A good G3 Mathematics answer makes a decision only after checking units, rates and constraints. This independent guide works through a fictional moisture-control scheduling problem with two devices, measured collection rates, energy charges and a limited operating window. It teaches how to calculate, model, test and explain feasibility without treating a laboratory rate as a guaranteed daily output.
The scientific foundation is Volume 1349 on evaporation and condensation. Earlier Mathematics owners include Volume 1342 on water-recovery rates and inequalities and Volume 1346 on samples, variation and fair comparisons. The new owner here is the combined scheduling and least-cost decision under explicit whole-hour constraints.
Quick route: translate the words into a data table; find comparable unit rates; define the variables; write three inequalities; test whole-number schedules; calculate the tariff and cost; justify the least-cost feasible option; state the modelling assumptions.
1. The decision is not which machine has the largest number
After the Physics work in the preceding chapter, a fictional habitat team has two moisture-control units and one short operating window. A supplier offers a capacity label, a technician reports actual trial volumes, and the finance officer lists the energy used per running hour. Mathematics should tell the team which schedules are feasible under a stated model, but it cannot establish a machine’s real-life performance where conditions differ. The first question is therefore more careful than ‘Which is best?’ It asks, ‘Given these measured and assumed rates, which schedules can meet the stated collection target within eight operating hours and an energy-cost ceiling?’ This question has variables, units, constraints and an answer that can be checked. That is the central habit for a strong G3 Mathematics contextual problem.
2. Build the data table from the words
Use the following invented exercise dataset throughout this chapter. Unit A collected 0.18 L in three hours under one trial and used 0.20 kWh for each hour it ran. Unit B collected 0.32 L in four hours under a separate trial and used 0.35 kWh per running hour. The exercise assumes a common electricity tariff of $0.32 per kWh and an eight-hour maximum operating window. The minimum collection target is 0.50 L. An additional cost ceiling is $0.80 for the shift. These are not actual product specifications or recommended indoor conditions. The job of the learner is to translate this case to mathematics, test the assumption of constant average rate and reach a limited conclusion. Write all the units in the table before calculating anything.
3. Calculate each observed rate
For A, observed average collection rate is 0.18 ÷ 3 = 0.060 L/h. For B, it is 0.32 ÷ 4 = 0.080 L/h. Those are rates computed from the trials, not guarantees that collection must continue linearly. You may convert them to 60 mL/h and 80 mL/h respectively, but do not change the numerical value without converting its unit. A common error is to compare the raw volumes, 0.32 and 0.18, without accounting for the different observation times. Another error is to use B’s four-hour measurement as though it covered the full eight-hour shift. The first fix is to put all values on a common basis. The second is to say explicitly that the projection assumes comparable conditions and a stable rate.
4. Keep observed, projected and advertised values apart
There are at least three different kinds of number in a word problem. An observed number comes from a specified trial. A projected number comes from extending a model. An advertised figure comes from a label or claim and may use different conditions. For example, the projection of B’s collection over seven hours is 7 × 0.080 = 0.56 L, based on its measured average and a constant-rate assumption. It is not a seven-hour observation. A test answer should use words such as ‘if the measured rate remains constant’. This tiny condition can rescue a large piece of reasoning. When a task asks for a decision, show whether the answer is established by measurements or conditional on the model. Mathematically correct arithmetic cannot turn an assumption into an observation.
5. Introduce variables only when they help
Let x be the number of hours Unit A operates and y the number of hours Unit B operates. In a simplified scheduling model, x and y are non-negative. If the eight-hour cap refers to total unit-hours, then x + y ≤ 8. If both units are permitted to operate simultaneously, their total unit-hours can exceed elapsed clock time; the wording must decide which model applies. This chapter’s main exercises specify a cap of eight total unit-hours. That makes the constraint x + y ≤ 8 valid. An examiner may instead tell the learner that the units can each run for up to eight clock hours. Do not mechanically reuse an old inequality. Define the variable’s meaning before the mathematics and repeat the meaning when interpreting the result.
6. Translate the collection target to an inequality
With the assumed constant rates, predicted collection in litres is 0.06x + 0.08y. The target says ‘at least 0.50 L’, so the model is 0.06x + 0.08y ≥ 0.50. The inequality direction matters. Reversing it would select schedules that collect too little. A physical target may be exactly equal to the boundary or above it, so ≥ is appropriate. If the case instead said ‘no more than 0.50 L’, the relationship would be ≤. A robust student re-reads the plain-English sentence after writing the inequality. Substitute a simple test: zero operating hours produces zero collection, and zero should fail a minimum target. That test quickly reveals whether the sign is wrong.
