When 70% does not mean 70% of what you think
A mathematics problem can be easy to calculate and hard to understand. A question says that 70% of an ore sample is iron(III) oxide. A student multiplies the total mass by 0.70, sees a neat answer and calls it the mass of iron obtained. The multiplication was right; its interpretation was wrong. The result was mass of iron(III) oxide, a compound that also contains oxygen. A second proportion is needed before the learner reaches the theoretical mass of the element iron.
This is the central skill of the chapter: follow each quantity through a chain without losing the meaning of the whole. You will work from ore mass to oxide mass, from oxide to theoretical metal, from theoretical metal to actual recovery, and from actual recovery to practical comparisons. The quantities will be deliberately labelled at every step. Once you can keep their meanings apart, difficult-looking real-world Mathematics questions become manageable.
The short working rule is to draw a ‘mass path’ before calculating: ore → contained Fe2O3 → theoretical Fe → recovered Fe → usable output. Each arrow has its own ratio or condition. A number written above the wrong arrow may produce a reasonable-looking answer that belongs to a different problem. If you have only one minute to check your work, check the arrows and their units.
This is an invented teaching context, not a claim about an actual mine or production plant. The figures are selected to make the calculations transparent. Industrial ore assays, recovery methods and product specifications vary. You are learning Mathematics, not receiving a forecast of the economics of iron production.
How this volume fits the learner route
Read Volume 0950 first for the Chemistry of iron(III) oxide and why oxygen is removed during reduction. If ratios, percentages and units are still insecure, use the PSLE Mathematics readiness foundation before returning here. The integrated accuracy routine in Volume 0949 will help when a correct calculation is followed by a public claim that is too broad.
The Sengkang G3 SEC route explains subject levels and exam preparation. This chapter owns one narrower Mathematics intent: ore-grade composition and recovery. It does not replace the existing electroplating geometry and thresholds lesson, which concerns a different physical and mathematical task.
1. The real SEC Mathematics assessment
The official 2027 G3 Mathematics syllabus K310 has two papers. Paper 1 is 90 marks in two hours and fifteen minutes, worth 50% of the assessment. Paper 2 is also 90 marks in two hours and fifteen minutes, worth 50%; its final question applies Mathematics in a real-world scenario. The syllabus stresses showing essential working and interpreting solutions in context.
The syllabus includes number and algebra, geometry and measurement, and statistics and probability. A contextual question may combine percentages, ratios, algebra, tables, graphs, and a decision under stated constraints. This guide uses those core tools with an invented ore dataset. It does not predict a particular examination passage or give extra marks for learning technical mining vocabulary.
Unless a question specifies another level of accuracy, the official syllabus generally expects non-exact numerical answers to three significant figures, or one decimal place for angles in degrees. Always preserve enough precision through intermediate stages before rounding. In this article, we sometimes keep more digits while explaining a calculation so the effect of rounding is easy to see.
2. Name the denominator before the numerator
Consider the sentence: ‘A 2.00-tonne sample is 70% iron(III) oxide by mass.’ The word ‘sample’ supplies the denominator. The compound Fe2O3 supplies the numerator category. The contained oxide mass is 2.00 multiplied by 0.70, giving 1.40 tonnes. Nothing in that sentence yet says the entire 1.40 tonnes will become iron metal. The oxide also contains oxygen.
A percentage is not a free-standing property. Seventy percent of what? Of the ore, of the oxide, of the maximum theoretical iron, or of the delivered product? One way to prevent confusion is to attach a phrase every time you write a percentage: ‘70% of ore mass is Fe2O3’ rather than simply ‘70%’. That habit improves both accuracy and communication.
When a problem gives ore grade as elemental iron rather than iron(III) oxide, do not apply the oxide-composition fraction again. The definition changes the path. Mathematically, a percentage statement is an equation between two specified masses; the label tells you which one.
3. Convert a chemical formula to a mass fraction
For the original example, assume the relative atomic masses Fe = 56 and O = 16. Two iron atoms contribute 2 × 56 = 112 relative mass units. Three oxygen atoms contribute 3 × 16 = 48. The relative formula mass of Fe2O3 is 160. Therefore the fraction of pure Fe2O3 mass attributable to iron is 112/160 = 0.70.
