Direct and inverse proportion questions are easy only after the relationship is classified correctly. This workshop trains the learner to identify the invariant first, then scale, calculate, check units and reject false proportionality.
It continues from Vol 0118 and the G3 SEC learner route. For 2027 school candidates, use the official K310 Mathematics syllabus and SEAB G3 directory for current requirements. All contexts below are original teaching examples.
Proportion begins with a relationship, not a formula
A proportional question is asking how one quantity changes when another changes. Before choosing multiplication, division or a constant, identify the two quantities and decide what stays fixed in the relationship. This prevents the common error of applying a familiar ratio to a context where the proportional rule has changed.
Direct proportion means the ratio stays constant
If y is directly proportional to x, then y/x is constant and y = kx for some constant k. Doubling x doubles y, provided the same proportional relationship still applies. The learner should be able to explain what k means in context rather than treating it as a symbol introduced only to solve an equation.
Inverse proportion means the product stays constant
If y is inversely proportional to x, then xy is constant and y = k/x. Doubling x halves y when the stated model remains valid. The key question is not whether one number increases while the other decreases, but whether their product is the quantity that remains fixed.
Not every increase is direct proportion
A quantity can increase with another without being directly proportional. A fixed fee plus a per-unit charge gives a straight-line relationship with a non-zero intercept, not direct proportion. If the line does not pass through the origin, the ratio y/x is not constant across all x.
Not every decrease is inverse proportion
A quantity can decrease as another increases without satisfying xy = constant. For example, a linear decrease such as y = 100 – 2x is not inverse proportion. The learner should test the defining invariant rather than classify from direction alone.
Worked case 1: constant unit rate
A tap delivers 18 litres in 6 minutes at a constant rate. The rate is 3 litres per minute, so volume is directly proportional to time under this model. At 10 minutes, the volume is 30 litres. The reasoning depends on the constant rate, not on memorising cross-multiplication.
Worked case 2: direct proportion from a table
Suppose a table shows x values 2, 5 and 8 with y values 7, 17.5 and 28. The ratio y/x is 3.5 each time, so the data are consistent with direct proportion. If one row produced 29 instead of 28, that row would not fit the exact proportional model.
Worked case 3: inverse proportion from a table
Suppose x values are 2, 4 and 8 while y values are 24, 12 and 6. The products xy are all 48, so y is inversely proportional to x. The ratio y/x is not constant, which is exactly why a direct-proportion method would be wrong.
Choose the constant deliberately
In y = kx, k has units of y per x. In xy = k, k has the combined unit of x times y. Units help the learner identify whether a proposed constant makes sense. A constant with the wrong unit is evidence that the model may have been formed incorrectly.
Use the origin as a diagnostic for direct proportion
A graph of y against x for direct proportion is a straight line through the origin. A straight line with a positive intercept can still represent a linear model, but it is not direct proportion. This distinction matters when the task asks for proportional reasoning rather than only line interpretation.
Inverse-proportion graphs are curved
A graph of y = k/x does not form a straight line in ordinary x-y coordinates. The curve falls rapidly for small positive x and more slowly as x increases. The learner should not expect a straight line merely because the relationship has a simple formula.
The first exam decision is classification
Before calculating, ask whether the question states direct proportion, inverse proportion, a constant rate, a fixed total, or another relationship. Classification determines the invariant. Once the invariant is correct, the arithmetic is usually simpler.
Worked case 4: recipe scaling
A recipe uses 250 g of flour for 10 portions and the ingredients scale directly with portions. For 16 portions, flour = 250 × 16/10 = 400 g. The direct-proportion assumption comes from the problem statement; real recipes can have practical constraints, but the examination model controls the calculation.
Worked case 5: fixed work and number of workers
A simplified task requires 72 worker-hours and all workers operate at the same constant rate with no interference. If n workers share the work, time t satisfies nt = 72. This is inverse proportion under the model. Nine workers take 8 hours; twelve workers take 6 hours.
State the assumption in inverse-work models
Real teams may not scale perfectly because coordination and task structure matter. In a Mathematics question that specifies equal rates and fixed work, use the model given. If asked to evaluate, mention that the inverse relationship depends on those assumptions.
