Electricity becomes a mathematics problem the moment a person has to compare how long equipment runs, how much energy it uses, what a price actually covers and whether a proposed saving is worth the cost. This guide teaches those calculations through a fictional school energy exhibition. It begins with unit conversion and proportional reasoning, then moves to budget constraints, graph interpretation and a defensible final recommendation.
The work follows Volume 1129 on the a.c. generator, transformer and transmission losses, but this is an independent Mathematics lesson. You do not need to remember the Physics equations to answer a Mathematics examination question; the examination would need to provide any specialised context or information outside its own syllabus. What matters here is a set of K310 skills: percentages, rates, ratio and proportion, algebra, numerical accuracy, graph reading and interpretation of a real-world problem.
What to know first
- Convert minutes into hours, watts into kilowatts, and statements into consistent quantities before multiplying.
- Use a clear model for estimated electricity units: energy in kWh = power in kW multiplied by operating hours, when power is assumed constant.
- Distinguish the price per unit, the total variable energy charge and other bill components that may or may not be included.
- Compare alternatives using equal operating conditions, not mismatched days or times.
- Report uncertainty and a feasible recommendation, not a claimed saving the provided data cannot support.
Know the 2027 Mathematics paper you are preparing for
The official SEC G3 Mathematics subject is K310, previously referenced as O-Level 4052. Its Paper 1 lasts two hours 15 minutes, carries 90 marks, and contributes 50% of the subject assessment. It has about 26 short-answer questions. Paper 2 also lasts two hours 15 minutes, carries 90 marks and 50% weighting, and has nine or ten questions of varying lengths. The final Paper 2 question focuses on applying Mathematics in a real-world context, and candidates answer all questions. These are details of the published 2027 K310 syllabus.
An approved calculator may be used in both papers, but essential working must still be shown. The syllabus says non-exact numerical answers are normally given to three significant figures, and angles in degrees to one decimal place, unless a question specifies otherwise. Money amounts require contextual judgement, such as cents when quoting a payable price. An examination question may specify its own accuracy instructions. Always follow those first. An earlier PSLE Mathematics timing guide offers useful pacing habits, but the duration and structure of the SEC papers are different.
Read SEAB’s official 2027 G3 subject list if you are unsure which subject code applies. The national SEC will record the subjects and levels registered; there is no combined EMS written examination. The energy exhibition used here is original practice material.
First principle: name the quantity and its unit
A common mistake is to multiply three numbers merely because they appear near one another in a paragraph. Before calculating, name the required quantity. Power is a rate of energy transfer. Time records how long the equipment operates. Energy use combines the two. Electricity tariffs commonly quote a price per kilowatt-hour, written kWh. A kilowatt-hour is an energy unit, not a unit of power and not an hour of use by itself.
For a simple steady-power estimate, convert watts to kilowatts by dividing by 1,000, convert minutes to hours by dividing by 60, then multiply power by time. A 10 W lamp running for three hours uses 0.010 kW × 3 h = 0.030 kWh. The statement ‘it uses 30 kilowatts’ is wrong in both magnitude and unit. The useful habit is to carry the unit through each line, so that kW multiplied by h visibly becomes kWh.
In a Mathematics paper, the context might define energy consumption directly. Use the definition in the stem. Do not assume every unfamiliar technical relationship must be recalled from another subject. The exam evaluates how effectively you use the information provided, translate the words into a model and interpret the outcome. This is why the same unit-discipline skill transfers between everyday bills, transport rates and school project budgets.
Build a baseline estimate before comparing options
A fictional display board uses twelve 8 W lamps for three hours on each of 25 days. The total constant power is 12 × 8 = 96 W, which is 0.096 kW. Total operating time is 3 × 25 = 75 h. Energy is 0.096 × 75 = 7.2 kWh. At an entirely hypothetical energy price of $0.32 per kWh, the energy-only charge is 7.2 × $0.32 = $2.304, or about $2.30 rounded to cents. This is not a quotation of any actual Singapore electricity plan or school invoice.
It matters that the twelve lamps all run simultaneously. If the timetable says six operate on weekday afternoons and the other six operate only on weekends, the all-together model is no longer justified. Split the use into separate groups, calculate their contributions and add them. A neat answer using the wrong operating assumptions is still wrong. In a Paper 2 application question, the most difficult move may be identifying the actual schedule rather than the multiplication.
