SEC G3 Mathematics K310: A backup-power problem is solved by distinguishing rate from amount. Multiply each device’s stated power by the time it runs, add the energy used, then compare that total with the usable capacity rather than the nominal label. When a count of packs must be whole, state the inequality and round in the direction required by the constraint. This original learner guide continues the habitat-relay series at a more demanding decision stage.
G3 Mathematics for the first SEC examination in 2027 has two 90-mark papers, each lasting 2 hours 15 minutes. The last Paper 2 question emphasises mathematics in a real-world setting. A fictional energy schedule is useful practice for rates, percentages, algebra, inequalities, whole-number solutions and the interpretation of data; it is not a forecast of examination topics or advice about real backup supplies.
Quick answer: Write every rate and duration with units; calculate demand interval by interval; apply reserves to the right base; compare unrounded values; choose a feasible whole-number count; justify the final decision in one clear sentence.
A precise answer begins with the demand
The imaginary habitat has added a temperature-sensitive relay. The Physics chapter explained how a measured voltage can change when resistance changes. A separate decision now confronts the team: if its main supply is unavailable, how long can a backup pack run a set of low-power teaching devices? That is a mathematics question about quantities, units, rates, constraints and how to interpret a result. The supplied energy may be enough for one schedule but not another. A correct total is useful only after the student has checked which devices run together, for how long and under what assumed limit.
This guide works through the units, the total-versus-rate distinction and the whole-number choices that appear in practical planning. Nothing describes an actual emergency system or gives instructions for real equipment. All loads, capacities, prices and operating times are invented classroom figures. The purpose is to practise a kind of reasoning tested in G3 Mathematics: translating a written situation into a mathematical model, calculating accurately, interpreting the answer and explaining why a superficially attractive alternative may be infeasible.
Read the official mathematics format before practising
Singapore–Cambridge SEC begins in 2027. G3 Mathematics uses subject code K310, with 4052 shown by SEAB as the older reference code. There are two papers. Paper 1 lasts two hours 15 minutes and contains about 26 short-answer questions for 90 marks. Paper 2 also lasts two hours 15 minutes and carries 90 marks, with nine or ten questions. The final Paper 2 question specifically focuses on applying mathematics to a real-world scenario. Both papers count for half the subject result and require all questions to be attempted.
The syllabus expressly expects necessary working, provides relevant formulae and permits an approved calculator on both papers. It includes contexts such as household utility bills, transport schedules and real-world graphs. None of this means the fictional relay context will appear in the exam. It means the underlying habits—calculate with units, choose an appropriate method, check a bound and interpret a feasible solution—are legitimate preparation. Use the subject’s official syllabus rather than extrapolating from a favourite tuition example.
Distinguish power, energy and time
Power describes a rate of energy transfer. Energy used is that rate multiplied by time when the power is constant. In this practice case, the load is stated in watts and energy is expressed in watt-hours. A 20 W device operating for three hours uses 60 Wh; the same device running for half an hour uses 10 Wh. A result in watts cannot be compared directly with a capacity in watt-hours without adding a duration. That is a units error, not a small rounding slip.
The model has a limit: real devices may vary their demand and real storage has losses. A school problem can explicitly assume constant power and an ideal conversion to a usable capacity. Treat those as assumptions supplied by the task, not laws about every battery. When the problem supplies a usable-energy figure, do not quietly replace it with the pack’s nominal figure. A clear solution writes ‘usable energy’ and ‘scheduled energy demand’ as two separate lines before comparing them.
Create an inventory without double-counting
Suppose four classroom devices have constant stated powers: a monitoring module 6 W, a diagnostic unit 8 W, a light 12 W and a circulation fan 24 W. If all four run together, total load is 50 W. For 2.5 hours, demand is 50 × 2.5 = 125 Wh. That computation works because every device shares the full duration. If the fan runs for only one hour while the other devices run for 2.5 hours, its consumption must be calculated separately: (6+8+12) × 2.5 + 24 × 1 = 65 + 24 = 89 Wh.
