The answer is not always a single number
A capacity word problem is a small exercise in responsible decision-making. It gives a fixed demand, an amount used by each extra item and a limit that must not be passed. An equation would find a value at equality. An inequality identifies a set of possible values. The final instruction may then ask for the largest whole-number count in that set, the smallest count that meets a target, or every count that satisfies several rules. These are different jobs. A student who solves the algebra but ignores the meaning of the last line can still give the wrong answer.
This lesson extends Vol 0238’s foundations of linear inequalities and number lines with complete decisions involving mixed units, fixed and variable quantities, strict thresholds, alternative time windows and integer restrictions. It is not a duplicate of Vol 0336 on piecewise utility bills. Cost models ask what a resource costs after thresholds; this chapter asks whether a modelled quantity satisfies a constraint. It follows Vol 0339 on physical fault diagnosis, but all device demands here are invented mathematics data rather than an inspection or approval of any real electrical installation.
Know the official examination frame
The official 2027 G3 Mathematics K310 syllabus includes linear inequalities in one variable, simultaneous equations, algebraic formulae, ratio, percentages, rates, graphs and real-world application. According to the SEAB 2027 G3 school-candidate directory, K310 succeeds the earlier reference code 4052. The assessment scheme consists of Paper 1 and Paper 2, each lasting two hours fifteen minutes, each worth 90 marks and 50 percent. Paper 1 contains about twenty-six short-answer questions; Paper 2 contains nine to ten questions, with its final question directed at a real-world scenario. These details were checked against the current official syllabus rather than carried over from memory.
The syllabus names AO1 standard techniques, AO2 solving contextual problems and AO3 reasoning and communication. Together they explain the point of this chapter. Translating words correctly is AO2. Rearranging an inequality is AO1. Showing that seven works but eight fails is AO3. Workings matter; the specification warns that omission of essential working can cost marks. Every invented problem below is a teaching exercise, not a forecast of a particular examination question.
Define the object counted by the unknown
Write “Let n be the number of additional displays.” This fixes two important facts: n counts a specific kind of thing, and n is a non-negative whole number unless the question says otherwise. Never let n turn into total watts halfway through the solution. A typical expression is P = 460 + 230n, where P is power in watts, 460 W is a fixed demand, 230 W is the contribution per optional display, and n is a count. If the problem limits P to at most 2300 W, the mathematical sentence is 460 + 230n ≤ 2300.
This is a model, so it is necessary to name the assumptions. Each optional unit is identical, the fixed demand remains constant, and the named devices are being considered at the same time. These assumptions do not certify any actual circuit or product. They make the examination word problem solvable. Clear variables also help the reader audit your argument and identify what changes if a new condition is introduced.
Read the language before writing the sign
At most twelve includes twelve and becomes n ≤ 12. Less than twelve excludes twelve and becomes n 12. Do not scan the nearest keyword and guess. “Not fewer than twelve” means at least twelve; “no more than twelve” means at most twelve. The word not changes the statement that follows it.
Test equality to confirm the sign. Could exactly twelve be accepted by the sentence? If yes, equality belongs in the sign. If no, it does not. The boundary may determine whether the final whole-number answer is seven or eight. In a timed paper this reading decision matters as much as the arithmetic. A question with “below the cap” must be treated differently from one that says “not above the cap”.
Build the fixed-plus-variable expression
In a typical planning question the fixed contribution appears once. A 580 W fixed component plus n optional 220 W components gives 580 + 220n, not 580n + 220n. The mistake occurs because a student sees two wattages without asking which one repeats. Look for phrases such as “each”, “per”, “one-off”, “fixed” and “already in use”. They decide whether a number is multiplied by n.
The upper limit belongs on the opposite side of the inequality. For example, 580 + 220n ≤ 1900. Subtract 580 to obtain 220n ≤ 1320, then divide by positive 220 to obtain n ≤ 6. If a minimum of three units is also required, add n ≥ 3. The feasible whole-number counts are three, four, five and six. This is why interpreting the final condition is a separate step: the algebra alone would have admitted zero, one and two.
The sign-reversal rule is not optional
Adding or subtracting the same number from both sides preserves the order of an inequality. Multiplying or dividing by a positive number also preserves the direction. Dividing by a negative number reverses it. Consider 15 − 2n ≥ 5. Subtract 15: −2n ≥ −10. Divide both sides by −2, reversing the sign: n ≤ 5. Test n=4 in the original: 15−8=7, which is at least five. Test n=6: 15−12=3, which fails.
