G3 SEC Maths Ventilation Timetables | Learner’s Guide Vol 1187

Build a timetable that satisfies the mathematics—not just the energy total

A school greenhouse demonstration has twelve half-hour operating slots, a fixed energy budget and a maintenance interval in which the fan must remain off. The mathematics is straightforward only after the real-world words have been translated into units, variables, inequalities and clock times. This chapter works through one complete fictional timetable and then adds uncertainty, cost, optimisation and evidence checks.

The 2027 G3 SEC Mathematics syllabus is K310. Its problems test mathematical methods and reasoning, not a prediction of greenhouse examination themes. All fan energy figures, school rules and prices in this chapter are hypothetical teaching values. They must not be treated as actual Singapore tariffs, equipment specifications or plant-care advice.

Quick route: Count the slots correctly; define a whole-number choice variable; build the on/off energy formula; solve a cap inequality; translate mandatory and unavailable slots; check consecutive runs; construct a timetable; calculate prices and percentage changes; test a revised energy model and uncertainty bounds; finish with a decision that states its assumptions.

Why a temperature reading is not a schedule

A thermistor circuit can indicate a change in temperature, but a school demonstration still needs a timetable. The electrical response in the preceding Physics guide and the timetable in this Mathematics guide are different decisions. This chapter assumes an invented school greenhouse opens to visitors for six hours and divides that period into equal half-hour slots. A fictional planner must choose when a demonstration fan operates while respecting an energy budget, a maintenance restriction and a short viewing requirement.

The distinction matters in examination-style reasoning. A physical sensor can supply a reading; the timetable uses a planning model with constraints. We have not measured how much ventilation actual plants need, nor defined a commercial fan controller. The numbers are invented to practise equations, inequalities, schedules, integer choices and appropriate conclusions rather than to recommend a real greenhouse operating policy.

For examination practice, the useful check is to write down the meaning of each quantity before calculating and to test the result against the original wording. If the result is a count of complete slots, it must be a whole number. If it is an energy figure, it needs kWh. If it is a schedule, actual times and every restriction must be visible. A neat equation is only part of a complete answer.

Know what G3 Mathematics K310 actually assesses

The Singapore-Cambridge Secondary Education Certificate begins in 2027. G3 Mathematics is syllabus K310, replacing the older school-candidate O-Level naming and code 4052. The official syllabus emphasises concepts and skills, reasoning, communication, connections, applications and modelling. Paper 1 lasts two hours fifteen minutes for 90 marks. Paper 2 also lasts two hours fifteen minutes for 90 marks, including a real-world context question; each paper carries half the total weighting.

Approved calculators are permitted for both papers. The published syllabus, not an independent guide, governs the detailed rules for a particular cohort. The greenhouse example below is one constructed practice model, not an alleged SEC examination topic. The practical lesson is to read a question as a set of defined quantities and constraints, then provide a numerical answer that is meaningful in that setting.

Draw the clock before writing an equation

The fictional visitor session runs from 09:00 to 15:00, a period of six hours. At thirty minutes per slot, there are twelve slots. Slot 1 is 09:00–09:30; slot 2 is 09:30–10:00; slot 3 is 10:00–10:30; slot 4 is 10:30–11:00; slot 5 is 11:00–11:30; slot 6 is 11:30–12:00. Slots 7 through 12 continue in the same order until 15:00.

Students who treat six hours as six slots have doubled the allowable duration of each selection. One defensible conversion is six times sixty minutes divided by thirty minutes, giving twelve. The result is dimensionless because minutes cancel. A good timetable shows both slot numbers and clock times so that a symbolic answer is traceable to an actual interval. There is no need to draw an ornate timetable; a simple numbered row is enough.

Define a decision variable clearly

Let x_i equal 1 if the fan is on in slot i and 0 if it is off, for i from 1 to 12. Each x_i is therefore an integer with only two possible values. Let n be the total number of on-slots, so n=x_1+x_2+…+x_12. This is an elementary way to organise a discrete planning problem, not a requirement to learn a formal university optimisation algorithm.

Writing a definition first prevents an important ambiguity. If x_i represented minutes rather than a yes-or-no decision, the energy equation would have different units. The index i names a particular half-hour interval; the total n counts selected intervals. For all twelve off, n=0. For all twelve on, n=12. A physically possible plan must keep n between those values and meet any additional restrictions.

Build the energy model from all twelve slots

Suppose the fictional fan-and-display setup uses 0.12 kWh during an on-slot and 0.02 kWh during an off-slot. An off-slot still uses electricity because the monitoring display stays active. If the fan is on in n slots, it is off in 12−n slots. The total is E=0.12n+0.02(12−n) kWh, which simplifies to E=0.24+0.10n kWh.

