G3 SEC Mathematics Reverse Energy Budgets

When one number is a limit rather than an answer

A school science club has a fictional energy budget of 170,000 joules and an example pack that needs 38,650 joules to go from ice at minus five degrees to liquid water at ten degrees. The calculator reports 170,000 divided by 38,650 as about 4.398. Does that mean the club can complete 4.398 packs? If each pack must stay whole, it means four packs at most. The decimal is useful for analysing the constraint, but it is not a feasible decision.

This guide teaches the Mathematics behind that distinction. We will reverse equations to find unknown masses, build expressions for several thermal stages, compare exact requirements with available energy, use inequalities to decide feasibility and interpret whole-number constraints. The setting comes from Physics, but the essential skills are algebra, proportional reasoning, estimation, interpreting a mathematical model and checking that an answer fits the words of the question.

The scenarios are invented learning problems. No real school, medicine, food-storage service or safety certification is being assessed. A mathematical energy budget cannot on its own guarantee a real temperature outcome. We will keep physical assumptions explicit so that calculation becomes a tool for reasoning rather than an excuse to claim that every system behaves perfectly.

What the 2027 SEC Mathematics syllabus actually assesses

The first Singapore–Cambridge Secondary Education Certificate examinations will take place in 2027. At G3, Mathematics is syllabus K310, the continuation of the former 4052 standard. The official assessment has two written papers of two hours and fifteen minutes each, each worth 90 marks and 50 percent. The published syllabus develops number and algebra, geometry and measurement, and statistics and probability, with mathematical problem-solving and modelling across its questions.

The ice-pack calculations here are applications of algebra, rearrangement, inequalities, ratios, units and mathematical interpretation. They are not a new K310 syllabus unit called ‘latent heat’ or a guarantee of a Physics formula appearing in a Mathematics paper. For pure G3 Physics, the separate subject is K323. Check the official 2027 G3 list and K310 syllabus for the examinable Mathematical content. The context is a useful route into harder real-world problems, not a substitute for the actual syllabus.

Start by understanding what is fixed

In a forward calculation, mass is given and energy is requested. In a reverse calculation, energy is given and mass is requested. In an optimisation or feasibility question, both may appear as limits: only a certain amount of energy is available, and packs may come in whole units. Many learners can manipulate a formula yet fail to notice this change in the question’s purpose.

A reliable first sentence is: ‘Let m be the mass of ice in kilograms that completes the stated journey.’ Another might be: ‘Let n be the whole number of equal packs.’ These variables are not interchangeable. A mass can vary continuously in an ideal model, but a count of sealed packets takes integer values. The answer’s domain comes from the story, and it determines the rounding decision at the end.

A three-stage journey becomes one algebraic expression

Assume a hypothetical sample of ice initially at −5°C warms to 0°C, melts completely and then its resulting liquid water warms to 10°C. In the stated ideal model, use c_ice = 2100 J/(kg°C), latent heat of fusion ℓ = 334,000 J/kg and c_water = 4200 J/(kg°C). For an unknown mass m in kilograms, the first stage requires 2100 × 5 × m joules, the second requires 334,000 × m joules, and the third requires 4200 × 10 × m joules.

Adding produces E = (10,500 + 334,000 + 42,000)m = 386,500m. This is a linear relationship between the total useful energy and the mass under the stated assumptions. The coefficient has units of joules per kilogram; its value comes from adding stage-wise energy requirements for one kilogram. When the mass is 0.100 kg, E is 38,650 J. When m doubles, the ideal energy requirement doubles because every stage scales with mass.

Why the combined coefficient is useful

Algebra lets you compress a detailed physical sequence without losing its meaning. Once the stages have been justified, a single coefficient of 386,500 J/kg makes later comparisons quick. Yet compression must not erase conditions. If the starting temperature changes from −5°C to −10°C, the initial warming component changes. If the final liquid temperature changes, the last component changes. If melting is incomplete, the model must branch instead of using the full three-stage formula.

