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When one quantity increases and another decreases, they are not automatically inversely proportional. Inverse proportion has a stronger condition: multiplying one quantity by a factor divides the other by that same factor. Their product stays constant. That condition is what justifies the calculation.
For example, if x and y satisfy xy = 60, changing x from 3 to 6 changes y from 20 to 10. Doubling x halves y. But y = 30 − x also decreases as x increases, and its product with x does not generally remain constant. A downward trend alone does not identify the model.
The official 2027 G2 Mathematics syllabus includes direct and inverse proportion. This explainer develops the basic inverse relationship, its assumptions and its checks. Classroom examples are fictional idealised models; the student’s school determines topic sequence and the level of extension required.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–28
For simple inverse proportion, write y = k/x or xy = k, with x nonzero. In the positive-quantity situations used through most of this guide, x and y are positive and k is positive. Find k from one known pair, then use the same product to calculate another pair.
The comparison form is x₁y₁ = x₂y₂. If x changes by factor r, y changes by factor 1/r. An increase of 25% in x means multiplication by 1.25, so y is divided by 1.25 and becomes 80% of its original value. It does not automatically decrease by 25%.
Before applying the rule, check what remains fixed. A fixed distance gives speed × time = distance when speed is constant over the journey. A fixed shared total gives number of equal shares × amount per share = total. If the total, rate or conditions change, the same inverse model may no longer apply.
An invariant is something that remains unchanged while other quantities vary. For inverse proportion, the invariant is the product xy. This is the key information to look for in a question, whether it is stated algebraically or described through a fixed total.
Suppose a 48-metre ribbon is cut into equal pieces. If there are n pieces and each has length l metres, then nl = 48, assuming the whole ribbon is used and cutting losses are ignored. Six pieces have length eight metres each; eight pieces have length six metres each.
The two quantities move in opposite directions because the same total is being partitioned more or less finely. Their relationship is not established merely by observing that more pieces look shorter. It is established by the fixed total and the equal-piece condition.
Write the invariant before inserting numbers. The sentence number of pieces × length per piece = total length shows why multiplication is appropriate and what must stay fixed. Once the model is visible, the arithmetic becomes a consequence rather than a guessed operation.
Consider y = 20 − x for positive x less than 20. When x changes from 4 to 8, y changes from 16 to 12. One increases and the other decreases, but doubling x does not halve y. The products are 64 and 96, so they are not equal.
Now consider y = 64/x. At x = 4, y = 16. At x = 8, y = 8. This time the product remains 64 and doubling x does halve y. Both relationships decrease over the chosen positive domain, but only the second is simple inverse proportion.
One shared data point cannot distinguish these models. Both pass through x = 4, y = 16. You need the relationship statement, the model assumptions or additional relevant information to determine which law is intended. A convenient first pair is not a proof of the whole relationship.
This is a useful habit beyond proportion questions. A pattern’s direction tells you something, but not everything. Ask what quantitative rule the pattern obeys. Inverse proportion describes a specific multiplicative rule, not just the everyday phrase more means less.
In y = k/x, k is the constant of proportionality. It is fixed within the particular model. The variables x and y may change, but k does not change from row to row. Multiplying both sides by x gives xy = k, provided x is not zero.
The letter k does not always stand for the same number in different questions. One model may have k = 36, another k = 120. Find it from the stated pair rather than importing a remembered value from an earlier exercise.
For example, if y = 14 when x = 5, then k = 5 × 14 = 70. The relationship is y = 70/x. When x = 10, y = 7; when x = 2, y = 35. Each pair has product 70, which is the shared structural check.
Keep variable definitions visible. If x is speed and y is time, k has a distance interpretation under the fixed-distance model. If x is number of recipients and y is amount per recipient, k represents the fixed distributed total. The same algebra can describe different quantities, so the final units still matter.
The expression k/x is undefined at x = 0. For a positive nonzero total k, there is no finite value of y that makes 0 × y = k. This restriction belongs to the relationship; it is not an optional calculator warning.