The 0.24 kWh term is the energy use of twelve off-slots. Each on-slot substitutes 0.12 kWh for 0.02 kWh, adding 0.10 kWh. This offers a quick check against the original expression. When n=0 the model gives 0.24 kWh; when n=12 it gives 1.44 kWh. A student’s model that gives zero energy when the fan is off has ignored an explicit piece of the hypothetical scenario.

Distinguish power from energy

Power is measured in watts or kilowatts; electrical energy in this exercise is measured in kilowatt-hours. The supplied 0.12 and 0.02 kWh values are energy per half-hour slot. They do not need to be multiplied by thirty minutes again. If instead the question supplied a power of 0.24 kW during an on-slot, then energy for a half-hour would be 0.24×0.5=0.12 kWh.

This distinction is often more important than calculator speed. Multiplying a slot-energy figure by a duration a second time produces a number with the wrong meaning. The previous Physics guide explains voltage and current in a sensing circuit; neither a 4.0 V signal nor a 6.0 V supply states the entire fan system’s electrical energy without additional information. Keep each unit attached to the appropriate quantity.

Translate the energy cap into an integer inequality

The invented committee permits at most 1.04 kWh over the session. Substitute the model: 0.24+0.10n≤1.04. Subtract 0.24 to obtain 0.10n≤0.80, then divide by 0.10 to obtain n≤8. Because n counts complete half-hour slots, n must be a whole number between zero and eight under the energy cap.

The boundary 8 is included because the rule says ‘at most’. A plan using eight on-slots consumes exactly 1.04 kWh. If the wording had said ‘less than 1.04 kWh’, eight would fail and the maximum whole number would be seven. The distinction between a strict inequality and one that includes equality can change the final schedule.

Add a minimum-service requirement

Now imagine that staff need the fan operating in at least five complete slots during opening hours. This is a planning rule invented for the example, not a horticultural requirement. Mathematically it is n≥5. Combining the service requirement and the energy cap gives 5≤n≤8. The allowed integer totals are therefore n=5,6,7,8.

This answers which totals could be feasible under the two whole-day rules. It does not yet answer which particular times are feasible. A student who finds n=6 and stops has not handled the schedule restrictions. Inequality algebra gives the feasible range of totals; the next stage tests the actual placement of on-slots.

Interpret a maintenance prohibition

The maintenance notice says the fan must be off in slot 6, from 11:30 to 12:00. Write x_6=0. This condition does not reduce the total number of slots in the day from twelve to eleven. There are still twelve slots, but one is forced to be off. The basic energy formula still counts the standby energy in the maintenance interval.

A common error is to delete slot 6 from the calculation and accidentally count only eleven intervals of standby energy. Removing a time interval from the set of possible on-times is not the same as removing it from the clock. The mathematical model must preserve the difference between ‘unavailable for operation’ and ‘not present in the day’.

Encode the three-slot demonstration window

The school has set a viewing demonstration for the consecutive slots 3, 4 and 5. It requires the fan on in all three: x_3=x_4=x_5=1. That immediately accounts for three of the at-least-five on-slots. Since slot 6 must be off, the demonstration block ends with a natural break before any later block of operation.

This is a stronger requirement than ‘at least three slots somewhere’. It specifies where the operation must occur. Students must attend to words such as during, before, consecutive and exactly. Converting them to slot numbers is not clerical work: it is the reasoning that makes the final answer valid.

Apply a maximum-consecutive operation rule

A further fictional constraint allows no more than three consecutive on-slots. In any four neighbouring slots, at least one must be off. One could express this as x_i+x_(i+1)+x_(i+2)+x_(i+3)≤3 for each starting index i from 1 through 9. For a school-level explanation, it is enough to inspect every group of four consecutive slots and verify that not all four are selected.

Do not confuse a maximum run length of three with a maximum total of three on-slots. Two separate three-slot blocks are allowed if an off-slot separates them. The distinction is visible in a timetable and could be missed by a single inequality for n. Counting totals and checking local arrangements are complementary tasks.

Construct a feasible timetable deliberately

Consider on-slots 3,4,5,8,9,10 and off-slots 1,2,6,7,11,12. There are six on-slots, so n=6. The energy is E=0.24+0.10×6=0.84 kWh. The required demonstration slots 3–5 are on, slot 6 is off for maintenance and the later run 8–10 is also only three consecutive slots long.

This is a constructive proof of feasibility: the timetable exists and each constraint can be checked. Do not merely say ‘six is between five and eight’. That numerical range is necessary but not enough when placement restrictions are present. An answer earns credibility by showing the actual selected slots and walking through every requirement.

