The invented planetarium needs to coat a removable iron mounting plate, not its displayed metal-rich specimen. The plate has a rectangular front and back, narrow outer edges and a circular through-hole. Its true paintable area is about 0.822 square metres. With two coats, a stated coverage rate and a 15% allowance, it needs about 0.315 litres of paint. One fictional 0.50-litre tin at $20 meets that requirement and costs less than two 0.25-litre tins at $24.
This G3 Mathematics guide shows exactly how those results follow, how to change the model and how to write a reasoned decision in a SEC-style contextual problem. It makes no recommendation about painting genuine museum objects, nor any prediction about future examination questions.
- 2027 syllabus: Mathematics K310 (older code 4052)
- Geometry: two perforated faces, outer edges and an inner cylindrical wall
- Calculations: surface area, units, coating rate, repeated coats and percentage allowance
- Decision: whole-tin capacity, cost ceiling and cheapest feasible order
- Continuity: Chemistry Volume 1222 explains the protective mechanism; this chapter models coating quantity
1. Understand the actual task
The fictional planetarium’s team must estimate paint for a removable iron plate. They are not painting a meteorite and they are not being asked to restore a real artefact. The plate measures 0.80 m long, 0.50 m high and 0.020 m thick. A circular hole of radius 0.080 m passes completely through the plate. The front, back, four outer edges and the curved surface inside the hole all receive paint. Every surface needs two coats. Coverage is six square metres per litre for each coat; a 15% additional quantity is allowed for waste. A small 0.25 L tin costs $12, and a 0.50 L tin costs $20. This complete specification lets the learner distinguish the continuous quantity consumed from the integer purchase required. Write those two outputs separately at the start of the solution.
2. Connect the model with the 2027 G3 paper
SEAB assigns Mathematics K310 to 2027 G3 SEC school candidates, replacing the historical 4052 reference. The published examination specification has two 90-mark papers, each lasting two hours and fifteen minutes and contributing 50% of the subject grade. Paper 2 ends with a question applying mathematics in a real-world scenario. Calculators are permitted on both papers, yet essential working must be shown. This invented painting case practises the sorts of modelling and interpretation involved in that syllabus, not a forecast of an actual paper. Avoid claims that a particular coating question will appear. The exam value lies in recognizing the solid, choosing consistent units, constructing justified expressions and explaining a constrained decision in ordinary language. Those skills transfer to tanks, boxes, packaging, railings and other non-routine word problems.
3. Sketch three views before calculating
Draw the plate from the front, from the side and in a simple three-dimensional sketch. Label length and height on the front view, thickness on the side view and radius inside the circular opening. Shade the large front face, then mark the back face separately. Trace the outer perimeter to identify the four rectangular side surfaces. Finally draw a finger pointing to the wall inside the through-hole: that narrow curved surface is also painted. A through-hole removes material from the broad faces but creates an exposed inner wall. The diagram is not decorative. It is a ledger of surfaces that prevents both omission and double counting. If the question instead describes a shallow circular recess, do not force the through-hole formula onto it. The word identifying the physical feature determines the mathematical model.
4. Calculate the rectangle before subtracting anything
One full rectangular face has area 0.80 multiplied by 0.50, or 0.400 square metres. The dimensions are already in metres; no further conversion is required. If the original drawing gave 80 cm and 50 cm, converting to 0.80 m and 0.50 m would be essential before using a coating rate in square metres per litre. Alternatively, calculating 4,000 square centimetres and then dividing by 10,000 gives the same 0.400 square metres. Do not divide the area by 100; that is the length conversion factor, not the square-area factor. Label the area as a full rectangle before accounting for the opening. A clear solution distinguishes the geometry that exists on paper from the actual paintable surface after the hole is removed.
5. Calculate the circle that is missing
The area removed from a broad face is πr². With r = 0.080 m, π(0.080)² is approximately 0.020106 square metres. This is the empty opening as viewed from the front. The front paintable area is therefore approximately 0.400 − 0.020106 = 0.379894 square metres. A common mistake is to substitute the diameter into the radius formula, making the circular area four times too large. Another is to write πr instead of πr² and get a length rather than an area. Check the dimensions: the squared length is the clue. A calculator can evaluate π accurately, but it cannot decide whether the diagram labels the radius or diameter. That decision belongs to the student.