The coefficient is therefore not a universal ‘energy per ice pack’. It belongs to a specified material journey. One of the most valuable exam habits is to write the conditions beside any shorthand expression. If the question gives different constants, use the given values. A learner who remembers 386,500 without remembering how it was obtained is likely to misuse it on the next, slightly different problem.

Reverse the equation carefully

Suppose useful energy is limited to 170,000 J for the stated three-stage journey. Solve 386,500m = 170,000, giving m ≈ 0.4398 kg. This is the theoretical continuous mass, assuming all of the energy reaches the material as modelled. It is not yet a whole-pack answer. If the packs each contain 0.100 kg, the corresponding count is 4.398. Only four entire packs can be completed. The useful energy for those four is 154,600 J.

Five packs would need 193,250 J, which exceeds 170,000 J. That two-sided check gives the decision credibility: four work and five fail. You can express the logic as 38,650n ≤ 170,000, with n a nonnegative integer. The greatest valid n is four. Writing a correct inequality and interpreting its integer constraint often demonstrates deeper understanding than reporting a bare quotient from the calculator.

Reverse problems need a named unknown

Consider the expression E = 386,500m. If a question asks for the remaining energy after preparing three packs, do not solve for m again. Three packs of 0.100 kg represent 0.300 kg, so the required energy is 115,950 J. The remaining budget is 170,000 − 115,950 = 54,050 J. That is more than enough for one additional pack, but not for two more. The output of one calculation becomes the input to the next.

Write the variable meaning each time you change the question. A symbol can stand for a mass, a pack count, time, useful power or remaining budget. In long word problems, using the same unlabelled letter for different quantities is a common source of invisible mistakes. A well-chosen variable reduces the amount of explanation you need and clarifies what the answer can mean.

Compare margins, not just the final decision

A feasible answer may still be fragile. With a 170,000 J budget, four packs require 154,600 J, leaving a margin of 15,400 J. As a proportion of the available budget this is about 9.06 percent. That arithmetic margin measures how much room the ideal model leaves. It does not prove that environmental losses will be smaller than the margin; the model has not measured them.

Now reduce the budget to 150,000 J. The quotient 150,000/38,650 is about 3.88, so only three full packs fit. A difference of 20,000 J in the total budget changes the feasible whole-pack count. In an examination, this threshold effect is more informative than saying that the answer ‘changes a little’. A quantity can vary smoothly while a discrete decision changes in steps.

Connect whole-number constraints with inequalities

Suppose a club needs at least six complete 0.100 kg packs under the same energy model. Set up 38,650n ≤ B, with n = 6 and B the required total useful budget. This gives B ≥ 231,900 J. If the available source supplies only 220,000 J, the target is infeasible under the ideal assumptions; it is not acceptable to round six down to 5.69 while claiming the target has been met.

When a question asks for the smallest budget to complete a specified number of units, multiply the cost per unit by the count. When it asks for the largest number of units within a fixed budget, divide and take the greatest feasible integer. Both are algebraic, but their final logic differs. Identify ‘at least’, ‘at most’, ‘minimum’ and ‘maximum’ before choosing a rounding direction.

Never round up simply because the decimal is large

A quotient of 4.99 does not make five units feasible under a strict ‘at most’ capacity constraint. A quotient of 4.01 does not demand five units either. The correct whole-number operation is constrained by the question. For ‘the fewest buses required to transport 101 passengers if each holds 25’, round up to five. For ‘the most full packs possible with a given energy budget’, round down to four. The same arithmetic-looking fraction can call for opposite decisions.

Make the inequality explicit. A feasible count n satisfies the total requirement ≤ the available resource. An insufficient count for a minimum demand fails a different inequality. These are not arbitrary rounding rules memorised for school. They are consequences of what makes the decision valid. After choosing an integer, multiply back to verify it.

