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Direct inverse proportion and compound measures - Mathematics Year 9

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Introduction

If one quantity doubles, does the other double, halve, or do something else? Choose a model from its assumptions, then use units and a second check to test the result.

Solve four direct/inverse proportion contexts and three compound-measure problems. Distinguish density, speed and unit price, and explain why an increasing graph or a single data pair cannot establish proportionality on its own.

Choose direct models from a constant quotient

Direct proportion means multiplying one variable by a factor multiplies the other by the same factor. A model $y = k x$ has constant quotient y/x for nonzero x and passes through the origin if zero is in its domain. The constant k carries any required units; it is not necessarily a dimensionless number.

For the source assumption y proportional to x, with y=21 at x=7, the constant is k=21/7=3. Therefore y=3 x. At x=10, the predicted y is 30. Reverse-check 30/10=3 and use a unitary route: one x-unit corresponds to three y-units, so ten correspond to thirty. The assumption of direct proportion matters: a single observed pair 21, 7 would also fit many non-proportional rules.

An increasing relationship is not automatically directly proportional. The rule y=3 x+5 increases but does not have a constant y/x and does not pass through the origin. For x 1 and 2, it gives 8 and 11, whose quotients 8 and 5.5 differ. A fixed starting charge breaks a pure price-per-unit model even if each extra item has the same additional cost.

Four complete proportion contexts. All quantities are invented mathematical examples. State which quotient or product remains fixed and retain the stated conditions.

Task	Given context and assumption	Model	Required result	Independent check
1	Assumed direct: y 21 when x 7	y=3 x	x 10 gives y 30	30/10=21/7=3
2	Fabric £4.80 per metre, no fixed charge	C=4.8 L	7.5 m costs £36	36/7.5=4.8
3	Six equal-rate workers, 8 h, fixed divisible work	t=48/w	12 workers take 4 h	12×4=6×8=48 worker-hours
4	Rectangle area 60 cm², width x, length y	y=60/x	Width 5 cm gives length 12 cm	5×12=3×20=60 cm²

For task 2, let L be fabric length in metres and C cost in pounds. A constant £4.80 per metre with no fixed charge gives C=4.8 L. For 7.5 m, this is£36. Split the length into 5 m and 2.5 m: £24+£12=£36 provides a second calculation. A £5 handling charge would instead give C=4.8 L+5 and would not be direct proportion.

For direct models, a scale-factor method is efficient when the new input is an easy multiple. A unitary method or explicit constant is clearer for an awkward factor. Both preserve a ratio, not a difference. Adding 3 to both entries of 21: 7 gives 24: 10, which has quotient 2.4 rather than 3; that additive change does not preserve the original direct relationship.

Plot an assumed direct model using at least two checked points and the origin where permitted. Interpret a straight line through the origin as consistent with the rule, not as proof of an unstated real-world law. Keep contextual limits such as nonnegative lengths and valid purchase quantities.

Choose inverse models from a constant product

Inverse proportion has a constant product: as one positive variable multiplies by a factor, the other divides by that factor. The basic model is $y = \frac{k}{x} (x \neq 0)$ . Its graph is curved, not a downward straight line. Decreasing alone is insufficient: the rule y=20−x decreases but its product xy is not fixed.

In task 3, the fixed workload takes 6 equal-rate workers 8 h, so the workload model is 48 worker-hours. With 12 workers, time is 48/12=4 h; with 4 workers, it is 48/4=12 h. A scale-factor check doubles the workers and halves the time. The model assumes identical constant rates, perfect sharing, no setup delay and no interference. Real teamwork can violate these assumptions; this arithmetic is not a staffing or safety instruction.

In task 4, the fixed rectangle area is 60 cm². Width 3 cm requires length 20 cm, while width 5 cm requires length 12 cm. The product remains 60 cm². If width doubled from 3 to 6, length would halve from 20 to 10. Here k has area units and x, y must be positive physical lengths, not merely nonzero real numbers.

The worker and rectangle contexts use the same mathematical product structure for different reasons. In a width-length problem the fixed product is area; in a simple work-rate model it is assumed worker-hours. Always name the invariant rather than selecting “inverse” merely because one number becomes smaller.

The original Corbettmaths preview explicitly includes an inverse-square extension: y is inversely proportional to x squared, with y 2 when x 3. That model is y=18/x², not the basic y=k/x in our two tasks. Compare the invariant x² y=18 with xy=k; do not transfer a square that the problem has not stated. Original: corbettmaths, Inverse Proportion - Corbettmaths. The written explanations and native tasks work without sound or the external player.