Convert the timetable into clock times

The first on-block, slots 3–5, runs from 10:00 to 11:30. The second, slots 8–10, runs from 12:30 to 14:00. The fan is off during the 11:30–12:00 maintenance slot and during the other off-slots. This translation is useful because an answer listing slot indices may be mathematically correct yet hard for a reader to use.

Make sure endpoints are not counted as additional half-hour periods. The interval 10:00–11:30 is ninety minutes, equal to three half-hour slots, not four. When a problem gives a schedule by starting and ending times, draw a number line or tally the intervals. It is safer than counting the printed clock labels.

Check an apparently attractive but invalid schedule

Suppose a student selects slots 3,4,5,6,8,9,10. That gives n=7 and energy 0.94 kWh, which is below the 1.04 kWh cap. Nevertheless the plan fails: slot 6 must be off and the run 3–6 contains four consecutive on-slots. Energy compliance does not excuse breaches of two other conditions.

This is a powerful examination habit: evaluate each constraint independently. A real-world scenario may include several rules; satisfying one does not imply satisfying the rest. When showing that a proposed plan is impossible, identify the exact violated condition rather than making a vague statement that the total is ‘too many’.

Use a table of totals to expose the pattern

For n=5,6,7,8 the total energies are 0.74,0.84,0.94,1.04 kWh respectively. The values form an arithmetic progression with common difference 0.10 kWh because each extra on-slot adds that amount to the total. A simple table gives the planner an immediate sense of the feasible energy range before a detailed timetable is chosen.

The progression alone cannot tell us whether a particular configuration of eight on-slots satisfies a maximum-three-consecutive rule. Tables are strong at expressing the model’s overall relationship, while detailed slot diagrams handle the local restrictions. Use the right representation for the part of the problem being answered.

Test a fictional electricity price accurately

Suppose an exercise gives an illustrative electricity charge of S$0.32 per kWh. It is not an actual Singapore tariff. For the six-on-slot schedule, the modelled energy charge is 0.84×S$0.32=S$0.2688, or about S$0.27 to the nearest cent. The model does not include fixed fees, demand charges or other components unless the question specifies them.

It is important to label the result ‘modelled energy charge’. Do not claim this is the complete real household or school electricity bill. If a question asks for savings relative to another timetable, calculate both with the same price and interval basis. When comparing alternatives, keep unrounded values until the last step to avoid an artificial discrepancy.

Calculate a difference against a longer operation plan

An eight-on-slot plan would use 1.04 kWh, while the six-on-slot plan uses 0.84 kWh. The difference is 0.20 kWh. Relative to the eight-on-slot baseline, the energy reduction is 0.20/1.04×100%, approximately 19.2%. Relative to the six-on-slot plan instead, the same absolute difference represents 0.20/0.84×100%, about 23.8%.

Percentages need named denominators. A student should write ‘19.2% less than the eight-slot plan’ rather than the floating statement ‘19.2% saving’ if the reference is unclear. The two percentages describe different comparison questions; they are not two possible answers to one precisely stated problem.

Deal with a changed consumption estimate

Suppose new illustrative information revises on-slot use to 0.14 kWh while the off-slot remains 0.02 kWh. The revised model is E=0.14n+0.02(12−n)=0.24+0.12n. Under the same 1.04 kWh cap, 0.12n≤0.80, so n≤6.666…, and n must be a whole number. The new maximum is six complete on-slots.

A student who rounds 6.666… up to seven ignores the inequality. Seven revised on-slots would use 1.08 kWh and exceed the cap. Conversely, six use 0.96 kWh and satisfy it. Always test the proposed integer boundary in the original equation if the direction of rounding is important.

Distinguish exact totals from measured estimates

The per-slot energies in this exercise are assigned model values. If they had come from a meter rounded to two decimal places, the true totals might need uncertainty handling. A displayed 0.12 kWh cannot support an unlimited number of decimal places in a real measurement claim. The problem should tell you whether to treat numbers as exact model inputs or rounded measurements.

When no uncertainty information is provided in an ordinary examination question, use the given figures as instructed and do not invent a confidence interval. When uncertainty is part of the question, identify its source and test the worst case against the cap. Precision is a tool for honest decisions, not a way to make a simple model artificially complicated.

Build an upper-bound model for robust budgeting

For a separate extension, suppose actual on-slot use may be anywhere from 0.12 to 0.13 kWh and off-slot use anywhere from 0.02 to 0.025 kWh. These are hypothetical bounds. A conservative upper-energy calculation is Emax=0.13n+0.025(12−n)=0.30+0.105n. If the cap remains 1.04 kWh, the condition 0.30+0.105n≤1.04 gives n≤7.0476…, hence at most seven whole slots.