This 70% is not the ore grade. It is the composition fraction of a chemical compound. The two figures happen to be numerically equal in our example, but they have different denominators. A particularly robust student writes ‘Fe fraction of Fe2O3 = 112/160’, not just another 70%, so that the source of the result remains visible.
The theoretical iron mass in the 1.40 tonnes of contained oxide is 1.40 × 112/160 = 0.980 tonnes. If your calculator gives 0.980, read it as 980 kilograms, because one tonne is 1,000 kilograms. This is the maximum iron mass attributable to that Fe2O3 in the simplified model assuming complete conversion and no relevant losses. It is not the mass of final steel.
4. Keep a working trail with units
A well-presented solution might be written in three lines. Contained Fe2O3 = 2.00 t × 0.70 = 1.40 t. Theoretical Fe = 1.40 t × 112/160 = 0.980 t. Theoretical Fe = 980 kg. Each line has a clear object and unit. An examiner can follow the reasoning even if the final arithmetic is later affected by an error.
Avoid writing ‘2 × 70 × 70 = 980’ without explaining the percentage conversions. The calculation hides which quantities are fractions and which are percentages. In general, to multiply by 70%, use 0.70 or 70/100, not 70. A well-named working line is often faster than repairing a chain of ambiguous numbers.
The process can also be expressed as a single compound fraction: 2.00 × 0.70 × 0.70 = 0.980 tonnes. This is perfectly valid once both factors are explicitly defined. The detailed method, however, is safer when a later question modifies only one stage, such as changing the assay or recovery rate.
5. Introduce recovery without changing the theoretical maximum
Suppose the processing worksheet says 80% of the theoretical iron is recovered as iron metal. Then recovered iron = 0.980 × 0.80 = 0.784 tonnes, or 784 kilograms. The gap between theoretical iron and recovered iron is 0.980 − 0.784 = 0.196 tonnes. This is an invented recovery assumption, not a universal industry value.
Notice that the 80% is measured against the theoretical iron mass, not the original 2.00 tonnes of ore. If you multiply ore mass directly by 80%, you obtain 1.60 tonnes, which exceeds the theoretical quantity of iron available in the described sample. A simple upper-bound check catches the error without needing to repeat every arithmetic step.
The ratio of recovered iron to the original ore mass in this example is 0.784/2.00 = 0.392, or 39.2%. That is a valid whole-sample yield description. It must not be renamed 80% recovery, because 80% was already the fraction of theoretical iron recovered. Both numbers are correct for different denominators.
6. Distinguish metal content from product mass
Suppose the collected output is a 0.800-tonne material that contains 98% iron by mass. The actual iron content is 0.800 × 0.98 = 0.784 tonnes. If the theoretical iron was 0.980 tonnes, the iron recovery is 0.784/0.980 = 80%. This is deliberately designed to agree with the previous example while revealing a new distinction: recovered product can include non-iron material.
If a question asks for mass of recovered iron, use the mass of the element in the product. If it asks for total product mass, use the full output, including the stated non-iron fraction. If it asks for purity, divide the iron mass by product mass. Readers may casually call a partly impure product ‘iron’, but a mathematically careful answer follows the defined quantities.
This reasoning is also useful outside Chemistry. A food bag may be 80% usable contents, a manufactured part may have a packaging mass, and a mixture may contain a known concentration of an active ingredient. The same denominators matter in each case. Learn the structure rather than memorising the ore story.
7. Reverse a target
A planning question may give a required recovered output and ask for the starting ore mass. Imagine the original ore grade and 80% recovery remain unchanged, but the target is at least 1.18 tonnes of recovered iron. The recoverable fraction of ore is 0.70 × 0.70 × 0.80 = 0.392. Therefore starting ore mass must be at least 1.18/0.392 ≈ 3.0102 tonnes.
