Quick answer: the denominator is time
When electricity-use measurements cover intervals of different lengths, add the energy contributed by each interval. Do not take a simple average of the recorded power readings unless the intervals are equal. For a steady value within an interval, energy in watt-hours equals power in watts multiplied by time in hours. The time-weighted average power equals total energy divided by total elapsed time. If part of the log is missing, report what is known and show a defensible range rather than invent a value. These ideas apply beyond electricity, but the worked setting here is a fictional school monitor.
Why this Mathematics chapter follows Physics
Volume 0625 showed why electrical power belongs to a specified source or branch and why a one-second reading cannot establish full-day energy. The mathematical task now begins after those physical quantities have been identified. We receive power values from a simplified, time-stamped log and reconstruct the total. We will also solve budget inequalities, bound unknown intervals and test misleading percentage claims. This is not a second owner for electrical circuit theory: return to the Physics chapter for where the values came from.
Official G3 Mathematics requirements
The Singapore–Cambridge SEC starts in 2027. The SEAB 2027 G3 subject directory lists Mathematics as K310, with 4052 as its pre-2027 reference. The official K310 syllabus specifies two papers of two hours fifteen minutes each, 90 marks each, each worth 50%. Candidates answer all questions. Paper 2 ends with an extended real-world application. The syllabus includes rate, percentage, ratio, algebra, inequalities and interpretation of data. It explicitly permits utility bills as a context. Essential working matters; non-exact answers are normally to three significant figures unless another accuracy is specified. The power-log dataset below is original learning material, not a SEAB specimen paper.
The first clean calculation
Consider a fictional monitor operating at 12 W for 10 minutes, 18 W for 20 minutes and 6 W for 30 minutes. Convert durations to hours: 1/6, 1/3 and 1/2. The three energy amounts are 2 Wh, 6 Wh and 3 Wh. Total energy is 11 Wh. The whole observation period is one hour, so its time-weighted average power is 11 W. The ordinary average of readings, (12 + 18 + 6)/3 = 12 W, is incorrect because a 10-minute reading and a 30-minute reading were given equal influence. The distinction is mathematical rather than rhetorical.
Read the boundaries of intervals
If a log is labelled 09:00–09:10, 09:10–09:30 and 09:30–10:00, the durations are 10, 20 and 30 minutes. No interval overlaps another if each ends precisely where the next begins. A new label recorded at 09:10 might represent the reading at that instant rather than the average of the next 20 minutes; the question must say what the entries mean. When a problem explicitly states each listed power was constant over its interval, multiplication is valid. Otherwise, do not quietly treat instantaneous readings as interval averages. This assumption is the hidden hinge of many tempting answers.
One denominator, several quantities
Power is a rate. Energy is a total. A cost per kWh applies to an energy quantity. A percentage compares a chosen part with a defined whole. An average can be unweighted or time-weighted depending on the question. These are related, but they do not share a single denominator. The learner should write units beside the given numbers before choosing an operation. For example, 6 W × 0.5 h = 3 Wh, while 6 W ÷ 0.5 h is a different quantity and usually does not answer the question.
Convert minutes, seconds and hours deliberately
At the start of every question, choose a unit system. For Wh, use W and hours. For J, use W and seconds. A ten-minute interval is 1/6 hour, not 0.10 hour. A 45-minute interval is 0.75 hour, not 0.45 hour. These errors can produce answers that look credible when presented with three significant figures. Show the conversion once and use it consistently. Where the requested output is kWh, divide the final Wh result by 1000.
Totals from tables and graphs
Suppose a table has columns for starting time, ending time, power and operating state. Add a new derived column for elapsed time, then one for interval energy. The energy column is calculated row by row. Summing powers alone does not create energy. On a piecewise-constant power–time graph, the rectangle width is duration and its height is power; the rectangle represents energy. This is an intuitive area interpretation, not a requirement to use advanced integration. It is especially useful when a diagram mixes narrow high-power intervals with long low-power periods.
