A low initial price does not settle a long-term decision
A school group is comparing two educational demonstration devices. One costs less to buy but more to maintain. A second costs more at the beginning yet may need fewer replacements. The advertisement for the cheaper device looks persuasive because the first number is smaller. But a useful mathematical answer asks which costs occur once, which recur, what they depend on, and how much evidence supports the forecast. A quotation is not a lifetime guarantee.
This chapter develops a precise G3 Mathematics habit: apply successive percentages, interpret cost functions over time, reverse a known change, and decide whether a financial conclusion follows from the arithmetic. It is not a generic list of formulae. The primary task is to build and test a budget over a stated period while preserving all of its assumptions. That turns ‘which is cheaper?’ from a guess into an argument.
From Vol 0331, the Physics generator-voltage waveform guide, we know that a demonstration may produce electrical output when mechanical motion changes magnetic linkage. That scientific result alone does not establish whether the equipment is cost-effective. Vol 0328 works on mathematical intersections and feasible roots, while Vol 0330 practises reporting figures accurately in an email. Here the new, distinct ownership is successive cost changes and honest long-term comparisons.
We will use fictional school purchasing data and original worked examples. They describe no actual school purchase, vendor quotation, energy bill or government subsidy. The figures exist so that a learner can practise a real calculation without confusing a classroom model with a real recommendation.
What the official 2027 K310 syllabus supports
The SEAB 2027 G3 school-candidate syllabus directory, updated 26 October 2026, identifies Mathematics as K310, replacing the former reference code 4052. The official K310 Mathematics syllabus includes percentages and reverse percentages under Number and Algebra, direct and inverse proportion, use of algebraic formulae, and real-world problems involving personal or household finance. Its supplied formulae include compound interest. This guide uses those published mathematical tools, not a new examination format invented for the topic.
K310 Paper 1 has about 26 short-answer questions, lasts two hours 15 minutes, carries 90 marks, and contributes 50% of the total. Paper 2 also lasts two hours 15 minutes, carries 90 marks and contributes 50%. It has nine to ten questions, and its final question focuses specifically on a real-world scenario. A cost-comparison narrative is therefore appropriate practice, though no particular generator or school-budget case is promised in the actual examination.
The syllabus says essential working matters and specifies that non-exact numerical answers are generally given to three significant figures, or angles to one decimal place, unless the question specifies otherwise. Money calculations may require a sensible final cent or dollar rounding according to the task. It is good practice to retain intermediate precision and round only at the end. Premature rounding is especially damaging when a cost is multiplied across several years.
The wider G3 SEC EMS Learner’s Guide explains how the examination subjects connect. A learner who needs to repair percent-of-a-quantity from primary school should use the PSLE Learning Guide before advanced forecasts. The Mathematics Hub remains the owner of general number and algebra practice, while the present volume is a worked examination-performance chapter.
Make four separate boxes before calculating
Every real-world money question becomes clearer when you divide information into four groups. The first group is a one-off cost, such as purchase and installation. The second is a recurring cost, such as a service charge each year. The third is a variable amount, such as operating hours, number of units or electricity consumption. The fourth is a condition: a cap, deadline, discount rule, replacement interval or uncertainty.
This separation prevents a common error: adding a percentage of purchase price to a percentage of running cost as though the bases were identical. If a part’s annual maintenance is 4% of its original price, that is a particular statement. If maintenance rises by 4% each year, that describes a changing yearly charge. The words contain the same number and sign but require different mathematics.
A table is often enough. Write columns for year, opening cost, recurring cost, expected saving, and cumulative difference. If the question states a budget cap of $600, check whether the cap refers to initial purchasing or the entire period. A plan could pass the purchase cap and still exceed the total-cost cap; a mathematically correct total can be the wrong answer to an incorrectly interpreted condition.
Costs should be stated as costs. Avoid treating a hypothetical saving as cash already received. If a school has not measured energy use before and after an installation, any claimed saving is at best a forecast based on assumptions. Mathematics can compare models precisely, but it cannot turn an unsupported assumption into an observed fact.
Percentage multipliers keep successive changes honest
A 10% increase turns an amount P into 1.10P. A 10% decrease turns it into 0.90P. These multiplier forms are easier to combine than repeated verbal instructions. For a $200 charge, one rise of 10% gives $220. A following decrease of 10% gives $220 × 0.90 = $198, not $200. Returning by the same numerical percentage does not reverse the first step because the second percentage uses a different base.
