Fast answer: Energy used during an interval is the later cumulative reading minus the earlier one. A constant-rate segment is modelled by energy = rate × time, while stopped or changing operation needs separate intervals. G3 Mathematics tests equations, graphs, inequalities and real-world judgement—not just pressing the calculator buttons.
This original guide continues the motorised-ramp sequence without repeating the physics explanation. Start with the worked examples, then attempt the twelve practice cases and check each conclusion against the stated constraints.
Contents
- The graph that turns a physical experiment into a mathematics problem
- Fast answer and the most useful six checks
- Know the actual G3 Mathematics paper
- Find the change, not the final reading
- Build a linear equation from two data points
- Understand a piecewise schedule
- Why the domain belongs next to the formula
- Converting units without changing the question
- A complete operating-cost calculation
- Compare fixed-charge and usage-charge plans
- Thresholds: ‘at most’ and ‘at least’ are not decoration
- Rounding near a decision boundary
- Working with measurement intervals
- Percentages and a duty-cycle calculation
- Reading an irregular table
- Scatter plots and cautious fitting
- A decision can depend on two constraints
- How a small graphing mistake changes a conclusion
- The four-week route from foundation to examination transfer
- Practice 1 — energy across a change
- Practice 2 — find an intercept correctly
- Practice 3 — translate a stop into mathematics
- Practice 4 — compare a stepped tariff
- Practice 5 — a fixed fee versus no fixed fee
- Practice 6 — check a budget boundary
- Practice 7 — a duty cycle with idle consumption
- Practice 8 — interpret irregular times
- Practice 9 — an inequality with a fixed start-up amount
- Practice 10 — numerical bounds
- Practice 11 — compare two eligible options
- Practice 12 — a full real-world question
- A worked examination answer layout
- Frequently asked questions and next steps
The graph that turns a physical experiment into a mathematics problem
A motor lifts a small load on a track. A sensor records electrical energy supplied after 1, 2, 3 and 4 seconds. The numbers are not themselves a Mathematics answer. The examiner may want you to organise the observations, find an interval rate, build a linear relationship, compare operating plans, or decide whether a constraint is satisfied. This volume is about choosing and defending the mathematical representation. It does not repeat the physics derivation of a motor’s useful output.
Begin with Volume 0685, Physics work, power and efficiency if the meaning of the quantities is unclear. Return to Volume 0684, speed-squared and graph gradients if fitting a straight-line model is unfamiliar. Here we move from a continuous smooth trial to practical time intervals, switches and costs. The mathematical work is transferable to water meters, transport schedules, production planning and bills.
The main lesson is simple: a context adds detail, but mathematical precision comes from knowing which variables are related. A table can report accumulated energy, yet the question asks for energy used during a particular interval. A graph can show a straight segment, yet the question asks about an interval across a change of operating mode. Read the interval before finding the slope.
Fast answer and the most useful six checks
When a device uses energy at a constant rate of P joules per second for t seconds, E = Pt, where E is energy in joules. In a real-world mathematics question, this is a given relationship that you can rearrange. A straight energy-time graph with a constant gradient represents a fixed rate during that period. A device that stops, restarts or changes operating rate requires a piecewise account; one overall straight-line equation may conceal important changes.
First identify the x- and y-variables. Second copy the units exactly. Third decide whether the table shows accumulated totals or per-interval amounts. Fourth select the relevant interval, not necessarily one starting at zero. Fifth keep full precision until the stated final rounding. Sixth ask whether the calculated result meets the actual condition: ‘at most’, ‘at least’, ‘greater than’ and ‘no later than’ lead to different inequalities.
A quick error test is to draw a tiny number line or sketch. If energy is cumulative and the meter never runs backwards, the total should not decrease over an interval of positive consumption. If a plan has an initial fixed charge, the graph may have a positive intercept even when usage is zero. That intercept is a real part of the model, not automatically a plotting mistake.
Know the actual G3 Mathematics paper
SEAB identifies K310 as 2027 SEC G3 Mathematics, with 4052 listed as the earlier O-Level reference code. Paper 1 is 2 hours 15 minutes, 90 marks and 50% of the subject; it has about 26 compulsory short-answer questions. Paper 2 is also 2 hours 15 minutes, 90 marks and 50%, with nine or ten compulsory questions. The final Paper 2 question focuses on applying mathematics to a real-world scenario. Relevant working matters: the official syllabus warns that omission of essential working can lose marks.