An exam question may hide the schedule in a table rather than a paragraph. Draw a time line and list which devices are active in each interval. The same device should not be counted twice in overlapping rows. The word ‘total’ refers to a sum of energy over the stated interval, while ‘maximum simultaneous load’ refers to the highest combined power at one instant. Those two quantities can lead to different decisions, so the examiner’s actual question matters.
Turn a reserve percentage into usable capacity
Imagine one pack is labelled as providing 240 Wh under the simplified classroom model. The task directs the team to keep 20% unused as a reserve. That means the planned usable amount is 80% of 240 Wh: 192 Wh. The 125 Wh full-load schedule fits with 67 Wh of planned usable capacity still available. A learner who subtracts 20 rather than 20% would obtain the wrong result. A learner who calls 240 Wh usable would ignore an explicit constraint.
Use an inequality when the problem asks whether the plan is feasible: scheduled demand must be less than or equal to usable capacity. The sign is not decorative. A plan that exactly equals 192 Wh passes the stated numerical rule but leaves no extra allowance beyond the reserve already built into the model. If the question asks for a more cautious margin, that must be modelled separately. Do not quietly invent a second safety margin and present it as official policy.
Whole packs create a discrete decision
A second case offers small educational packs with 100 Wh nominal capacity each and the same 20% reserve rule. Each pack therefore contributes 80 Wh to the plan. For a demand of 125 Wh, write 80n ≥ 125, where n is the number of packs. Dividing gives n ≥ 1.5625. Since a fractional physical pack is impossible, the least integer choice is two. The arithmetic quotient is a lower bound; the final answer is an integer decision.
Rounding always upward is correct only when the inequality calls for a minimum whole count and the objects cannot be divided. If the question instead asks the greatest number of completed 25 Wh activities that can fit into 80 Wh, use the largest integer not exceeding 80/25, which is three. The right direction of rounding comes from the story and the inequality. Teach yourself to name that direction before pressing the calculator.
Different intervals, different rates
Suppose the relay needs its 6 W monitor for five hours, but an 18 W display runs only for the first two hours and an 8 W diagnostic unit only for the last hour. The total demand is 6 × 5 + 18 × 2 + 8 × 1 = 74 Wh. It would be wrong to multiply the sum of all three powers by five hours, because the devices do not share one common duration. A timeline is often clearer than a large equation.
When intervals overlap, break the schedule into non-overlapping pieces. From hour zero to two, the monitor and display run at 24 W. From hour two to four, only the monitor runs at 6 W. From hour four to five, monitor and diagnostic unit run at 14 W. Energy is 24 × 2 + 6 × 2 + 14 × 1 = 74 Wh. Two valid methods agree, giving a useful consistency check.
Translate tables into a linear expression
If a fan draws 24 W for x hours and a lamp draws 12 W for y hours, total demand in Wh is E = 24x + 12y, assuming neither term changes with time. The equation does not tell us x and y separately. One total-energy observation supplies one relation, not a unique schedule. A second independent condition, such as total device-hours x + y = 5, may make the unknowns solvable when both equations are consistent.
Consider E = 84 Wh and x + y = 5. Substituting y = 5 − x gives 24x + 12(5 − x) = 84. Hence 12x = 24, x = 2 and y = 3. Check: 48 + 36 = 84. The mathematical solution makes sense only because both durations are non-negative. In some word problems, a negative or fractional answer would signal a wrong model or an infeasible physical interpretation, even if the algebra were internally correct.
Budget is not the same as energy
A fictional charging tariff of S$0.32 per kilowatt-hour is used solely for arithmetic. One kilowatt-hour is 1000 watt-hours. A 50 W load running for four hours consumes 200 Wh, or 0.2 kWh. At the invented tariff, the calculated electricity cost is 0.2 × S$0.32 = S$0.064. A student who multiplies 200 by 0.32 obtains S$64 and has missed the unit conversion.