7. Compute the energy-cost expression
Unit A uses 0.20 kWh per running hour, so x hours uses 0.20x kWh. Unit B uses 0.35y kWh. Together the modelled energy consumption is 0.20x + 0.35y kWh. At an illustrative tariff of $0.32 per kWh, cost in dollars is C = 0.32(0.20x + 0.35y), or C = 0.064x + 0.112y. The $0.80 ceiling becomes 0.064x + 0.112y ≤ 0.80. These are monetary costs, not power ratings or collection rates. The calculation has two distinct layers: energy use and price per unit energy. Keeping the layers visible makes it much easier to repair a mistake. Do not multiply litres by electricity cost unless the question asks for cost per litre.
8. Find a feasible single-unit schedule
For B alone, y must satisfy 0.08y ≥ 0.50, giving y ≥ 6.25 hours. If full-hour shifts are required, the smallest whole number is seven. Seven hours yields a projected 0.56 L and costs 7 × $0.112 = $0.784. The operating-time condition and $0.80 cost ceiling are both met. For A alone, 0.06x ≥ 0.50 implies x ≥ 8.333… hours, which exceeds the eight-hour limit. A is therefore infeasible as a sole unit under the stated model. This does not mean Unit A is inferior in every respect. It is a statement about a chosen set of constraints and given rates. A decision is meaningful only with its scope attached.
9. Check an apparent cheapest option against time
Unit A has a lower hourly energy cost than B, but ‘cheaper per hour’ is not the same as ‘cheaper way to reach the target in time’. Under a hypothetical longer window, A would need nine whole hours and its energy cost would be 9 × 0.064 = $0.576. That is below the budget but outside this case’s eight-hour cap. A student who recommends A simply because $0.576 is less than $0.784 has ignored a binding constraint. The correct response distinguishes economic efficiency from feasibility. In contextual mathematics, rejected options deserve a mathematical reason. Write the rejected option’s violation in one line rather than burying it in a paragraph. The examiner can then see that you considered the complete problem.
10. Explore mixed schedules without random guessing
The model allows combinations such as x = 2, y = 5, giving a projected collection of 0.06(2) + 0.08(5) = 0.52 L, seven total unit-hours and cost 0.064(2) + 0.112(5) = $0.688. This is feasible under all three conditions. The mixed schedule is cheaper than B alone at seven hours in this simplified model, although the collection target is reached with a smaller predicted margin. Another combination x = 5, y = 3 produces 0.54 L in eight total unit-hours, costing $0.656. The aim is not to choose a random attractive answer but to compare viable possibilities. If the question demands the least cost across integer schedules, the student must search systematically rather than stop at the first feasible pair.
11. Find a least-cost whole-hour mix carefully
For integer, non-negative x and y with x + y ≤ 8, a systematic comparison can be made by checking allowed y values and the smallest x needed for the target. One candidate is x = 7 and y = 1: collection = 0.42 + 0.08 = 0.50 L, operating time eight hours, cost = 7 × 0.064 + 1 × 0.112 = $0.560. This meets the target exactly, and it costs less than x = 5, y = 3 or x = 2, y = 5. With y = 0 the target cannot be met within eight hours; with y = 1 at least seven A-hours are needed; larger y values replace some lower-cost A-hours with higher-cost B-hours. Under the stated conditions, x = 7, y = 1 is the least-cost integer schedule. State that the result belongs to this simple model, not to actual equipment performance.
12. Show your search method, not a long unstructured table
One method is to let y take integer values from zero to eight. For each y, calculate the minimum non-negative integer x satisfying the collection inequality, then reject any pair violating x + y ≤ 8 or the cost ceiling. You do not need to print every rejected combination in a short examination answer, but show enough working to explain how you found the optimum. Another approach is to examine the boundary 0.06x + 0.08y = 0.50 and compare feasible integer points. If the problem does not explicitly require integer hours, do not impose them silently; fractional operating durations may change the mathematical optimum. The important habit is to read whether the outcome is continuous, discrete, rounded up, or constrained by practical units.
13. Use graphs to make constraints visible
The line 0.06x + 0.08y = 0.50 separates combinations that meet the target from combinations that do not. The line x + y = 8 marks the time cap. The cost line 0.064x + 0.112y = 0.80 marks the budget cap. Draw a clear scale with x-hours on one axis and y-hours on the other, label intercepts, and test a known point to determine each permitted half-plane. Only the intersection of all permissible regions is feasible. The graph is a decision aid, not a substitute for numerical checking when exact integer choices matter. A sketch with unlabeled axes or a shaded region based on the wrong inequality can hide a conceptual mistake. The meaning of every boundary should be stated in words.
