Multistage transformer mathematics: the method in two paragraphs
For a two-transformer context, solve each ratio at the correct stage, carry only supported values to the next stage and reverse operations when a target is given at the far end. Percentages and power losses are separate from voltage ratios unless the problem explicitly provides a model connecting them. Finish with the stated constraints: a plausible calculation is not necessarily a feasible solution.
The reliable method is to use a labelled table, not a memorised story. Identify the starting quantity, name each operation and write its inverse for reverse problems. Check units, decide whether a number is an amount or a percentage, and delay rounding until the answer. This volume builds advanced quantitative control from basic proportional reasoning.
Quick route: ratios; forward products; reverse operations; fixed versus percentage losses; successive percentages; inequalities; integer constraints; interpretation and error checks.
Read Volume 0830 for the physics boundaries. Volume 0738 covers inverse transformer ratios more generally; here the ownership is explicitly multistage algebra and mixed feasibility conditions. For foundations, use the PSLE word-problem route.
Official examination references: the 2027 SEAB subject list and Mathematics K310 syllabus. This is independent teaching practice, not a specimen paper.
The central mathematical problem is a chain, not a single ratio
A textbook ratio may fit on one line: one quantity is five times another. A multistage transformer context is harder because the question places two changes in sequence, perhaps adds a loss in between, and then asks for a value at the beginning rather than the end. The temptation is to multiply every displayed number. Better work begins with a map. Draw boxes for the first transformer, the cable, the second transformer and the load. Write each quantity under its box. The model then becomes a sequence of ratios, percentages and constraints, all of which belong to the G3 Mathematics toolkit.
This volume is a mathematics owner, not a replacement for the physics explanation. We assume the stated physical relations are provided or known, then practise using them as algebraic data. The heart of the chapter is interpreting paired quantities, changing direction when solving backwards, and distinguishing a result that is mathematically possible from one that also meets the stated constraints. That last step is what separates routine calculation from stronger examination performance.
Keep the quantities and units in one column each
A power-transfer example may involve volts, amperes, watts, number of turns and percentages. These are not interchangeable quantities. The ratio of two voltages can be compared with a turns ratio for a single ideal transformer. A ratio of power delivered to power input can express an efficiency. A current calculation uses power divided by a corresponding voltage. Before working, write a small unit key: V for voltage, A for current, W for power, and a whole number for turns. If a value is written in kW, convert it before combining it with W.
An apparently harmless units error can change a sensible answer by a factor of one thousand. In a worked example, 0.72 kW is 720 W, not 0.72 W. The first is plausible as an invented school model’s input; the second is a completely different quantity. The question is not asking you to have engineering intuition. It is asking you to preserve the mathematical meaning of every number. Place units beside intermediate results, not only beside the final answer.
Construct a two-row ratio map
Let transformer A have 120 primary turns and 1,800 secondary turns. The output-to-input voltage factor is 1,800/120 = 15. Let B have 1,500 primary turns and 150 secondary turns. Its factor is 150/1,500 = 0.1. If the connecting cable has no stipulated voltage drop, the combined voltage factor is 15 × 0.1 = 1.5. Thus a 20 V source would correspond to a 30 V final output in the idealised voltage chain. This is a mathematical mapping between specified boundaries.
The word combined is not permission to ignore context. If the receiving-end voltage differs from the sending-end voltage, the combined factor cannot simply carry that value unchanged across the cable. First apply A’s factor; next apply the stated cable relation; finally apply B’s factor. In algebraic form, Vfinal = Vsource × rA × rCable × rB, but the rCable factor exists only if the question actually provides or supports one. Never invent it to make an answer look complete.
Fractions are often clearer than decimal guesses
The second transformer’s factor might be 3/25 rather than a terminating-looking decimal chosen from memory. Keeping a fraction can reduce premature rounding. For example, if the first stage multiplies voltage by 25/3 and the second multiplies by 3/5, their exact product is 5. The shared factor of 3 cancels in the algebra, even though the two transformers remain distinct physical stages. Write the fractions beside their names so the cancellation does not hide which stage created which factor.
This approach also avoids confusing input-to-output and output-to-input ratios. If a question gives Vp/Vs = Np/Ns, then Vs/Vp = Ns/Np. Both relations are valid; they express opposite directions. Instead of remembering a sequence of letters, ask whether the stage increases or decreases the stated voltage. Your numerical factor should agree with that conclusion. That simple directional check catches many inversions before they spread through the rest of the work.