For that explicitly stated extension, substituting x 3 gives 18/9=2. Doubling x would quarter y, rather than halve it. The course's basic inverse models instead halve y when x doubles. The contrast shows why precise wording and the actual power of x matter. The text and native tasks remain usable without sound or the optional external player.

Retain all four context solutions with the chosen invariant, units, model conditions and a second method. A numerical fit at one pair alone is insufficient to infer a universal direct, inverse or inverse-square law.

Interpret three compound measures and consistent units

A compound measure combines different quantities. Division does not make their units disappear: density is mass per volume, average speed is distance per time, and unit price is cost per stated amount. Decide which quantity belongs in the numerator and which in the denominator before calculating.

Three complete compound-measure problems.

Measure	Given data	Calculation	Result	Original-quantity check
Density	540 g, 200 cm³	540/200	2.7 g/cm³	2.7×200=540 g
Average speed	150 km, 2.5 h	150/2.5	60 km/h	60×2.5=150 km
Unit price	£3.60, 750 g	3.60/0.750	£4.80/kg	4.80×0.750=£3.60

For source density 540 g in 200 cm³, the result is 2.7 g/cm³. Multiply density by volume to reconstruct 540 g. This is an average over the stated sample; uniformity is an additional assumption if using the same density for every smaller part. No actual weighing or laboratory measurement is claimed by this fictional calculation.

To express that density in kg/m³, one gram is 0.001 kg and one cubic centimetre is 0.000001 m³. Therefore 2.7 g/cm³=0.0027/0.000001=2700 kg/m³. A centimetre-to-metre conversion factor must be cubed for volume. Using 100 rather than 1000000 for cm³ to m³ is the first error, even if a calculator performs the later division correctly.

For average speed 150 km in 2.5 h, the result is 60 km/h. In SI units 150000 m/9000 s=50/3 m/s, approximately 16.67 m/s. This average does not prove a constant instantaneous speed throughout the trip. The check 60 km/h×2.5 h=150 km keeps distance units and avoids confusing a speed with a time.

For unit price, convert 750 g to 0.750 kg before dividing£3.60 by mass. The result is£4.80/kg, not£4.80 per gram. A comparison pack costing£5.20 for 1.2 kg has rate£13/3 per kg, approximately£4.33/kg, and is cheaper per kilogram. The larger total price alone does not decide value; assume comparable goods and no other charges. Retain exact values until the final rounded money comparison.

Unfamiliar transfer. A new direct model has y=28 at x=8, so k=3.5 and x 12 gives y 42. A fixed workload of 72 worker-hours with 9 equal-rate workers predicts 8 h. A new sample 840 g in 300 cm³ has average density 2.8 g/cm³. Rebuild 28, 72 and 840 from each proposed rule to check the three different contexts independently.

Keep four proportion models, three compound answers and units, the inverse-square contrast, a corrected first error and new-value transfer. A unit check can expose a reversed division even when its numerical value looks plausible. Record actual later retrieval separately; these computations do not certify measurement competence or the truth of an untested real-world model.

Interactive Tasks

Quiz: Test Your Knowledge

Assumed direct y 21 at x 7 gives which k? (3) (!14) (!147) (!1/3)

Then x 10 gives which y? (30) (!24) (!70) (!210)

Which model is directly proportional? (y=3 x) (!y=3 x+5) (!y=20−x) (!y=x²)

Six equal-rate workers take 8 h; 12 workers in the same ideal model take? (4 h) (!16 h) (!8 h) (!48 h)

Fixed area 60 cm², width 5 cm: length? (12 cm) (!55 cm) (!300 cm) (!65 cm)

Density 540 g in 200 cm³? (2.7 g/cm³) (!0.37037 g/cm³) (!340 g/cm³) (!108000 g/cm³)

That density in kg/m³? (2700) (!27) (!0.0027) (!2700000)

Average speed 150 km in 2.5 h? (60 km/h) (!375 km/h) (!147.5 km/h) (!0.01667 km/h)

£3.60 for 750 g: unit price? (£4.80/kg) (!£0.0048/kg) (!£2.70/kg) (!£2700/kg)

For an explicitly inverse-square model, doubling x does what to positive y? (Quarters y) (!Halves y) (!Doubles y) (!Adds two to y)

