Big picture
Two quantities may change together in a predictable way. In direct proportion, their ratio remains constant. In inverse proportion, their product remains constant. Identifying the correct relationship matters more than choosing a multiplication trick.
More identical notebooks cost more money at a fixed unit price. More equally efficient workers take less time to finish a fixed job. Both examples involve changing quantities, but the mathematical models are different. This lesson teaches how to state assumptions, recognise the model and solve problems using a transparent method.
Learning goals
Define direct and inverse proportion using constant ratios or products.
Test numerical tables and distinguish proportional from non-proportional change.
Solve direct-proportion problems by the unitary and ratio methods.
Solve inverse-proportion problems involving work, speed, supplies and distribution.
Explain why fixed charges and changing conditions may break a model.
Handle multi-step problems with consistent units and meaningful checks.
Core ideas
1. Ratios compare corresponding quantities
A ratio compares quantities by division. Before comparing values, keep units consistent and align each quantity with its matching observation. If x denotes notebook count and y denotes total cost, pair each count with the cost of that count, not a different row.
Worked example. Three identical notebooks cost ₹54 and five cost ₹90. Cost per notebook is 54 ÷ 3 = ₹18, and 90 ÷ 5 = ₹18. The equal ratio y/x suggests direct proportion if the price remains fixed and there are no extra charges.
A ratio need not be dimensionless when it relates different quantities: ₹18 per notebook is a rate. When comparing two lengths as a pure ratio, convert them to the same unit first. For example, 2 m:50 cm = 200 cm:50 cm = 4:1, not 2:50.
2. Define and recognise direct proportion
Quantities x and y are in direct proportion when y/x = k is constant, for nonzero x. Write y = kx. If x is multiplied by a positive factor, y is multiplied by the same factor. In typical count-cost situations, zero items correspond to zero cost under the stated assumptions.
Notebooks x | Cost y, ₹ | y/x, ₹ per notebook |
|---|---|---|
2 | 36 | 18 |
4 | 72 | 18 |
7 | 126 | 18 |
Worked example. Find the cost of 9 notebooks. Since k = 18, y = 18 × 9 = ₹162. Doubling the count doubles cost; adding one notebook adds ₹18. The decisive property is constant ratio, not merely that both quantities increase.
3. Unitary and proportion methods
The unitary method first finds the value for one unit, then the required number of units. It is especially useful when you want to explain what each operation means.
Worked example. Six metres of cloth cost ₹810 at a fixed rate. Cost of 1 m = 810 ÷ 6 = ₹135. Cost of 11 m = 135 × 11 = ₹1485. This assumes the same cloth quality and no bulk discount.
The ratio method uses corresponding pairs. If x₁, y₁ and x₂, y₂ are direct pairs, y₁/x₁ = y₂/x₂. Thus 810/6 = y₂/11, giving 6y₂ = 810 × 11 and y₂ = 1485. Cross-multiplication is justified by the direct relationship; it is not a substitute for identifying that relationship.
4. More direct-proportion applications
At constant speed, distance is directly proportional to time. Worked example: a cyclist covers 18 km in 1.5 hours. Speed = 18 ÷ 1.5 = 12 km/h. In 2.25 hours, distance = 12 × 2.25 = 27 km. The assumption excludes stops or speed changes.
A recipe scaled without changing its proportions gives another example. If 4 servings require 300 g of flour, one serving needs 300 ÷ 4 = 75 g. Ten servings need 75 × 10 = 750 g. Scaling ingredient amounts does not imply that cooking time must scale in the same way.
For a map with scale 1 cm representing 8 km, a measured route of 6.5 cm corresponds to 6.5 × 8 = 52 km. Use the actual route length on the map if the question concerns a winding route; a straight-line measurement would answer a different question.
5. Define and recognise inverse proportion
Quantities x and y are in inverse proportion when xy = k is constant. Write y = k/x for nonzero x. If x is multiplied by a positive factor, y is divided by that factor. Doubling x halves y.
Workers x | Days y | Worker-days xy |
|---|---|---|
3 | 24 | 72 |
6 | 12 | 72 |
9 | 8 | 72 |
Worked example. For the same job, 8 workers require y = 72 ÷ 8 = 9 days. The result assumes equal worker efficiency, equal working hours, a divisible job and no interference. More workers do not always make real tasks faster if they cannot work independently.
6. Inverse problems by fixed total
Identify what remains fixed. In worker-time questions it is the job; in speed-time questions it is distance; in supply questions it is total stock. Then express the constant product in meaningful units.
Worked example: workers. Twelve workers complete a job in 15 days. Total work = 12 × 15 = 180 worker-days. With 20 equally efficient workers, time = 180 ÷ 20 = 9 days. Since there are more workers, the required time should be less than 15 days.