Two-sided testing guards against error

The quickest check for a maximum whole count n is to verify that n units are feasible and n+1 units are not. For four packs, 4 × 38,650 = 154,600 ≤ 170,000 and 5 × 38,650 = 193,250 > 170,000. The pair proves maximality under the model. For a minimum count needed to meet a target, verify that n units meet the target while n−1 do not.

The logic extends beyond ice packs to seating, transport, inventory, packaging and scheduling. Students often solve a real-world problem by finding a decimal and applying a familiar rounding rule. The two-sided test turns intuition into evidence. It takes only one more line and is especially valuable when a high-stakes question says ‘justify your answer’.

A reverse mass question with a target temperature

Imagine a fixed 0.300 kg of warm liquid water initially at 28°C. A variable mass x of ice at 0°C is added, and the problem asks what ice mass would produce a final all-liquid mixture at 10°C, assuming an isolated system and the stated values c = 4200 J/(kg°C) and ℓ = 334,000 J/kg. The warm water loses 0.300 × 4200 × (28 − 10) = 22,680 J.

Each kilogram of ice at 0°C requires 334,000 J to melt and a further 4200 × 10 = 42,000 J to warm to the final temperature. Hence 376,000x = 22,680 and x ≈ 0.0603 kg. This is about 60.3 g. The final temperature is above zero, so the assumed complete-melting branch is consistent with the found mass. This calculation is a different inverse problem from simply dividing a fixed heater budget by energy per pack.

Check the branch before solving a mixed system

Take 0.250 kg of water at 32°C and 0.090 kg of ice at 0°C. The warm water can release 0.250 × 4200 × 32 = 33,600 J by reaching zero. The ice needs 0.090 × 334,000 = 30,060 J to melt. There is enough energy for complete melting, and the remaining 3540 J warms the combined 0.340 kg of water. The temperature rise is 3540/(0.340 × 4200) ≈ 2.48°C.

Now increase the ice mass to 0.120 kg. Melting would require 40,080 J, more than the 33,600 J that the warm water can provide to zero. In the ideal model, the final mixture remains at zero degrees with some ice present. The amount of melted ice is 33,600/334,000 ≈ 0.1006 kg, leaving about 0.0194 kg ice. One changed value forces a different equation branch. This is precisely the kind of transfer test that distinguishes understanding from memorised working.

Express an energy budget as a linear function

Let E(m) = 386,500m for mass m ≥ 0 under the example’s fixed three-stage conditions. The graph of E against m is a straight line through the origin. Its gradient is 386,500 joules per kilogram. If the budget is 170,000 J, the horizontal line E = 170,000 intersects the linear model at about 0.4398 kg. Graphically, that intersection is the largest theoretical continuous mass.

The pack-count condition adds a discrete grid of allowed masses: 0, 0.1, 0.2, 0.3, 0.4, and so on. The final feasible mass is 0.4 kg, not the continuous intersection. This illustrates a broad mathematical idea: an equation can describe a continuous resource relationship while the real decision restricts values to a discrete set. If the question does not impose whole packs, do not invent that constraint.

Use a table to show the decision boundary

List the number of complete 0.100 kg packs, followed by required energy: one, 38,650 J; two, 77,300 J; three, 115,950 J; four, 154,600 J; five, 193,250 J. This table makes the step boundary visible without a graph. It also helps detect arithmetic slips: each additional pack adds exactly 38,650 J in the ideal model.

A table is not automatically better than an equation. Choose it when the question asks for a range of options or a clear practical recommendation. Use an equation when you need to find a boundary value, and then a small table or inequality to prove the whole-number decision. In a timed examination, structure the work to give the reader the key result quickly without concealing the reasoning.

Time changes energy only under a stated rate model

A useful constant power of 40 W corresponds to 40 joules transferred each second. If the source operates for 50 minutes, the ideal useful energy is 40 × 3000 = 120,000 J. Under the 386,500 J/kg three-stage coefficient, this supports a continuous mass of about 0.3105 kg, or at most three complete 0.100 kg packs. Confusing minutes with seconds would undercount energy by a factor of sixty.