Memorising “flip the arrow” without knowing why is fragile. On the number line, −6 is smaller than −4 even though 6 is greater than 4. Multiplication by −1 reverses order. The rule can be remembered through that relationship. If you are unsure, pick a value from your proposed solution and substitute it into the original. A contradiction often exposes the wrong sign immediately.
Whole-number decisions are not ordinary rounding
Suppose the algebra produces n ≤ 7.6 and n counts complete objects. The largest valid n is seven, not eight. Rounding 7.6 to the nearest integer would cross the limit. If the inequality is n ≥ 7.6, the smallest complete count is eight. If the condition is n < 8, the maximum is seven even though 8 is a whole number. Think about membership of a set, not rounding as a formatting operation.
Always check the proposed value and the next integer. If n≤6 follows from 460+230n≤1840, substitute six: 1840 W fits. Substitute seven: 2070 W fails. The two calculations defend the maximum. If the question asks for all possibilities, write the set or a correct interval restricted to integers instead of announcing only six. A clean conclusion follows the command word.
Mixed units must be made consistent
An expression such as 0.46 + 230n ≤ 2.3 is meaningless if the first and last numbers are kilowatts while the middle number is watts. Either change everything to watts, 460 + 230n ≤ 2300, or change everything to kilowatts, 0.46 + 0.230n ≤ 2.30. Both solve to n≤8. Use whichever form makes the arithmetic and explanation clearer.
The same discipline applies to time, distance, money and mass. A half-hour is thirty minutes; a kilowatt-hour is an energy unit rather than a power rating. If a science scenario gives current I and voltage V, a simplified relationship P=VI may establish a model power. But a modelled numerical limit is not permission to plug in real devices. The only legitimate conclusion here is about the figures the question supplies.
The percentage base is a hidden trap
Suppose a purely hypothetical mathematical policy holds back twenty percent of a nominal 2300 W allowance. The remaining usable amount is eighty percent: 0.8×2300=1840 W. If fixed demand is 460 W and optional units use 230 W, write 460+230n≤1840. Then n≤6. If you instead apply twenty percent only to the fixed 460 W, you have changed the whole on which the percentage is defined.
Questions sometimes say “leave 300 W unused” instead of stating a percentage. With a 2100 W total, write 2100−P≥300, or P≤1800. Notice that the quantity which must be at least 300 is the spare allowance, not the used power. Translate exactly what the sentence describes. There is no universal twenty-percent electrical safety rule in this lesson; the percentage belongs solely to the invented problem.
Several constraints require their intersection
A programme may require at least four displays, permit at most nine by space, and permit at most seven by a model demand limit. The valid counts satisfy n≥4, n≤9 and n≤7 simultaneously. Therefore n can be four, five, six or seven. Taking a union of these conditions would allow values that break one of them. The word and is crucial: all conditions must be satisfied.
If an activity requires at least eight units but its cap allows at most six, there is no feasible integer solution. Do not choose seven as a compromise. Seven would break both instructions. A strong answer states that the conditions conflict, and if asked for an explanation, names the lower and upper bounds responsible. Mathematically impossible is a valid conclusion when warranted by the model.
Different time windows need different equations
Imagine a fictional morning exhibition with 500 W fixed demand and an evening event with 800 W fixed demand. Each shared optional panel uses 150 W, and the model threshold is 1700 W during either interval. Morning: 500+150n≤1700, so n≤8. Evening: 800+150n≤1700, so n≤6. If the same n panels remain in both sessions, the common maximum is six. Averages are not justified because the question limits instantaneous demand in each interval.
This approach generalises to rooms with separate capacity limits, delivery trucks with different weight restrictions and timetables with two session durations. Create one condition for each interval. Then identify whether the problem asks for an arrangement that works throughout or allows different arrangements at different times. The time wording decides whether one shared n or several variables are needed.
Reverse the unknown when the question changes
Instead of asking how many displays fit, a problem may ask the maximum power p of each of five displays. If the fixed base is 350 W and the total bound is 1750 W, write 350+5p≤1750. Then 5p≤1400 and p≤280 W. The variable is watts per display, not an integer count. Power can be represented as a continuous quantity in the problem, so do not apply the whole-number rounding rule without a reason.
Likewise, a time problem may ask for the duration of a session, not the number of sessions. If six sessions plus 18 minutes of setup must fit in 90 minutes, 18+6t≤90 gives t≤12 minutes. The algebraic pattern looks familiar, but a meaningful answer changes with the identity of the unknown.