When a value increases by r% each period for n periods, its modelled amount is P(1 + r/100)^n. If it decreases by d% each period, use P(1 – d/100)^n when the simplifying model makes sense. This looks like the structure of compound interest, but the subject need not be a bank account. It can describe a fictional maintenance quote, operating rate or replacement allowance; the mathematical idea is repeated multiplication.
Keep the meaning of a percentage point separate from a relative percentage change. Moving a predicted failure proportion from 20% to 25% is a rise of five percentage points, and a 25% relative rise, because 5 divided by the original 20 is one quarter. Students often write ‘the rate increased by 5%’ when the intended claim is ‘five percentage points’. In a recommendation, that mistake can understate or overstate the change.
To reverse a percentage increase, divide the final amount by the multiplier. A displayed price of $216 after an 8% rise began at $216/1.08 = $200. Subtracting 8% of $216 does not recover the original because the final price is not the original base. Write the relationship ‘final = original × multiplier’ first, then rearrange.
Cumulative cost is not the last year’s bill
Suppose a demonstration kit costs $480 to buy and its annual service charge begins at $60, rising by 5% each subsequent year. The first three annual services cost $60, $63, and $66.15. Their three-year total is $189.15, and purchase plus these three services is $669.15. It would be wrong to add only the third year’s $66.15 to the purchase price and call that the lifetime cost.
A projected annual saving can be placed in a different column. Suppose an original exercise gives an illustrative $35 saving in each of those three years. The total saving is $105 under the forecast, leaving a net three-year cost of $564.15 when compared against a zero-purchase baseline that otherwise has identical costs. This is arithmetic on a stated hypothetical model, not evidence that a real generator produces that saving.
If a device is bought halfway through a year, read the question’s convention rather than automatically charging a full annual maintenance amount. If the task says service is paid at the beginning of each year, that affects which payments lie in the period. If it says a warranty includes the first year’s service, that line must not be charged again. The most frequent long-word-problem errors are omissions and double counts, not advanced algebra mistakes.
Useful tables distinguish a total, a net difference, and a break-even point. A total is the sum of specified components. A net difference subtracts a stated benefit or alternative. Break-even occurs when two cumulative models are equal. They are three different questions; answer only the one asked, with the precise horizon.
Break-even is a conditional crossing, not a guaranteed future event
Imagine option A costs $120 initially and $45 per year. Option B costs $300 initially and $15 per year. After n whole years, the simple models are A(n) = 120 + 45n and B(n) = 300 + 15n. Set them equal: 120 + 45n = 300 + 15n, giving 30n = 180 and n = 6. In this simplified equal-use scenario, their cumulative costs are equal after six years.
Before six years, the cheaper purchase A has a lower cumulative cost; after six years, B becomes cheaper within this model. But the model assumes constant yearly charges, equivalent service, and no unexpected replacement or financing. A question might add a three-year school project horizon. Then the six-year break-even is mathematically correct yet irrelevant to the decision: the more expensive option does not pay back during the project.
When the running charge rises by a percentage each year, the calculation is not the same straight line. Build a year-by-year table and compare totals at the required dates. An equation may still be possible, but the exam usually rewards clear justified mathematical working. Do not force a linear intersection onto a genuinely compounded process.
If a cost estimate comes with a range rather than a point, check more than one scenario. For example, an annual operating saving between $20 and $50 leads to a much wider range of possible break-even times than a single $35 estimate suggests. The mathematics shows sensitivity to assumptions; it does not certify which future value will occur.
Units, rounding and feasibility
Never subtract watts from dollars or compare an energy quantity with a power rating. A power rating of 10 W sustained for one hour corresponds to 10 Wh of energy, or 0.010 kWh. To estimate an electricity bill, first find energy in kWh, then multiply by the price per kWh specified in the problem. A generator rated at a particular power is not automatically producing that power all day.
Suppose a fictional light uses 20 W for five hours each school day, across 180 days. Annual energy is 20 × 5 × 180 = 18,000 Wh, or 18 kWh. At an illustrative rate of $0.30 per kWh the energy cost is $5.40. That does not include purchase, maintenance or taxes unless stated. A saving claim of $100 on this usage alone would require other savings or assumptions not present in the model.