The expression E = Pt connects two rates and quantities: power is energy per time. If useful power is variable, a single multiplication using one arbitrary reading may be inappropriate. The question might provide a piecewise schedule or a graph from which total transferred energy can be obtained. Read the given rate information before choosing the arithmetic; do not silently assume a constant rate.

A two-rate schedule needs piecewise reasoning

Suppose a fictional source transfers a useful 50 W for ten minutes and then 25 W for twenty minutes. The first period delivers 50 × 600 = 30,000 J. The second delivers 25 × 1200 = 30,000 J. The combined useful energy is 60,000 J. It would be incorrect to multiply 50 W by the entire thirty minutes, because the rate changes after the first stage.

This illustrates the same reasoning as a piecewise price tariff or an interval-gradient problem. Break the description into sections where the rule is fixed, calculate each part and combine them. The fact that two periods deliver equal energy despite different powers and durations is a useful conceptual check. A high rate over a short period can contribute as much as a low rate over a longer one.

Ratios and sensitivity: what happens if the mass doubles?

Under fixed starting and finishing states and the same material constants, E = 386,500m. Double m and E doubles. Halve m and E halves. But doubling the number of complete packs does not necessarily double a budget’s unused margin, because the available total may remain fixed. The linear relationship describes the physical requirement, not every feature of the resource decision.

Suppose each pack contains 0.080 kg instead of 0.100 kg. Its model energy requirement is 30,920 J. A 170,000 J budget would permit five complete packs, using 154,600 J, with six packs requiring 185,520 J. Although five packets now fit, their combined ice mass is 0.400 kg, exactly the same total as four 0.100 kg packs. This illustrates why packet count and material mass must not be mixed up.

Do not compare percentages without the base

With 170,000 J available, four 0.100 kg packs leave 15,400 J unused. The unused share of the budget is 15,400/170,000 × 100 ≈ 9.06 percent. The spare fraction relative to the four-pack requirement is instead 15,400/154,600 × 100 ≈ 9.96 percent. Both computations can be legitimate, but they answer different percentage questions.

State the denominator in words before you compute. The distinction connects with our earlier guide on percentage points, relative changes and denominators. Percentages are not self-explanatory. They become evidence only when the reader knows what the stated quantity is a proportion of.

Precision and error bounds

A mass reported as 0.440 kg to three significant figures is a rounded continuous answer. It does not make the limit exactly 0.440000 kg, nor guarantee that five 0.100 kg packs fit. In the ideal worked numbers, the exact quotient is about 0.439844… kg, so a report rounded to 0.440 kg could appear to sit exactly on a boundary at a coarser scale. For feasibility, preserve unrounded numbers until you have checked the inequality.

A stated energy budget might also be an estimate. If uncertainty information is supplied, evaluate whether the worst relevant bound changes the number of whole packs. If no uncertainty information is supplied, do not invent a formal confidence interval. The precision and reverse-check guide Vol 0411 explains why intermediate rounding can change a borderline decision.

Break-even thresholds for pack counts

To complete four packs in the standard model, the budget must be at least 154,600 J. To complete five, it must be at least 193,250 J. Thus any budget B satisfying 154,600 ≤ B < 193,250 permits exactly four as its maximum whole-pack count, assuming nonnegative B and no other constraints. That is an interval statement, stronger than a result calculated for one budget.

This kind of reasoning helps with questions that ask when a decision changes. Rather than calculate a separate quotient for every possible budget, find the thresholds and describe the ranges. Notice that the upper bound is excluded: at 193,250 J, five become feasible. Inequality signs carry real meaning. A candidate who writes ≤ where < is needed can accidentally include a case that changes the answer.

Check dimensions and units as part of algebra

The coefficient 386,500 J/kg multiplied by 0.100 kg produces 38,650 J. Dividing 170,000 J by 38,650 J/pack produces about 4.398 packs. The unit cancellation reflects the mathematical meaning of the operation. If you divide a total energy in kilojoules by energy per pack in joules, the result will be wrong by a factor of a thousand unless you first align the units.