Why you should check the original expression
A calculator can give a numerically plausible solution to a wrongly copied equation. Substituting into the original catches missing fixed charges, misplaced signs and unit errors. If 420+175n≤1820, subtract 420 to obtain n≤8. A student who divides 1820 by 175 first and claims ten units has ignored the fixed portion. Check eight: 420+1400=1820. Check nine: 420+1575=1995, above the bound.
Direct verification is particularly valuable when the result is an integer boundary. It does not require you to trust every earlier algebraic line. It also proves the decision in language a reader understands. The final sentence can state that eight fits exactly, whereas nine exceeds the limit. That is stronger than simply writing n=8 without saying what n means.
Avoid premature numerical rounding
If each specified model load is 451 W at an illustrative 230 V, its current from P=VI is 451/230 ≈1.96087 A. Seven loads require 3157/230 ≈13.726 A in the simplified model. Rounding the current of each load to 2 A before multiplication would give 14 A. If a numerical threshold were 13.8 A, premature rounding would change the feasibility decision. Keep the original expression as long as possible and round only when the calculation is complete.
K310 ordinarily uses three significant figures for non-exact answers unless a question specifies another accuracy. That convention does not mean intermediate work should be rounded at every step. The count of units is exact, not a measurement to three significant figures. State the exact integer first and the relevant comparison separately.
Do not confuse a mathematical bound with safety approval
These worked problems sometimes use electrical terms because this volume continues an English–Mathematics–Science progression. The source of the model is supplied data, not testing of live circuits. An algebraic inequality can show that a fictional combination falls below a fictional cap, but it cannot assess conductor condition, protective-device suitability, legal requirements or the safety of a physical installation. Outside the textbook, those matters belong to responsible personnel and appropriate expertise.
This limit of inference is a worthwhile examination habit beyond electricity. A budget model does not prove a product is fit for purpose. A population graph does not prove causation. A mathematical capacity count does not prove a venue is authorised for that arrangement. Say precisely what the numbers establish and do not attach extra conclusions.
The learning route and canonical owners
For sign reversals and number lines, return to Vol 0238. For equations in two unknowns and graph intersections, Vol 0234 has the primary method. Vol 0167 covers Physics electrical power and energy. Vol 0336 owns tiered tariffs. These links support learning without competing for the same search intent. The Mathematics Hub is the subject owner and the PSLE Learning Guide provides elementary percentage and unit repair.
Readers who need to express mathematical reasoning in prose can use the Secondary English Learning Hub. Readers revising cause and effect in electrical models can use the Primary 6 Science Learning Hub and Vol 0339. For the whole route, follow the G3 SEC EMS Learner’s Guide. This lesson ends with a new ability: defending a bound through a real number, a condition and a clear interpretation.
Twenty-eight worked problems: move from a formula to a defensible decision
The following numbers are invented practice data. Every example demonstrates a different reading or reasoning demand. Work each from the original wording before reading its solution.
Problem 1: Maximum including equality
Question. A 2300 W model has a 460 W fixed component and identical 230 W additional units.
Worked solution. 460+230n≤2300, hence n≤8. Eight units fit exactly; nine would reach 2530 W.
What this tests. At most includes equality. If the word were strictly below, the maximum would drop to seven.
Problem 2: Maximum with strict inequality
Question. The same fixed and unit demands apply, but the demand must remain below 2300 W.
Worked solution. 460+230n<2300 gives n<8. The maximum non-negative integer is seven.
What this tests. A decimal or boundary value cannot be accepted merely because the equality version used it.
Problem 3: Rounding toward feasibility
Question. A 1600 W cap includes 450 W of fixed demand and 100 W per unit.
Worked solution. 450+100n≤1600 gives n≤11.5. The largest integer is eleven.
What this tests. Twelve units would demand 1650 W, violating the cap. Conventional rounding is inappropriate.
Problem 4: Minimum target
Question. A demonstration must reach at least 740 W, with 260 W fixed and 80 W per added unit.
Worked solution. 260+80n≥740 gives n≥6. Six is the smallest qualifying count.
What this tests. This is a lower bound; reporting a maximum would answer a different question.
Problem 5: A time-slot model
Question. Setup takes 18 minutes and every session takes 12 minutes; the total available period is 90 minutes.
Worked solution. 18+12n≤90 gives n≤6. Six sessions use exactly 90 minutes.
What this tests. Setup occurs once. Do not multiply the fixed setup time by n.
Problem 6: Reserved margin in a given model
Question. The stated exercise reserves twenty percent of a 2300 W numerical capacity; fixed use is 460 W.