If ore can be supplied only in whole 0.5-tonne loads, the mathematical minimum of about 3.01 tonnes is not an available purchase. Three tonnes would provide only 3 × 0.392 = 1.176 tonnes of recovered iron, slightly below 1.18. The next available 3.5-tonne load would be required under those assumptions. Rounding the starting mass to 3.0 tonnes would cause a false ‘feasible’ decision.
This is a key difference between a calculation and a planning decision. The question might demand a minimum number of containers, batches or deliveries. The model’s continuous solution is a starting point. The practical answer must respect whole-number or step-size constraints. When the target is a minimum, do not round down merely because it looks closer.
8. Compare two grades before comparing rates
Suppose Ore A contains 70% Fe2O3 and Ore B contains 60% Fe2O3 by mass. If the theoretical iron fraction within Fe2O3 remains 70%, Ore A contains a theoretical 49% elemental iron by mass while Ore B contains 42%. Under equal recovery conditions, A yields more iron per tonne of starting ore. The difference is seven percentage points of theoretical iron content, not merely a seven-percent relative advantage.
The relative increase from 42% to 49% is (49 − 42)/42 × 100 ≈ 16.7%. That percentage is a different question from the seven-percentage-point absolute difference. Learn to say which comparison you are making. People often use the word ‘percent’ for both, and the resulting public claims can mislead even when the source numbers are correct.
Ore B might still be preferred in a broader cost problem if its price, transport or processing circumstances differ. A higher concentration is not always the whole decision. In a Mathematics examination, use the extra constraints the question gives rather than assuming that the highest-grade input must be the most economical.
9. Compare recovery paths fairly
Imagine two invented processing plans applied to the same 2.00-tonne ore sample. Plan C has 80% recovery of theoretical iron and yields 0.784 tonnes. Plan H has 85% recovery and yields 0.980 × 0.85 = 0.833 tonnes. The absolute output difference is 0.049 tonne, or 49 kilograms. The recovery-rate difference is five percentage points.
Relative to Plan C’s 0.784 tonnes, Plan H yields 0.049/0.784 × 100 = 6.25% more iron. The latter is not the same as saying recovery rises by 6.25 percentage points. If the question asks which plan produces more output from equal input, both can be useful descriptions, but each must carry its own denominator.
Do not infer that Plan H is environmentally or economically superior merely from recovery percentage. The planning data may omit energy, reagent manufacture, labour and material quality. Mathematics tells you exactly what the given quantities imply; it cannot manufacture missing criteria.
10. Graph the process without confusing cumulative quantities
A line graph could show recovered mass over processing time for a specified run. A separate bar chart might show total iron recovered by different runs. They are not the same representation. When a graph’s vertical axis gives cumulative mass, its gradient can suggest rate of increase over an interval. When it gives mass per hour, the height already represents a rate. The wording on the axis determines the operation.
A plot that begins above zero can exaggerate differences between two outputs. If the purpose is to compare 0.784 and 0.833 tonnes, a bar chart from zero gives a clearer visual sense of the change. A detailed table can still show exact values. In an examination, decide what the graph actually supports rather than being persuaded by its design.
If no measurements were made between two times, a straight connecting segment is a convenient representation, not proof that recovery was perfectly uniform throughout that interval. The same principle appeared in the solar-panel evidence review. Contexts differ; the logic of interpreting a small dataset remains the same.
11. Means and spread for repeated runs
Three fictional runs with equal theoretical iron input report recovered iron masses of 0.760, 0.784 and 0.812 tonnes. Their mean is (0.760 + 0.784 + 0.812)/3 ≈ 0.785 tonnes. The median is 0.784 tonnes. The range is 0.812 − 0.760 = 0.052 tonne. Those statistics answer different questions: typical centre by one measure, central ordered value by another and the extent of the observed spread.
A mean tells you something about average output across the three runs, but with so few observations it does not prove long-run reliability. A median can be less affected by an extreme value. The range is simple but uses only the minimum and maximum. If a paper provides more observations, use the measure of spread and the comparison method it requests, such as the standard deviation, carefully.
Do not add the three run outputs to the same batch’s mass unless the runs truly represent separate input batches. A problem might describe repeated measurements on one sample instead. Repetition and production are different things. Identify what each record represents before summing.