The syllabus emphasises Number and Algebra, Geometry and Measurement, and Statistics and Probability, as well as reasoning and communication. A motor-data task belongs to a mathematical modelling context. It does not mean students need to learn new university engineering formulae. Read any physical relationships supplied, translate into manageable arithmetic or algebra, and explain decisions using the quantities in the question.
Unless stated otherwise, the syllabus specifies three significant figures for non-exact numerical answers, with a different convention for angles in degrees. In extended contexts, it is wise to show enough intermediate precision to avoid changing an inequality decision through early rounding. A calculator result is a number; a mathematical answer also identifies what the number measures and whether it is suitable for the context.
Find the change, not the final reading
Suppose a cumulative energy meter reads 12 J at 3 s and 47 J at 8 s. The energy consumed between those readings is 47 − 12 = 35 J. The interval is 8 − 3 = 5 s. The average rate for that interval is therefore 35 ÷ 5 = 7 J/s. Simply dividing 47 by 8 treats the whole history from time zero as though the problem asked for it; that gives the wrong interval.
Students often handle ‘difference in readings’ correctly with distance travelled but forget it for energy, water use or money. The general method is the same: later total minus earlier total. If the question asks how much was used from 14:00 to 14:30, take the change between those clock times; do not interpret the 14:30 meter value as the half-hour usage.
Write the difference explicitly. ‘Energy from 3 to 8 seconds = 47 − 12’ is a small line that makes the meaning of the arithmetic visible. If the graph is not smooth, this still gives the interval total. The average rate may be computed without claiming the rate remained constant at every moment.
Build a linear equation from two data points
Suppose an experiment reports 18 J at 2 s and 42 J at 6 s during one fixed operating mode. If a linear model is justified, gradient m = (42 − 18)/(6 − 2) = 6 J/s. Substitute a point into E = 6t + c. Using 18 = 6 × 2 + c gives c = 6. The fitted relation is E = 6t + 6 for the interval represented by the model.
The intercept here is 6 J. It can represent cumulative energy already recorded at t = 0, or a baseline of the chosen time origin; the information given is not enough to assign a unique physical cause. This matters because a student can obtain a perfect line equation and then invent a story that the data do not support. Graph interpretation is a separate skill from the algebra of finding m and c.
To predict the reading at 5 s, substitute t = 5 and obtain E = 36 J. Because 5 s lies within the observed 2-to-6-second range, this is interpolation within the chosen linear model. Predicting at 30 s would be extrapolation: mathematically possible, but the device might change mode, stop or heat up before that time. Identify the difference when the question asks about reliability.
Understand a piecewise schedule
A device operates at 8 J/s for its first 3 s, stops for 2 s, then runs at 5 J/s for another 4 s. Its cumulative energy after the first stage is 8 × 3 = 24 J. While stopped, its cumulative total remains 24 J if the model assumes zero consumption in that idle state. During the last four seconds it adds 5 × 4 = 20 J. The final total at 9 s is 44 J.
On a cumulative graph, the first part is a line with gradient 8; the stop is horizontal; the final running part rises with gradient 5. Draw the segments in chronological order and join them at the correct cumulative energy readings. A graph that drops back to zero every time the device changes mode would be a graph of separate interval energy, not cumulative consumption.
Piecewise descriptions can be written in words when notation becomes cumbersome. If the question specifically requires equations, state the relevant time domain with each expression. For 0 ≤ t ≤ 3, E = 8t. For 3 ≤ t ≤ 5, E = 24. For 5 ≤ t ≤ 9, E = 24 + 5(t − 5). This is more accurate than using E = 8t for the whole nine seconds.
Why the domain belongs next to the formula
A formula can be algebraically correct for one interval and misleading elsewhere. In the last stage above, E = 24 + 5(t − 5) describes cumulative energy after the restart at t = 5 s. Substituting t = 2 produces a value that has no interpretation under that stage because the motor had not yet reached it. The calculation is possible; the model use is invalid.