6. Double the broad faces, not the opening alone
Because the plate has a front and a back, each has the same circular opening. Two paintable broad faces together measure 2(0.400 − 0.020106) = approximately 0.759788 square metres. The brackets matter. Writing 2(0.400) − 0.020106 wrongly subtracts the hole only once, even though the opening removes an area from both faces. If the back face were hidden and unpainted, only one broad-face term would be used; this problem says both are coated. Do not import a real-life assumption about surfaces against a wall. Read the question’s definition of painted surfaces as the rule of the mathematical case. A candidate who draws and labels each face usually notices the factor of two without needing to memorise an elaborate composite-solid formula.
7. Unwrap the four outside edges
The plate has thickness 0.020 m. Its four outside edges form narrow rectangles: two of dimensions 0.80 × 0.020 and two of dimensions 0.50 × 0.020. Add them to obtain 2(0.80 + 0.50)(0.020) = 0.0520 square metres. Another way is to imagine unwrapping the outer edge into a strip whose length is the front rectangle’s perimeter, 2.60 m, and whose width is the plate thickness. Both arguments agree. A perimeter has units of metres; multiply by thickness in metres to obtain square metres. Never add the bare perimeter directly to a face area. The geometric meaning of each term should stay visible so that the marker can follow the student’s method even when a later arithmetic entry is wrong.
8. Account for the wall inside the hole
The hole is a cylinder passing through the plate. Its internal wall area is circumference times depth, or 2πrt. With radius 0.080 m and depth 0.020 m, this is 2π(0.080)(0.020) = approximately 0.010053 square metres. The term is small, but omitting it shows a misunderstanding of the physical surface. There is no circular base painted at the bottom because the hole passes right through. At the same time, the cylindrical wall exists and is exposed. The learner should be able to name that wall while pointing to the side of a paper cylinder or cardboard tube. This tactile description gives the formula meaning: the curved surface becomes a rectangle when unwrapped, with width equal to circumference and height equal to thickness.
9. Assemble the complete surface-area expression
Write the model before entering it: A = 2(LH − πr²) + 2(L+H)t + 2πrt. Here L = 0.80, H = 0.50, r = 0.080, and t = 0.020, all in metres. Substitution produces A = 2(0.80×0.50 − π×0.080²) + 2(0.80+0.50)×0.020 + 2π×0.080×0.020 = approximately 0.8218407105 square metres. To three significant figures this is 0.822 square metres. Do not round to 0.822 before using the figure in a multi-stage coverage calculation; keep the full calculator value. The final expression is a readable argument: the first term counts the two broad faces, the second the four outer edges, and the third the inner wall. Every term has a physical referent.
10. Check the result with approximate mental arithmetic
Two uncut rectangular faces would measure 0.80 square metres. The two circular openings subtract about 0.040 square metres. The outside and inside edges together add about 0.062 square metres. A final result a little above 0.80 is therefore sensible. This check does not repeat every calculator step. It tests whether the size and signs of the terms fit the diagram. If a candidate obtains 8.2 square metres, the answer fails this rough check because the plate measures less than a metre in each principal direction. If a candidate obtains only 0.38, one broad face probably dominates the calculation and other surfaces may have been missed. Reasonableness helps catch a model error before the student carries it through a percentage and purchasing calculation.
11. Read coverage as a rate with units
Coverage is six square metres per litre for one coat. This means one litre is estimated to coat six square metres once. Required litres for one coat equal surface area divided by coverage: 0.8218407105 square metres ÷ (6 square metres per litre) = approximately 0.136973 L. Notice the unit cancellation. Dividing an area by an area-per-volume rate leaves a volume. Multiplying area by coverage would have the wrong dimensions and would produce an implausible number. A learner should write the rate as 6 m²/L in the margin, not as a naked 6. When a case changes to 8 m²/L, the needed litres should decrease, not increase. That directional check is another fast way to detect an inverted operation.