Try a quick reasonableness check. One 8 W lamp used for 75 h requires 0.008 × 75 = 0.60 kWh. Twelve such lamps require 12 × 0.60 = 7.2 kWh. The second route confirms the first. Two valid calculation routes are valuable, not because the examiner needs both, but because an independent cross-check catches copied digits, misplaced decimals and mistaken time conversions.
Rates and percentages are not the same calculation
A rate compares a quantity with a unit of another quantity. The cost per kWh is a unit rate. A percentage expresses a comparison with a base. If a model’s energy usage falls from 100 kWh to 80 kWh, the reduction is 20 kWh and the percentage reduction relative to the original 100 kWh is 20%. If the usage later rises from 80 to 100 kWh, the percentage increase relative to 80 kWh is 25%, not 20%. The direction changes the denominator.
This denominator issue causes many avoidable errors in real-world questions. Always write ‘percentage of what?’ before using a formula. For a saving, compare the difference to the original cost unless the question defines another base. For a revised cost, multiply the original by 1 minus the reduction as a decimal. A 15% reduction on $240 gives $240 × 0.85 = $204. To reverse a known final $204 after a 15% reduction, calculate $204 ÷ 0.85 = $240. Dividing by 0.85 and multiplying by 1.15 are not the same operation.
Separate a quoted rate from a complete bill
Suppose a worksheet gives an invented energy-only rate of $0.30 per kWh, a fixed monthly service amount of $3.00 and a flat 10% surcharge for the exercise. These are fictional conditions, not statements about Singapore tariffs or taxes. If the model uses 50 kWh in a month, the energy portion is $15.00. Add the fixed $3.00 to obtain $18.00. If the fictional 10% charge applies to that $18.00 subtotal, the total becomes $19.80. The order matters because the surcharge base has been specified.
A weaker response would compute 50 × 0.30 = $15.00 and stop, ignoring the fixed charge. Another weak response adds the surcharge only to the energy portion without checking the instructions. In a real-world question, each line of the given policy is a mathematical constraint. Mark the base for each fee explicitly and do not invent a real-world billing convention. If the worksheet says the rates already include all fees, do not add another invented surcharge.
Students often remember percentages but overlook whether an amount is tax-inclusive, applies per day or applies once per billing period. Take five seconds to annotate ‘per kWh’, ‘per month’ and ‘percentage of subtotal’. The same habit works for delivery charges, booking fees, instalment purchases and discounts.
Direct and inverse proportion in an energy context
At constant power, energy use is directly proportional to time: doubling hours doubles energy. At constant duration, using twice as many identical lamps doubles energy use. These are valid direct proportional relationships when the other relevant quantities remain fixed. They may not hold if the number of hours changes or if the equipment adjusts its power automatically. State the fixed conditions before invoking direct proportion.
Inverse proportion appears in a different kind of problem. For a fixed quantity of work performed by equal-rate teams, more teams may reduce required time, under a simplified assumption. Do not force inverse proportion onto electricity bills just because two quantities move in opposite directions. Higher efficiency equipment could reduce use, but the relationship between an arbitrary price and energy consumed is not automatically inverse. Know the mathematical structure before choosing the method.
Example: five identical exhibition lamps consume 2.5 kWh over a specified identical interval. Under equal operating conditions, eight lamps use 2.5 × 8/5 = 4.0 kWh. Explain the scaling factor. If the problem also says the eight-lamp display runs for only half as long, account for both changes: 2.5 × 8/5 × 1/2 = 2.0 kWh. The order of scaling factors does not matter, but overlooking one condition does.
Use algebra when the unknown is embedded in the schedule
Suppose six identical 15 W lamps run for h hours, and the estimated total energy must be no more than 1.8 kWh. Total power is 90 W or 0.09 kW. The inequality is 0.09h ≤ 1.8. Divide by the positive 0.09 to obtain h ≤ 20. If the event lasts four days with equal use each day, maximum running time per day is 5 h. A learner can solve the algebra quickly; the tougher part is explaining what the 20 means and recognising the four-day requirement.
Now change the problem. The lamps must run at least 3 h daily for five days. The total minimum is 15 h, which satisfies the 20 h limit. There is a feasible interval from 15 to 20 h total under the simplified model. If the organiser instead insists on six hours each day for five days, the demand is 30 h and the plan is infeasible. A correct examination conclusion should say ‘does not meet the given limit’, not simply report ’30’. Context decides whether the number represents a solution or a warning.