This bound assumes every on-slot and off-slot can reach its stated maximum together; it is a deliberately cautious independent bound, not a statistical prediction. Eight on-slots could use up to 1.14 kWh under these example limits. A robust choice protects against a specified range, whereas the earlier nominal choice uses a single pair of given values.

Find where nominal and robust answers differ

The nominal calculation allowed eight on-slots. The conservative upper-bound calculation allows seven. That does not mean eight nominal slots have suddenly become ‘wrong’. It means the question has changed from ‘does this plan meet the modelled figure?’ to ‘does it remain under the cap for every value in the supplied uncertainty ranges?’ The distinction between an estimate and a guarantee is central.

Students can practise this contrast with two adjacent columns labelled ‘nominal’ and ‘worst case’. Each column should state its assumptions, equation, integer bound and interpretation. That structure prevents the frequent mistake of quoting the stricter number without explaining which question it answers.

Separate the fan timetable from its sensor trigger

The previous Physics chapter showed that an NTC potential divider could produce an output that varies with temperature, and that a fictional indicator could respond to a voltage threshold. The mathematical timetable here does not automatically infer temperatures for its slots. Without a time series of observed sensor readings, the model cannot say which slot would trigger an automatic fan.

A question might provide both readings and a schedule. Then connect them only through the explicitly stated rule. If a proposed timetable is manually programmed, say so. If an automatic controller overrides the timetable, the scenario must specify which constraint takes precedence. The strongest mathematical model begins by explaining its decision rule instead of quietly importing facts that were never given.

Decide whether to minimise cost or maximise operation

A problem must name its objective. If the only goal is minimising energy subject to at least five on-slots and the other constraints, a feasible five-slot plan may be preferred to the six-slot example. If the goal is providing more operation while respecting a cap, the best objective may select a larger n. Neither is automatically the real-world ‘best’ plan for plants or visitors.

The distinction between constraints and objectives is essential. A limit says what cannot be exceeded; an objective says what to maximise or minimise among feasible possibilities. Write both in words before solving. Many incorrect conclusions arise from choosing the cheapest plan when the task actually asks for the maximum number of demonstration intervals.

Consider tie-breakers when two plans cost the same

Under the simple E=0.24+0.10n model, two schedules with the same n consume the same energy. They may still differ in timing. A fictional school might prefer an on-block near visitor arrivals, or require supervision during particular slots. Such preferences must be stated and translated into constraints or a separate ranking rule.

Do not claim that one six-slot schedule is financially superior to another if the model gives identical energy for both. Instead note that the plans tie on energy and explain what extra information would be needed to distinguish their suitability. This habit is useful well beyond electricity questions: a correct calculation can leave a decision unresolved.

Read a graph of energy against number of slots

Plot n on the horizontal axis and E on the vertical axis. The nominal rule E=0.24+0.10n is a straight-line relationship, but physically allowed n-values are integers. It is appropriate to mark discrete plotted points rather than implying that a quarter of a half-hour slot has been authorised by the planning rules. The gradient is 0.10 kWh per extra on-slot; the E-intercept is 0.24 kWh.

If a question asks you to interpret a gradient, give its units and context. ‘0.10’ alone is incomplete. It represents the additional modelled energy caused by replacing one off-slot with an on-slot. The intercept does not mean the fan itself uses electricity while switched off; it represents the stated baseline system consumption in all twelve slots.

Explain how to check a proposed solution systematically

Build an audit row with five columns: total on-slots; energy; mandatory slots 3,4,5; prohibited slot 6; longest consecutive run. Evaluate every candidate against every column. For the six-slot plan, the results are n=6, E=0.84 kWh, required slots present, slot 6 absent, longest run 3. It passes the stated rules.

This final check is more reliable than asking whether a plan ‘looks fine’. It also trains students to show enough working for a marker or teammate to follow. A list of numbers is not yet a solution until the conditions are explicitly verified. In a timed examination, the compact audit often catches errors faster than redoing the entire algebra.

Train the transition from Primary to G3 reasoning

At PSLE level, a student may already solve problems involving rates, percentages, elapsed time and units. G3 adds richer algebraic notation and simultaneous restrictions. The practical bridge is to make one statement at a time: what is the unknown, which operations are allowed, and what does a solution mean in ordinary language? The timetable is an excellent setting for that discipline.