12. Respect the two-coat instruction
Every paintable surface requires two coats, so the idealised volume doubles. V = 2×0.8218407105÷6 = approximately 0.2739469035 L. This operation is separate from the earlier use of thickness. Plate thickness controls the size of exposed sidewalls; two coats controls how many times the total area is covered. Do not double thickness again when the instruction says two coats. Nor should students assume the second coat has a different coverage rate unless the case provides one. The original problem defines equal coverage to keep the mathematical reasoning focused. The resulting theoretical volume is smaller than half a litre, consistent with a plate of less than a square metre and a stated product covering several square metres per litre.
13. Interpret the fifteen percent allowance
The words “allow an additional 15% for waste” mean multiply theoretical volume by 115%, or 1.15. The adjusted requirement is 0.2739469035 × 1.15 = approximately 0.315038939 L. Report 0.315 L to three significant figures but use the unrounded figure when comparing tin sizes. Fifteen percent is not fifteen litres, nor a fixed 0.15 L. The extra amount is approximately 0.0410920 L, equal to 15% of the theoretical consumption. A different wording, such as “15% of purchased paint is lost,” would imply a different equation. The exam candidate must notice which quantity the percentage describes. Putting a short phrase under the calculation — “extra 15% of theoretical use” — prevents a surprisingly common misunderstanding.
14. Separate paint required from paint purchased
The model predicts roughly 0.315 L needs to be available. Shops sell whole tins. One 0.25 L tin is insufficient because it holds less than 0.315 L, even though its price is lower. Two 0.25 L tins provide 0.50 L at a cost of $24. One 0.50 L tin also provides 0.50 L, but costs $20. The latter meets the volume constraint and costs less. This is an integer decision based on continuous calculations. Students should not round 0.315 down to 0.25 to make a cheaper choice fit. A practical decision can require a different kind of rounding from a numerical answer given to three significant figures. Always check whether a requirement is a minimum threshold before choosing a discrete package.
15. Show why the purchase is optimal
A candidate could state the cheapest feasible option without proving it, but a stronger solution shows the relevant alternatives. Zero tins provide nothing. One small tin supplies only 0.25 L and fails the volume condition. Two small tins cost $24. One large tin is enough and costs $20. Adding any positive-priced extra tins cannot improve on the $20 cost under the stated menu. Therefore one large tin is cheapest among the offered choices. If the problem introduces another tin size or a special discount, the list of candidates changes and the comparison must be repeated. Do not claim a universal rule that bigger tins always cost less per litre. The mathematical conclusion belongs to this particular price list and consumption model. Real-world application questions reward this discipline of checking the available choices.
16. Work out the nominal remainder
The selected 0.50 L tin exceeds the adjusted requirement by 0.50 − 0.315038939 = approximately 0.184961061 L. This is a nominal unallocated quantity in the mathematical model. It is not identical to the 15% waste allowance, which has already been added to the estimate. Nor does the model show whether the remainder will be usable after storage. Define each figure before subtracting: theoretical consumption, allowance-adjusted requirement and purchased volume. A learner who writes “0.185 L is wasted” has made an unsupported interpretation. The question describes an allowance and package sizes, not an observed disposal record. Similar distinctions arise in budget models: the estimated cost, authorised budget and money left in an envelope are not interchangeable.
17. Test a changed coating specification
Suppose a later fictional instruction excludes the back broad face but still includes the front, outer edges and inner hole wall. Remove one perforated broad face, approximately 0.379894 square metres, from the complete area. The new paintable area becomes approximately 0.441947 square metres. Two coats at six square metres per litre and an additional 15% would require approximately 0.169413 L. Under the unchanged tin list, one 0.25 L tin is now sufficient and cheaper at $12. The change in specification alters the optimal purchase. That is exactly why the physical description must come before calculations. A student who remembers “buy a 0.50 L tin” from the original worked example without reading the new condition is not transferring mathematical understanding.
18. Explore a changed thickness
If thickness rises to 0.030 m while length, height and hole radius remain the same, the planar areas do not change, but both kinds of wall become wider. The outside edges become 2.60×0.030 = 0.0780 square metres, and the inside wall becomes 2π(0.080)(0.030) ≈ 0.015080 square metres. Add those to the unchanged broad-face total ≈0.759788 to obtain approximately 0.852868 square metres. The increase is modest, not 50% of the entire area. Multiplying the original total area by 1.5 would incorrectly assume every dimension and every surface responds equally to the thicker plate. A good candidate asks which terms contain the changed variable. That question is a bridge from arithmetic to algebraic modelling.