Many real-world Paper 2 problems present formulas as ‘Plan A applies for the first two hours; Plan B applies afterwards’. Copying an equation without its condition loses the most important information. Before substituting a number, ask whether that number lies in the defined interval or price band.
Use small boundary checks. At t = 5, the last-stage equation gives E = 24, agreeing with the earlier plateau. At t = 9, it gives E = 44, agreeing with the stage arithmetic. Those two checks prove continuity at the join and correct total addition. They do not prove the physical assumptions, but they protect the mathematical stitching.
Converting units without changing the question
Energy may be described in joules, kilojoules or kilowatt-hours. In a supplied real-world context, remember that 1 kJ = 1000 J and 1 kWh = 1000 Wh. Since 1 W = 1 J/s, a device drawing 200 W for 3 hours uses 200 × 3 = 600 Wh = 0.6 kWh. Do not multiply 200 by 3 and call the result joules; the time was given in hours, not seconds.
A 200 W input over 180 seconds gives 200 × 180 = 36,000 J. The number is different from the three-hour example because the durations are different. If a question supplies a conversion factor, use that factor rather than guessing a relation between watt-hours and joules. For example, 1 Wh = 3600 J follows from one watt maintained for one hour, which contains 3600 seconds.
Dimensional checking helps when a unit feels unfamiliar. W × h gives Wh; W × s gives J. J ÷ s gives W. A price per kWh multiplied by kWh gives money. Energy divided by price per kWh does not give the cost. Write units beside the first line of working until the operations become reliable.
A complete operating-cost calculation
Suppose a demonstration motor consumes a steady 0.45 kW when operating, and it runs for 40 minutes a day over 20 days. Treat 40 minutes as 2/3 of an hour. Total operating time is 20 × 2/3 = 13 1/3 hours. Energy used is 0.45 × 13 1/3 = 6.0 kWh. At an illustrative energy price of $0.30 per kWh, the usage-based cost is $1.80. This is a mathematical exercise price, not a statement of Singapore’s current electricity tariff.
The order can be changed if every factor is consistent: daily energy is 0.45 × 2/3 = 0.30 kWh, then twenty days give 6.0 kWh. Either route works. The important step is converting minutes to hours before multiplying by kilowatts. If the question also states a fixed charge, add that separately; do not silently include or exclude it.
A useful final sentence is ‘The estimated usage-based electricity cost for twenty operating days is $1.80 at the stated illustrative rate.’ It distinguishes cost from energy and declares the assumed price. A learner who returns only 6.0 without a unit has not yet answered the money question.
Compare fixed-charge and usage-charge plans
Assume Plan A charges $2.00 fixed plus $0.25 per kWh, while Plan B charges no fixed amount but $0.40 per kWh. Let x represent monthly energy use in kWh. Then A(x) = 2 + 0.25x and B(x) = 0.40x. Equality occurs when 2 + 0.25x = 0.40x, so 2 = 0.15x and x = 13 1/3 kWh.
Below that use level, the larger usage charge of Plan B can still be cheaper because it has no fixed amount. At 10 kWh, Plan A costs $4.50 and B costs $4.00. At 20 kWh, A costs $7.00 and B costs $8.00. This pair of substitutions is an essential sanity check: the preferred plan should change across the break-even point.
The inequality must follow the question. If the family wants Plan A not to cost more than B, solve 2 + 0.25x ≤ 0.40x, giving x ≥ 13 1/3. If usage occurs in whole kWh increments in a particular model, the first whole-number level satisfying the condition is 14 kWh. But consumption is often not constrained to whole numbers, so do not round a threshold without reading the context.
Thresholds: ‘at most’ and ‘at least’ are not decoration
A model device must use no more than 90 J during a test. Suppose a fixed start-up cost is 18 J and it then consumes 6 J every second. The model is E = 18 + 6t. The constraint is 18 + 6t ≤ 90. Subtract 18, divide by 6 and obtain t ≤ 12 seconds. Writing t = 12 alone misses the fact that every shorter non-negative run is also within this energy cap.
If the question instead requires at least 90 J of supplied energy, the inequality is 18 + 6t ≥ 90 and the condition becomes t ≥ 12 seconds. The algebra is nearly identical, but the real-world conclusion reverses. This is why it helps to restate the condition in plain English immediately before solving.