In a fixed-distance journey, zero speed does not produce a finite completion time. In an equal-sharing problem, dividing a fixed positive quantity among zero recipients is not a meaningful allocation. The algebraic restriction and the contextual restriction point to the same boundary.
Some mathematical discussions allow negative x and y with a nonzero domain. If k is positive, their signs must match. That extension is valid algebraically, but it does not make a negative number of recipients meaningful in an ordinary sharing problem. Context can restrict a mathematical domain further.
For this guide’s practical examples, use positive inputs unless stated otherwise. Do not substitute zero simply because a table begins at zero in many other topics. Inverse proportion has a different structure from a straight-line model with an intercept, and its domain must be read accordingly.
A fictional vehicle travels 180 km at a constant speed. At 60 km/h it takes three hours. How long would the same journey take at a constant speed of 75 km/h, ignoring stops and assuming the same route length?
Let v be speed in kilometres per hour and t be time in hours. The fixed-distance relationship is vt = 180. Therefore t = 180/75 = 12/5 = 2.4 hours. This is two hours and 24 minutes because 0.4 hour is 24 minutes.
The comparison route gives the same result. Speed changes by factor 75/60 = 5/4, so time changes by reciprocal factor 4/5. Three hours multiplied by 4/5 is 12/5 hours. The speed increase is 25%, while the time decrease is 20%.
Check using the original invariant: 75 × 2.4 = 180 km. Also check direction: the faster constant speed should require less time for the same distance. The direction check alone cannot prove the number, but the product check confirms that the calculated pair fits the stated model.
The distinction is about the quantity held constant. Direct proportion preserves a ratio y/x. Simple inverse proportion preserves a product xy. A relationship can preserve neither, even if its values form a neat pattern or move consistently in one direction.
| Relationship | Example rule | What stays constant |
|---|---|---|
| Direct proportion | y = 4x | y/x = 4 |
| Inverse proportion | y = 36/x | xy = 36 |
| Neither | y = 20 − x | Neither ratio nor product in general |
| Neither | y = 5 + 36/x | A shifted reciprocal rule, not simple inverse proportion |
The last row is especially useful. Its graph decreases for positive x, and it contains a reciprocal term. However, xy = 5x + 36, which changes with x. It is not correct to call y inversely proportional to x merely because 36/x appears somewhere in its formula.
Test a relationship using its stated definition, not visual resemblance. If a question says y is inversely proportional to x, use y = k/x. If it says y − 5 is inversely proportional to x, the constant-product statement is x(y − 5) = k. Those are different models with different predictions.
Suppose a table lists x values 2, 3, 6 and 9, with corresponding y values 18, 12, 6 and 4. The products are 36, 36, 36 and 36. These observations are consistent with y = 36/x at the listed inputs.
Do not compare differences instead. The x increments are one, three and three, while the y decreases are six, six and two. Constant additive changes are not the defining feature of inverse proportion. The important comparison is multiplicative, through products or reciprocal factors.
A table can refute an exact inverse relationship when a stated exact pair has a different product. If the last y value were five, its product with nine would be 45, so the listed exact table would not fit one constant-product rule.
However, several matching products do not logically prove that every unlisted input must obey that law. In a school question, the model may be given or inferred under the exercise’s stated conditions. In a real investigation, a table offers evidence to evaluate, not an unrestricted guarantee about every possible input.
If y is inversely proportional to x and y = 18 when x = 7, write xy = k. Substitution gives k = 126. When x = 12, the new y is 126/12 = 10.5. Keep the original pair together so that the constant is formed from corresponding values.
A common mistake is to calculate 18/7 and use that as k. That would be the constant ratio for direct proportion, not the constant product for inverse proportion. The error happens before the second calculation, so repeating division practice will not fix the model choice.
Another mistake uses one old value and one new value to find the constant: 18 × 12. Those values do not form a corresponding pair in the supplied data. Label the two pairs as old and new if the question contains several numbers.
The constant-first method takes a little space but keeps the logic auditable. It is particularly useful when the unknown is the input rather than the output. Once k is known, either variable can be found by dividing that same constant by the other.