Worked case 6: speed and time for fixed distance
For a fixed journey distance, speed and time are inversely proportional when speed is treated as constant over the trip. If 60 km/h takes 3 hours for the same distance, 90 km/h takes 2 hours. The product speed × time remains 180 km.
Do not confuse average speed with fixed-speed models
A multi-stage journey with stops or changing speeds may not fit a simple inverse-proportion rule. First determine whether the model actually treats the whole journey as a fixed distance travelled at one constant speed. Context decides whether the invariant applies.
Worked case 7: intensity over area
Suppose a fixed quantity is spread uniformly over area A, giving amount per unit area I. Then IA is constant and I is inversely proportional to A. Doubling the area halves the amount per unit area. The model is about a fixed total distributed across a changing area.
Percentage change under direct proportion
If y is directly proportional to x, a 12% increase in x produces a 12% increase in y, provided k remains constant. This follows because both are multiplied by the same factor 1.12. The learner can often predict the percentage result before using actual numbers.
Percentage change under inverse proportion
If y is inversely proportional to x, increasing x by 25% multiplies x by 1.25 and y by 1/1.25 = 0.8. Therefore y decreases by 20%, not 25%. Inverse percentage changes are not symmetrical.
Worked case 8: a 20% decrease in x
If y is inversely proportional to x and x falls by 20%, x is multiplied by 0.8. Then y is multiplied by 1/0.8 = 1.25, so y increases by 25%. This is a useful check against the false idea that inverse proportion simply reverses the same percentage.
Use scale factors instead of repeated unitary steps when useful
If the relationship is direct, multiply by the same scale factor. If inverse, divide by that factor or multiply by its reciprocal. This gives a compact way to reason about large or non-integer changes while preserving the structure.
Cross-multiplication is a consequence, not the concept
For direct proportion, y1/x1 = y2/x2 can be rearranged by cross-multiplication. For inverse proportion, x1y1 = x2y2. Mixing these two templates is a common error. The learner should identify which invariant applies before moving symbols.
Worked case 9: comparing two direct-proportion models
Model A has y = 3x and Model B has y = 4.5x. Both are directly proportional, but B has the larger constant of proportionality. For the same x, B gives 1.5 times A. This comparison is about k, not merely about one sampled value.
Worked case 10: comparing two inverse models
Model C has xy = 60 and Model D has xy = 90. For the same positive x, D gives a larger y because its constant product is larger. The curves share the same general shape but represent different fixed totals.
Units can distinguish direct and inverse models
If y is dollars and x is kilograms, direct proportionality may have k in dollars per kilogram. If x is workers and y is hours for fixed work, k may be worker-hours. Reading the unit of k can reveal whether the model’s meaning is coherent.
Zero values need interpretation
In y = kx, x = 0 gives y = 0. In y = k/x, x = 0 is not allowed. This domain difference matters. An inverse-proportion model cannot be extended through zero merely because the graph appears to approach an axis.
Negative values need context
Algebraically, inverse and direct proportion can include negative values where the context permits. Many physical quantities such as time, number of workers or mass are restricted to non-negative values. Respect the domain of the model rather than applying pure algebra blindly.
Worked case 11: finding k from one pair
If y is directly proportional to x and y = 27 when x = 6, then k = 27/6 = 4.5. Therefore y = 4.5x. If x = 14, y = 63. The first pair identifies the whole proportional model under the stated assumption.
Worked case 12: inverse k from one pair
If y is inversely proportional to x and y = 15 when x = 4, then k = xy = 60. Therefore y = 60/x. If x = 12, y = 5. The constant product is the easiest way to rebuild the model.
Recognise proportionality from equations
y = 7x is direct proportion. y = 7/x is inverse proportion. y = 7x + 3 is linear but not direct proportion. y = 7/x + 3 is neither the standard inverse-proportion form nor direct proportion.
Recognise proportionality from language
Phrases such as “at a constant rate per unit” can signal direct proportion. Phrases such as “fixed total shared equally” can signal inverse proportion. Do not depend on trigger words alone; translate the meaning into an invariant.