12. Rounding and false pass results
Return to the 1.18-tonne target. An exact predicted output of 1.176 tonnes is below target even though it rounds to 1.18 tonnes to three significant figures. You must test feasibility against the unrounded or suitably precise value, not against a prematurely rounded display. State that the prediction falls short by 0.004 tonne, equivalent to four kilograms.
A practical statement can be: ‘Three tonnes would give 1.176 tonnes under the assumed fractions, so it does not meet the 1.18-tonne requirement; a larger permitted batch is necessary.’ This is better than silently rounding and declaring success. Context and inequalities often matter more than the final digit of a calculation.
When measurements themselves have limits of accuracy, another layer of caution is needed. The word ‘at least’ makes a boundary decisive. If a measured or reported mass has uncertainty near the target, the student should not invent a guarantee. Use the stated precision and any given bounds to reason about what can be concluded.
13. Translate word problems into algebra
Let x be the starting mass of ore in tonnes. If 70% of the ore is Fe2O3, 70% of Fe2O3 mass is iron and recovery is 80%, then y, the recovered iron mass, is y = 0.392x. The model’s gradient is 0.392 tonnes of recovered iron per tonne of ore, under the specified assumptions. If x increases by one tonne, y increases by 0.392 tonne in this simplified linear model.
The equation does not claim that a real production process operates with identical recovery at every scale. It encodes the assumed constant fractions in the exercise. A good learner knows how to use the model and where it could fail. If recovery decreases for larger batches, the linear equation would need revisiting when new data are supplied.
An inequality turns the target into 0.392x ≥ 1.18. Dividing by 0.392 gives x ≥ about 3.01. If the available batches are whole half-tonne loads, x must satisfy both the numerical inequality and the discrete loading rule. A real-world answer needs both conditions.
14. Twenty-four short worked cases
These original cases vary the denominator, assumption, unit and target. Solve each in a notebook, write the meaning of every intermediate value, then read the worked correction. Do not use the numbers as real mine recovery rates. They are invented to teach transfer across unfamiliar examination contexts.
Practice 1: A 5.00 t sample is 60% Fe2O3. Find oxide mass.
Worked answer: 3.00 t, because 5.00 × 0.60 = 3.00. This is not yet iron mass. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Name the substance before reporting the number. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 2: The previous 3.00 t Fe2O3 is completely reduced. Find theoretical iron.
Worked answer: 3.00 × 112/160 = 2.10 t Fe. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Distinguish compound mass from elemental contribution. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 3: The 2.10 t theoretical iron has an actual recovery of 75%.
Worked answer: Actual recovered iron is 1.575 t, or about 1.58 t to three significant figures. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Keep the unrounded mass available for later thresholds. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 4: An output product is 2.00 t and 95% pure Fe.
Worked answer: Iron content is 1.90 t, while product mass remains 2.00 t. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Purity has the product mass, not original ore mass, as denominator. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 5: A sample contains 0.60 t theoretical iron and yields 0.48 t Fe.
Worked answer: Recovery is 0.48/0.60 ×100 = 80%. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Never divide by total ore unless that is the specified yield definition. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 6: A student uses 70 instead of 0.70 in a mass formula.
Worked answer: Multiply by 70/100 or 0.70. Using 70 exaggerates the result one hundredfold. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Percentage notation must be converted correctly before multiplication. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 7: Theoretical iron is 0.980 t and measured iron is 1.10 t.
Worked answer: The measurement exceeds the theoretical limit under the stated assumptions. Recheck mass definition, purity or data. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: An impossible-looking result is a diagnostic clue, not a reason to ignore theory. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 8: Two plans recover 80% and 85% of theoretical iron.
Worked answer: The rate difference is 5 percentage points, not 5% of the first rate. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Absolute and relative comparisons have different interpretations. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 9: Plan A yields 0.784 t and Plan B 0.833 t.
Worked answer: Difference is 0.049 t = 49 kg. Relative to A, B is 6.25% higher. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: State the baseline when reporting relative change. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 10: Ore A has 49% theoretical iron; Ore B has 42%.