Use the endpoint check. At 12 s the expression gives 90 J, so equality is included for ‘no more than’ and ‘at least’. At 13 s it gives 96 J, which violates the cap but meets a minimum. The test number is not a replacement for algebra; it is a simple way to catch a reversed inequality or a mistaken threshold.
Rounding near a decision boundary
A model predicts 49.96 J, and the task has a maximum permitted energy of 50 J. Rounded to one decimal place, this becomes 50.0 J. The decision should not be based on a prematurely rounded value if the original model provides a more precise estimate. Likewise, a prediction of 50.04 J rounds to 50.0 J but exceeds the stated threshold.
When an inequality decision is close, keep the calculator’s extra precision internally and report a conclusion consistent with that information. If the measurements themselves are only accurate to a broad interval, do not create a false claim that the device definitely passes because the computed central estimate is 49.96. You may need to consider measurement or model uncertainty when asked to evaluate the result.
Compare the amount with the cap before writing a rounded display figure. A clear answer reads: ‘The unrounded model estimate exceeds the 50 J limit by 0.04 J, so it does not meet the limit under the stated model.’ This is not unnecessary fuss. It protects the meaning of ‘at most’ from the arithmetic convenience of rounding.
Working with measurement intervals
Suppose the input rate is estimated as between 5.8 and 6.2 J/s and the operating duration is 10 s, ignoring start-up energy. With non-negative values and a simple constant-rate assumption, the predicted energy lies between 58 J and 62 J. A student who reports exactly 60 J as a certainty has hidden the supplied range. The midpoint can be a useful estimate, but it should not erase the interval.
If the task says the energy must not exceed 60 J, the entire range does not satisfy the condition. Some possible values meet the cap and others do not. A conservative design would account for the upper value of 62 J. That judgement arises from the inequality and supplied uncertainty, not from a new physical law.
Now include a known fixed 12 J start-up contribution. The interval becomes 70 J to 74 J. This example shows why separate terms should be added after carrying each stated range through the calculation. Unit conversion, interval arithmetic and final interpretation are different steps; keeping them separate makes them easy to check.
Percentages and a duty-cycle calculation
A motor operates for 18 minutes in every 30-minute cycle. Its operating duty fraction is 18/30 = 0.6, or 60%. If the operating input is 0.50 kW and the stopped consumption is assumed zero, each cycle uses 0.50 × (18/60) = 0.15 kWh. Four complete cycles use 0.60 kWh.
The tempting shortcut ‘60% of 0.50 kW equals 0.30 kW’ describes an average power over the full cycle under these assumptions, not the energy itself. Multiply that average by the full two-hour duration of four cycles to obtain 0.60 kWh. Alternatively, add operating time first: 72 minutes = 1.2 hours, and 0.50 × 1.2 = 0.60 kWh.
If the stopped mode uses 0.05 kW rather than zero, add the energy from the 12 idle minutes per cycle. Each cycle then uses 0.50 × 0.3 + 0.05 × 0.2 = 0.16 kWh. Four cycles use 0.64 kWh. The change looks small but matters when comparing plans or interpreting a requirement over many cycles.
Reading an irregular table
A results table may give times 0, 2, 5, 9 and 12 seconds with cumulative energies 0, 12, 27, 51 and 51 J. The adjacent interval rates are 12/2 = 6, 15/3 = 5, 24/4 = 6 and 0/3 = 0 J/s. It would be incorrect to divide every energy change by a fixed two seconds because the times are unevenly spaced.
The overall average rate from 0 to 12 seconds is 51/12 = 4.25 J/s. It is not the unweighted mean of the four interval rates; the intervals have unequal durations. A weighted approach, using each energy change and its own time length, gives the same total-over-total result. This problem pattern is useful beyond motors: growth records, travel logs and water collection may all contain irregular intervals.
If the final plateau means the device is off, that explanation may be plausible under the stated setup. A mathematics solution should at minimum report that no cumulative energy increase was recorded over those three seconds. Avoid claiming a precise cause without the context. A flat graph can arise from a switched-off motor or from the scope and precision of the measurements.