14. Do not mistake a decimal answer for a whole unit
If 6.25 hours is allowed, it means six hours and fifteen minutes, not six hours and twenty-five minutes. If the machine can only be booked in complete hours, round upward to seven hours because six hours would miss the minimum collection target. Ordinary rounding to the nearest integer would be inappropriate here. A requirement to purchase packs, hire buses, select people or schedule whole sessions often produces the same issue. Learn to distinguish mathematical rounding from practical integer constraints. In a final answer, the phrase ‘seven whole operating hours’ tells the reader why the number was rounded up. That short explanation can be more valuable than another line of arithmetic.
15. Convert energy units before pricing
Electricity charges are typically calculated from energy in kilowatt-hours. A student may confuse W, kW, Wh and kWh. In a simple example, a device drawing 250 W has a power of 0.250 kW. Run for three hours at that constant power, it uses 0.750 kWh. At $0.32/kWh, the modelled energy charge is $0.24. Do not write that 250 W means 250 kWh in one hour. The relationship between power and energy helps explain the arithmetic, but the G3 Mathematics question primarily tests correct unit handling and contextual interpretation. If the problem supplies energy used directly in kWh, do not invent an additional conversion. Read the unit printed beside the number and use the same base across alternatives.
16. Compare percentages against the same baseline
Suppose a second trial of B collects 0.36 L in four hours instead of 0.32 L. The increase is 0.04 L, and the percentage increase relative to the first trial is 0.04 ÷ 0.32 × 100% = 12.5%. It is not 0.04 ÷ 0.36. The denominator must match the phrase ‘increase from’. If the later result drops back to 0.32 L, the percentage decrease from 0.36 L is 0.04 ÷ 0.36 × 100%, about 11.1%; equal absolute changes need not have equal percentage sizes in opposite directions. In examination writing, identify the original value before using a percentage formula. The skill transfers directly from primary-school fractions and percentage comparisons to G3 context questions.
17. Read an average without assuming uniform performance
The average collection rate of B over four hours is 0.080 L/h. Its first two hours could produce relatively little and the next two relatively more. A mean is a summary across the observation period, not a minute-by-minute description. If the question gives cumulative collections at hours one through four, calculate interval changes by subtraction before comparing hourly performance. A graph that increases steadily does not necessarily mean its slope is constant. The slope over a specified interval gives the rate over that interval in the simplified model. A pupil who is accurate with division but ignores the graph’s shape can still misread an examination question. State explicitly whether you are discussing average rate, instantaneous trend or projected rate.
18. Separate uncertainty from arithmetic mistakes
Two trials may differ because conditions change, because instruments have limits or because the process genuinely varies. Mathematics cannot erase these causes merely by reporting the mean to three significant figures. For a practical decision, consider plausible lower and higher observed rates. Suppose B’s measured rate across comparable tests falls between 0.065 and 0.080 L/h. At seven hours its projected collection would be between 0.455 and 0.560 L under the constant-rate assumption. The lower bound fails the 0.50 L target. A responsible answer says that seven hours meets the target under the original 0.080 L/h model but is not guaranteed under the stated range. This is not a reason to abandon calculation; it is a reason to interpret calculation honestly.
19. Use a sensitivity check before recommending a schedule
At B’s original modelled 0.080 L/h, the seven-hour plan gives 0.56 L and a margin of 0.06 L above the target. If the modelled rate were 10% lower, it would become 0.072 L/h, and seven hours would yield 0.504 L, a margin of only 0.004 L. At 15% lower, the rate becomes 0.068 L/h, and the same plan yields 0.476 L, below target. The arithmetic shows which assumption matters. It does not predict the machine’s actual failure probability. In a complex case, a short sensitivity calculation can make a recommendation defensible. Specify the hypothetical change, calculate its effect, and compare it with the threshold. The explanation should not promise reliability where no probability distribution has been supplied.
20. Apply the official accuracy convention
For 2027 G3 Mathematics K310, the syllabus says that non-exact numerical answers should generally be given to three significant figures, or angles in degrees to one decimal place, unless the question specifies another accuracy. In financial contexts, the question may reasonably specify cents, and the answer should follow that instruction. Keep intermediate values unrounded when possible so that the final decision does not change from premature rounding. For example, 0.50 ÷ 0.08 is exactly 6.25; replacing 6.25 with 6.3 hours too early could affect a later cost estimate. An examination answer should be both numerically accurate and contextually appropriate. State final monetary costs to two decimal places when the question asks for dollars and cents, while preserving exact comparisons where needed.
