Algebra · Eqora guide
Inverse proportion: what stays fixed?
Test the product, reverse the scale factor and check the model before deciding that more must mean less.
Twice as many people can sometimes finish a job in half the time. But sometimes they cannot: there may be only one usable tool, or everyone may need the same safety briefing. Inverse proportion describes a particular mathematical relationship, not every situation where one quantity increases while another decreases. The useful question is what remains fixed when the two quantities change together.
You will build a constant-product model, find a missing value, interpret a graph and test the assumptions behind a result. The calculations use positive quantities in practical examples. Paper is enough, so the method is free to practise without an app or an answer service. We will also distinguish an exact classroom model from measurements that only approximately fit it.
Eqora publishes this independent learning guide. It is a learning aid, not a substitute for exam work or a guarantee of correctness or grades. Check the task, notation, assumptions and result, and follow your course rules. The workshop photographs suggest sharing a fixed amount; they are not counted mathematical diagrams and do not encode the numerical examples.
A fixed product is stronger than an opposite trend
Two positive quantities x and y are inversely proportional when xy = k for one fixed positive constant k. Equivalently, y = k/x. The same k must work throughout the stated domain, not change from row to row. OpenStax’s treatment of inverse variation gives this reciprocal form and the method of finding the constant from a known pair. Our numbers below are original worked examples, not measurements taken from a source or a photograph.
Consider a fixed shared cost of 240 currency units, split equally among n people. If p is each person’s contribution, np = 240, so p = 240/n. Eight people pay 30 each; twelve pay 20 each; fifteen pay 16 each. The product is the total cost, with units of people times currency units per person. This model requires the same bill and equal shares, with no extra charge for adding participants. State those conditions before using the formula.
Use the reciprocal factor, not the same difference
Suppose x changes by a positive factor q: the new input is qx. The new output is k/(qx) = (k/x)/q = y/q. Multiplying one quantity by q therefore divides the other by q. Doubling gives half; tripling gives one third. The argument works for any positive factor, not only whole-number multipliers. If the input is halved, q = 1/2 and the output is divided by 1/2, which means doubled.
In the shared-cost example, going from eight people to twelve multiplies the number by 12/8 = 1.5. Divide the contribution by 1.5: 30/1.5 = 20. Do not subtract four from the contribution because four people were added. Those quantities have different units, and inverse proportion is multiplicative rather than an equal-difference rule. Writing the input ratio above the output calculation makes the direction of the change visible and prevents a common mistaken multiplication by 1.5.
Check every paired row in a table
Take paired values x = 2, 3, 6, 9 and y = 18, 12, 6, 4. Their products are 36, 36, 36 and 36. Thus these rows are consistent with y = 36/x. The input gaps are unequal, which does not matter: product equality uses each matching pair, not the spacing between successive inputs. Do not multiply all the inputs together or pair one row’s input with the next row’s output.
A decreasing table alone is insufficient. For y = 20 − x, the inputs 1, 2 and 3 give outputs 19, 18 and 17. Both quantities move in opposite directions, but the products are 19, 36 and 51, so this is not inverse proportion. Also, a few matching observations do not prove a universal physical law. They support a proposed model over the observed range. A classroom statement may explicitly specify inverse proportionality; an experimental claim needs an explanation of the fixed quantity and its limitations.
Find the constant before solving the unknown
A steady inlet supplies water at 3 litres per minute and fills a tank in 24 minutes. Assume the tank starts empty, its target volume is fixed and there is no outflow. With rate r and time t, rt is the delivered volume: k = 3 × 24 = 72 litres. At 6 litres per minute, t = 72/6 = 12 minutes; at 8 litres per minute, t = 9 minutes. Both products return 72, checking the calculation against the model.
If the question instead gives a target of 18 minutes, solve 18r = 72, so r = 4 litres per minute. Decide which variable is missing before dividing. An answer of 4 minutes would have the right numeral but the wrong meaning. Our guide to converting length, area and volume helps keep unit changes separate from model changes. Converting minutes to seconds changes the numerical constant when the rate unit is unchanged, so use one consistent system before comparing products.