Scatter plots and cautious fitting
Not every pair of points lies perfectly on a line. Measurements might show a roughly linear rise in cumulative energy with small deviations due to reading precision, fluctuations in operation or noise. A reasonable straight-line model aims to capture the general trend, rather than joining every data point with a jagged zigzag and declaring each segment a different operating law.
A good choice of axes places time horizontally and cumulative energy vertically when time is the input. Label both quantities and units. Use an appropriate scale that permits the observed range to be displayed clearly. When interpreting a fitted gradient, calculate the rise over the run using two well-separated points on the line rather than choosing two points pressed together where reading error overwhelms the difference.
If a question asks for a prediction between observed times, identify it as an interpolation from the model. If it asks well beyond the observed span, acknowledge that the underlying operating mode may not continue. A linear equation can be mathematically manipulated outside its measured domain, but that does not make it a reliable forecast.
A decision can depend on two constraints
Suppose a demonstration must finish within 12 s and consume at most 65 J. Option A takes 10 s and consumes 62 J. Option B takes 9 s but consumes 71 J. Option C takes 13 s and consumes 58 J. Only A meets both conditions. Students sometimes select B because it is fastest or C because it uses least energy. They have optimised one measure while ignoring the other.
Write the constraints separately: time ≤ 12 and energy ≤ 65. Evaluate each option against both inequalities. An option that fails one required condition is ineligible even if it excels on the other. This is the logic behind many extended real-world mathematics tasks involving cost, capacity, journey time or safety.
If the question asks which eligible option has the lowest energy use, first screen for eligibility, then compare energy among the survivors. If none meets both conditions, say so and support it with values. Do not invent a compromise by averaging the options unless the problem explicitly allows a mixed strategy.
How a small graphing mistake changes a conclusion
Consider a graph whose vertical scale increases by 10 J per large square, but the student assumes 5 J. Every reading and gradient will be halved. A slope can look neat and still be wrong because the axis was not checked. Similarly, if the horizontal axis uses minutes but the answer is presented as J/s, a conversion is missing by a factor of 60.
Build the habit of writing a three-part gradient: difference in vertical values, difference in horizontal values, and units of vertical over horizontal. For example, (54 − 18) J ÷ (9 − 3) s = 6 J/s. The numerical calculation is ordinary, but the notation prevents the conclusion from losing its meaning.
When a graph includes a break in the axis or an origin that is not displayed, do not assume a line begins at (0,0). Use observed points to find the gradient, and calculate the intercept only when it is required or meaningful. The diagram is evidence for the mathematical method; it is not decoration.
The four-week route from foundation to examination transfer
Week one: secure cumulative totals, interval differences and unit conversions. Use short questions until a learner automatically subtracts the earlier reading from the later one. Week two: find gradients, straight-line equations and intercepts, then introduce one piecewise timeline. Week three: solve cost-plan intersections, consumption thresholds and combined inequalities, checking results with substitutions. Week four: attempt mixed real-world questions under a sensible time limit and annotate each uncertainty or assumption.
Keep a correction log with four headings: quantity, operation, unit, and interpretation. A wrong ratio belongs under operation; forgetting hours-to-minutes belongs under unit; selecting the wrong plan after correct calculations belongs under interpretation. Reviewing error types is more useful than simply redoing a whole worksheet from the beginning.
A learner who still struggles with ratio and signed arithmetic should return to an earlier step before taking on complex cost comparisons. The aim of this route is not to force everyone through the same schedule. It is to create a reliable bridge from basic operations to judgement in longer questions.
Practice 1 — energy across a change
A meter rises from 25 J at 4 s to 73 J at 12 s. Calculate the change in energy and average rate over that interval. The change is 73 − 25 = 48 J; the time is 12 − 4 = 8 s; the average is 6 J/s. These three lines should be visible in the working, not compressed into an unexplained 73/12.
Variation: the recorded 73 J is revised to 69 J. The change becomes 44 J and the average becomes 5.5 J/s. The interval duration did not change. This is a good self-check because it isolates the effect of one reading rather than changing the whole problem.