21. Know where each official Mathematics paper assesses you
The 2027 G3 Mathematics syllabus K310 specifies two papers, each worth 90 marks and 50%. Paper 1 lasts two hours and fifteen minutes and contains about 26 short-answer questions. Paper 2 also lasts two hours and fifteen minutes, contains nine to ten questions, and ends with a real-world scenario problem. All questions are compulsory. A strong learner therefore needs reliable basic techniques as well as the ability to translate longer information into mathematical constraints. The official assessment objectives cover standard techniques, solving problems in varied contexts, and reasoning or communication. The moisture-control exercise is not an official specimen question; it is a rehearsal for the combination of modelling, calculation and interpretation called for by the syllabus.
22. Write the final decision in one disciplined paragraph
An exam response might say: ‘Let x and y be the whole hours assigned to Units A and B. Under the measured constant-rate model, the constraints are x + y ≤ 8, 0.06x + 0.08y ≥ 0.50 and 0.064x + 0.112y ≤ 0.80. The schedule x = 7, y = 1 meets all three: eight unit-hours, 0.50 L predicted collection and $0.56 modelled electricity cost. A systematic integer comparison gives this as the least-cost schedule under the stated assumptions. Actual collection may vary with operating conditions.’ This paragraph combines equations, verification and a properly limited conclusion. It avoids a familiar failure: saying ‘choose A’ without giving an operating schedule or showing that the target is met.
23. Repair the weak link in a timed paper
If you lose marks, diagnose the exact point where the solution diverged. A unit-rate error needs fraction and unit practice. An inequality error needs language-to-symbol translation. A rounding error needs whole-number constraint practice. A graphing error needs boundary and half-plane tests. A decision error needs a habit of checking every requirement after finding a candidate. Do not respond to all these mistakes by doing fifty more unrelated algebra questions. Pick three variations of the same failure and solve them with your notes closed. Then write a two-sentence explanation of why the method is correct. On examination day this reduces reliance on rote pattern recognition and helps with the unfamiliar final question.
24. The distinction between a mathematics answer and a factual promise
This plan lives in a fictional learning case. A real device’s water collection may depend on temperature, humidity, maintenance and how the equipment operates. The supplied rates are idealised averages and the unit-hour assumption is a modelling choice. The numbers can prove that a schedule meets the constraints of that model; they cannot certify indoor air quality or building safety. This is exactly why careful mathematical writing uses conditional language. Mathematics gives a valid result from stated premises. Responsible communication keeps the premises visible when the result leaves the page. The next English chapter will test how to explain those premises to people who need a recommendation rather than pages of algebra.
Practice paper: fourteen checks with complete reasoning
These are original learning exercises, not SEAB specimen questions. Try each before viewing its answer, then change the condition in the follow-up. One carefully checked variant teaches more than copying the result from a worked example.
Practice 1: Which machine has the greater observed collection rate?
Reasoning and answer: A: 0.18 ÷ 3 = 0.060 L/h; B: 0.32 ÷ 4 = 0.080 L/h. B has the greater observed average rate, by 0.020 L/h. Do not compare the raw total volumes from different-length trials.
Extension: Would the ranking necessarily hold in a warmer or drier room? Explain why a mathematical comparison has a scope.
Practice 2: How much would Unit A collect in six hours under the constant-rate model?
Reasoning and answer: 0.060 × 6 = 0.36 L. The answer must be labelled projected, because the original observation covered only three hours.
Extension: How does the projection change if the observed rate were 0.055 L/h instead?
Practice 3: What is the modelled electricity cost for B in seven hours?
Reasoning and answer: Energy use is 0.35 × 7 = 2.45 kWh. Cost is 2.45 × $0.32 = $0.784. If asked to the nearest cent, report $0.78, but retain $0.784 for checking against a precise $0.80 cap.
Extension: Why should an intermediate amount not always be rounded before the final comparison?
Practice 4: Can A alone meet 0.50 L in eight hours?
Reasoning and answer: At 0.060 L/h, its eight-hour projection is 0.48 L. That is less than the required 0.50 L. A alone fails the target under the model.
Extension: What is the smallest whole number of hours A would need if the time cap were relaxed?
Practice 5: Does B alone meet the target at six whole hours?
Reasoning and answer: Six hours gives 6 × 0.080 = 0.48 L, below the target. Seven whole hours gives 0.56 L and is sufficient in the stated model.
Extension: Explain why 6.25 is not rounded down when the unit must operate for whole hours.
Practice 6: Test x = 2, y = 5.
Reasoning and answer: The total is seven unit-hours; collection is 0.12 + 0.40 = 0.52 L; cost is 0.128 + 0.560 = $0.688. All conditions are satisfied. A feasible plan is not automatically the cheapest.