Bounds: when a decimal is not an exact measurement
Measurements are not always exact. If a lamp is labelled as using 12 W to the nearest watt, that label represents a rounded value in a simplified mathematical exercise, though a manufacturer’s rating should not automatically be treated as a measured rounding interval in real life. A question that expressly says ‘measured as 12 W to the nearest watt’ implies a true value from 11.5 W inclusive up to 12.5 W exclusive. If runtime is exactly 10 h, the consumption lies from 0.115 kWh up to but not including 0.125 kWh under the constant-power model.
Be careful about the distinction between a measurement interval and device uncertainty. An instrument marked ±0.5 W is not necessarily identical to a number rounded to the nearest watt in every experimental sense. For a Mathematics bounds question, use the stated rounding precision and the lower and upper bounds it implies. If the paper gives uncertainty differently, use that wording. This protects students from applying a memorised half-unit rule when it does not fit the stem.
A graph may show cumulative consumption with plotted values rounded to one decimal place. You cannot claim a difference of 0.003 kWh from that presentation. Report only the precision the data supports, or explicitly calculate from the original unrounded figures if provided.
A decision problem with purchase price and operating cost
The school considers two fictional lighting packages for a display room. Option A costs $80 more to purchase but consumes 20 W while operating; Option B consumes 60 W. The room operates for 2,500 h in a year, the assumed energy charge is $0.34 per kWh, and the question tells us to ignore maintenance, changes in price and other costs. Option A’s running-energy difference is 40 W, or 0.040 kW. Over 2,500 h the energy saving is 100 kWh. At $0.34 per kWh, that is $34 per year.
The extra initial price is $80. A simple payback estimate is $80 ÷ $34 ≈ 2.35 years. This is not a guarantee of any actual product lifetime or real-world savings. It is the mathematical result under the explicitly provided assumptions. If the school plans to keep the display for only 18 months, Option A will not recover the extra cost in that interval through this simplified energy saving alone. If the decision period is five years and all assumptions remain fixed, the five-year energy saving is $170, which exceeds the $80 additional price by $90.
Notice that a ‘best buy’ answer changes with the decision horizon. You are not choosing a universally superior product. You are evaluating a stated objective. The question may ask for the lower immediate cost, lower annual running cost or lower total cost over a chosen duration. Circle the objective before comparing.
Express total cost as an algebraic function
Let t be the number of years. Ignore changes in annual usage and price for this exercise. If Option B costs $200 initially and consumes $51 a year, its modelled total cost is B(t) = 200 + 51t. If Option A costs $280 and consumes $17 a year, A(t) = 280 + 17t. To find the break-even point, set the totals equal: 280 + 17t = 200 + 51t. This gives 80 = 34t, or t = 80/34 ≈ 2.35 years.
The graphs of these functions are straight lines with different vertical intercepts and gradients. The more expensive purchase has the higher intercept but lower gradient. The point of intersection represents equal modelled accumulated total cost. Before that point B is cheaper; after that point A is cheaper. This connection between algebra and graphs is more useful than memorising a rule for ‘payback questions’. It shows what the variables mean and how the conclusion changes with time.
A graph with t measured in years on the horizontal axis and total dollars on the vertical axis should use the same scales for both options. Plotting one line with months and the other with years would make the picture misleading. When reading a supplied graph, identify what its origin means. An initial purchase cost is not automatically zero at time zero; it is the vertical intercept.
The difference between a model and a forecast
An equation such as A(t) = 280 + 17t assumes that annual consumption and energy price do not change. That makes it a simple comparison model. A forecast would require evidence about future prices, usage and maintenance. Students should not turn the model into a promise. If the question asks whether the package definitely saves money over the next decade, the correct answer may be conditional: it does so under the supplied assumptions, but the assumptions may not hold.
This is especially important when explaining numbers to people outside the Mathematics classroom. ‘The model projects a $90 saving over five years if runtime and rates remain constant’ is stronger than ‘We will save $90’. The first sentence makes the logical condition visible. Good mathematical communication includes its own limitations.