Worked example: speed. A bus takes 4 hours at 45 km/h for a fixed journey. Distance = 45 × 4 = 180 km. At 60 km/h, time = 180 ÷ 60 = 3 hours. This assumes the same route and no change in stoppage time. Speed and time are inverse only while distance is fixed.
7. Stocks and equal distribution
A fixed quantity shared equally produces inverse variation between number of recipients and each person's share. Worked example: ₹960 shared among 8 pupils gives 960 ÷ 8 = ₹120 each. Shared among 12 pupils it gives 960 ÷ 12 = ₹80 each. Number × share remains ₹960.
For supplies, consumption per person must stay constant. Food sufficient for 18 people for 20 days represents 18 × 20 = 360 person-days. For 24 people, duration = 360 ÷ 24 = 15 days. Different appetites, spoilage or new supplies would change the model.
Do not confuse units: person-days measure stock relative to a daily allowance, not a number of people by itself. The answer sought is days, so divide person-days by people.
8. Why opposite movement is insufficient
Two quantities are not inverse merely because one rises while the other falls. Test their product. Ages of a parent and child both rise together but are not usually in direct proportion. The number of pages remaining decreases as pages read increases, but their sum, not product, is fixed.
Worked test. In pairs (2, 18), (3, 12), (6, 6), the products are 36, 36, 36, so inverse proportion fits. In pairs (1, 10), (2, 9), (3, 8), the products are 10, 18, 24, so the relationship is not inverse proportion, even though y decreases as x increases.
Likewise, pairs (1, 5), (2, 8), (3, 11) increase together, but y/x equals 5, 4 and 11/3. Their constant difference pattern y = 3x + 2 is not direct proportion because the fixed extra 2 prevents a constant ratio.
9. Fixed charges and graphs
Suppose delivery cost is ₹30 plus ₹12 per item. For 2 items, cost = 30 + 12 × 2 = ₹54. For 4, cost = 30 + 12 × 4 = ₹78. The ratios 54/2 = 27 and 78/4 = 19.5 differ. Total cost is not directly proportional to item count, although the item-only charge is.
A direct-proportion graph y = kx is a straight line through the origin. A fixed-charge model is a straight line with a nonzero vertical intercept. An inverse model y = k/x is curved in the positive quadrant; joining a few points by straight segments only approximates its shape.
These graphs offer a visual check, but the table or formula supplies the exact relationship. Not every straight-line graph represents direct proportion.
10. Multi-step changes and remaining work
Worked example. Eight workers can finish a job in 18 days. They work for 6 days, then 4 more equally efficient workers join. Total work = 8 × 18 = 144 worker-days. Completed work = 8 × 6 = 48 worker-days. Remaining = 144 − 48 = 96 worker-days. New workforce = 8 + 4 = 12. Further time = 96 ÷ 12 = 8 days. Total elapsed time = 6 + 8 = 14 days.
Do not apply the larger workforce to the whole job, because the first part is already complete. State whether the question asks for additional time or total time; the answers here are different.
11. Check proportional reasoning
Estimate the direction before calculating. Greater quantity at a fixed price should cost more; greater speed over a fixed distance should take less time. Then check the invariant: equal ratios for direct variation, equal products for inverse variation.
For direct proportion, x₁/x₂ = y₁/y₂. For inverse proportion, x₁/x₂ = y₂/y₁. Keeping a labelled table prevents reversal. Avoid mixing minutes with hours, grams with kilograms or workers with working hours per worker. Write assumptions briefly when the context requires them.
A reliable approach to questions
Identify the two changing quantities and the condition held fixed.
Decide whether constant ratio, constant product or neither applies.
Convert all relevant units to consistent forms.
Use a unitary calculation or a labelled proportion equation.
In multi-step problems, remove completed work or consumed stock first.
Check direction, invariant, units and whether total or additional time is requested.
Connections that matter
Ratios, percentages and unitary reasoning provide the computational foundation.
Algebra expresses direct and inverse relationships as y = kx and xy = k.
Graphs distinguish proportional lines through the origin from other linear or curved models.
Common mistakes
Using “both increase” as proof of direct proportion: verify the ratio.
Calling any opposite movement inverse: verify the product.
Ignoring fixed charges: separate the constant charge from the proportional component.
Recalculating a whole job after workers join: subtract work already completed first.
Revision checklist
I can recognise direct and inverse proportion from assumptions and data.
I can identify the constant ratio or product.
I can solve cost, recipe, speed, work and supply problems.
I can explain why a fixed-charge model is not direct proportion.
I can solve a change-of-workforce problem in stages.
I can check units, direction and the meaning of my answer.