Missing measurements are not zero
A log may contain a blank interval because the recorder was disconnected. If the unknown load is bounded between 0 W and 9 W for 20 minutes, its energy contribution is between 0 and 3 Wh. Add this range to the known total instead of inserting zero without evidence. An interval bound gives a mathematically honest estimate; a single chosen value without justification gives invented precision. If a paper explicitly asks for an estimate based on a stated model, you may calculate that model and label it as an assumption.
Percentage comparisons need the same period
Imagine one day with 12 Wh and a later day with 9 Wh over comparable observation windows. The reduction is 3 Wh, which is 25% of the first day’s 12 Wh. But if the second day contains only half as many operating hours, the comparison is not necessarily a reduction in power or improved efficiency. Separate change in consumption from change in exposure. A percentage is not a physical explanation; it is a ratio whose reference value must be clear.
Budget inequalities and feasibility
An imagined activity allocates a maximum of 0.080 kWh to an exhibit. A fixed 0.020 kWh is already needed for setup, and every extra one-hour session uses 0.015 kWh. Then 0.020 + 0.015n ≤ 0.080, so n ≤ 4. If n counts complete sessions, its values are whole numbers and the largest feasible choice is four. A mathematical solution of 4.6 sessions may be useful for a continuous variable, but not for an indivisible completed event. Identify whether the result is a duration, a count or an amount before rounding.
Costs are conditional on a stated rate
A sample task may state an invented rate of $0.30 per kWh. For a measured 0.011 kWh, the energy-only cost is $0.0033. Do not label this as a real Singapore electricity tariff or a complete bill. Household bills can involve additional components, different billing periods and taxes. The K310 real-world mathematics is in applying a stated rate accurately. The existing Volume 0336 remains the canonical route for tiered utility charges; this volume owns unequal-time logs and their derived energy.
Check your result with a bound
A simple reasonableness check can catch errors before detailed recomputation. If a one-hour log never drops below 5 W and never rises above 15 W, its energy must be between 5 Wh and 15 Wh, assuming those are interval-average or constant powers. A calculated 42 Wh violates the given range. Similarly, average power must lie between the minimum and maximum interval powers if the intervals cover the entire period and all weights are positive. These tests often expose a misplaced conversion or double-counted interval.
Present essential working
Write a short table or a clearly ordered calculation: duration in hours, interval energy, total, and required interpretation. Avoid writing only the final total because the official K310 notes warn that omission of essential working loses marks. If a question asks whether the budget is met, finish with an explicit comparison in the same units and a yes-or-no conclusion. If a missing interval prevents a definite decision, say so and give the range of possible results.
A full multi-stage example
A fictional school monitor has a 6 W baseline for 40 minutes and a 24 W communication state for 20 minutes. Its hour uses 6×(40/60) + 24×(20/60) = 12 Wh. Over six identical hours the model uses 72 Wh = 0.072 kWh. At an invented energy-only rate of $0.30 per kWh the model charge is $0.0216. Compare a proposed variant that uses 8 W for 50 minutes and 12 W for 10 minutes: its one-hour energy is 8×(50/60)+12×(10/60) = 8⅔ Wh. The reduction per hour is 3⅓ Wh. The original is higher despite the variant’s increased baseline because the full operating pattern differs.
Why a stronger answer separates measured, modelled and unknown
There are three different forms of evidence in a data question. A measurement is a number actually recorded. A model is a rule imposed on those numbers, such as treating each 10-minute reading as constant over a whole interval. An unknown is a quantity the data do not determine. A confident learner can say which category an answer belongs to. For example, a computed weekly consumption is a projection when based on one representative day; it is not automatically a measured weekly total. Precision of arithmetic must never hide weakness in evidence.