Table comparison: keep denominators equal
A fictional school tests three display settings. Mode A uses 0.8 kWh across four hours. Mode B uses 1.5 kWh across five hours. Mode C uses 1.2 kWh across eight hours. To compare energy per hour, divide each use by its runtime. A is 0.20 kWh/h, B is 0.30 kWh/h and C is 0.15 kWh/h. C has the lowest average energy use per hour in this simplified set. The raw total energy of Mode A is smallest, but its operating period is also shortest. Those are different comparisons.
If the question asks which mode consumed the least energy during its own observation window, choose A. If the question asks which would use least energy per hour under a stable-rate assumption, choose C. If the question asks which achieves a required light output or satisfies an access standard, the table is insufficient because it lacks that information. The best examination answer depends on the exact question, not the first ranking the student notices.
Statistics and the danger of a single attractive average
Suppose four daily runtime measurements are 2 h, 2 h, 3 h and 9 h. The mean is 4 h. The median is 2.5 h. The unusually long fourth day pulls the mean above the typical shorter days. If a decision requires enough battery time for every day in this small sample, designing for the mean alone may be inappropriate. If a question asks for mean, calculate mean. If it asks for a suitable summary of typical use, consider the distribution and justify your choice.
Do not declare the 9-hour day an ‘error’ merely because it is large. It may be a genuine long event. Investigate the context before excluding it. The point of statistics is not to make data look tidy; it is to describe it accurately and choose a sensible measure for the question. If the data is skewed or small, acknowledge that limitation rather than making a false population claim.
Composite problems: one calculation feeds another
A school plans an energy exhibition for 15 days. Each day, eight lamps operate for 3.5 h at 18 W each. The invented electricity price is $0.36/kWh. First calculate the combined power: 8 × 18 = 144 W = 0.144 kW. Next calculate total time: 15 × 3.5 = 52.5 h. Energy is 0.144 × 52.5 = 7.56 kWh. The energy-only charge is 7.56 × 0.36 = $2.7216, approximately $2.72.
The organiser says a new system will reduce the estimated energy use by 25% under the same schedule. The revised energy is 7.56 × 0.75 = 5.67 kWh. Estimated energy charge is 5.67 × 0.36 = $2.0412, or about $2.04. The saving is about $0.68 at this scale. It would be misleading to call that a large financial saving without considering the number of displays, their operating schedule and any purchase cost. The context matters after the arithmetic ends.
Alternative check: 25% of 7.56 is 1.89 kWh, and 1.89 × $0.36 = $0.6804. This agrees with the charge difference before rounding. Students who round too early may produce an avoidable one-cent discrepancy. Keep extra precision through the intermediate calculations and round the final amount in the context required.
Problem set A: short-answer precision practice
Question A1: A 75 W device runs for 40 minutes. What energy does it use in kWh under constant power? Convert 75 W to 0.075 kW and 40 min to 2/3 h. Multiply to obtain 0.050 kWh. The most common wrong answer is 3 kWh, from treating minutes as hours. Write both conversions before multiplying.
Question A2: Six identical devices use 3.6 kWh in a fixed period. How much do ten identical devices use in the same period? Direct proportion gives 3.6 × 10/6 = 6.0 kWh. The value is not 2.16 kWh; that would reverse the ratio. More devices under identical time and operating conditions should not reduce total usage in this simplified model.
Question A3: A display’s modelled usage falls from 48 to 36 kWh. Find the percentage reduction. Difference = 12 kWh, base = 48 kWh, reduction = 12/48 × 100% = 25%. If usage instead rises from 36 to 48, the increase is 12/36 × 100% = 33⅓%. Explain why the percentage bases differ.
Question A4: A price is increased by 8%, then decreased by 8% of the new value. Is it back to the original? No. Starting at $100 gives $108, then $108 × 0.92 = $99.36. Equal percentage increase and reduction do not cancel when the base changes. This is useful beyond bills; it applies to repeated price changes and discounts.
Problem set B: rate, graph and algebra
Question B1: A cumulative-use graph is a straight line through (0, 0) and (5, 1.5), with time in hours and energy in kWh. The gradient is 1.5/5 = 0.30 kWh per hour. Under a constant-rate model, eight hours gives 2.4 kWh. Note the unit on the gradient. A gradient is not merely ‘0.30’; it expresses the rate at which consumption accumulates.
Question B2: A revised graph intersects the vertical axis at 0.4 kWh. Can that be explained by the same model starting from zero usage? Not without an initial offset or a change in what is being recorded. The intercept may represent previous consumption before the observation period. Read the caption and the defined origin. An apparent graph anomaly may come from the chosen baseline rather than the equipment itself.