If a purchase cap is $500 and the total including mandatory delivery is $520, a ‘small’ overrun remains an overrun. Likewise a proposal can have the lowest three-year projected cost but fail a nonfinancial constraint such as equipment safety or permitted use. The mathematical answer should state both the ranking and whether each candidate satisfies the full task.
Round final money appropriately. A computation yielding $237.645 may need $237.65 to the nearest cent; a whole-dollar budget statement may state about $238. If the exam requires three significant figures, follow that instruction. The model’s precision must not exceed the confidence in the assumptions: reporting a 30-year payback of 7.382914 years from highly uncertain costs suggests more certainty than the evidence provides.
Read the context before choosing a formula
A useful written solution starts with one sentence that names the decision. ‘We need the three-year net cost of the two devices, including purchase and annual service’ is more helpful than immediately copying a percentage formula. Then define variables and list known values. Work in chronological order, state units, and return to the decision sentence.
Use estimation as a guardrail. Three years of roughly $60 maintenance is around $180, so a total near $670 for the $480 kit is plausible; $480 + $60 × 1.05^3 ≈ $550 is not a valid three-year total, because it represents one service term rather than their sum. The estimate catches the conceptual mistake before detailed arithmetic hides it.
Show one line per operation when bases change. For instance, ‘original price × 0.90 × 1.08’ shows a 10% discount followed by an 8% charge on the discounted amount. Simply writing ‘minus 2%’ would not in general equal the product, because percentage changes compound. Transparent working is easier to mark and easier for the student to audit.
Close by interpreting a result: ‘Option B costs $30 less over four years under the stated constant-service model, provided the equipment gives equivalent service.’ That is stronger than a bare number because it preserves horizon and assumptions. It is also better than claiming that the option is definitely better for a real school.
Twenty-two original worked decisions
These short problems are deliberately mixed. Some ask for arithmetic, some for interpretation, and some for a decision that the data cannot yet justify. Solve each before opening the answer mentally. In Paper 2, the hardest part may be deciding which information belongs in the calculation rather than performing the final multiplication.
Decision 1: A $400 part rises by 5% in one year.
Multiply 400 by 1.05 to obtain $420. The $20 difference is five percent of the original $400. If a second 5% rise follows, the new base is $420, not $400. This is the distinction between a single change and successive changes.
Exam check: Check whether the problem says ‘per year’ or ‘once’. The multiplier is the same in one step, but the model is different over time.
Decision 2: A maintenance fee of $80 rises by 5% for two years.
The first rise gives $84.00 and the second gives $88.20. Directly, 80 × 1.05² = 88.20. Adding 5% twice to the original $80 gives $88 and loses twenty cents because it ignores compounding. A small monetary difference is still a different mathematical answer.
Exam check: Preserve intermediate cents and state the two-year horizon.
Decision 3: A quotation falls by 15% and then rises by 15%.
Begin with an illustrative $200. The fall gives 200 × 0.85 = $170; the rise gives 170 × 1.15 = $195.50. The result is $4.50 below the starting price. Equal percentage labels do not cancel because the second percentage is applied to a smaller base.
Exam check: Write both multipliers; do not replace them with a verbal ‘cancel out’.
Decision 4: An item costs $270 after a 10% price rise.
Let P be the original. Then 1.10P = 270, so P = 270/1.10 ≈ $245.45. Subtracting $27 from $270 would mistakenly use ten percent of the final figure. If a question asks for exact working, keep 270/1.10 until the final currency rounding.
Exam check: Reverse a known multiplier by division, not by applying an opposite percentage to the final base.
Decision 5: A promotional price is $189 after a 10% discount.
The listed original price was $189/0.90 = $210. The discount amount is $21. Check the result forwards: 10% of $210 is $21 and $210 – $21 = $189. The reverse calculation is a good diagnostic because learners often multiply by 1.10 instead.
Exam check: Verify an inverse calculation by reapplying the original operation.
Decision 6: Two options have equal purchase cost but different service schedules.
If each costs $300, option A has constant $25 annual service and B starts at $20 but increases by 10% per year. Over three years A adds $75, whereas B adds 20 + 22 + 24.20 = $66.20. B is cheaper over the stated period in the simplified model. It may not stay cheaper indefinitely.
Exam check: Do not rank options from their third-year charge alone; add the complete stated period.
Decision 7: A device costs $150 plus $30 each year.
After n years, total cost is C = 150 + 30n. For five years the total is $300. The gradient of a cost-against-years line is $30 per year and the intercept is the initial $150. These units help interpret the equation. They also explain why a graph can support a decision without complicated algebra.