Separate inverse proportion from direct proportion
With direct proportion, y = ax and the ratio y/x is constant. For example, at a fixed price of 6 currency units per item, buying x items costs y = 6x. Doubling the number doubles the cost. With inverse proportion, y = k/x and the product xy is constant. Sharing a fixed bill among twice as many people halves each contribution. The word fixed matters in both cases, but the fixed quantity and the operation are different.
Some relationships are neither. The decreasing rule y = 1/x² has xy = 1/x, which varies with x. Doubling a positive input divides the output by four, not two. It is an inverse-square relationship, not the basic inverse proportion xy = k. Likewise, y = 20 − x is a decreasing straight line rather than a reciprocal rule. Choose a model from its defining equation or justified invariant, not just from a keyword such as less, faster or shared.
Identify the assumptions behind workers and time
Suppose a batch requires 24 station-hours of processing. If n identical stations work independently at a constant rate, with enough work to share evenly and no delays, the processing time is t = 24/n hours. Four stations need six hours; six stations need four. Station-hours are an idealized workload measure here, not a promise about real staffing. Unequal skill, a shared bottleneck or work that cannot be divided can invalidate the equal-rate assumption.
Now add a fixed one-hour setup before processing begins. The total duration is T = 1 + 24/n. Four stations need seven hours, while eight need four hours. More stations still mean less time, but nT equals n + 24, not a constant: the products are 28 and 32. Doubling the stations does not halve the total duration. You may apply inverse proportion to the processing part T − 1, provided the setup really stays fixed, but not silently to the entire job.
Percentage changes are not mirror images
An input increase of 25% multiplies x by 1.25. In an inverse proportion, the output factor is 1/1.25 = 0.8, so y decreases by 20%, not 25%. For example, raising a steady filling rate from 4 to 5 litres per minute changes the time for 72 litres from 18 to 14.4 minutes. The time difference is 3.6 minutes, which is 20% of the original 18 minutes.
More generally, an input increase of a decimal fraction u gives output factor 1/(1 + u). Its fractional decrease is u/(1 + u). This statement assumes 1 + u is positive; interpret the sign carefully if describing a decrease instead. A 50% decrease in the positive input halves it and doubles the output, a 100% increase. Always identify the starting quantity used as the percentage base. Opposite directions do not imply equal percentages or equal absolute differences.
Read the curve and respect the domain
For k > 0 and x > 0, the graph y = k/x is a decreasing curved branch, not a straight line. It approaches the positive axes without reaching either at a finite positive input. You cannot substitute x = 0, because division by zero is undefined. In a physical model, indefinitely increasing the input may also be impossible: tank capacity, available equipment or a maximum safe flow restrict the useful range long before a mathematical limit is reached.
Plotting y against 1/x instead produces y = k(1/x), a straight line through the origin in the transformed coordinates. An arbitrary straight trend is not enough; its intercept must be zero for this exact model. Algebraically, allowing negative x adds another branch when k is positive, but negative people or negative filling times do not belong to our practical examples. Whole people also require integer n. A formula’s algebraic domain and the admissible values of a word problem are related but not identical.
Test a new case without copying the first calculation
Try this independently: a fixed 90-litre amount takes 15 minutes to deliver at a steady rate. What rate delivers it in 12 minutes? First recover the original rate, 90/15 = 6 litres per minute. The shorter time is multiplied by 12/15 = 0.8, so the rate is divided by 0.8: 7.5 litres per minute. A separate calculation, 90/12 = 7.5, agrees. The product check is 7.5 × 12 = 90 litres, and the direction is sensible: less time requires a higher rate.
For a different setting, split 210 currency units equally among fourteen people, then ten. The contributions are 15 and 21, with the same product 210. If the bill changes too, stop and build a new equation instead of preserving the old constant. Our guide to checking answers without an answer key explains why a second representation is more useful than repeating identical arithmetic. Finish by naming the invariant, the domain and the real-world assumptions; those explain why the answer applies, not merely how it was calculated.