Memory Game

Direct proportion	Constant quotient and model y=kx
Inverse proportion	Constant product and basic model y=k/x
Density	Mass divided by volume
Average speed	Total distance divided by total time
Unit price	Cost per agreed unit amount
Invariant	Quantity kept constant under the model

Drag and Drop

30	Predicted y at x 10 in the source direct model
4 h	Twelve-worker time in the ideal workload model
12 cm	Rectangle length at width 5 cm and area 60 cm²
2.7 g/cm³	Source average density
£4.80/kg	Source unit price after converting grams

Crossword Puzzle

DIRECT	Proportion with constant quotient
INVERSE	Basic proportion with constant product
DENSITY	Mass per volume
AVERAGE	Describes total-distance over total-time speed
INVARIANT	Value preserved by the model
CONVERSION	Change to consistent measurement units

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Find the source direct constant: Use 21/7 and rebuild 21.
Check inverse products: Verify 6×8=12×4 and 3×20=5×12.
Rebuild density mass: Use 2.7×200=540 with units.
Convert a purchase amount: Write 750 g as 0.750 kg before comparing rates.

Standard

Solve all four proportion contexts: Retain the invariant, units, conditions and a second method.
Solve three compound problems: Calculate density, speed and unit price with original checks.
Compare equivalent units: Derive 2700 kg/m³ and 50/3 m/s without early rounding.
Repair an additive-ratio error: Explain why 21: 7 and 24: 10 are not equivalent.

Advanced

Transfer three new contexts: Find direct 42, work time 8 h and density 2.8 g/cm³.
Challenge model assumptions: Explain setup delays or a fixed price charge without inventing measured results.
Distinguish inverse powers: Contrast xy constant with x² y constant and doubling effects.
Evaluate value and evidence: Compare£4.80/kg with£13/3 per kg, retain exact arithmetic and actual later retrieval.

Learning Check

Four proportion contexts: Solve each with its invariant and original check.
Three compound results: Retain numerator, denominator and complete units.
Independent unit conversions: Derive the same density and speed in SI units.
Model and first-error analysis: Reject increasing-only and additive-ratio shortcuts.
Unfamiliar transfer and value: Check new models and compare pack unit costs.
Portfolio and model limits: Keep all required work and distinguish assumptions from observation.

Worked checks and assessment guidance

Direct y 21/x 7 gives k 3 and y 30 at x 10. Fabric£4.80/m gives£36 for 7.5 m. Fixed workload 48 worker-hours gives 4 h for 12 workers and 12 h for 4. Fixed rectangle 60 cm² gives 12 cm length at 5 cm width.
540/200=2.7 g/cm³; 150/2.5=60 km/h; £3.60/0.750 kg=£4.80/kg. Multiplying each by its denominator reconstructs 540 g, 150 km and£3.60.
2.7 g/cm³=0.0027 kg/0.000001 m³=2700 kg/m³.150000 m/9000 s=50/3 m/s. Cubic units need a cubed conversion factor; average speed does not prove constant motion.
y 3 x+5 increases but its ratio differs at x 1/2.21: 7→24: 10 changes 3 to 2.4. Basic inverse yk/x halves output when x doubles; the explicitly inverse-square video model y 18/x² quarters it.
New direct 28/8=3.5 gives 42 at 12; 72 worker-hours/9 gives 8 h; 840 g/300 cm³=2.8 g/cm³.£5.20/1.2 kg=£13/3 per kg is below£4.80/kg under comparable-goods assumptions.
Retain four proportion contexts, three compound answers, units, independent checks, first-error repair and unfamiliar transfer. Equal worker rates, uniform density and zero fixed charges are model assumptions; no actual work completion, measurement or later mastery is invented.

Learning evidence

Retain four direct/inverse proportion solutions, three compound measures with units and independent checks, a corrected first error and unfamiliar transfer.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, number, ratio, fluency, reasoning and problem solving. Year 9 allocation is editorial within KS3. Numerical models, tasks and feedback are original; images retain actual Commons attribution and videos name their original publisher. Reviewed 12 November 2026.

Cuisenaire-Rods.jpg, Celcom at en.wikipedia, CC BY 3.0. Unchanged original; caption and learning activities are original.

Cuboid.abc.png, Cuboid.png: en:User:Tomruen, CC BY-SA 4.0. Unchanged original; caption and learning activities are original.