19. Write area as a function of hole radius
Let r be variable, still measured in metres, while length, height and thickness stay fixed. Then A(r) = 2(0.40 − πr²) + 0.052 + 2πr(0.020). Simplifying gives A(r) = 0.852 − 2πr² + 0.040πr. Two effects compete: enlarging the hole removes more material from the broad faces, but it lengthens the cylindrical inner wall. The expression belongs to a physical domain where the circular hole fits within the rectangle. With a central hole and shorter side 0.50 m, for instance, a radius above 0.25 m would not be feasible. A formula can still produce a numerical value outside that range, but the model would no longer describe the stated object. Examination modelling requires attention to the domain, not only symbol manipulation.
20. Understand when similarity can simplify a calculation
Imagine enlarging the entire plate uniformly by a factor of 1.25: length, height, thickness and hole radius all scale together. Every surface area then scales by the square of the linear factor, 1.25² = 1.5625. The new total is approximately 0.8218407105×1.5625 = 1.28413 square metres. This shortcut is valid because the solid remains geometrically similar. If only height and length change while thickness remains fixed, it is not generally correct to apply the same factor to all exposed surfaces. The learner must identify whether “scale up” means every linear dimension or just selected features. In the SEC, similarity and mensuration questions often test interpretation of which quantities scale by a factor, a squared factor or a cubed factor.
21. Make units do the checking
Every major line should carry a unit. L×H yields square metres; perimeter×thickness also yields square metres; circumference×thickness yields square metres. Surface area divided by square metres per litre yields litres. Multiplying litres by a dimensionless 1.15 leaves litres. The number of tins is a whole count, and count×dollars per tin yields dollars. These relationships form a chain of checks. If an answer to “How many tins?” is written as 0.63 m², the unit warns that the learner is answering a different question. If a student adds 0.5 L and $12, the addition is meaningless. Dimensional thinking is more dependable than guessing whether to multiply or divide based on keywords such as “per” and “each.”
22. Respect numerical accuracy but avoid premature rounding
The K310 syllabus’s general guidance is to present non-exact numerical answers to three significant figures unless the problem specifies something else, with angles in degrees normally to one decimal place. Intermediate rounding can compromise a multistep result, especially when a volume lies near a package-size threshold. In this example keep 0.8218407105 in the calculator while computing the two coats and allowance. Report 0.315 L at the end, then compare the underlying requirement with available tins. The whole-number purchase should not be expressed as 0.630 of a tin when the customer must buy sealed packages. A final decision can be exact even though the quantity used in obtaining it is approximate. Recognising that difference is part of interpreting answers.
23. Find a missing surface through a second method
A student can recompute the area by grouping it differently: planar faces contribute approximately 0.759788 m²; all wall surfaces contribute 0.0520 + 0.010053 = approximately 0.062053 m². The sum remains approximately 0.821841 m². This second route helps identify missing components because it forces the learner to consider all flat and curved faces separately. Under exam time pressure, an independent check need not be long. A rough estimate that the result should be just over 0.80 m² may be enough. Use a full second method when the model is especially easy to misread and the later answers depend on it. Verification is not time-wasting when it prevents the same error from propagating through several marks.
24. Diagnose the student who subtracted one circle
Suppose a script reads 2(0.80×0.50)−π(0.080)² for the broad faces. The calculation removes the circular opening only once. The correct expression is 2(0.80×0.50−π(0.080)²), because the hole appears on both the front and back. Rather than simply saying “use brackets,” explain what the brackets mean: each broad face has its own missing circular region. This concept-first correction is more durable than memorising an isolated formula. In a different problem the number of affected faces might not be two, so the student must be able to inspect the solid. When marking practice work, highlight the diagram next to the erroneous expression and let the learner redraw the exposed regions.
25. Diagnose the student who omitted the inner wall
Another answer uses the correct two broad faces and all outer edges but gives no term for the inside of the circular opening. The result is about 0.812 m² rather than 0.822 m². The numerical gap may seem small, yet the omission reveals an incomplete surface model. Ask the learner to imagine touching the inside edge of the hole. That surface has physical area and the question says all exposed surfaces are coated. The appropriate extra term is circumference times thickness, 2πrt. The correct response is not just to append that formula: the candidate should explain why it exists. The same principle applies to holes in boxes, cut-out channels and other composite geometry questions. Naming the omitted face repairs the underlying mistake.