In a real bill, taxes, fixed charges and changing tariffs may matter. In an examination problem, follow the charges that the stem actually supplies; do not guess a live Singapore price. Money often requires rounding to cents only at the requested final stage. Carry full precision during intermediate calculations, and make clear whether an answer is a theoretical energy cost, a purchase price or a total budget including other items.
Percentage growth and reduction
A revised device uses 15% less power than a 40 W earlier model. Its new stated power is 40 × 0.85 = 34 W. Over three hours, it consumes 102 Wh rather than 120 Wh, saving 18 Wh. Note that the percentage applies to power at the same duration; it does not automatically describe a battery’s usable capacity. Readers must identify the base of each percentage.
If a future change raises demand from 100 Wh to 115 Wh, that is a 15% increase based on 100 Wh. Returning from 115 to 100 Wh requires a reduction of 15/115, or approximately 13.0%. Equal-looking percentage changes are not usually reversible by the same rate. This becomes important when a question asks which upgrade recovers an earlier budget after two successive changes.
Thresholds and boundary cases
Suppose a plan is considered feasible only when energy demand is at most 160 Wh. A calculated result of 159.8 Wh is numerically within the limit; 160.2 Wh is not. A result printed to the nearest whole watt-hour could display 160 Wh in both cases and hide the difference. When the stem requires a decision at a threshold, compare unrounded numbers before rounding for presentation.
A boundary case helps test the model. What happens if duration is zero? The calculated energy must be zero for a constant-power device. What happens if a power value doubles while duration stays fixed? Energy doubles. If an equation predicts otherwise, inspect your units and coefficients. These checks are faster than rebuilding an entire solution and often expose a sign or conversion mistake in seconds.
Average versus peak demand
The average power over an interval is total energy divided by total time. A schedule that uses 90 Wh in three hours has average power 30 W. That does not prove demand was 30 W at every instant. It might have been 50 W for one hour and 20 W for two hours. The average describes one aggregate, not the shape of the entire schedule.
If a task mentions a maximum allowed simultaneous load, use the actual powers active in each segment, not the average alone. In the example, peak load is 50 W. A student who checks only a three-hour average of 30 W could miss a 40 W peak limit. State which constraint you tested: energy capacity across time or maximum power at a point. A robust answer addresses each named restriction.
Designing a reasoned final Paper 2 response
The final G3 Mathematics Paper 2 task may combine several strands in a real-world setting. Read the decision question first and mark every supplied restriction. Build a table with columns for activity, rate, duration, amount and unit. Translate hard limits into inequalities. Calculate the candidates, choose only feasible whole-number options, then explain the conclusion using the figures. An explanation such as ‘Plan B is better because it needs two packs rather than three while satisfying the stated reserve’ is stronger than ‘Plan B is best’.
Do not invent extra sophistication when the data are simple. An integer count and a total-energy comparison may be enough. Conversely, do not stop at an arithmetic total when the question asks whether the plan meets a threshold. The mathematics is complete only when the numerical result is interpreted. Where multiple choices satisfy the constraints, identify the given objective—for instance least number of packs, smallest cost or longest scheduled duration—before ranking them.
How to review the work without wasting another hour
Make a four-column correction log: what I read, what I calculated, what I concluded and what the question actually asked. If the error was a watt-to-kilowatt conversion, revise that skill rather than repeating an entire chapter on probability. If the error was rounding down a required pack count, practise integer inequalities. If the error was assuming all devices operated for the same duration, practise interval tables. Each correction should name a cause that can be tested tomorrow.
Then attempt one fresh problem without referring to the model answer. Annotate the units before typing into a calculator. Estimate the magnitude first: 50 W for two hours should be near 100 Wh, not 0.1 Wh and not 100 kWh. Finish by writing a single sentence that answers the decision question in its original terms. The transition from number to plain language is exactly where many otherwise competent learners lose clarity.
Thirty-five original mathematics exercises and reasoning checks
Each prompt practises an error students really make: wrong denominator, wrong time interval, wrong unit, inappropriate rounding or an answer that ignores the question’s decision. Work it independently, then read the reasoned solution. These are original examples, not reproduced SEAB papers.