How to use this in revision
Start by doing five very short conversions correctly. Then solve three equal-interval logs, three unequal-interval logs and two missing-interval problems. Only after those are stable should you add costs, percentages and whole-number decisions. When correcting an error, identify whether it arose from reading the interval, selecting the operation, choosing a denominator or interpreting the answer. This narrow diagnosis is more efficient than repeating an entire worksheet. Once the process becomes reliable, change the story while keeping the mathematical structure and test whether the skill transfers.
Worked question bank: unequal time, energy and decisions
These original, self-contained questions develop from conversions into missing-data judgements. Answers use the stated fictional figures. The calculation should be attempted before its explanation; the final check should always name whether a figure is measured, calculated or projected.
Worked question 1: A fair first log
Question. Power is 10 W for 15 minutes and 4 W for 45 minutes.
Working. Energy is 10×0.25 + 4×0.75 = 5.5 Wh. Over one hour the average power is 5.5 W.
Reader-safe conclusion. An unweighted mean of 7 W is wrong because the 4 W state lasts three times longer. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 2: Three unequal intervals
Question. A display runs at 9 W for 10 min, 3 W for 20 min and 12 W for 30 min.
Working. The contributions are 1.5 Wh, 1 Wh and 6 Wh, so total is 8.5 Wh.
Reader-safe conclusion. The largest interval dominates the total even though its power reading is not the largest possible. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 3: Translate time stamps
Question. Readings cover 08:05–08:17 at 5 W and 08:17–08:47 at 8 W.
Working. Durations are 12 and 30 minutes, or 0.2 and 0.5 h. Energy = 1 + 4 = 5 Wh.
Reader-safe conclusion. The time stamps are a boundary description; do not mistake clock times for durations. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 4: An hour and a half
Question. A model sensor uses 7 W steadily for 90 minutes.
Working. 90 min is 1.5 h, so energy is 10.5 Wh = 0.0105 kWh.
Reader-safe conclusion. Dividing by 90 instead of converting minutes loses the physical meaning of the units. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 5: Time-weighted average
Question. A logger reads 2 W for 1 h and 8 W for 3 h.
Working. Energy is 2 + 24 = 26 Wh over 4 h. The correct average is 6.5 W.
Reader-safe conclusion. A simple reading average of 5 W is not justified with unequal durations. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 6: Equal durations as a check
Question. Two 20-minute periods show 4 W and 10 W.
Working. Energy = 4×1/3 + 10×1/3 = 14/3 Wh. Mean power for 40 min is 7 W.
Reader-safe conclusion. For equal durations, the simple arithmetic mean agrees with the time-weighted result. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 7: One-second spike
Question. A log reports 50 W for one second and 2 W for 59 seconds.
Working. Energy in the minute is (50×1 + 2×59)/3600 = 0.0467 Wh approximately.
Reader-safe conclusion. One second of high demand does not describe a minute of constant high demand. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 8: Watt-hour versus joule
Question. A steady 6 W load lasts 10 minutes.
Working. Energy in joules is 6×600 = 3600 J. It is 1 Wh, equivalent to 3600 J.
Reader-safe conclusion. Both routes agree; if they do not, recheck the 60-second and 60-minute conversions. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 9: Missing interval lower bound
Question. The first 20 minutes use 4 W; the next 20 are unrecorded; the final 20 use 7 W.
Working. Known energy = 4/3 + 7/3 = 11/3 Wh. If the missing power is non-negative, total is at least 11/3 Wh.
Reader-safe conclusion. Do not treat the missing interval as a measured 0 W value. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 10: Missing interval upper bound
Question. A 30-minute gap is guaranteed not to exceed 12 W.
Working. Its unknown energy is between 0 and 6 Wh. If known energy is 8 Wh, total lies between 8 and 14 Wh.
Reader-safe conclusion. This bound is useful even when no exact estimate can be defended. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 11: Average from total
Question. An hour-long model consumed 13.5 Wh.
Working. Average power is 13.5 Wh ÷ 1 h = 13.5 W.