The equation x₁y₁ = x₂y₂ does not introduce a new rule. Both sides equal the same k. It is simply a shorter way to connect two corresponding pairs without writing the constant separately.
For example, if x changes from 8 to 20 while y starts at 15, write 8 × 15 = 20 × y₂. Therefore y₂ = 120/20 = 6. The factor route also works: x is multiplied by 20/8 = 5/2, so y is multiplied by 2/5.
Avoid arranging numbers by visual position alone. Write the product of the old pair equal to the product of the new pair. If you instead set 8/15 = 20/y₂, you are imposing a constant ratio and will obtain an inappropriate direct-proportion result.
Use the form you can explain. A concise correct comparison equation is better than a longer unexplained manipulation, but speed should follow understanding. If you forget which numbers go where, return to xy = k and label the quantities before attempting a shortcut.
If x increases by 20%, its multiplier is 1.2 = 6/5. Under simple inverse proportion, y is divided by 6/5, so its multiplier is 5/6. The percentage decrease in y is 1 − 5/6 = 1/6, or 16 2/3%.
The percentages are not symmetric because each change is measured against its own original value. Multiplicative inverses restore the product, but equal positive and negative percentage changes do not generally cancel. Increasing one quantity by 20% and decreasing the other by 20% multiplies the product by 1.2 × 0.8 = 0.96.
For a numerical example, start with x = 10 and y = 30, product 300. Increasing x to 12 requires y = 25, not 24. The value 25 preserves 12 × 25 = 300, whereas 12 × 24 = 288.
When percentage language becomes confusing, temporarily replace it with a factor sentence. Write x is multiplied by 6/5, therefore y is multiplied by 5/6. Only convert back to a percentage after the reciprocal relationship is secure.
If x decreases by 40%, it becomes 60% of its original value, with multiplier 0.6 = 3/5. The inverse multiplier for y is 5/3. Therefore y increases by 2/3 of its original value, or 66 2/3%.
For instance, x = 20 and y = 9 have product 180. Reducing x to 12 gives y = 180/12 = 15. The increase from nine to fifteen is six, which is two thirds of nine. It is not a 40% increase.
This also explains why values become large as a positive input becomes very small. With the same positive constant product, a smaller positive factor needs a larger partner. But x cannot reach zero within y = k/x, and a practical context may impose limits well before that boundary.
Do not treat the algebra as a promise that any real system can operate at extreme inputs. A model of ideal equal sharing may be exact within its definition; a model of physical production may stop being useful when capacity, minimum sizes or other constraints matter.
A fixed 84-litre quantity is shared equally among n identical containers, with no loss and sufficient container capacity. If each receives V litres, then nV = 84. Seven containers receive 12 litres each, while twelve containers receive seven litres each.
The assumptions are doing real mathematical work. Equal sharing makes one common per-container amount meaningful. A fixed total keeps the product unchanged. No loss ensures that the entire 84 litres is accounted for. If any of these conditions changes, revisit the model.
If four additional litres are added before the second distribution, the new total is 88 litres, so the old constant 84 cannot be retained. If some containers receive different amounts, a single per-container V no longer describes every container. The arithmetic formula must match the situation being represented.
For count-based contexts, the domain may also be discrete. You can allocate among seven or eight recipients, but an ordinary number of recipients is not 7.4. A calculated non-integer input might indicate that an exact equal-share requirement cannot be met under the stated whole-number constraints.
A textbook may model a fixed job completed by n identical machines, each operating at the same constant independent rate. Under those assumptions, n × completion time is constant. Doubling the number of machines halves the time because total production per hour doubles.
Suppose four ideal machines complete a job in nine hours. The model gives 36 machine-hours of work, so six such machines would take six hours. The product 4 × 9 = 6 × 6 checks the calculation under the stated conditions.
Do not transfer this rule automatically to people or real production systems. Workers may need coordination, machines may share a bottleneck, and some tasks cannot be divided into independent parallel pieces. More resources can reduce time without producing exact inverse proportion.