Question B3: A budget allows at most $12 in energy-only costs, and the given rate is $0.30/kWh. Maximum consumption is 40 kWh. If each daily session uses 2.5 kWh, the number of complete sessions must satisfy 2.5n ≤ 40, so n ≤ 16. If n counts whole events, report the integer maximum of 16 rather than a fractional session. An answer has to fit the object being counted.
Question B4: A planned schedule has 18 sessions at 2.5 kWh each. Required energy is 45 kWh, beyond the 40 kWh limit. The organiser cannot repair the plan by rounding 18 down to 16 without acknowledging two removed sessions. The correct decision is to revise the schedule, usage or budget and state the trade-off explicitly.
Problem set C: the longer Paper 2 scenario
School X is deciding between a portable exhibit costing $450 to purchase and a fixed exhibit costing $620. The portable exhibit is predicted to consume 80 kWh per year; the fixed exhibit 25 kWh per year. Use a fictional energy rate of $0.40/kWh. Assume an eight-year comparison, identical educational usefulness, no maintenance costs and constant energy prices. Portable annual energy cost is $32; fixed is $10. The eight-year totals are portable $450 + 8 × $32 = $706 and fixed $620 + 8 × $10 = $700. Under the assumptions, fixed costs $6 less over eight years.
The answer is incomplete if it simply says ‘fixed is better’. State that the cost advantage is small relative to the model’s uncertainty and that usefulness is assumed equal, not measured. A different rate, maintenance schedule or decision horizon can reverse the result. In an exam, the command may ask for the cheaper option under the given model, in which case fixed is the supported choice. If it asks for a critique, acknowledge missing information.
Find the break-even year by equating 450 + 32t and 620 + 10t. Then 22t = 170, so t ≈ 7.73 years. Before that, portable has the lower cumulative cost; after that, fixed does. Check the intersection against the eight-year calculation: eight years is just beyond break-even, so the fixed display’s advantage should be positive but small. The two methods agree. That mutual consistency is a strong check.
Problem set D: a constraint the brochure hides
Two fictional lamps have identical purchase prices. Lamp L consumes 10 W but provides only 60 arbitrary brightness units. Lamp M consumes 16 W and provides 120 brightness units. The school needs at least 100 units for a fair display. Which is suitable if there is only room for one lamp? L has the lower power use, but it does not satisfy the brightness constraint. M is feasible. The correct conclusion is not ‘L saves energy’; energy saving alone does not meet the complete objective.
Now suppose the school may install two Lamp L units. Their combined light output would be 120 units, and combined power is 20 W, greater than M’s 16 W. In this simplified problem M provides the required output with lower power. Real lighting measures and practical suitability are more complex, but the mathematical structure is clear: define the constraint first, then compare feasible choices.
Problem set E: interpret uncertain reporting
A student’s final claim says, ‘This plan will always cut the electricity bill by 30%.’ The calculations used two sample days and assumed fixed operating hours, a uniform energy-only rate and no other charges. A defensible revision is: ‘Under the stated operating schedule and constant energy rate, the model predicts a 30% reduction in the energy portion of the bill.’ That statement is less dramatic but mathematically sound. It names the model’s scope.
Another claim says, ‘The new system is twice as efficient because it costs half as much to run.’ That may not follow. Energy costs depend on use, price and sometimes service terms; efficiency is a separate physical idea. In a Mathematics answer, do not use a technical label unless the defined quantities support it. Explain the measurable difference instead.
A practical method for reading dense case-study questions
On the first reading, underline the decision being asked for. On the second, mark quantities with units and the period to which each applies. On the third, list assumptions: same hours, constant unit price, initial price included, maintenance excluded, number of devices fixed. Only then choose arithmetic, proportion, algebra or a graph. The order prevents a common failure in long questions: doing perfectly correct calculations on a model that does not answer the question.
Keep a small working table with columns for ‘quantity’, ‘unit’, ‘source’ and ‘use in calculation’. If the source says power is in watts, write the converted kilowatts before computing kWh. If the table says a rate is monthly, do not mix it with annual totals without conversion. If the source gives a duration range rather than a single figure, consider whether a range of answers is required. These procedures are not formal requirements in every paper; they are ways of protecting accuracy.