Practice 2 — find an intercept correctly
A steady mode passes through (t,E) = (1,14) and (5,38). Gradient is (38 − 14)/(5 − 1) = 6. Use E = 6t + c; at t = 1, 14 = 6 + c, hence c = 8. The equation is E = 6t + 8 on the modelled interval. At 3 s the predicted total is 26 J.
One tempting error is to use E = 6t because the gradient is 6. That line gives 6 J at 1 s, contradicting the observed 14 J. A gradient describes change per unit time; it is not automatically the whole equation. Always test the final expression against one original point.
Practice 3 — translate a stop into mathematics
A device runs at 4 J/s for 5 s, then is off for 3 s. With zero starting cumulative energy and zero off-mode consumption, energy at t = 5 is 20 J and at t = 8 remains 20 J. The first segment rises, then the graph is flat. The overall average across eight seconds is 20/8 = 2.5 J/s.
Do not confuse the overall average with the operating rate. Four J/s characterises the first stage, but the three-second stop lowers the full-period average. If the task asks for energy consumed rather than average rate, the answer is simply 20 J; adding an extra division would change the requested quantity.
Practice 4 — compare a stepped tariff
An illustrative billing rule charges $0.20 per kWh for the first 10 kWh and $0.35 per kWh for any additional usage. What is the usage charge for 14 kWh? The first block costs 10 × $0.20 = $2.00, and the next four units cost 4 × $0.35 = $1.40. Total $3.40. Multiplying all 14 kWh by $0.35 would incorrectly apply the second-band price to the first band.
At exactly 10 kWh the charge is $2.00. At 10.5 kWh it is $2.175 before any currency-rounding instruction. State the boundary rule rather than inventing a discontinuous jump in cost. This exercise is about a piecewise mathematical rule, not a current real electricity tariff.
Practice 5 — a fixed fee versus no fixed fee
Plan M costs $1.50 plus $0.18 per unit, while Plan N costs $0.30 per unit. Set 1.50 + 0.18x = 0.30x. The difference in usage rates is $0.12 per unit, giving break-even at x = 12.5 units. For x = 10, M costs $3.30 and N $3.00, so N is cheaper. For x = 20, M costs $5.10 and N $6.00, so M is cheaper.
The crossing point is not the answer to every comparison question. If the task asks for the cheapest plan at 16 units, substitute x = 16 and report prices. If it asks for the usage interval where M is at least as cheap as N, solve the inequality and state x ≥ 12.5. Read the command before deciding whether to produce a point or a range.
Practice 6 — check a budget boundary
A club has a usage-cost budget of $12. An activity consumes 0.8 kWh per session and the given rate is $0.25 per kWh. Cost per session is 0.8 × 0.25 = $0.20. Let n be the number of sessions. Then 0.20n ≤ 12, so n ≤ 60. If n must be a non-negative whole number, the maximum is 60 sessions.
If the equipment must be booked in blocks of eight sessions, the maximum whole number of complete blocks is 7, or 56 sessions, rather than the theoretical 60. This second condition changes the practical result. State the number of blocks and the corresponding number of sessions so the decision can be checked.
Practice 7 — a duty cycle with idle consumption
A device operates for 12 minutes at 300 W and idles for 18 minutes at 30 W during each half-hour. In hours, 12 minutes is 0.2 h and 18 minutes is 0.3 h. Energy per cycle is 300 × 0.2 + 30 × 0.3 = 69 Wh. Across six cycles, the total is 414 Wh = 0.414 kWh.
If a student assumes idle means zero, the result becomes 360 Wh across six cycles, which understates usage by 54 Wh. An idle power reading is a relevant term, not a footnote. The method is to account for each operating state separately, then add the stage energies.
Practice 8 — interpret irregular times
At times 0, 3, 7 and 10 seconds, a meter reports 0, 21, 37 and 61 joules. Interval rates are 7, 4 and 8 J/s respectively. Over all ten seconds, the mean rate is 61/10 = 6.1 J/s. The arithmetic mean of 7, 4 and 8 is about 6.33, which is not the full-period average because the intervals last different lengths.
To explain the difference, point to the duration of each stage. The 4 J/s interval lasts four seconds, whereas the 7 J/s interval lasts three. An unweighted mean treats them as equal contributions. The correct total-energy method weights every period naturally through its measured consumption.