Reader-safe conclusion. The average is a rate derived from total energy, not necessarily any instant’s recorded value. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 12: Average over 2.5 hours
Question. A log totals 30 Wh over 150 minutes.
Working. Time is 2.5 h; average power is 12 W.
Reader-safe conclusion. Dividing by 150 without reconciling units would produce Wh/min, not W. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 13: Reverse a duration
Question. A constant 8 W load must use 12 Wh.
Working. Time needed is 12/8 = 1.5 h, or 90 minutes.
Reader-safe conclusion. Check the direction of the equation: energy divided by power gives time. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 14: Reverse a power
Question. A fixed load transfers 9 Wh in three hours.
Working. Its constant power is 9/3 = 3 W.
Reader-safe conclusion. The given wording says constant; without it, this is only a time-average power. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 15: Energy-only cost
Question. A model consumes 0.050 kWh and the question states $0.30 per kWh.
Working. Cost is $0.015, or 1.5 cents. Keep sufficient decimal places.
Reader-safe conclusion. It is a fictional energy-only calculation and says nothing about standing fees or current tariffs. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 16: Two equal monthly scenarios
Question. A fictional month is modelled with 40 identical days, each using 0.08 kWh.
Working. Projected consumption is 3.2 kWh. The model assumption is deliberately artificial.
Reader-safe conclusion. An arithmetic projection is not a measurement of actual monthly consumption. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 17: Percent reduction
Question. Usage changes from 20 Wh to 15 Wh for the same duration.
Working. The reduction is 5 Wh, and 5/20×100% = 25%.
Reader-safe conclusion. The original 20 Wh is the denominator because the comparison asks how much usage fell. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 18: Percent increase
Question. An equal-length trial changes from 12 Wh to 15 Wh.
Working. The increase is 3 Wh, or 25% of the initial 12 Wh.
Reader-safe conclusion. Do not use the later value of 15 Wh as the base unless the question requests it. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 19: Reverse percentages
Question. A modified model’s 18 Wh is 75% of the former consumption.
Working. The earlier amount was 18÷0.75 = 24 Wh.
Reader-safe conclusion. Dividing rather than subtracting 25 percentage points recovers the original. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 20: A misleading half-day comparison
Question. A first day’s 12-hour log uses 30 Wh, while a later six-hour log uses 18 Wh.
Working. The later total is smaller, but average powers are 2.5 W and 3 W respectively.
Reader-safe conclusion. The raw totals do not support a claim that the second period operated at lower power. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 21: Maximum sessions
Question. The energy allocation is 70 Wh, fixed setup takes 10 Wh and each session 12 Wh.
Working. 10+12n≤70 gives n≤5, hence at most five whole sessions.
Reader-safe conclusion. Six sessions exceed the cap; rounding a non-integer upward can violate a constraint. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 22: Unknown fixed cost
Question. Total modeled energy is at most 100 Wh for four sessions of 18 Wh each.
Working. If setup energy is x, x+4×18≤100 gives x≤28 Wh.
Reader-safe conclusion. A negative setup energy is not physically meaningful, so also require x≥0. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 23: Multiple operating states
Question. A fictional plan runs 15 W for 20 minutes, 9 W for 40 minutes, then 6 W for one hour.
Working. Energy is 5 + 6 + 6 = 17 Wh over two hours; average is 8.5 W.
Reader-safe conclusion. Be careful not to apply a two-hour total to each row. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 24: Graph area check
Question. A step graph shows 5 W for 0.4 h and 10 W for 0.6 h.
Working. The two rectangle areas are 2 Wh and 6 Wh, giving 8 Wh.
Reader-safe conclusion. Width represents elapsed time, not the order of observations. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 25: A missing endpoint
Question. Readings are given at noon and 2 pm, with no statement about what happened between.
Working. You cannot determine energy by multiplying either value by the two-hour period. A curve or interval rule is needed.
Reader-safe conclusion. A line drawn between the dots is a modelling choice, not an observed continuous history. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 26: Bounds on mean power
Question. A full 60-minute log contains only readings between 3 W and 11 W.