Reverse the whole chain with inverse operations
Suppose the combined ideal voltage factor is 3/2 and the desired final load voltage is 36 V. The source must be 36 ÷ (3/2) = 24 V. Students who multiply 36 by 3/2 again are moving forward even though the problem asks them to go backwards. One reliable layout shows an arrow from source to load labelled ×3/2. A second arrow from load to source is labelled ÷3/2, or ×2/3. It is a reversible mathematical map.
If the chain includes a fixed subtraction, such as a stipulated 4 V drop in a simplified numerical model, reverse the operations in reverse order. For instance, if Vfinal = (5Vsource − 4) ÷ 2, a final value of 28 gives 56 = 5Vsource − 4, then 60 = 5Vsource, then Vsource = 12. The physical model must authorise that fixed drop, but the algebraic method is transferable to many real-world contextual problems.
Separate constant power loss from percentage efficiency
If a link receives 500 W and loses a fixed 25 W, its output is 475 W. If instead it delivers 95% of its input, its output is also 475 W for this one input. The two rules produce the same number at 500 W, but they are not the same function. At 800 W, a constant 25 W loss gives 775 W, while 95% delivery gives 760 W. A question may deliberately compare models that agree at one point and diverge at another.
The important mathematics is the difference between a fixed amount and a multiplicative percentage. Read the wording closely: “loses 25 W” and “loses 5%” have different units and different implications. Do not carry a percentage from one operating point to a new input unless the question says the percentage model continues to apply. In an evidence question, one reading cannot generally prove a universal efficiency law.
Compute a percentage with the correct baseline
Efficiency is normally expressed as useful output divided by input, multiplied by one hundred. If 450 W enters a specified device and 405 W leaves it usefully, the ratio is 405/450 = 0.90, or 90%. The loss is 45 W. A common wrong answer divides 45 by 405 and calls that the loss percentage. That quotient compares loss with output, not loss with input; it answers a different mathematical question.
Write the denominator in words before calculating. For every claim of the form “x percent of something”, identify the something. If a report says that an upgrade saved 12% of the original 500 W loss, the reduction is 60 W, and the new loss is 440 W. If the same phrase refers instead to 12% of final loss, an inverse-percentage calculation is needed. Percentage questions often test reading more than arithmetic.
Treat successive percentages as successive operations
Suppose one hypothetical stage delivers 96% of its input, and the next stage delivers 92% of what it receives. The combined delivery fraction is 0.96 × 0.92 = 0.8832, or 88.32%. It is not 96% + 92%, and it is not automatically 100% − (4% + 8%) = 88%. The second percentage acts on a smaller base. The distinction is modest numerically in this example but essential to the correct model.
An elegant solution keeps a symbolic starting value, perhaps P. The first stage gives 0.96P; the second gives 0.92(0.96P) = 0.8832P. You can then substitute any authorised starting power. This also makes reverse calculation easy: a desired final 441.6 W means an input of 441.6/0.8832 = 500 W. The method generalises to successive discounts, population changes and mixed-stage percentage models.
Use inequalities to test the design, not just solve it
A school exhibition may impose an output limit: the theoretical final voltage must lie between 18 V and 30 V inclusive. If a model gives Vfinal = 1.5x, where x is the adjustable input voltage, solve 18 ≤ 1.5x ≤ 30. Dividing through by 1.5 yields 12 ≤ x ≤ 20. The answer is an interval of inputs, not a single preferred value. Check the endpoints by substitution: 12 produces 18 and 20 produces 30.
A learner may solve the two boundary equations correctly and then forget the inequality signs. That loses the meaning of the task. A feasible interval describes every input within the specified domain, provided the model is valid over that range. If the variable represents whole turns or an integer number of equipment units, the domain may require whole-number answers. Always combine the algebraic interval with any stated real-world restriction.
Integer turns and whole-number feasibility
A winding count is a whole number. A model that gives 127.5 turns as an exact design instruction may need further treatment. If the question asks for the smallest whole-number value meeting a minimum, round upward only after considering the inequality. For example, if 8n ≥ 1,004, then n ≥ 125.5, so the smallest allowed whole number is 126. Rounding to 125 would violate the minimum even if it looked close.