What examiners can see in the working
In K310, omission of essential working can lose marks. Show the mathematical link between a figure and the conclusion. For a straightforward power-time calculation in a supplied context, write the unit conversions and multiplication. For an algebraic cost comparison, write the two expressions and the equality that defines break-even. For a percentage, show the correct base in the denominator. For a graph, identify the meaningful coordinates and gradient. A reader should be able to follow the chain without guessing why a number appeared.
The final answer should reflect the real object. A fractional number of staff members, lamps or whole sessions may need to be rounded according to feasibility, not automatically to three significant figures. If six whole devices are needed to meet a minimum capacity, a mathematical estimate of 5.2 devices means at least six, not five. Explain that integer decision in words.
Recovery from common errors
If your answer is 1,000 times too large, check watts and kilowatts. If it is 60 times too large or small, check minutes and hours. If a percentage change looks backwards, check the original base. If a payback period seems impossibly short, check whether you have compared annual cost with purchase cost rather than monthly cost. If a graph gives a negative running cost for ordinary positive time, recheck the intercept and gradient. If two options are compared over different observation periods, normalise by time before claiming one is inherently lower-use.
When correcting an error, do more than write the right number. Write why the original approach failed, then solve a slightly changed question without looking at the model. That demonstrates repair rather than temporary recognition. A useful error ledger separates unit conversion, proportional model, arithmetic, graph interpretation, assumption and final recommendation. Each entry points to a different kind of practice.
A seven-day improvement plan
Day one: practise watts-to-kilowatts and minutes-to-hours with ten short examples. Day two: work on percentages and reverse percentages using bill-like data. Day three: compare rates under unequal durations. Day four: express cumulative cost as linear functions and locate a break-even point. Day five: solve inequality and integer-feasibility questions. Day six: complete one full mixed case under timed conditions. Day seven: redo only the mistakes with different numbers, then explain aloud how the conclusion depends on the assumptions.
Give yourself credit for correctly rejecting an unsupported conclusion. Some questions are designed to test whether you can say that data is insufficient. If the setup cost is missing, do not make up a cost to choose the cheaper package. If prices change across periods but no rate is given, state which comparison cannot be completed. Mathematical discipline includes knowing what cannot be calculated.
Connecting earlier learning and later performance
The route from PSLE to G3 Mathematics builds steadily. Primary-school arithmetic provides multiplicative reasoning, ratios, fractions and basic percentage ideas. Secondary Mathematics adds formal algebra, inequalities, graph interpretation and multi-condition problems. The PSLE Mathematics timing guide is a useful foundation for checking and pacing, but K310’s papers and expectations are distinct. For wider study structure, consult the G3 SEC EMS learner hub.
The Physics transformer guide in Volume 1129 explains why electricity transmission can involve different voltages and currents. This Mathematics guide is the owner of the cost-and-constraint method. The next English guide will help students turn a defensible numerical conclusion into a situational message suited to a real audience. Do not merge the subject-specific papers simply because the scenarios connect.
Frequently asked questions
Do I need to know electricity physics for K310?
No separate Physics syllabus should be presumed. An application question must provide the context needed for its mathematical work. This article deliberately supplies its energy-use model so you can practise proportion, percentage, algebra and decision-making.
Is a kWh a measure of power?
No. It measures energy. kW is a power unit; kWh combines a constant power rate with time.
Should I round money to three significant figures?
Use the question’s instructions and an appropriate money representation, often cents. The syllabus gives a default accuracy for non-exact numerical answers, but exact and context-specific instructions take priority. Preserve unrounded intermediate values where possible.
Why is the cheaper unit not always the cheaper total option?
Total cost may include initial price, operating use, fixed fees and the number of periods. You must compare the full quantity the question defines.
What if two options are almost equal?
State the calculated comparison and its assumptions. If the difference is small and the task asks for evaluation, recognise that excluded costs or uncertain inputs could alter the recommendation.
Sources and the next step
The examination details in this guide were checked against the SEC overview, 2027 G3 school-candidate list and official K310 Mathematics syllabus. Rates, devices, purchase costs, running times and cases are fictional, invented for practice; they are not current retailer quotations, regulated tariffs or official examination questions. Continue with Volume 1131 for K300 English Paper 1 situational writing, and then use the EMS case to test whether scientific, numerical and verbal claims remain aligned.