Extension: Compare it with x = 7, y = 1 using one common objective.
Practice 7: Write the minimum collection condition using x and y.
Reasoning and answer: The inequality is 0.06x + 0.08y ≥ 0.50. The coefficients are litres collected per running hour, while x and y are hours. The product has units of litres.
Extension: Why would the origin (0, 0) lie outside the feasible side of this boundary?
Practice 8: Explain why x+y≤8 may be invalid in a different version of the question.
Reasoning and answer: It assumes the cap is on combined unit-hours. If both machines may operate for eight elapsed hours simultaneously, the individual constraints x≤8 and y≤8 may be appropriate instead. The words define the mathematics.
Extension: Give a practical reason that the model’s scheduling assumption must be specified.
Practice 9: Convert 175 W for four hours into kWh.
Reasoning and answer: 175 W = 0.175 kW. Over four hours the energy is 0.175 × 4 = 0.700 kWh. A watt is a power unit, not an energy unit.
Extension: At $0.32 per kWh, what is the modelled charge for that run?
Practice 10: Calculate an increase from 0.32 L to 0.36 L.
Reasoning and answer: The increase is 0.04 L. Percentage increase = 0.04 ÷ 0.32 × 100% = 12.5%. The initial value is the denominator.
Extension: Why is the reverse percentage decrease from 0.36 L not also 12.5%?
Practice 11: What does a seven-hour prediction of 0.56 L say about a guaranteed minimum of 0.50 L?
Reasoning and answer: It exceeds the target under the constant-rate model, but the margin is not a guarantee. Without evidence that the rate will remain at least as large, the conclusion is conditional.
Extension: Recalculate the margin if the rate falls to 0.068 L/h.
Practice 12: Identify the main mark-losing error in ‘B is best because 0.32 is bigger than 0.18’.
Reasoning and answer: The statement ignores the unequal trial durations, energy use, schedule cap and cost ceiling. It also confuses an observation with a decision. Convert to unit rates and apply every constraint before recommending.
Extension: Write a replacement conclusion in fewer than 40 words.
Practice 13: Why might a least-cost result still be unsuitable?
Reasoning and answer: The model may omit practical constraints, or the observed rates may vary under new conditions. Within the examination, identify any specified unmodelled condition and state its effect without inventing evidence. Mathematical optimum means optimum for a stated objective and feasible set.
Extension: Name one extra quantitative constraint a question might reasonably add.
Practice 14: What would be wrong with claiming a 0.08 L/h rate proves a daily output of exactly 1.92 L?
Reasoning and answer: 1.92 L is a 24-hour projection at a fixed rate, not a measured daily result. A question may allow the projection, but it must be described as conditional and checked against any daily operating limit.
Extension: What is the 12-hour projection if all conditions and the model remain unchanged?
Useful questions from parents and learners
Will all SEC Mathematics questions be word problems?
No. The official 2027 K310 scheme has a wide range of short-answer and longer questions. This guide focuses on the higher-transfer skill of reading and modelling a context; it does not replace fluency in algebra, number, geometry, statistics and probability.
Is the lowest cost automatically the right answer?
Only when cost is the specified objective and every constraint is satisfied. A cheaper schedule that misses the volume target or time limit must be rejected. Do not remove an inconvenient constraint simply to produce a pleasing price.
How does a Primary School percentage lesson connect to this?
The idea of comparing a change with the original quantity remains essential. At G3 the figures may appear inside a multi-step modelling problem, where students must first decide what the original value means.
Do I need a different formula for every new story?
No. Many unfamiliar cases use familiar skills: ratios, rates, algebraic expressions, inequalities, graphs, accuracy and interpretation. The challenging step is often translating the information into the right mathematical relationship.
What can I do the evening before a timed practice?
Rehearse reading units, defining variables and checking the final decision. Then do one compact task without notes, mark the error type and correct it. Tired repetition of every possible question type is less useful than targeted repair.
Related owners and verified 2027 requirements
The connected science mechanism is Volume 1349 Physics. For background work on scheduling, compare Volume 1294’s integer runtime constraints; for statistics and honest variation, return to Volume 1346. These links support particular weak skills, not a second duplicate owner for this moisture-plan decision.
Official reference: 2027 G3 SEC syllabus index and SEAB/MOE G3 Mathematics syllabus K310. The two 90-mark papers, two-hour-and-fifteen-minute durations and final Paper 2 real-world question are taken from that official syllabus. All device figures, prices, results and assumptions above are invented teaching data, not actual product claims or a recommendation about controlling a building.