Working. Its average power must lie between 3 W and 11 W if readings describe every constant segment.
Reader-safe conclusion. An answer of 14 W violates the stated range and exposes an arithmetic or unit error. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 27: A billing threshold
Question. A hypothetical energy-only budget is $0.09 at a constant $0.30 per kWh.
Working. Permissible energy is at most 0.30 kWh, found by dividing budget by price per kWh.
Reader-safe conclusion. This is a simple linear charge; do not import the tiered tariff structure owned by Volume 0336. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 28: Two devices not equally observed
Question. Device A consumes 6 Wh in one hour and device B 9 Wh in three hours.
Working. Their average powers are 6 W and 3 W respectively. B uses more energy because it operates longer.
Reader-safe conclusion. Name whether the question compares total energy or average power. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 29: Estimating within an interval
Question. A question assumes a constant 8 W for a missing 15-minute block.
Working. Under that explicit model, estimated energy is 2 Wh. The estimate depends on the 8 W assumption.
Reader-safe conclusion. Write ‘estimated’ rather than reporting the derived value as a direct measurement. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 30: Uncertainty range
Question. A 20-minute block is described as between 6 W and 9 W.
Working. Its energy lies between 2 Wh and 3 Wh. If the rest totals 5 Wh, whole-log energy lies between 7 Wh and 8 Wh.
Reader-safe conclusion. An interval can support a decision even when a single exact value cannot be determined. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 31: Feasibility with uncertainty
Question. A 10 Wh budget faces known energy 8.5 Wh and a missing block between 1 and 2 Wh.
Working. Total lies between 9.5 and 10.5 Wh. The plan may exceed budget; permission is not guaranteed.
Reader-safe conclusion. Report the uncertain feasibility rather than forcing a yes or no. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 32: Rounding a small bill
Question. A calculation yields $0.000675 under a stipulated price.
Working. This amount is less than one cent, so premature cent-rounding to $0.00 would lose information in later sums.
Reader-safe conclusion. Carry sufficient precision through intermediate calculations and round according to final-question requirements. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 33: Rounding hours wrongly
Question. A 17-minute interval occurs at 12 W.
Working. Energy is 12×17/60 = 3.4 Wh. Do not round 17/60 to 0.3 h first and call the answer 3.6 Wh.
Reader-safe conclusion. Preserve the exact fraction or enough calculator digits until final reporting. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 34: Whole-number purchase
Question. A monitor consumes 5 Wh per session and 23 Wh remains in the allocation.
Working. 4 complete sessions fit, leaving 3 Wh. Five would require 25 Wh.
Reader-safe conclusion. The mathematical quotient 4.6 is not the number of complete sessions possible. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 35: Alternating modes
Question. A device runs at 12 W for 5 minutes and 3 W for 5 minutes, repeating this 10-minute cycle six times.
Working. One cycle uses 1 Wh + 0.25 Wh = 1.25 Wh. Six cycles use 7.5 Wh in an hour.
Reader-safe conclusion. Count cycles and durations, not merely the distinct power levels. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 36: Compare a new schedule
Question. Old plan: 8 W for 30 minutes and 2 W for 30 minutes. New plan: 6 W for 40 minutes and 3 W for 20.
Working. Old uses 5 Wh; new uses 4 Wh + 1 Wh = 5 Wh. Consumption is equal.
Reader-safe conclusion. A changed operating pattern need not change total energy. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 37: A label versus measured history
Question. A printed label says maximum 30 W, while a complete modelled hour is logged at 6 W.
Working. The logged hour uses 6 Wh under its assumptions. The rating is a limit or condition, not a time series.
Reader-safe conclusion. Do not replace observed interval values with the largest printed number. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 38: From kWh back to Wh
Question. A fictional calculation gives 0.0075 kWh.
Working. Multiply by 1000 to get 7.5 Wh, allowing a direct comparison to a table expressed in Wh.