Worked answer: Absolute difference is 7 percentage points; relative to B, A is about 16.7% greater. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Select the comparison requested, not the one with a bigger numeral. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 11: A 3.00 t ore batch gives 39.2% recovered Fe.
Worked answer: Output is 3.00 ×0.392 =1.176 t. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Do not treat 39.2% as oxide grade: it is a combined output fraction. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 12: The 1.176 t predicted output is compared with a 1.18 t target.
Worked answer: It fails by 0.004 t or 4 kg, although rounding could show 1.18. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Test an inequality before presenting the rounded answer. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 13: A target of 1.18 t uses recovery model y=0.392x.
Worked answer: x≥1.18/0.392≈3.01 t of ore. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Reverse a multiplicative model by dividing by the effective fraction. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 14: Ore arrives only in 0.5 t loads; mathematical minimum is 3.01 t.
Worked answer: Choose 3.5 t, not 3.0 t. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Minimum targets and discrete choices demand upward adjustment. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 15: A graph shows recovered mass from 0.40 to 0.70 t over 10 hours.
Worked answer: Average gradient over that interval is 0.030 t/h, or 30 kg/h. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: A gradient needs both change in output and change in time. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 16: A chart starts its vertical scale at 0.75 t to compare 0.784 and 0.833.
Worked answer: The cropped scale magnifies apparent difference; read the marked values and note the truncation. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Visual emphasis does not replace numerical comparison. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 17: Three run outputs are 0.760, 0.784 and 0.812 t.
Worked answer: Mean≈0.785 t; median=0.784 t; range=0.052 t. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: These statistics answer different questions about the data. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 18: The middle of three observations is reported as their mean.
Worked answer: The middle ordered value is the median. Calculate the mean by adding and dividing by the number of runs. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Centre has more than one useful definition. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 19: The reported ore mass is in tonnes, but iron demand is in kilograms.
Worked answer: Multiply tonnes by 1,000 before comparing: 0.784 t is 784 kg. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Use consistent units before drawing a pass/fail conclusion. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 20: The overall recovery is 0.392 of ore and a student reports 392%.
Worked answer: 0.392 is 39.2%. Multiplying by 100 gives the percentage. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Fractions and percentages require a deliberate conversion. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 21: A batch is 60% oxide and has 80% recovery; iron fraction within oxide is 70%.
Worked answer: Recovered iron fraction of ore is 0.60×0.70×0.80=0.336, or 33.6%. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Write each factor’s meaning to prevent applying the wrong percentage. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 22: One supplier has higher ore grade but higher transport cost.
Worked answer: Higher grade alone proves more theoretical iron per tonne, not that the total plan is cheaper. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: A feasible decision needs all criteria named in the question. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 23: A reported output includes a 2% non-iron component.
Worked answer: If product mass is 0.800 t, actual Fe mass is 0.800×0.98=0.784 t. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Do not count impurities as recovered iron when asked for elemental content. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 24: A class subtracts oxygen mass from total ore instead of contained Fe2O3.
Worked answer: First isolate the oxide mass, then apply its composition. Ore may contain other materials. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Atom fractions belong to a specified compound, not an unanalysed mixture. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 25: Two runs have unequal input masses and recovered iron masses of 0.6 and 0.8 t.
Worked answer: Normalise each output by the relevant input or theoretical maximum before comparing efficiency. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: More total output may reflect more input rather than a better process. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 26: A model assumes constant recovery but the data show it decreases with batch size.
Worked answer: The relation y=0.392x would no longer be justified across those sizes. Use the new information or restrict the conclusion. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: A mathematical model is conditional on its assumptions. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 27: A product purity increases from 90% to 95%.
Worked answer: The increase is 5 percentage points; relative to 90%, it is approximately 5.56%. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Percentage-point language prevents common interpretation errors. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 28: A result is 1.575 t with a requirement of 1.58 t.
Worked answer: It is below the requirement by 0.005 t before rounding. Report the threshold decision using the unrounded value. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Display precision and decision precision are not always the same. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 29: A spreadsheet computes 1.40×(112/160) but labels output ‘Fe2O3 recovered’.