Worked solution. Usable allowance 1840 W. Then 460+230n≤1840 and n≤6.
What this tests. The twenty percent belongs to the total model capacity, not only the fixed load.
Problem 7: Two operating sessions
Question. Morning base 500 W, evening base 800 W, optional panels 150 W, common limit 1700 W.
Worked solution. Morning n≤8 and evening n≤6, so a shared count cannot exceed six.
What this tests. Use both sessions, not their average. The strictest condition controls a shared decision.
Problem 8: The unknown is power, not count
Question. Five identical units with unknown power p join a 350 W base under a 1750 W threshold.
Worked solution. 350+5p≤1750 gives p≤280 W per unit.
What this tests. A watt value is not a number of whole devices. Name the unknown before solving.
Problem 9: No more versus less than
Question. Two statements impose n≤8 and n<8 respectively. Find each maximum for integer n.
Worked solution. The first permits eight; the second permits seven.
What this tests. The presence or absence of equality can shift the final feasible count.
Problem 10: Three constraints
Question. The activity needs n≥4, room planning says n≤9, and an independent numerical model says n≤7.
Worked solution. The common integer solutions are 4,5,6,7.
What this tests. Taking all numbers allowed by any one condition is wrong; each condition must hold.
Problem 11: An impossible set of requirements
Question. An activity states n≥8 while another condition requires n≤6.
Worked solution. No feasible integer n exists.
What this tests. Do not invent a compromise at seven; it would fail both conditions.
Problem 12: Kilowatts and watts
Question. Fixed 0.5 kW, optional 250 W each, total maximum 2.0 kW.
Worked solution. 500+250n≤2000 gives n≤6, using watts throughout.
What this tests. Writing 0.5+250n≤2.0 mixes incompatible units.
Problem 13: Strict lower bound
Question. An exhibition needs more than five complete practice stations.
Worked solution. n>5, so at least six whole stations are needed.
What this tests. More than excludes equality. At least five would include five.
Problem 14: Dividing by a negative
Question. Solve 18−3n≥6 with n non-negative.
Worked solution. −3n≥−12, and dividing by −3 reverses the sign: n≤4.
What this tests. Test n=5: 18−15=3 fails. The wrong sign would admit it.
Problem 15: A fixed booking time
Question. A 100-minute period includes 22 minutes of fixed briefing and 13 minutes per station.
Worked solution. 22+13n≤100 gives n≤6. Six uses 100 minutes; seven takes 113.
What this tests. The answer is a whole station count; do not round up a fractional maximum.
Problem 16: A modelled current bound
Question. Seven 451 W units operate at an illustrative 230 V. Find total demand.
Worked solution. P=3157 W and I=3157/230≈13.726 A.
What this tests. Rounding 451/230 to 2 A before multiplying would inflate the result to 14 A.
Problem 17: Inconsistent arithmetic in a table
Question. A table claims five 180 W units plus a 740 W base total 1540 W.
Worked solution. Correct calculation is 5×180+740=1640 W.
What this tests. A derived cell can be checked against original numbers instead of accepted automatically.
Problem 18: Reserve stated as unused amount
Question. At least 300 W of a 2100 W numerical cap must remain unused.
Worked solution. 2100−P≥300, or P≤1800 W.
What this tests. The phrase at least applies to unused capacity, not consumed power.
Problem 19: Two categories with known total
Question. Six devices are Type A at 120 W or Type B at 240 W; combined power is 1080 W.
Worked solution. a+b=6 and 120a+240b=1080. Divide by120 and subtract: b=3, a=3.
What this tests. This is the separate simultaneous-equations method, not a two-variable inequality; the canonical owner is Vol 0234.
Problem 20: Different days, one common count
Question. Day 1 has base 500 W under 1400 W; Day 2 has base 650 W under 1700 W. Each optional unit is 150 W.
Worked solution. Day 1 n≤6; Day 2 n≤7. A count used both days must satisfy n≤6.
What this tests. Choosing seven ignores the restriction on the first day.
Problem 21: Non-overlapping loads
Question. A 1200 W exhibit runs in the first hour, and a 1300 W exhibit runs only in the second, under a 1800 W demand limit.
Worked solution. They are not simultaneous; each period individually stays below 1800 W in the model.
What this tests. Adding 2500 W assumes simultaneous operation, which contradicts the timetable.
Problem 22: A compulsory starter device
Question. A calculation yields n≤5, and the prompt says at least one device must be present.