Reader-safe conclusion. The decimal conversion is separate from any price multiplication. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 39: Distinguish ratio and percentage
Question. Trial A uses 12 Wh and B uses 9 Wh over comparable periods.
Working. Their ratio A:B is 4:3. B is 75% of A and 25% less than A.
Reader-safe conclusion. A ratio, share and percentage reduction describe related but different comparisons. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
Worked question 40: Unnecessary precision
Question. A log calculation yields 11.333333 Wh using nominal values.
Working. For an ordinary K310 non-exact answer the three-significant-figure convention would give 11.3 Wh unless told otherwise.
Reader-safe conclusion. Use precise working internally while presenting a rounded final value to the requested standard. For a complete examination answer, keep the chosen time denominator visible and attach the correct unit to each intermediate total. If the wording asks whether a plan is feasible, use your result to answer that question rather than stopping at a number.
The complete exam-style question: phase A
A monitor records four operating intervals: 07:00–07:15 at 6 W, 07:15–07:35 at 18 W, 07:35–08:00 at 3 W, and 08:00–08:30 at 12 W. The durations are 15, 20, 25 and 30 minutes. Their contributions are 1.5 Wh, 6 Wh, 1.25 Wh and 6 Wh. The full 90-minute observation contains 14.75 Wh. Therefore average power equals 14.75 ÷ 1.5 ≈ 9.83 W to three significant figures. Notice that adding the four powers gives 39 W, which is not the answer to any energy question here. The intermediate columns provide the audit trail.
The complete exam-style question: phase B
Suppose the same monitor is operated for ten days according to precisely the same 90-minute model. Energy is projected at 147.5 Wh, or 0.1475 kWh. At the question’s hypothetical $0.30 per kWh, the energy-only charge is $0.04425. If an unknown extra half-hour block might draw between 0 W and 9 W daily, add between 0 and 4.5 Wh each day: the ten-day total becomes a range of 147.5 to 192.5 Wh. The cost range is $0.04425 to $0.05775. The point is not the tiny monetary amount; it is the difference between an exact model and the incomplete record.
The complete exam-style question: phase C
A proposed activity has a cap of 0.160 kWh over ten days. Under the known 90-minute log alone, its projection of 0.1475 kWh is below that cap. With a possible additional half-hour block on every day, its upper projection is 0.1925 kWh and it may exceed the cap. The correct judgement is therefore conditional: the plan fits on the recorded schedule but is not guaranteed to fit when missing operation is considered. A mathematically sound recommendation explains the range and proposes obtaining the missing operating data. This kind of conclusion is stronger than merely announcing that the cheapest-looking number wins.
Where this topic sits in the learning ecosystem
The PSLE Reality Lab article about an annual kWh label is a useful bridge: it teaches the difference between a stated standard-use figure and an actual household outcome. Volume 0622 is the owner for sensor threshold inequalities, and Volume 0614 is the owner for current–voltage gradient models. This chapter does not duplicate their purpose. It teaches how to reconstruct energy totals from unequal operating periods, then decide whether a conclusion is supported.
A 25-minute practice plan
Spend five minutes on unit conversions, five on identifying time intervals in a table, seven calculating interval energies, five on bounds or budget inequalities and three reviewing the written conclusion. If you made a mistake, choose the first operation that went wrong and redo that step with the units displayed. A strong correction looks like a short explanation, not an overwritten calculator result. Before the next English chapter, write a two-sentence description of the evidence: what was actually measured, what was calculated and what remains unknown. That distinction becomes the backbone of honest argument.
Authoritative references
The official K310 syllabus for 2027 G3 Mathematics controls examination rules and content. The SEAB SEC overview explains the 2027 transition. This chapter contains original paper-based application exercises, not a guaranteed forecast of a future paper, nor a statement of the current electricity tariff. The next English guide addresses how to present an argument when these calculations support some claims but not others.