This is a general G3 Mathematics habit. Do not round a capacity, item count or turns requirement using ordinary nearest-integer rules without checking what the inequality means. A minimum requirement and a maximum limit can demand different treatments. The final sentence should explain why a chosen integer satisfies the constraint rather than merely listing a rounded number.
Distinguish estimates from precise results
Suppose a calculation ends with 17.3642 V. If the question specifies three significant figures, the result is 17.4 V. If it asks for a whole-number limit, do not round first without checking feasibility. When intermediate steps are needed, keep sufficient calculator digits until the final rounding. This prevents accumulated error, especially in a chain of two fractional ratios and an efficiency factor.
The 2027 G3 Mathematics K310 syllabus asks for non-exact numerical answers to three significant figures unless a different accuracy is stated, with a separate convention for angles. Examination questions may specify other forms, such as exact fractions or a stated number of decimal places. The instruction on the particular question takes precedence. A calculation can be mathematically correct yet incomplete if the final answer ignores the requested accuracy and unit.
Explain what a missing parameter does
Consider Vfinal = x × 12 × rB, where x is source voltage and rB is an unspecified positive ratio. You cannot determine a unique Vfinal from x alone. You may write a symbolic expression, identify the missing factor or, if bounds on rB are given, give a range. A student who makes up rB = 0.1 has not solved the question; they have changed it. The distinction matters when interpreting tables and schematics whose labels are incomplete.
A high-quality response can state a conditional result: “If the connecting stage has no voltage drop and B’s factor is one tenth, the output would be 1.2x.” The word if does real work. It separates a hypothetical model from an established fact. This is mathematical precision and good English at the same time; both are part of an integrated examination skill.
Use a table to avoid swapping stage values
A four-column working table can be more reliable than a single chain of equal signs. Name the stage, input, operation and output. For example: A: 16 V, ×8, 128 V; cable: 128 V, ×0.95 under a stipulated multiplicative model, 121.6 V; B: 121.6 V, ×1/4, 30.4 V. The table makes clear that 121.6 V is B’s input, not its output.
If the cable statement instead describes power loss, the voltage row cannot simply multiply by 0.95. You would need the appropriate voltage data or a model that connects power and current. One purpose of the table is to expose this missing relationship. Good working does not merely lead to an answer. It also reveals when the givens are insufficient for the kind of answer requested.
A word problem can conceal the order
Some questions state the final load requirement first, then describe the source and both transformer turn counts in an order that is not the physical sequence. Start by reading once for the story and a second time for the relationships. Rewrite the facts by location rather than by sentence order. If a target appears at the start of the paragraph, that does not mean it is the first operation to perform.
This is the same skill that makes PSLE word problems manageable and becomes even more important at SEC. The student has to build a mathematical representation that is independent of the order in which the prose was written. Once the route is visual, the calculation is ordinary ratio and algebra. Without that representation, even a strong calculator user may combine unrelated values.
Why the K310 paper structure matters
For the 2027 G3 SEC Mathematics syllabus K310, Paper 1 lasts two hours fifteen minutes, carries 90 marks and has a 50% weighting. Paper 2 also lasts two hours fifteen minutes, carries 90 marks and contributes 50%. Paper 2 includes an extended real-world problem, while both papers require essential working. The syllabus includes ratio, direct and inverse proportion, percentages, algebra and equations, inequalities, interpretation of data and other topics. The transformer story is an illustrative context, not a separate prescribed engineering unit in Mathematics.
The best preparation is therefore broader than memorising these exact numbers. Practise translating a situation into a relation, showing a logical sequence, testing the answer and interpreting it in the original context. A real-world problem may combine topics, and its wording may introduce assumptions that differ from the ones used here. Your working should make it easy for a marker to see each justified step.
Twenty-eight short examinations of the method
Write each answer before viewing the correction. Every numerical example is a teaching model. The goal is to learn when to multiply, when to invert, and when a conclusion requires more information, not to memorise a particular transformer diagram.
Practice 01 — A multiplies voltage by 12; B multiplies by 1/4. What is the total factor?
Answer and working. 12 × 1/4 = 3. The combined ideal voltage factor is three, if nothing changes the voltage between the stages.