Exam check: A straight-line model assumes the annual fee stays constant.
Decision 8: Option A costs $100 plus $50 annually; option B costs $250 plus $20.
The difference in initial cost is $150 and the difference in annual running cost is $30. Under constant costs, B catches up after 150/30 = 5 years. At five years each is $350. Before five years A is cheaper; beyond that B is cheaper. A six-month project cannot claim a five-year payback.
Exam check: Give the threshold and compare it with the permitted time horizon.
Decision 9: A project lasts three years, but a model breaks even in four.
The numerical crossing at four years does not meet the project’s condition. State that no payback occurs within the three-year horizon under the supplied assumptions. An answer that says simply ‘the project pays back in four years’ misses the actual decision, even though its algebra is correct.
Exam check: Finish every break-even problem by checking the horizon named in the question.
Decision 10: An annual payment begins at $50 and increases by 20% yearly.
The first four yearly payments are $50, $60, $72 and $86.40. Their total is $268.40. The fourth-year fee alone is not the four-year expenditure. A simple sum of the four terms is sufficient, although more advanced learners may recognise the repeated-multiplier structure.
Exam check: Label each year explicitly before summing; this prevents skipped payments.
Decision 11: A student reports a 5% reduction from 20% to 15%.
The difference is five percentage points. Relative to the original 20% rate, the reduction is 5/20 = 25%. The sentence ‘a 5% reduction’ is ambiguous or misleading. In an examination report, choose a precise term and show the base used. Small wording changes can carry large numerical differences.
Exam check: Distinguish the change in a rate from the percentage change of the rate.
Decision 12: A bill for 40 kWh uses an invented tariff of $0.25/kWh.
Multiply 40 kWh by $0.25 per kWh to get $10.00. The kWh unit cancels, leaving dollars. The result is not a price per hour; the energy has already combined power with usage time. If another fixed fee is supplied, add it separately. Do not charge the tariff per watt.
Exam check: Units are a quick test of whether the correct quantities have been multiplied.
Decision 13: A 15 W lamp runs four hours daily for 200 days.
Energy is 15 × 4 × 200 = 12,000 Wh = 12 kWh. At an illustrative $0.30/kWh, the electricity component is $3.60. A claim that the lamp ‘uses 15 kWh per day’ would confuse W and kWh. This calculation assumes the rated power is sustained over all stated hours.
Exam check: Write power × time = energy before introducing a tariff.
Decision 14: A plan saves a modelled $18 per year but costs $120 to install.
The simple payback under constant savings and zero additional expense would be 120/18 ≈ 6.67 years. However, if complete years are required, cumulative savings first exceed $120 after seven years. This model excludes maintenance and the value of time, and the $18 saving is assumed rather than observed.
Exam check: Say exactly what the model includes and whether the threshold must be crossed in whole years.
Decision 15: A quotation gives $250 for a kit plus $30 mandatory delivery.
Total initial spending is $280, not $250. Against a $270 cap the offer is infeasible even if it is the cheapest kit. If delivery were optional, the treatment could differ, but the question has called it mandatory. Finding a low base price is not enough to satisfy a fully stated budget.
Exam check: Check every mandatory charge before announcing that a plan is under budget.
Decision 16: An estimate is $400 with a possible 10% overrun.
The upper illustrative amount is $400 × 1.10 = $440. If the approved cap is $420, the estimate sits below the cap but its stated upper case exceeds it. The correct recommendation may be to seek a better quotation or reduce scope rather than treat $400 as certain.
Exam check: Present a range as a range; do not replace it with a guaranteed outcome.
Decision 17: An item is discounted by 20%, then charged 8% of the discounted price.
From $500, discount to $400; then add 8% of $400 to obtain $432. The combined multiplier is 0.80 × 1.08 = 0.864. It is not correct to say simply that 20% off and 8% added means 12% off: the bases differ. The final reduction relative to $500 is 13.6%.
Exam check: Order and base are essential in sequential percentage calculations.
Decision 18: A budget uses a rounded rate of $0.28/kWh rather than an exact quotation.
With 50 kWh, the modelled energy charge is $14.00. The arithmetic is exact for the assumed $0.28 rate, but the real bill is not guaranteed to be exactly $14 if the actual rate or fee structure differs. The mathematics supports a conditional estimate, not an invoice.