A learner does not need to leap straight into a complicated twelve-variable expression. Begin by counting slots, building the one-variable energy model, then adding the special location rules. This order moves from a secure foundation to advanced decision-making. When a mistake occurs, diagnose whether it arose in the clock conversion, the algebra or the final feasibility check.

Present a Paper 2-style final answer

An effective response might read: ‘Let n be the integer number of half-hour on-slots. E=0.12n+0.02(12−n)=0.24+0.10n kWh. From E≤1.04, n≤8. Selecting slots 3,4,5,8,9,10 gives n=6 and E=0.84 kWh. Slot 6 remains off and no on-run exceeds three slots, so the schedule is feasible under the given assumptions.’

Notice the order: define, calculate, restrict, construct, verify, conclude. This is longer than a one-number answer, but it avoids an unsupported decision. When a question is marked for explanation, the logic is part of the result. In other questions, avoid needless repetition; follow the command words and marks given.

Worked practice: short questions with explanations

Use the invented six-hour scenario and the nominal equation E=0.24+0.10n kWh unless a question explicitly changes a value. Try each question before opening the answer in the following paragraph. These are original practice cases, not actual SEAB past-paper questions.

Convert 7.5 hours into half-hour slots.

7.5×60/30=15 slots. Use minutes in both numerator and denominator to avoid mixing units.

To check this answer, identify the given quantities, preserve the meaning of the selected comparison, and inspect any boundary carefully. For numerical results, substitute back into the original model. For statements about what a plan proves, separate the mathematical assumptions from unmeasured real-world effects.

Evaluate E for five on-slots.

E=0.24+0.10×5=0.74 kWh. This fits the nominal 1.04 kWh cap and satisfies the whole-day minimum count.

Check seven on-slots against the nominal cap.

E=0.94 kWh, which does not exceed 1.04 kWh. A specific seven-slot timetable must still satisfy placement rules.

Find the greatest integer n if E must be strictly below 1.04 kWh.

0.24+0.10n<1.04 gives n<8, so the greatest integer n is seven.

What happens if slot 6 is unavailable?

Set x_6=0, but still count its 0.02 kWh off-slot use. The entire six-hour session remains twelve slots.

Check a run of 3,4,5,6.

It fails the maintenance rule at slot 6 and the maximum-three-consecutive rule. A correct energy total cannot rescue it.

Calculate the price of 0.94 kWh at the invented tariff.

0.94×S$0.32=S$0.3008, or S$0.30 to the nearest cent. This excludes any unspecified fees.

Calculate the added energy of one more on-slot.

Replacing an off-slot by an on-slot adds 0.12−0.02=0.10 kWh in the nominal model.

Find the maximum revised on-slot count at 0.14 kWh per on-slot.

The revised upper bound is six whole slots, since n≤6.666… and seven would exceed the cap.

What is the energy of six revised on-slots?

E=0.24+0.12×6=0.96 kWh, below the 1.04 kWh cap.

What does the energy graph’s intercept mean?

0.24 kWh is the baseline for twelve off-slots. It is not an initial supply voltage or a fixed electricity price.

Is 5.5 on-slots a feasible solution?

Not under the model’s requirement that every slot is either fully on or fully off. The variable n is an integer.

Can an automatic fan’s exact switching times be deduced?

No. The slot-by-slot temperature or voltage record and a clear controller rule would be needed.

Does the least-energy timetable always give the best result?

No. A minimum-energy objective differs from maximising operation or meeting visitor requirements. The objective must be stated.

Give one way to prove feasibility.

Provide a full schedule and test its energy, minimum operation, mandatory slots, prohibited slot and consecutive-run restrictions.

Which plan is guaranteed within the illustrated worst-case cap?

Any feasible schedule with at most seven on-slots satisfies the conservative energy upper bound, provided the described uncertainty assumptions hold.

Keep reading along the learner route

This scheduling problem is a distinct continuation of Learner’s Guide Vol 1186 on the Physics thermistor controller. For earlier sensor mathematics, study Vol 1183 on humidity calibration and bounds. The broader G3 Mathematics foundations and algebra owner explains where equations and problem solving fit into secondary learning. Students moving from primary school can begin at the PSLE-to-G3 preparation route. Work backwards whenever a weak foundation blocks a later calculation.

Check the official source before planning examinations

Consult SEAB’s official 2027 G3 Mathematics K310 syllabus, the G3 subject-syllabus directory and the SEC overview. An independent worked example can make a topic easier to grasp, but the official documents govern syllabus coverage and examination conditions. The fictional ventilation planning rules in this article are not operational instructions for an actual greenhouse.

Small Group Tutorials

How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 1187 | Mathematics Greenhouse Ventilation Plan — Integer Time Slots, Energy Budgets and Feasible Decisions