Diagnostic. Multiplying by four in B would reverse a step-down ratio.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 02 — Source is 9 V, factor A = 8 and factor B = 1/3. Find final voltage.
Answer and working. 9 × 8 × 1/3 = 24 V under the stated ideal connection.
Diagnostic. Rounding one third to 0.3 too early gives a needless error.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 03 — Final output is 45 V with total factor 5/2. Find input.
Answer and working. 45 × 2/5 = 18 V. Reverse the mapping with the reciprocal.
Diagnostic. Do not multiply by 5/2 again; that walks in the wrong direction.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 04 — A has 210 primary turns and 1,050 secondary turns. Find its factor.
Answer and working. 1,050/210 = 5. This is the output-to-input voltage factor.
Diagnostic. The number of visible loops in a schematic is irrelevant if turns counts are labelled.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 05 — B has 1,200 primary turns and 240 secondary turns. Find its factor.
Answer and working. 240/1,200 = 0.2, a step-down factor of one fifth.
Diagnostic. Always form the output-to-input factor when working forward.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 06 — A factor 5 and B factor 1/5. What happens to the final voltage?
Answer and working. It returns to the starting numerical voltage in an ideal voltage chain.
Diagnostic. The ratio cancellation does not prove zero power loss in a cable not modelled as ideal.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 07 — The first stage outputs 200 V; a stipulated model removes a fixed 10 V before B, whose factor is 1/2. Final?
Answer and working. (200 − 10)/2 = 95 V in that explicitly stated model.
Diagnostic. Apply the fixed loss before B, not after the multiplication.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 08 — The final is 95 V from the preceding model. Recover first-stage output.
Answer and working. Reverse: 95 × 2 = 190, then add 10 = 200 V.
Diagnostic. The order of inverse operations is the opposite of the forward sequence.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 09 — 500 W input, 94% delivery. Find output.
Answer and working. 0.94 × 500 = 470 W; loss is 30 W.
Diagnostic. A loss of six percent is not a subtraction of six watts.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 10 — 470 W is 94% of input. Recover input.
Answer and working. 470/0.94 = 500 W.
Diagnostic. Reverse-percentage questions divide by the decimal multiplier.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 11 — 400 W enters a cable; 376 W leaves. Find delivery percentage.
Answer and working. 376/400 × 100 = 94%.
Diagnostic. Always identify input as the denominator of a delivery percentage.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 12 — A fixed loss of 24 W at 400 W input: output at 600 W input under same fixed-loss model?
Answer and working. 600 − 24 = 576 W.
Diagnostic. Do not reuse the original 94% delivery rate without the question authorising that percentage model.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 13 — A 94% delivery rule at 600 W: output?
Answer and working. 0.94 × 600 = 564 W.
Diagnostic. The fixed-loss and fixed-percentage models differ away from their shared reference point.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 14 — Stage 1 delivers 95%; stage 2 delivers 90%. Combined?
Answer and working. 0.95 × 0.90 = 0.855, or 85.5%.
Diagnostic. Do not simply subtract 5% and 10% from 100%; the second rate acts on the smaller output.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 15 — Final is 342 W after factors 0.95 and 0.90. Find initial power.
Answer and working. 342/0.855 = 400 W.
Diagnostic. Write the combined multiplier before dividing, and retain sufficient precision.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 16 — Solve 20 ≤ 2.5x ≤ 45.
Answer and working. 8 ≤ x ≤ 18.
Diagnostic. Keep both boundary inequalities and interpret the result as a feasible interval.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 17 — Solve 16x ≥ 1,005 for the minimum positive integer x.
Answer and working. x ≥ 62.8125, so the smallest integer is 63.
Diagnostic. Do not round down to 62, which would violate the minimum.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 18 — Output must not exceed 100 W and is 8n W. Largest integer n?
Answer and working. 8n ≤ 100 gives n ≤ 12.5, so the largest whole number is 12.
Diagnostic. A maximum capacity condition demands a different final whole-number choice from a minimum requirement.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 19 — Convert 1.25 kW to watts.
Answer and working. 1.25 × 1,000 = 1,250 W.
Diagnostic. Do not perform a power subtraction across kilowatts and watts without conversion.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 20 — Give 17.3642 V to three significant figures.
Answer and working. 17.4 V.
Diagnostic. Do not round the intermediate figures during a multistage calculation.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 21 — A meter is recorded to 0.1 V. Does 10.0 V mean mathematically exact ten?