Exam check: A correct answer separates arithmetic certainty from uncertainty in inputs.
Decision 19: A team compares two devices but ignores replacement after year three.
If the stated project lasts five years and one device must be replaced after three, a comparison of purchase plus five-year service that omits replacement is incomplete. Insert the relevant year-three expense before ranking alternatives. No amount can be invented unless the question supplies a price or asks for a variable expression.
Exam check: Missing data should be named instead of silently filled with a guessed number.
Decision 20: A student rounds each of five annual charges before adding.
If yearly charges carry fractional cents in intermediate model calculations, early whole-dollar rounding can accumulate avoidable errors. Keep the calculator values or enough decimal places, add the charges, and follow the final rounding instruction. The point of showing working is to keep the calculation reproducible, not to display an implausible number of digits in the final recommendation.
Exam check: Retain precision during computation and round once at the decision boundary.
Decision 21: Two costs are equal after six years, but a new warranty covers year one.
The initial equation assumed service payments in all six years. A warranty that removes year-one service changes the sequence and may shift the break-even year. Revise the model from the actual wording instead of keeping the previous answer. A small condition in a long question can invalidate otherwise correct equations.
Exam check: Recompute when the conditions change; never treat a calculated crossing as a permanent fact.
Decision 22: A model predicts a ‘saving’ of $200 because the device outputs 200 W.
Watts measure power, dollars measure money. Without operating time, usable energy, an electricity price, and a legitimate way of valuing output, a 200 W rating does not become a $200 saving. The answer should identify the missing conversion information before any payback calculation. This is not pedantry; it is dimensional reasoning.
Exam check: Never borrow a number from another unit merely because its digits look convenient.
An integrated mock question with a complete solution
Here is an original, fictional task. A school’s learning team has a budget of $650 for the first three years of a small teaching device. Option P costs $390, including installation, and has annual service of $70, $73.50 and $77.18 respectively. Option Q costs $480, including installation, and has a fixed annual service of $50. The two options are assumed to be equally useful for the stated three-year project. Which option is feasible, and which has the lower total cost?
First add every stated payment. P = 390 + 70 + 73.50 + 77.18 = $610.68. Q = 480 + 50 + 50 + 50 = $630.00. Both lie within the $650 three-year cap. P is $19.32 cheaper over the stated period. The answer is not that P is cheaper in every conceivable future year, and it does not establish that either actual device is scientifically better. It is a comparison under the supplied fictional financial and service assumptions.
Suppose the question now adds a requirement that a safety inspection costs $45 in year two only for P. Its updated three-year total becomes $655.68, so P fails the $650 cap while Q remains feasible at $630. The numerical ranking has changed in decision terms even though P’s original purchase price remains lower. A good student reads the addition, updates the correct term and explicitly revisits feasibility.
One more condition: if the $650 cap were an initial-purchase-only cap, both would be within it regardless of later service costs; ranking a three-year total would still be useful but would answer a different financial question. This is why understanding the wording precedes computing.
Independent transfer questions and answer route
Try five new tasks without referring to the worked examples: (1) recover a price from $345 after a 15% rise; (2) find the price of a $600 item after successive 10% and 5% reductions; (3) compare A(n)=80+25n and B(n)=200+10n at n=4 and find break-even; (4) compute energy from a 25 W load running two hours per day for 120 days; (5) explain why a modelled electricity saving is not automatically a verified annual saving.
Checks: (1) 345/1.15 = $300. (2) 600 × 0.90 × 0.95 = $513. (3) at four years A is $180 and B is $240; equality requires 80 + 25n = 200 + 10n, so n = 8 years. (4) 25 × 2 × 120 = 6000 Wh = 6 kWh. (5) A savings model requires actual comparable use, prices and conditions to be measured or justified. A claimed saving may change with usage or maintenance.
When correcting an error, label its cause: incorrect percentage base, missed recurring charge, unit conversion, time horizon, or unsupported assumption. Each calls for a different repair. Repeating a formula a hundred times does not fix a habit of reading ‘annual’ as ‘one-off’. A good improvement is to draw a timeline of payment dates before calculating.
This mathematical judgement belongs in the Mathematics Hub as a continuing skill, and it helps make scientific claims in the Science Hub more responsible. For clear evidence statements, use the Secondary English Learning Hub. The lasting rule is simple: state the base, show every period, identify the cap, and keep a projection separate from a fact.