Exercise 1: A 15 W lamp runs for four hours.
Solution: It consumes 15 × 4 = 60 Wh. The multiplication joins a power rate with its duration. Calling 60 W the answer would confuse the total amount of energy with the rate at which energy was transferred.
Check your method: Label the result as Wh before comparing with any pack.
Exercise 2: Convert 0.45 kWh to Wh.
Solution: 0.45 × 1000 = 450 Wh. The prefix kilo means 1000, so a numerical cost rate per kWh cannot be multiplied directly by a Wh figure. First align the units, then substitute.
Check your method: Use 450 Wh as a reasonableness reference for 0.45 kWh.
Exercise 3: Three identical 8 W modules run for two hours.
Solution: Total power is 3 × 8 = 24 W. Total energy is 24 × 2 = 48 Wh. The count of modules changes the rate before the duration is applied.
Check your method: Do not accidentally multiply by three twice.
Exercise 4: A 36 W fan runs half an hour.
Solution: Half an hour is 0.5 h, so demand is 36 × 0.5 = 18 Wh. Students often write 36 × 30 and obtain a value in watt-minutes without converting.
Check your method: Time must match the hour in Wh.
Exercise 5: What is a 20% reserve on 250 Wh?
Solution: Reserve is 0.20 × 250 = 50 Wh, leaving 200 Wh within the plan. The label ‘reserve’ describes the portion withheld, not the remaining usable portion.
Check your method: State both amounts when the question is ambiguous.
Exercise 6: Does a 205 Wh demand fit a 200 Wh usable limit?
Solution: No. It exceeds the stated cap by 5 Wh. The fact that it is close does not make it compliant. If the task gives a strict numerical threshold, use the correct inequality without rounding away the difference.
Check your method: Compare unrounded demand with usable capacity.
Exercise 7: What is the minimum number of 80 Wh usable packs for 190 Wh?
Solution: 80n ≥ 190 gives n ≥ 2.375, so three whole packs are needed. Two supply only 160 Wh and fail the condition. The integer result comes from the unit being indivisible.
Check your method: Check the adjacent smaller integer explicitly.
Exercise 8: How many complete 18 Wh tests fit in 100 Wh?
Solution: 100/18 ≈ 5.56; only five complete tests fit. Here the question asks for the maximum whole count not exceeding capacity, so round down rather than up.
Check your method: The inequality is 18n ≤ 100.
Exercise 9: Find total energy for 6 W for five hours plus 18 W for two.
Solution: The contributions are 30 Wh and 36 Wh, totalling 66 Wh. Do not add 6+18 and multiply by five because the 18 W device operates for only two hours.
Check your method: Show each duration beside the correct device.
Exercise 10: A 20 W unit works two hours, then 10 W three hours.
Solution: Energy is 20 × 2 + 10 × 3 = 70 Wh. The stages have different rates, so a time-weighted calculation is necessary. Average power is 70/5 = 14 W.
Check your method: This example separates total and average.
Exercise 11: Find the cost of 0.3 kWh at invented S$0.32/kWh.
Solution: The arithmetic result is S$0.096, about S$0.10 to the nearest cent. This is an invented tariff, not a claim about today’s electricity prices. Keep the calculation precise before rounding.
Check your method: Do not label S$0.096 as an actual household bill.
Exercise 12: A load uses 240 Wh over six hours. Average power?
Solution: 240/6 = 40 W. An average does not mean the load never exceeded 40 W. If the question sets a peak power limit, a separate timetable is needed.
Check your method: Specify ‘average power’ in the answer.
Exercise 13: A pack has 120 Wh nominal energy with 25% withheld.
Solution: Usable energy is 120 × 0.75 = 90 Wh. A plan needing 92 Wh is not feasible under this stated rule, even though 92 is below the nominal 120.
Check your method: Reserve is applied before comparing schedules.
Exercise 14: A 12 W device uses 90 Wh. Find operating time.
Solution: t = E/P = 90/12 = 7.5 h, assuming constant power. That is seven hours and 30 minutes. Do not confuse decimal hours with clock minutes; 0.5 h is 30 minutes.