Units are not a cosmetic addition after calculation. They are a quick diagnostic tool. In your rough working, attach J, kJ, kg, g, W and s to the relevant values. Once the units have been checked and converted consistently, the algebra can be quite simple. K310 questions often reward clarity about what a computed quantity represents.

A multi-condition feasibility problem

Suppose there are at most six containers available, each able to hold one complete 0.100 kg pack, but only 170,000 J of useful energy is available. The count n must satisfy both n ≤ 6 and 38,650n ≤ 170,000. The energy condition is tighter, so the maximum count is four. If the containers were limited to three instead, the storage condition would become the binding constraint.

Real problems frequently have more than one rule. If you solve only the numerical equation tied to one resource, you may overlook a stricter condition. A useful technique is to write each constraint on its own line and find the intersection of allowed values. State which condition actually limits the result. This turns a single-division exercise into mathematical modelling.

Mathematics is not a licence for a safety promise

A fictional worksheet may ask whether a pack has enough latent-energy capacity for a simulated demonstration. It does not ask you to certify food or medicine storage, decide real safe temperatures, or conduct an experiment. Actual cooling outcomes depend on heat transfer rates, geometry, insulation, external temperature, contact and other factors not captured by one energy-capacity calculation.

Keep the final recommendation within the model. ‘Four packs fit within the stated ideal useful-energy budget’ is justified. ‘Four packs will keep every item safe for two hours’ is not justified by the arithmetic. This is a lesson in interpretation as much as in calculation. A numerically correct answer should still be rejected if the final sentence introduces a new claim that the problem did not support.

What to do when the question changes

After solving a standard worked example, change one input rather than doing another copy. Increase the initial ice temperature from −5°C to −2°C. The solid-warming term becomes 2100 × 2 = 4200 J/kg, so the total coefficient falls to 380,200 J/kg for melting and warming to 10°C. Now change the final temperature to 8°C. The liquid-warming term becomes 4200 × 8 = 33,600 J/kg. Calculate the new coefficient before you calculate any count.

Transfer practice is not random difficulty. It teaches the model’s structure. The ice-warming term responds to the starting temperature; the melting term depends on mass and specific latent heat; the liquid-warming term responds to the desired final temperature. When you can identify which term changes, you can solve a new problem without fearing that it looks unfamiliar.

A compact timed-exam procedure

Read the question and write what is being maximised or minimised. Define the variable and its unit. Translate the physical journey into separate stages if necessary. Form the combined algebraic expression and solve the continuous model. Apply any domain restriction, such as a whole-number pack count, then test the proposed integer against the original inequality. Finally, label the conclusion as modelled, measured or conditional.

This approach builds on Vol 0419 on graph gradients and threshold times and Vol 0422 on phase-change energy. The next English guide turns our careful numerical conditions into sentences for different readers. The final integrated guide checks that those sentences still agree with the calculations.

Primary foundations worth keeping

Primary-school learners already encounter units, fractions, proportional relationships and word problems with conditions. The PSLE learner guide on interpreting diagrams as relationships is a useful stepping stone: a picture is not automatically a scale drawing, just as a context is not automatically an instruction to average temperatures. At secondary level, variables and inequalities make the same habit more powerful.

The important change is not that Mathematics has become less intuitive. It is that the hidden assumptions must be stated precisely. A diagram, table, equation and verbal conclusion should all describe the same constraints. If one of them gives a different story, return to the model. This simple check prevents attractive but unsupported answers.

Questions that decide whether you understand

Why is the continuous result 4.398 not the final answer for sealed whole packs? Because pack count is restricted to integers, and the energy budget allows at most four. Why must we compare available energy with the melting requirement in a mixture? Because complete melting determines whether a later liquid-temperature calculation is even valid. Why is a rounded 0.440 kg not a sufficient feasibility check? Because rounding can cross a decision threshold.