Recognise when a fixed term breaks direct proportion
A taxi fare with a fixed booking charge plus a per-kilometre rate is not directly proportional to distance because the fixed term remains when distance is zero. A learner who uses a direct ratio will understate or overstate the total depending on distance.
Worked case 13: a not-direct example
A service costs $12 fixed plus $3 per hour. At 2 hours the cost is $18; at 4 hours it is $24. Doubling hours does not double cost. The ratio cost/time changes, showing that direct proportion is not valid.
Worked case 14: a not-inverse example
A tank contains 100 litres and drains at 5 litres per minute. Remaining volume after t minutes is 100 – 5t, a linear decrease. The product t × volume is not constant. Decreasing with time does not imply inverse proportion.
Graph choice can reveal the invariant
For direct proportion, plotting y against x gives a line through the origin. For inverse proportion, plotting y against 1/x can produce a straight line through the origin if the model is exact. This transformation can help test the relationship when appropriate.
Do not over-transform in an examination
Use transformed graphs only when the syllabus context or question makes the method relevant. A direct calculation may be faster. The purpose of representation is to clarify the model, not to display every technique known.
Proportional reasoning and similarity
Similar figures use constant scale factors. Corresponding lengths are directly proportional. Areas scale with the square of the linear factor and volumes with the cube. This is not a contradiction; the proportional relationship changes because the measured quantity changes dimension.
Worked case 15: length to area scaling
If all lengths of a similar figure are multiplied by 1.4, corresponding lengths increase by 40%, while area is multiplied by 1.4² = 1.96, an increase of 96%. The area is not directly proportional to the linear scale factor.
Proportional reasoning and density
Density = mass/volume. If density is fixed for the same material, mass is directly proportional to volume. If mass is fixed while volume changes in a model, density is inversely proportional to volume. Which pair is proportional depends on what is held constant.
Proportional reasoning and frequency
Frequency and period can be inversely related in periodic contexts because one is cycles per unit time and the other is time per cycle. The product can be one when consistent units are used. This is a structural example of reciprocal quantities.
Do not import a proportional rule from a nearby topic
Seeing a ratio in a formula does not mean every pair of variables is directly or inversely proportional when other variables also change. Hold the other relevant quantities fixed before making a proportional statement.
Independent task A
y is directly proportional to x. When x = 8, y = 30. Find k, form the equation and find y when x = 14. Then state what k means in terms of y per unit x.
Independent task B
y is inversely proportional to x. When x = 5, y = 18. Find y when x = 12. Then predict whether y should be above or below 18 before calculating.
Independent task C
A fixed 240 km journey is modelled at constant speeds. Compare the travel time at 80 km/h and 96 km/h. Explain the proportional relationship and the assumption used.
Independent task D
A recipe uses 360 g of rice for 6 portions. Assuming direct scaling, find the mass for 15 portions. Then state why a fixed setup fee in a catering bill would make total cost not directly proportional to portions.
Independent task E
If y is inversely proportional to x and x rises by 60%, find the percentage change in y. Use scale factors rather than choosing the same percentage in the opposite direction.
Independent task F
Decide whether each equation shows direct proportion, inverse proportion or neither: y = 5x, y = 5/x, y = 5x + 2, y = 5/(x + 2). Explain one decision using an invariant.
Worked feedback A
k = 30/8 = 3.75, so y = 3.75x. At x = 14, y = 52.5. The constant 3.75 is the amount of y per unit x under this proportional model.
Worked feedback B
k = 5 × 18 = 90, so y = 90/12 = 7.5. It should be below 18 because x increased and the product stays fixed.
Worked feedback C
Time = distance/speed. At 80 km/h, time = 3 hours. At 96 km/h, time = 2.5 hours. For fixed distance and constant speed, speed × time remains 240 km, so speed and time are inversely proportional.
Worked feedback D
360/6 = 60 g per portion, so 15 portions need 900 g. A total bill with a fixed setup charge has a non-zero intercept, so total cost per portion is not constant across all quantities.