Worked solution. Feasible integers are n∈{1,2,3,4,5}.
What this tests. Zero meets the cap alone but violates the explicit minimum condition.
Problem 23: The missing fixed part
Question. A worksheet asks for maximum n when 420+175n≤1820.
Worked solution. Subtract 420 first: n≤8. Eight gives 1820; nine gives 1995.
What this tests. Dividing 1820 by 175 before subtracting fixed demand gives an invalid value.
Problem 24: A small increase that does not change the count
Question. A 460 W base and 230 W additional units face a cap rising from 1840 to 1850 W.
Worked solution. Both caps permit six units; seven would need 2070 W.
What this tests. A slightly higher real-number cap need not change a whole-number maximum.
Problem 25: Minimum and upper bound at once
Question. A programme needs at least three modules, each of 220 W, alongside a 580 W base under 1900 W.
Worked solution. 580+220n≤1900 gives n≤6; with n≥3, feasible integers are 3,4,5,6.
What this tests. A correct maximum alone does not answer a question asking for all feasible values.
Problem 26: Interpret a number line
Question. An inequality result is n>2.4 and n≤7, with n representing complete stations.
Worked solution. The solutions are integers 3,4,5,6,7.
What this tests. A continuous interval is useful for algebra but the final count set must respect integrality.
Problem 27: A rate problem uses the same structure
Question. One volunteer checks a fixed set of 15 forms and then 8 forms per cycle, with at most 95 forms processed.
Worked solution. 15+8n≤95 gives n≤10 complete cycles.
What this tests. The same algebra works without watts; translate the units rather than memorising a story.
Problem 28: A final boundary defence
Question. An answer gives n≤9 after solving a capacity inequality. What should a rigorous final line contain?
Worked solution. Define n, state that nine is the maximum whole count, show substitution at nine meets the original, and show ten fails.
What this tests. The number nine alone does not demonstrate that the story, units and inequality were interpreted correctly.
Model response one: the full reasoning chain
A fictional school exhibition has 1900 W model allowance and a fixed component of 580 W. Each additional panel uses 220 W. At least three panels are required. Let n be the number of added panels. The model gives 580+220n≤1900, so 220n≤1320 and n≤6. The minimum requirement gives n≥3. Therefore n is 3,4,5 or 6. At n=6, the model demand is exactly 1900 W; at n=7 it becomes 2120 W and fails.
This answer works because it does not stop after solving the upper bound. It identifies the complete solution set, checks the boundary and describes what the variable counts. It does not turn the arithmetic into an unsupported claim that a real exhibition is safe to operate. One clear sentence of interpretation is enough.
Model response two: explain why no solution exists
A second problem allows at most 1500 W and has a fixed 600 W demand with 180 W per optional module. The activity requires at least six optional modules. The inequality 600+180n≤1500 gives n≤5, but the activity requires n≥6. Since no n satisfies both, the model has no feasible solution.
The correct response is not “use five and a half”. Modules are complete objects. Nor is it “choose six because the activity needs six”; that would exceed the given mathematical cap. Name the incompatible conditions and stop unless the question explicitly asks which assumptions might be reconsidered.
Plan your revision for three sessions
First session: translate boundary words such as at least, fewer than, not above and strictly below. Test exact boundary values. Second session: practise mixed-unit fixed-plus-variable models and sign reversal, keeping exact arithmetic until the end. Third session: combine minimum counts, caps, reserves and different time windows. Recheck answers by substitution in the original statement.
Record the cause of each mistake: language, variable, unit, algebra, integer rounding, or interpretation. The aim is to make these checks automatic before the examination. Students who need a foundation can use the PSLE learning route, then resume the Mathematics Hub and the G3 SEC Learner’s Guide.
The final answer must match the job
If the question asks for an inequality, give the model and define the variable. If it asks for all possible whole numbers, show the full set. If it asks for a maximum, give one complete whole-number count and check the next value. If it asks whether the plan is feasible, consider every condition. And if it asks what the data prove, do not claim safety or operational permission from a classroom arithmetic model.
This discipline is valuable across English, Mathematics and Science. English establishes the precise meaning of the conditions, Mathematics identifies the possible values, and Science checks whether a causal claim follows from the evidence. The next volume combines all three in a controlled, fictional exhibition audit.
Official references: SEAB 2027 school-candidate directory and K310 Mathematics syllabus. Route: G3 SEC EMS Learner’s Guide, Vol 0238, Vol 0339, Mathematics Hub, Secondary English Hub.