The honest solution states the assumption rather than presenting an ideal model as an observed law. For a school exercise, follow the conditions supplied. For an open-ended application, explain what would need to stay fixed and what limitations could prevent the simple relationship from describing the actual system.
The relationship t = d/v is inverse proportion between time and constant speed only when distance d is fixed. If the route becomes longer at the same time as speed increases, you cannot predict the time change from speed alone.
Compare two journeys: 120 km at 40 km/h takes three hours, while 180 km at 60 km/h also takes three hours. Speed increased, but distance increased too. The equal times do not contradict the fixed-distance rule because its fixed-distance condition was not preserved.
Stops create another distinction. If moving time is d/v and there is an additional fixed stop time s, total elapsed time is T = d/v + s. Then vT = d + sv, which is not constant when s is positive. Moving time may be inversely proportional to speed while total journey time is not.
This is why defining the output matters. The word time might mean travel time while moving, scheduled time, elapsed time including stops or something else. A precise variable definition prevents a correct formula from being applied to the wrong quantity.
Decide whether each relationship is simple inverse proportion over positive inputs. First, y = 72/x. Second, y = 72 − x. Third, y = 4 + 72/x. Fourth, xy = 72. Explain the decision using a product, not only the direction of change.
The first and fourth are inverse proportion because multiplying x by y gives the constant 72. The second gives xy = 72x − x², which varies. The third gives xy = 4x + 72, which also varies. All three y formulas can decrease over suitable positive domains, but that observation does not make them equivalent.
If expanding the second product feels unfamiliar, test two inputs instead to refute constancy. At x = 2, y = 70 and xy = 140; at x = 4, y = 68 and xy = 272. One disagreement is enough to reject an exact universal constant-product claim.
The symbolic explanation establishes the defining relationship for the first and fourth. The numerical counterexample rejects the second. Keep these logical jobs separate: a few agreeing values suggest a model, while one valid disagreeing value can disprove an exact proposed law.
It is given that y is inversely proportional to x. When x = 9, y = 16. Find y when x = 24, and find x when y = 18. Start by writing the known corresponding pair and calculating its product.
The constant is 9 × 16 = 144. Therefore y = 144/24 = 6 for the first question. For the second, x = 144/18 = 8. Check both pairs: 24 × 6 = 144 and 8 × 18 = 144.
The direction checks differ. Changing x from nine to 24 should reduce y below sixteen, which six does. Changing y from sixteen to eighteen should reduce x below nine, which eight does. These comparisons give a quick warning if a direct-proportion method has been used accidentally.
Now explain why x = 0 is not another possible case. The rule would require division by zero, and the product equation would require zero times a finite y to equal 144. The domain condition belongs beside the method, not only beside unusual extension questions.
Under simple inverse proportion, x increases by 50%. What happens to y? In a second situation, x decreases by 25%. Find the percentage change in y. Work with exact multipliers before converting to percentages.
For the first, x is multiplied by 3/2, so y is multiplied by 2/3. The decrease is 1/3, or 33 1/3%. For the second, x is multiplied by 3/4, so y is multiplied by 4/3. The increase is 1/3, or 33 1/3%.
The equal numerical percentage changes in these two answers arise from these particular reciprocal factors, not from a general rule that every input percentage has a matching output percentage. Each question must be converted to its own factor.
Use x = 12 and y = 10 as a concrete check, with product 120. A 50% increase gives x = 18 and y = 20/3. A 25% decrease gives x = 9 and y = 40/3. Both new pairs preserve 120 exactly, even though their output values are fractional.
An ideal machine setup completes a fixed job using five machines in twelve hours. Someone predicts that ten machines will take six hours. Then the job size is increased by 20%. Under the same independent constant machine-rate assumptions, what completion time does the enlarged job require with ten machines?
The original work amount is 5 × 12 = 60 machine-hours. Increasing the job by 20% gives 72 machine-hours. With ten machines, the required time is 72/10 = 7.2 hours. The original six-hour prediction applied to the original job, not the enlarged one.
