics case explains what the readings can and cannot mean. Mathematics examines the arithmetic and visual representation of those readings. Together they teach how to stop a true numerical statement from becoming an unsupported scientific conclusion.

Is this a real filter specification?

No. Every reading, policy and comparison is invented for examination practice. The ratios should never be presented as measured product data or a recommendation for a commercial filter.

Why use both a mean and an aggregate ratio?

They describe different questions. The mean of outputs describes detector level; the aggregate ratio describes total output relative to total input. Separate statistics clarify rather than compete.

Where should a student begin if fractions are weak?

Return to the relationship part/whole, equivalent fractions, decimal conversion and percentage before tackling multiple trials. A short sequence of correct independent problems is more useful than attempting complex data questions prematurely.

What is the next learning step?

Read the related English guide about visual captions and unsupported claims. Then attempt the integrated EMS review without looking at the worked examples, to see whether the three subjects transfer into one coherent decision.

Small Group Tutorials

How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 1203 | Mathematics Planetarium Filter Audit — Background-Corrected Ratios, Percentages and Honest Comparisons