Check your method: Write the time in the form requested.
Exercise 15: A 12 W lamp and 24 W fan run together for three hours.
Solution: Their combined load is 36 W and energy is 108 Wh. If a 100 Wh limit applies, the schedule fails by 8 Wh. The answer must reach the feasibility decision, not stop at 108.
Check your method: Use the exact stated limit rather than a guessed margin.
Exercise 16: Two schedules each consume 120 Wh. Are their peaks equal?
Solution: Not necessarily. One might be 40 W for three hours, another 60 W for two hours. Equal energy can result from different power-time combinations and different peak demands.
Check your method: A total is not a complete time history.
Exercise 17: Explain why a 5% energy saving is not 5 Wh.
Solution: A percentage needs a base quantity. Five per cent of 80 Wh is 4 Wh; five per cent of 200 Wh is 10 Wh. Writing 5 Wh without identifying the base is unsupported.
Check your method: Start from the quantity to which the rate applies.
Exercise 18: A 40 W unit becomes 15% more efficient. Can power demand be assumed 34 W?
Solution: Only if the question explicitly models the change as a 15% reduction in power or gives a relationship that permits it. Efficiency is a ratio and is not automatically interchangeable with percentage power saving.
Check your method: Do not translate a descriptive label into an unprovided formula.
Exercise 19: Power rises from 20 W to 25 W. Percentage increase?
Solution: Increase is 5 W; relative to the original 20 W, that is 25%. The denominator is the initial value. Using 25 W as denominator would answer a different question.
Check your method: Write change/original × 100%.
Exercise 20: A 50 W load runs from 14:15 to 15:45.
Solution: Duration is one hour 30 minutes, or 1.5 h. Energy is 50 × 1.5 = 75 Wh. Clock arithmetic should be completed before multiplication.
Check your method: Convert all times to a shared unit.
Exercise 21: Solve 24x + 12y = 84 and x+y = 5.
Solution: Substitute y = 5−x: 24x + 60 − 12x = 84, so x=2 and y=3. Substitution back gives 84 Wh and five total hours. Both durations are non-negative.
Check your method: Check both equations, not only the first.
Exercise 22: What if the algebra produces x = −1 hour?
Solution: The negative value is not a physically valid duration in this context. It may indicate inconsistent constraints or that the mathematical line of solutions lies outside the feasible region. Report the infeasibility when asked; do not silently force x to zero.
Check your method: Distinguish an algebraic root from an acceptable contextual answer.
Exercise 23: A 160 Wh limit, demand computed as 159.96 Wh. Is it allowed?
Solution: Numerically yes, if the rule is demand ≤160 Wh and the model is exact as stated. Rounding 159.96 to 160 before comparing would not change this decision here but could hide important differences in a nearby case.
Check your method: Compare raw values before presenting the rounded result.
Exercise 24: Same limit, demand is 160.04 Wh.
Solution: It fails the strict limit by 0.04 Wh, even if shown as 160 Wh to three significant figures. A rounded display cannot rescue an infeasible plan. Keep sufficient precision for the comparison.
Check your method: A boundary decision uses the best available figures.
Exercise 25: A plan’s energy use doubles when its duration doubles. What assumption is implied?
Solution: That power stays constant over the compared interval. If devices switch on and off or power changes with time, a single proportionality may not apply. State the condition alongside the mathematical relationship.
Check your method: Proportional reasoning needs stable variables.
Exercise 26: Why is ‘the cheaper plan’ an incomplete final sentence?
Solution: The question may require feasibility as well as cost. A low-cost plan that fails its reserve condition is not an admissible choice. First filter options by hard constraints, then rank the feasible ones using the requested criterion.
Check your method: State the decision and the evidence together.
Exercise 27: A 90 Wh plan plus a 20% contingency on demand.
Solution: The required allowance is 90 × 1.20 = 108 Wh. Do not confuse increasing demand by 20% with withholding 20% of a pack’s capacity; these are different operations and can give different answers.