Worked feedback E
x is multiplied by 1.6, so y is multiplied by 1/1.6 = 0.625. Therefore y falls by 37.5%. The inverse percentage change is not 60%.
Worked feedback F
y = 5x is direct; y = 5/x is inverse; y = 5x + 2 is neither direct nor inverse; y = 5/(x + 2) is not inverse proportion between y and x in the standard form because xy is not constant.
How to review a proportion solution
Check the relationship before the arithmetic. Name the invariant, verify units, respect the domain and test whether a simple scale change behaves as expected. If doubling x does not produce the predicted response, inspect the model.
Repair route
Start with tables where the constant ratio or constant product is easy to see. Ask the learner to classify before solving. Keep numbers simple until direct and inverse invariants are stable.
Stabilisation route
Mix proportional and non-proportional examples. Include linear models with intercepts and decreasing linear models so the learner cannot classify from shape direction or keywords alone.
Extension route
Use percentage changes, similarity, compound units and contexts with one hidden fixed term. Ask the learner to predict direction and relative size before calculating.
What progress should look like
Progress is visible when the learner explains why a model is direct or inverse, uses the correct invariant, keeps units meaningful, rejects false proportionality and interprets percentage changes correctly.
Final operating rule
Before scaling, state what remains constant. If the ratio y/x stays fixed, use direct proportion. If the product xy stays fixed, use inverse proportion. If neither invariant holds, do not force a proportional template onto the problem.
K310 proportion checklist
- name the two quantities
- state what stays constant
- test constant ratio for direct proportion
- test constant product for inverse proportion
- check units and domain
- predict direction before calculating
- reject fixed-intercept and other non-proportional models
- return the result to context
Advanced proportion and model-audit laboratory
Advanced case: direct proportion hidden inside a compound unit
Suppose fuel used is directly proportional to distance when the vehicle’s consumption rate is fixed. The constant may be litres per kilometre. If the question instead gives kilometres per litre, that is the reciprocal rate. Translate the unit before choosing the multiplication or division.
Advanced case: inverse proportion hidden inside equal sharing
A fixed prize fund divided equally among n winners gives share s = k/n. Increasing the number of winners lowers each share. The model is inverse because the total fund stays fixed, not because division appears in the story.
Advanced case: direct proportion can fail after a threshold
A delivery service might charge a constant amount per kilometre only after a fixed minimum fee or within one distance band. Once a fixed term or tier appears, the whole model is not globally direct proportion. Use the rule that applies to the stated interval.
Advanced case: inverse proportion can fail under capacity limits
If workers share fixed work, inverse proportion assumes added workers can contribute at the same rate. Once the number of workers exceeds available workstations, the model can fail. A mathematical model is valid over the conditions that support its invariant.
Advanced case: transformed variables
If y is proportional to x squared, y/x² is constant. This is not direct proportion between y and x, even though it is direct proportion between y and x². Identify which transformed quantity makes the ratio constant.
Advanced case: reciprocal linearisation
If y is inversely proportional to x, plotting y against 1/x should give a straight line through the origin under the exact model. This can be a useful diagnostic in data questions because it tests the invariant through a different representation.
Advanced case: proportionality with powers
If y = kx³, tripling x multiplies y by 27. The learner should not call this simple direct proportion between y and x. The proportional relationship is between y and x³. Powers change sensitivity dramatically.
Advanced case: percentage sensitivity
In y = kx², a 10% rise in x multiplies y by 1.1² = 1.21, so y rises by 21%. In y = k/x², the same 10% rise multiplies y by 1/1.21 ≈ 0.8264, a decrease of about 17.36%. Predict direction before calculating.
Advanced case: dimensional interpretation
For similar solids, volume is proportional to the cube of a linear dimension when shape remains similar. A learner who scales volume directly with length has ignored dimension. State the transformed quantity that preserves proportionality.
Advanced case: density revisited
For the same material at fixed density, mass is directly proportional to volume. But if mass is fixed and volume changes, density is inversely proportional to volume. The proportional statement must name what is held constant; otherwise the same formula can support different relationships.