Practice 9 — an inequality with a fixed start-up amount
A process uses 9 J at start-up, then consumes 4.5 J each second. It must not use more than 81 J. Model E = 9 + 4.5t and solve 9 + 4.5t ≤ 81. Subtracting 9 and dividing by 4.5 gives t ≤ 16 seconds. Both the intercept and the direction of the inequality matter.
At t = 16 the model reaches 81 J exactly. At t = 16.1 it exceeds the limit. If the question requires time in complete whole seconds, the maximum is 16, but the mathematical solution already contains every real t from zero up to 16 unless an integer condition is given.
Practice 10 — numerical bounds
Suppose operating power is known to lie between 0.38 and 0.42 kW and operating time is exactly 2.5 h. Energy use lies between 0.38 × 2.5 = 0.95 kWh and 0.42 × 2.5 = 1.05 kWh. At a stated illustrative price of $0.28 per kWh, the charge lies between $0.266 and $0.294 before currency-rounding rules.
If the budget is $0.29, the interval straddles it. The model does not establish a guaranteed pass. A central estimate of 1.00 kWh and $0.28 is useful for planning but not the same as a worst-case bound. Write the conclusion that follows from the full interval.
Practice 11 — compare two eligible options
A school demonstration requires energy use at most 80 J and completion time at most 15 s. Trial A uses 75 J in 14 s. Trial B uses 69 J in 17 s. Trial C uses 82 J in 12 s. Only A meets both constraints. Selecting B because it has the lowest energy or C because it is fastest ignores the other condition.
If the question instead asks for the fastest option regardless of energy, C is the correct response. Mathematical decision-making is not a universal ranking: it is a ranking under the stated objective and constraints. Copy those conditions into a small two-column checklist before comparing outcomes.
Practice 12 — a full real-world question
A club plans three demonstrations each day for twelve days. Each demonstration includes 5 minutes at 0.60 kW and 3 minutes at 0.10 kW. The energy per demonstration is 0.60 × 5/60 + 0.10 × 3/60 = 0.055 kWh. There are 36 demonstrations, so total energy is 1.98 kWh. At a given price of $0.32 per kWh, the usage-based cost is $0.6336, or $0.63 to the nearest cent.
Now suppose the club must remain under 2.0 kWh. The planned use meets that cap by 0.02 kWh under the stated rates. That margin is small: if the operating times or power readings are approximate, a careful evaluation should not claim a substantial reserve. The conclusion is valid for the model, and its fragility should be reported when the question asks for judgement.
A worked examination answer layout
Write the variables and units on the first line. Under that, define the relationship or construct the table of intervals. Show each main substitution, especially where a piecewise condition or a conversion is used. When you solve an inequality, display the inequality sign on the final statement and carry over the variable’s relevant domain. Finish with a sentence that names the actual choice, limit or cost.
With a long real-world problem, place numerical facts in a concise table before writing the equation. This prevents double counting and helps you detect a missing operating period. The calculation can be short when the preparation is clear. A student does not earn clarity by producing pages of arithmetic that obscure what each figure represents.
Keep a final five-second audit: Did I use the requested time interval? Is this total or per stage? Did I convert units? Are the constraints all met? Have I rounded only after comparing the unrounded model result with the threshold? These checks are both faster and more reliable than rebuilding the whole solution at the end.
Frequently asked questions and next steps
Can I assume the meter begins at zero? No. Use the given initial reading and the stated time origin. A positive starting value may be relevant to a model intercept or previous consumption. Is a straight-line graph always the best model? Only when the data and operating assumptions justify it. Changes in mode are better handled by intervals.
Do I need to learn Physics to solve this mathematics problem? You need to understand the quantities given, but the primary skills here are ratios, rates, linear equations, graph reading, percentages, inequalities and model interpretation. For the physics meaning, consult the related guide instead of pretending that mathematics proves the mechanics.
Where do I go next? Review the Physics owner, Volume 0685, graph foundation in Volume 0684, Mathematics Hub and Examination Craft. For earlier foundations, Primary 6 and the Secondary transition offers a separate route. The 2027 SEAB K310 syllabus is the authority for the actual paper requirements.