Check your method: Identify which quantity the percentage modifies.
Exercise 28: Three modules draw 12 W, 8 W, 6 W but only two run together. Maximum possible pair?
Solution: Choose 12 W and 8 W to obtain 20 W. The full sum 26 W is impossible under the stated two-module rule. If a selection question asks for maximum load, enumerate valid combinations.
Check your method: Use the constraint before combining all the numbers.
Exercise 29: A plan has 24 W for two hours and 6 W for three hours. Average power?
Solution: Energy is 48 + 18 = 66 Wh over five hours, giving 13.2 W average. Averaging 24 and 6 directly gives 15 W, which ignores unequal durations.
Check your method: Use a time-weighted average when intervals differ.
Exercise 30: How does an inequality express a one-pack plan?
Solution: If usable capacity is 80 Wh, write E ≤ 80. Where E is a sum such as 24x + 12y, the feasible schedules satisfy 24x + 12y ≤ 80 with non-negative durations. Interpret the region before choosing a specific point.
Check your method: Write the equality boundary and the permitted side.
Exercise 31: Why should a candidate show working on K310 Paper 2?
Solution: The official syllabus warns that omitting essential working loses marks. A bare answer hides whether the student translated the situation correctly, applied the right unit conversion and selected a feasible integer choice. A few clean lines often suffice.
Check your method: Show the model, substitution, units and concluding comparison.
Exercise 32: What is one plausible estimate for 50 W over two hours?
Solution: Approximately 100 Wh. If a calculator shows 0.1 or 100,000 Wh, revisit the typed duration or unit conversion. Estimation is not decoration; it catches order-of-magnitude mistakes before a decision is made.
Check your method: Sense-check against 1 W × 1 h = 1 Wh.
Exercise 33: Does 0.5 hours mean five minutes?
Solution: No. It is half an hour, or 30 minutes. Similarly, 2.25 hours is two hours 15 minutes, not two hours 25 minutes. Clock and decimal conversions are common sources of hidden errors.
Check your method: Multiply decimal hours by 60 to find minutes.
Exercise 34: Can one energy total prove the battery is safe in the real world?
Solution: No. The total supports only the simplified mathematical comparison under stated assumptions. Actual equipment has electrical, thermal, environmental and operational requirements that a school arithmetic case omits. Write a bounded mathematical conclusion.
Check your method: Do not turn a school model into real operating approval.
Exercise 35: Summarise a correct final recommendation.
Solution: ‘Using the stated 20% reserve and constant-power model, two 100 Wh nominal packs provide 160 Wh usable, which exceeds the 125 Wh scheduled demand by 35 Wh. One pack would provide only 80 Wh, so two is the minimum whole count.’ This includes assumptions, method and decision.
Check your method: Prefer transparent evidence to an unsupported ‘best’.
Link the Primary foundation to the SEC problem
The arithmetic began in Primary Mathematics: multiplication, fractions, percentage and rates. The difference in G3 is that those operations are placed inside an explicit model with time intervals, units and constraints. If you struggle to identify the right operation from a word problem, practise the PSLE Mathematics diagnostic method or the PSLE unit-and-model checking route before returning to this extended task. A complex context does not make its foundation expendable.
Use the Mathematics owner for topic repair, especially rate versus total amount, bounds and intervals and constraint reasoning. Continue with the PSLE EMS learning route and the SEC EMS learner guide.
Where this volume sits
Vol 1290: relay delay mathematics; Vol 1292: earlier EMS relay evidence; Vol 1293: NTC thermistor and divider. The next English chapter asks how a student should describe the conditional result to a reader without presenting a fictional arithmetic estimate as a safety guarantee. A final EMS case then brings the strands together while preserving their separate examination standards.
Official syllabus verification
For codes and formats, consult SEAB’s 2027 G3 syllabuses for school candidates (index updated 26 October 2026) and G3 Mathematics K310, 2027. The planned power figures and tariff in this guide are entirely invented for mathematics practice, and no physical backup system or emergency operation should be decided from these examples alone.