What changes if useful power varies over time? The energy must be accumulated according to the separate rate intervals rather than multiplying one power by the full duration. Which quantity does the gradient of E against m represent? The useful energy required per kilogram for the specified state journey. These answers should become immediate; then longer examination questions are less about guessing and more about choosing a reliable order of operations.

Challenge 01: An exact budget with no surplus

A fictional source offers exactly 154,600 J. At 38,650 J per complete standard pack, four packs fit with zero surplus. Five do not. The boundary is inclusive for four because the inequality allows equality. Explain why a result at the precise threshold is feasible in the stated ideal model, but not a certificate of robust performance in practice. Change the budget to 154,599 J and re-evaluate: only three complete packs are possible, even though the missing energy is only one joule.

Write the defining equation or inequality before checking the worked discussion. Identify the units, state whether you are finding a continuous mass or a discrete count, and explain why the last mathematical step answers the actual question. Then make a one-number change and redo the decision. If the altered case requires a different formula branch or a different rounding direction, describe that change in a sentence. This way the practice builds flexible reasoning rather than recognition of familiar figures.

Challenge 02: A different initial temperature

A 0.100 kg ice pack starts at −2°C instead of −5°C but ends as water at 10°C. Using the supplied constants, the per-kilogram requirement is 2100×2 + 334,000 + 4200×10 = 380,200 J/kg. One pack needs 38,020 J. Divide a 170,000 J budget by that per-pack figure to get about 4.47, so four complete packs fit. Identify precisely why the coefficient changed: only the warming-of-ice stage became smaller.

Challenge 03: A different final liquid temperature

Keep the starting ice temperature at −5°C but change the required final liquid temperature from 10°C to 6°C. The per-kilogram coefficient becomes 2100×5 + 334,000 + 4200×6 = 369,700 J/kg. A 0.100 kg pack needs 36,970 J. Five packs would need 184,850 J and exceed a 170,000 J budget, so only four fit. The fact that the final temperature is lower does not remove the latent melting stage.

Challenge 04: Mass given in grams

A question supplies 75 g of ice per packet. Convert the mass to 0.075 kg, then multiply by 386,500 J/kg to get 28,987.5 J per pack. With 170,000 J available, five complete packs require 144,937.5 J, while six require 173,925 J. Therefore five are possible under the ideal conditions. State the remaining margin after five and why rounding the per-pack requirement too early could affect a borderline case.

Challenge 05: A mixed sample with just enough energy

Warm water has exactly 33,400 J available on cooling to the melting point. A 0.100 kg ice sample at zero degrees needs exactly 33,400 J to melt. In the ideal model all the ice just melts and the resulting water remains at zero degrees. There is no leftover energy to warm the liquid. This boundary example distinguishes ‘melts completely’ from ‘ends above the melting temperature’. State why a solution that predicts a positive temperature has introduced energy that is not present.

Challenge 06: A mixed sample that cannot finish melting

Use 0.250 kg warm water at 32°C and 0.120 kg ice at zero degrees. The warm water can release 33,600 J by reaching zero. Complete melting would need 40,080 J. Since the first value is smaller, no further temperature rise is possible while solid remains in the ideal model. Compute 33,600/334,000 ≈ 0.1006 kg melted and about 0.0194 kg left. Explain why it would be mathematically misleading to divide the remaining negative energy balance by combined heat capacity.

Challenge 07: Reverse the ice mass for an all-liquid target

A 0.300 kg water sample initially at 28°C must reach 10°C after mixing with ice at zero degrees. Write 0.300×4200×18 = x(334000+4200×10). The solution x≈0.0603 kg is the ice mass that can melt and warm to 10°C under the isolated ideal model. Why is the warm water’s cooling interval 18 degrees rather than 28 degrees? Because the final state is specified as 10°C, not zero. Name the unknown and its units before solving.

Challenge 08: Combine two useful-power intervals

A transfer source supplies 60 W usefully for eight minutes and then 30 W for twelve minutes. Energy is 60×480 + 30×720 = 50,400 J. A standard 0.100 kg pack needs 38,650 J, so one complete pack fits and a second does not. The average power over twenty minutes would be 42 W, but that must be obtained from total energy divided by total time, not by simply averaging 60 and 30 with equal weights.