26. Diagnose the student who used 15 as a volume
A rushed candidate takes the calculated theoretical 0.274 L and adds 0.15 L, claiming that an additional 15% means an additional 0.15 L. This misreads a percentage as a fixed quantity. Fifteen percent of 0.2739469035 L is about 0.0410920 L. The adjusted total is therefore about 0.315039 L. Give the learner two new bases — say 0.2 L and 1.0 L — and ask for 15% of each. They should obtain different absolute amounts. That short exercise makes the meaning of a percentage concrete. In a word problem always name the base before applying the factor. Some waste models are framed differently; interpret the exact phrase instead of forcing every percentage problem into one template.
27. Diagnose the cheapest-but-insufficient purchase
A candidate may choose the $12 tin because it is the lowest listed price. The constraint is not “spend the least regardless of outcome.” It is “buy enough paint at the lowest cost.” One 0.25 L tin fails because the required quantity exceeds 0.25 L. A correct solution first filters out infeasible packages, then compares prices among the survivors. This two-stage process applies to school transport, seating, printing and packaging. If an exam asks for a justification, state both the quantity and cost comparisons. The cheapest displayed item is not necessarily the cheapest feasible solution. Never round a demand downward simply to make it fit the budget. If no listed option satisfies every condition, the honest answer is that the stated constraints are incompatible.
28. Work with a tight budget
Assume the fictional supervisor allocates at most $22. One large tin costs $20 and supplies 0.50 L, so it satisfies the cost ceiling and covers the required 0.315 L. Two small tins supply sufficient volume but cost $24 and exceed the ceiling. One small tin meets the cost ceiling yet fails the paint-volume condition. The student should write the conditions separately: purchased volume must be at least the required amount, and total price must be no more than $22. Only after both inequalities are checked should a purchase be approved. This small case teaches why real-world application questions cannot always be solved by a single calculation. The candidate must hold several conditions in view and explain the choice rather than merely state a price.
29. Scale to several identical mounts
Five identical plates would require roughly five times the allowance-adjusted volume, or 1.57519 L. Suppose the shop also offers a 1.00 L tin for $35. Two such tins provide 2.00 L at $70. One 1.00 L tin plus two 0.50 L tins also provides 2.00 L but costs $75. Four 0.50 L tins cost $80. The two one-litre tins are cheapest among these particular feasible combinations. A student should also ask whether a combination of one one-litre tin and one half-litre tin is sufficient: it gives 1.50 L, which is short. This small inequality decides the case. The cheapest mix for one plate need not remain the cheapest mix for five. Recalculate the integer purchasing decision whenever scale or package sizes change.
30. Distinguish estimation from evidence
All dimensions, coverage and prices in this case are stipulated. The arithmetic is exact within its assumptions, but the model is not a measured conservation report. A real surface can have hidden fasteners, roughness and regions that should not be coated. Product labels may use different rates for different substrates, and paints may have preparation and application requirements. None of those details can be invented when the question does not provide them. For a mathematics examination, state a meaningful limitation such as “the coverage rate is assumed constant across the broad faces and the narrow hole wall.” Avoid the vague “real life is different” without explaining which assumption would matter. This scientific honesty links the maths chapter to the chemistry reasoning in the preceding volume.
31. Create a readable full-mark solution
A compact final presentation can contain three calculations and one decision. First, write the exposed-surface formula and obtain 0.8218407105 m². Second, calculate 0.8218407105×2÷6×1.15 ≈0.315 L. Third, compare whole-tin quantities: 0.25 L is too little; 0.50 L is enough. Fourth, compare cost: $20 for one large tin versus $24 for two small tins. Therefore choose the one 0.50 L tin at $20. The structure shows all necessary reasoning without reproducing every calculator keystroke. If the exam question requests only area, the purchasing work is unnecessary; if it asks for a minimum-cost order, both volume and price must appear. Let the question’s command word determine the scope.