Advanced case: pressure-volume model
In a simplified fixed-temperature gas model, pressure can be inversely proportional to volume for a fixed amount of gas. The learner should use this only when the problem establishes the relevant conditions. A formula remembered without conditions is not a complete model.
Advanced case: graph slope and proportional constant
For y = kx, the gradient of a graph of y against x is k. Therefore a steeper line through the origin represents a larger constant of proportionality. If the line has a non-zero intercept, the gradient may still be meaningful, but the relation is not direct proportion.
Advanced case: compare constants, not isolated values
When two proportional models use different k values, compare the constants to understand their structure. One sampled point can mislead if x values differ. Normalising by x or using the constant gives the fair comparison.
Advanced case: error from using the wrong pair
In a table, one value may be an outlier or measurement error. If the task states an exact proportional model, any valid pair should give the same k. If calculated k values differ, inspect the data rather than averaging automatically.
Advanced case: unit conversion before ratio
Suppose one row uses centimetres and another metres. Convert to consistent units before testing proportionality. A ratio can appear to change simply because the unit scale changed.
Advanced case: zero-intercept test with data
If an empirical line appears roughly linear but does not pass through the origin, it may represent y = a + bx rather than y = kx. The learner should distinguish approximate trend from exact proportional structure.
Advanced case: proportionality and feasibility
A direct model may predict 2.4 buses, 0.7 people or a negative time when extended carelessly. Interpretation must respect discrete and physical constraints. Mathematical scaling is not the final step when the output has a practical domain.
Advanced case: rounding after proportional calculation
Keep the exact or sufficiently precise result until the context requires rounding. If a quantity counts whole units, apply the decision rule after the proportional calculation. Premature rounding can shift a threshold or capacity decision.
Advanced case: inverse percentage trap
If y is inversely proportional to x, a 50% increase in x multiplies y by 2/3, which is a 33⅓% decrease. Returning x to its original value requires a 50% increase from the new y? No—the reverse percentage must be recalculated from the new base.
Advanced case: two-step scaling
If x first increases by 20% and then decreases by 20%, it does not return to the original value. In a proportional model, y follows the same multiplicative factors. Multiplicative scale factors should be composed, not added.
Advanced case: checking with one simple value
After forming a proportional model, test it with a convenient value. If doubling x in a claimed direct model does not double y, or doubling x in an inverse model does not halve y, inspect the equation before doing more arithmetic.
Workshop drill: classify without calculating
Take ten short contexts and label them direct, inverse or neither. Give one sentence naming the invariant or explaining why no invariant applies. This builds model selection separately from arithmetic fluency.
Workshop drill: explain k in words
For each direct or inverse model, state the meaning and units of k. If the learner cannot explain k, the symbolic model may not yet be attached to the context.
Workshop drill: convert representation
Move from words to equation, equation to table and table to graph. A proportional idea is stronger when the learner can recognise it across representations rather than only in one format.
Workshop drill: create a counterexample
For a claimed proportional relationship, invent a value that would violate the invariant. Counterexamples help distinguish a true proportional rule from a pattern observed at only one pair of values.
Workshop drill: compare percentage effects
Change x by 10%, 20% and 50% under direct and inverse models. Record the resulting percentage change in y. This makes the asymmetry of inverse percentage changes visible.
Workshop drill: detect hidden fixed terms
Given several cost formulas, identify which include a fixed charge. Explain why those cannot be directly proportional to quantity over the full domain.
Workshop drill: inspect domain
For each model, state whether x can be zero, negative, fractional or only integer. Proportional algebra should not be separated from the quantities it represents.
Workshop drill: use units as a model test
Before solving, predict the unit of k and the final answer. If the algebra produces a different unit, pause. Unit analysis can expose a wrong ratio or reciprocal.
Proportion under examination pressure
When time is short, the safest sequence is classify, state invariant, calculate, check direction, interpret. Skipping the classification step saves seconds only when the learner was already correct; when wrong, it can cost the entire solution.
Final proportion standard
A strong K310 response does not merely obtain a scaled value. It shows that the learner knows why the scaling rule applies, what remains constant, what units and domain make sense, and how to reject a proportional template when the context does not support it.