Answer and working. No. It is a reported measurement with a stated resolution, not an infinitely precise physical quantity.
Diagnostic. Keep calculator precision separate from measurement precision.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 22 — An unknown cable factor r is given only as 0.92 ≤ r ≤ 0.98. A outputs 100 V, B factor 1/4. Range?
Answer and working. Final lies from 23 V to 24.5 V if all factors are non-negative and the given multiplicative model applies.
Diagnostic. A range is more honest than assigning r a convenient single value.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 23 — A factor 8, cable voltage factor unspecified, B factor 1/5. Is final voltage unique?
Answer and working. No. It is 8r/5 times source voltage, where r is the missing cable factor.
Diagnostic. A symbolic result is a valid solution when data are incomplete.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 24 — A report compares a 20 W loss to 400 W input. Loss percentage?
Answer and working. 20/400 × 100 = 5%.
Diagnostic. Do not divide by the 380 W useful output if the phrase means percentage of input.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 25 — A stage’s output falls from 480 W to 432 W. Percentage decrease?
Answer and working. 48/480 × 100 = 10%.
Diagnostic. The original 480 W is the baseline for the change.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 26 — A model predicts output 30 V but maximum permitted is 28 V. What should the conclusion say?
Answer and working. The design is mathematically calculable but infeasible under the stated 28 V ceiling.
Diagnostic. A correct computation is not automatically an acceptable practical decision.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 27 — An exam answer says final 0.024 kV when the target was in volts. Convert.
Answer and working. 0.024 kV = 24 V.
Diagnostic. A final comparison must be made in common units.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
Practice 28 — A line gives 12.5% loss on 640 W input. Find power delivered.
Answer and working. Loss = 80 W; useful output = 560 W.
Diagnostic. The question asks for delivered power, not merely the amount lost.
The transfer question is whether the same decision would still be valid if the examiner reversed the order of information, changed the unit or altered a constraint. To check, label the input and output, name the operation, then state which part of the prompt makes the operation permissible. This simple discipline helps a student turn a correct isolated calculation into a correct complete response.
One extended real-world problem to integrate the chapter
A school technology team presents a fictional two-stage supply model. The input voltage x is multiplied by eight in transformer A, by a stated cable voltage factor 0.95, then by one fourth in transformer B. The final voltage must be at least 19 V but no more than 28.5 V. Write Vfinal = x × 8 × 0.95 × 1/4 = 1.9x. Solve 19 ≤ 1.9x ≤ 28.5 to find 10 ≤ x ≤ 15. The calculation is valid only because the cable’s voltage factor was explicitly supplied as part of the hypothetical mathematical model; it must not be inferred merely from a power-loss percentage.
The team next gives a separate power record: 500 W entering the cable, 465 W leaving it, and transformer B delivering 90% of its input power. The cable delivery fraction is 465/500 = 93%. B delivers 0.90 × 465 = 418.5 W. The overall useful delivery from the 500 W cable-input baseline to the final load is 418.5/500 = 83.7%. These figures describe a specific snapshot. They do not establish a universal performance rule at every operating point. In the examination, a stronger learner not only calculates 83.7%; they say exactly what the denominator represents.
Finally, a supervisor asks whether setting x = 16 V would be acceptable. It would produce 30.4 V under the given model, exceeding the 28.5 V ceiling. Although all the algebra works, the decision fails the feasibility test. You should not quietly lower the answer to 28.5 V or round 30.4 into an allowed value. Mathematical feasibility is a conclusion drawn from the actual inequality, not a preference for convenient numbers.
How to revise after making a wrong turn
When the error is an inverted turns ratio, draw a small arrow from primary to secondary. When the error is a percentage denominator, write ‘of what?’ above the fraction. When the error is rounding, retain calculator precision until the last line. When the error is a wrong inequality endpoint, substitute a value immediately outside the proposed range and test it. These small diagnostics are more effective than repeating fifty almost identical calculations without examining why the first one went wrong.
For further independent practice, revisit Volume 0810 on energy and costs and the foundational Mathematics route. The next volume changes discipline: it asks how language can distort or faithfully preserve the same multi-step information in a comprehension passage. That transition matters because a learner can handle the arithmetic correctly and still misread the claim being assessed.