32. Build a mini-mock with six decisions
Begin a short self-test without looking at the worked examples. Question one: identify which surfaces are painted and sketch the solid. Question two: calculate the combined broad-face area after removing the holes. Question three: determine the total sidewall area, naming the outside and inside components. Question four: convert the complete area to litres for two coats. Question five: apply the 15% allowance and choose tins at the minimum price. Question six: decide whether the answer changes if the back face is not painted, and explain why. The sequence is deliberately cumulative. It tests representation, geometry, rate, percentage, discrete optimisation and transfer. These are original practice prompts, not claims about the distribution of official marks in an unseen examination.
33. Mark the mini-mock constructively
For the self-test, check these benchmarks: broad faces approximately 0.759788 m²; outer and inner walls approximately 0.062053 m² together; full area approximately 0.821841 m²; two-coat theoretical consumption approximately 0.273947 L; adjusted requirement approximately 0.315039 L; minimum permitted purchase one 0.50 L tin at $20. If the back face is excluded, the area falls to approximately 0.441947 m² and the adjusted requirement to approximately 0.169413 L, so one 0.25 L tin becomes sufficient. An answer can have a small rounding difference and still be methodologically sound. Conversely, an apparently close final figure may conceal a missing face. Diagnose errors by model, units, operation, rounding and interpretation rather than by numerical agreement alone.
34. Take the problem beyond the plate
The most transferable idea is that unfamiliar contexts often combine familiar mathematical steps. A packaging problem might replace paintable area with cardboard area. A water-tank problem might replace coating litres with volume capacity. A school-event problem might replace tins with buses or ticket bundles. The same workflow remains: define the physical or numerical requirement, keep units consistent, compute the continuous amount and then apply any discrete constraints. What changes is the meaning of the surfaces, rates and thresholds. Students should practise explaining why each operation is appropriate, not merely copying its position in a remembered sequence. That habit develops reliable performance on new questions because the candidate learns to reconstruct a method from first principles.
35. A five-day improvement routine
On day one, sketch the plate from three views and label every exposed surface. On day two, solve the component areas without looking at the final formula. On day three, practise rate and percentage calculations with units beside each line. On day four, compare tin combinations against a fixed requirement and budget. On day five, attempt the six-question mini-mock under a reasonable time limit, then mark the first mistaken decision in each response. A student who repeatedly forgets the inner wall needs geometry practice, not more percentage drills. A student who calculates litres correctly but buys too few tins needs constraint practice. Parents and tutors can help by asking “What does this term represent?” and “Does your chosen package satisfy the condition?” Those questions lead students toward independent checking.
36. Frequently asked questions
Why does a through-hole reduce one area but add another? It removes circles from the front and back while exposing a cylindrical inner sidewall. Why not just compute volume? Paint coverage depends on the area being coated, not the plate’s material volume. Why do we divide by six? The label states six square metres can be covered by one litre for one coat. Why is 15% multiplied by 1.15? The task asks for 15% more than the theoretical quantity. Why can the number of tins not be 0.63? Because whole tins must be purchased. Why is the 0.50 L tin cheapest here? It is the lowest-cost feasible option under the supplied choices. Does this authorise painting any real specimen? No; the problem describes only an invented detachable support plate.
37. Conclude with the decision chain
Return to the sequence that resolves the case: read the scope of painting; represent every surface; use a correct expression; keep units coherent; compute the theoretical volume; apply the stated allowance; compare whole-package choices; and justify the cheapest feasible outcome. The final answer is approximately 0.822 m² of surface and 0.315 L required including allowance, leading to one 0.50 L tin costing $20. What matters more than these particular numbers is being able to rebuild the reasoning when the dimensions, shape, number of coats or prices change. This guide continues the Chemistry rusting case from Volume 1222 and prepares the next English chapter, where a student will speak about a proposed preservation plan without turning a modelled estimate into a guarantee.
Official references and continued learning
Use the official 2027 G3 syllabus index and K310 Mathematics specification for the actual examination requirements. These invented questions are for deliberate practice. Continue through the G3 SEC EMS learner hub, the Mathematics Hub, and the Secondary learning pathway. Read the prior Volume 1222 rusting and protection guide and the earlier Volume 1219 weighted-percentage investigation for two different uses of percentage reasoning. For the primary-school foundation, see PSLE Mathematics time budgeting. The next guide applies English oral communication to explaining the exhibition plan responsibly.