Worked answer: The calculated output is theoretical elemental Fe, not the input Fe2O3 mass. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Correct labels are as necessary as correct multiplication. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
Practice 30: A problem asks for the smallest number of 2 t loads to yield at least 5 t Fe at 39.2%.
Worked answer: Each load yields 0.784 t, so 5/0.784≈6.38; choose 7 whole loads. Explain your calculation using words before substituting numbers. Then check whether the result is possible given the earlier quantities and the stated constraints.
What this case teaches: Count whole units after dividing; check 6 loads produce only 4.704 t. In a real question, underline the reference quantity and identify whether the answer is a mass, fraction, percentage, rate or count. This small habit prevents a correct arithmetic operation being attached to the wrong object.
15. A complete real-world application question
A fictional school assessment gives two ore sources. Source A has 70% Fe2O3 by mass and costs 120 units per tonne of ore. Source B has 60% Fe2O3 and costs 95 units per tonne. In both, iron makes up 70% of Fe2O3 mass; assume 80% of theoretical iron is recovered. The question asks which source supplies at least 1.00 tonne of recovered iron at lower raw-ore purchase cost if ore can be bought in whole one-tonne lots. Other costs are declared equal for this exercise.
For A, one tonne of ore yields 0.70×0.70×0.80=0.392 tonne iron. To reach 1.00 tonne requires ceiling(1/0.392)=3 whole tonnes. Raw-ore cost is 3×120=360 units. For B, one tonne yields 0.60×0.70×0.80=0.336 tonne iron. The required number of whole tonnes is ceiling(1/0.336)=3. Raw-ore cost is 3×95=285 units. Both choices reach the target under the model.
The lower listed purchase cost for this exact order is Source B. It may at first surprise students because B has lower oxide content. The discrete purchase rule means both require three tonnes, so price decides between them in this restricted question. If the target changes, the minimum lot counts may change. Do not generalise the outcome to every order, and do not infer overall operating cost from raw-ore price alone.
To check, A’s three tonnes yield 1.176 tonnes iron and B’s three tonnes yield 1.008 tonnes. Neither violates the theoretical composition limit. A provides more excess output but costs 75 more units under the stated assumptions. If the task values only meeting a target at minimum purchase cost, B is the appropriate recommendation. If it values maximum output or reserve, the answer may differ.
16. Why many students lose marks despite a correct calculator result
There are four common failures. The first is using a percentage against the wrong whole. The second is rounding a threshold too early. The third is losing units during conversion. The fourth is answering a more interesting question than the question asked. Each has a specific repair: write the mass path; compare before rounding; place units in every line; and reread the instruction before writing the final sentence.
On an examination page, give your reasoning in a tidy sequence. You do not need long prose for every step, but you need sufficient working to show where each quantity came from. A final answer such as ‘3.5 tonnes’ is meaningful only if it has been derived from the fractions, the minimum target and the available lot size. Where the question asks for an explanation, add the sentence that makes those conditions clear.
17. Practice plan
First session: solve ten direct fraction conversions. Second: work backward from three targets, including one whole-batch constraint. Third: compare two input grades and distinguish percentage differences from percentage-point differences. Fourth: interpret a small chart and identify a misleading axis. Fifth: attempt the full source-selection problem without viewing the model answer and explain your recommendation in one paragraph.
If mistakes recur, do not respond by completing ever more random questions. Name the point at which the path breaks. Is it the contained oxide fraction, the chemical composition ratio, the actual recovery factor, or the final constraint? Return to that stage with three simpler examples, then try the full problem again. The aim is reliable transfer rather than memorising one attractive chain of numbers.
18. The next reading
The following English volume looks at claims about material supply and environmental benefits. It asks whether a publicity paragraph faithfully explains the distinction between theoretical yield, observed output and the wider claims the evidence cannot establish. For exam administration and subject codes, refer to SEAB’s official 2027 G3 index. This guide teaches a transferable mathematics method, not an additional compulsory examination topic.