The product of number and time is no longer 60 because the fixed total has changed. This is not a failure of proportion; it is a different total being modelled. The model becomes n × t = W, with W selected for the particular job.
Finish by interpreting 7.2 hours as seven hours and twelve minutes if clock-style duration is requested. Keep the assumptions visible: this is a fictional idealised production calculation, not a claim that real machines always scale perfectly or that a work schedule can be predicted from machine counts alone.
A learner is given y inversely proportional to x, with x = 4 and y = 15. When x becomes ten, the learner calculates y = 15 × 10/4 = 37.5. The arithmetic is correct for the expression written, but the expression uses direct rather than inverse scaling.
The input multiplier is 10/4 = 5/2. Therefore the output multiplier must be 2/5, giving y = 15 × 2/5 = 6. Equivalently, the constant is 60 and y = 60/10 = 6. The repaired answer restores the defining product.
Ask the learner to name the decision error: I used the same multiplier when I needed its reciprocal. That is more useful than saying I forgot the formula. It connects the correction to the meaning of the relationship.
Then test an input decrease, such as x changing from four to two. The answer should rise to 30. This follow-up checks whether the learner understands both directions rather than memorising that inverse proportion always makes the final answer smaller. It makes the partner change oppositely, relative to the particular input change.
For k positive and x positive, y = k/x gives a decreasing curve. Equal increases in x do not produce equal decreases in y. With k = 24, increasing x from two to four reduces y from twelve to six, while increasing x from four to six reduces y from six to four.
The curve therefore is not a straight line with a constant negative gradient. A negative straight-line gradient describes a different kind of change. You can recognise that distinction from the table before drawing any graph.
For positive inputs, the curve gets closer to the horizontal axis as x grows, but it does not reach y = 0 for a nonzero k at a finite x. It also does not include x = 0. These features follow from the formula, not from how neatly the graph is drawn.
Where graph work is part of your school programme, use the curve to support the algebraic understanding. Do not decide that a rough sketch proves inverse proportion because it slopes downward. A table of products or a stated reciprocal rule gives the defining information more directly.
If speed is measured in kilometres per hour and time in hours, their product is kilometres. In the fixed-distance model, the constant has the unit and meaning of distance. Writing k = 180 km is more informative than writing a detached 180 with no interpretation.
If n is a count of equal pieces and l is length in metres, nl has the meaning of total length. If n is the number of identical ideal machines and t is hours, nt is often described in machine-hours within the ideal work model. The unit reflects what was held fixed.
Do not multiply values expressed in incompatible conventions without first understanding the conversion. If one time is in minutes and another in hours, their numerical products with the same speed values will not be directly comparable. Put both pairs into one consistent unit system.
For example, 60 km/h for 30 minutes covers 30 km, not 1800 km. Convert 30 minutes to half an hour before forming the product. A constant-product check is meaningful only when the quantities and units are consistently defined across the compared pairs.
A classroom problem may state the inverse law explicitly, allowing exact calculation within that model. In observed data, similar products can support a proposed relationship, but they do not automatically establish a causal mechanism or guarantee accuracy outside the observed range.
Imagine three measured pairs with products 99, 101 and 100. They are not exactly constant, although measurement uncertainty might make an approximate inverse model worth considering. You would need to know the measurement precision and purpose before deciding whether the differences matter.
Conversely, an exact school table with products 99, 101 and 100 does not satisfy exact simple inverse proportion as written. Do not silently treat all contexts as approximate just because the values are close. Read whether the task supplies exact mathematical data or measured observations.
The useful boundary is between a defined relationship and evidence for a model. School calculation often begins after the relationship has been specified. Modelling requires an additional question: why should this rule represent the situation, and over what domain is that assumption reasonable?
Ask the learner what stays constant. If they answer the product, ask them to show it using a known pair. Then change one input by a clear factor and ask what must happen to the output. This reveals whether the law is understood before a longer word problem is introduced.