Challenge 09: Optimise with two simultaneous limits

At most five standard packs can be stored, and the useful energy budget is 130,000 J. The storage constraint is n≤5; the energy constraint is 38,650n≤130,000. The energy condition allows only three whole packs because four require 154,600 J. State the binding constraint. Then imagine a budget of 250,000 J: the energy supports six packs, but the storage limit restricts the answer to five. Identifying the tightest condition matters more than the initial division.

Challenge 10: Find a minimum rather than a maximum

A fictional task requires at least 0.350 kg of material to complete a specified state journey, using standard 0.100 kg packs. How many packs are needed to meet the mass target? At least four, because three contain only 0.300 kg. That is a round-up decision, unlike the maximum count permitted by a fixed energy budget. Four packs also need a minimum ideal useful-energy budget of 154,600 J. State both constraints; meeting the mass requirement does not guarantee that the energy source can support it.

Challenge 11: Compare two packet sizes fairly

Pack A contains 0.100 kg, pack B 0.080 kg, and both follow the same journey with coefficient 386,500 J/kg. At 170,000 J, four A packs or five B packs are feasible. Both options represent 0.400 kg of material and require 154,600 J in the ideal model. A comparison based only on number of packets would make B appear greater without identifying the shared total mass. Say what you are comparing and why equal pack counts are not the right basis.

Challenge 12: Repair an incorrect percent calculation

A learner calculates 15,400/154,600 as 9.96 percent and announces that 9.96 percent of the available budget is unused. The arithmetic is correct but the sentence gives the wrong denominator. The available budget is 170,000 J; the unused share of that budget is 15,400/170,000≈9.06 percent. The original 9.96 percent is unused energy relative to the four-pack requirement. Rewrite both statements with explicit bases, then decide which one the question asked for.

Challenge 13: Write a stepwise energy function

Define f(m)=386,500m for nonnegative continuous mass m. Now restrict the chosen mass to multiples of 0.100 kg. The allowed outputs are 0, 38,650, 77,300, 115,950 and so on. Explain why the original graph is continuous but the real package choice is discrete. If the budget is 160,000 J, the greatest allowed output not exceeding the budget is 154,600 J, corresponding to four packs. A continuous intersection alone would not finish the decision.

Challenge 14: Test how rounding affects a border

A source offers 193,240 J, ten joules short of the five-pack threshold of 193,250 J. A rough calculation might round the individual requirement to 38.6 kJ and falsely conclude that five need 193 kJ, which appears feasible. Keep the original values for the boundary test. Four packs are possible, five are not. This example shows why sensible significant-figure reporting should follow, not precede, the inequality check. Round the presentation without moving the physical decision boundary.

Challenge 15: Identify an unearned guarantee

A calculation shows that three packs require 115,950 J, within a 120,000 J useful-energy budget. The writer concludes that the three packs will keep a real container safe for precisely one hour. Reject the last clause. Energy capacity and a single budget do not determine heat-transfer rate or verify safety; the statement adds claims not contained in the model. Replace it with: ‘Three packs satisfy the stated ideal useful-energy requirement, leaving a modelled margin of 4050 J.’ Preserve what is supported and discard the invented promise.

Challenge 16: A changed coefficient reverses the maximum count

A hypothetical improved design needs only 30,000 J per complete pack, still with 170,000 J available. The quotient is 5.666…, so five complete packs fit, six do not. The maximum changes compared with the 38,650 J original design. But a smaller calculated requirement does not by itself prove a product is better in every way. State the numerical criterion, the assumptions and the feasible count. This exercise checks whether you can keep algebra, discrete constraints and verbal interpretation aligned.

Small Group Tutorials

How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 0423 | Mathematics Reverse Energy Budgets — Ice Mass, Whole Packs and Feasibility