For x = 6 and y = 20, the constant is 120. If x triples, y becomes 20/3. If x halves, y doubles to 40. These questions are deliberately about factors, not percentage vocabulary, so they isolate the inverse relationship itself.
If the learner calculates the constant correctly but reverses the next factor, use the product equation again. If they choose a ratio rather than a product, compare one direct and one inverse example side by side. If the calculation works but the context is wrong, return to what is fixed.
A single label such as weak at proportion hides these different needs. A short explanation and a carefully varied pair of questions often reveal a more precise next step. The goal is not more working for its own sake, but working that makes the mathematical decision visible.
For the broad topic foundation, read Inverse Proportion in Secondary Mathematics. For a paired comparison of relationship types, use How to Be Good at Direct and Inverse Proportion. This article’s narrower job is to make the constant-product test and its conditions dependable.
For application judgement, continue to How to Test Whether a Mathematical Model Makes Sense. It places product checks alongside units, boundaries and assumptions, which matters when the relationship is not simply given by the question.
These links have different purposes. Use the foundation page to rebuild the definition, the comparison page to distinguish direct from inverse scaling, and the modelling page to examine whether a relationship is appropriate. Reading all three is not necessary for every learner or every correction.
The connected habit is to identify structure before computation. Once a constant is understood as a conserved total, reciprocal scaling becomes easier to explain, and conditions that break the model become easier to notice.
Define the two quantities and choose consistent units. State why inverse proportion applies, either because it is given or because the stated ideal conditions preserve a product. Write xy = k, find k from a corresponding pair, calculate the missing value and interpret it in context.
Then check the product and direction. A larger positive x should give a smaller positive y for a positive constant. The numerical product should match the original constant. Check domain restrictions and any whole-number requirement before writing the final sentence.
If the answer disagrees with the context, do not immediately change the arithmetic. First ask whether the total changed, whether a fixed extra term was present, whether one pair used different units or whether the quantities were never actually inversely proportional.
This sequence is short enough to become a habit, but each step has a distinct job. Model choice protects meaning, substitution protects correspondence, arithmetic finds the value, and checking tests whether the result still belongs to the original relationship. Leaving out the first step can make every later line look tidy while answering the wrong mathematical question.
If the definition is unclear, practise deciding which product stays fixed in an equal-sharing example. If the model is clear but the calculations fail, practise corresponding pairs with exact fractions. If percentage changes are the difficulty, translate them into multipliers before applying reciprocals.
If word problems are the main difficulty, highlight the fixed quantity and write a sentence describing the assumptions. A fixed job, fixed distance or fixed shared amount is not a decorative phrase. It is the reason that one product can remain unchanged.
Return to the G1/G2/G3 Mathematics explainer directory for related guides on rates, percentage bases and checking. The Clementi Secondary 1 Mathematics reference sets out the wider emphasis on understanding rules before trying to perform them quickly.
Keep one sentence available: inverse proportion means reciprocal factors because the product stays constant. It explains the formula, the direction of change, the percentage asymmetry and the assumption check. That is a stronger foundation than remembering only that one quantity goes up while the other goes down.
These questions separate the exact mathematical relationship from common shorthand descriptions.
Is every decreasing graph inverse proportion?
No. A decreasing graph shows a direction of change. Simple inverse proportion additionally requires y = k/x, equivalently a constant product xy. A straight line such as y = 20 − x decreases over a positive domain without satisfying that condition.
Should the percentage changes be equal and opposite?
No. The multipliers are reciprocal, not the percentage changes. If x is multiplied by 1.25, y is multiplied by 0.8. That is a 25% increase in x and a 20% decrease in y. Both changes preserve the same product.
Can the input be zero?
Not in y = k/x. Division by zero is undefined. For a nonzero constant, no finite partner can make a zero input produce the required product. A practical context may impose further restrictions, such as positive speeds or whole numbers of recipients.
Why do more workers not always halve the time?
Exact inverse scaling needs a fixed divisible job and equal independent constant work rates without bottlenecks or coordination effects. Real work may not meet those assumptions. Treat the textbook relationship as a stated ideal model, not a universal claim about people or production.