Direct Proportion — Scaling with a Constant Rate

Introduction

Direct proportion is the simplest, most common scaling relationship in GCSE maths: when one quantity changes, the other changes by the same factor. If you double one, you double the other; if you halve one, you halve the other. We summarise this idea with the statement \(y \propto x\) and the equation \(y = kx\), where \(k\) is the constant of proportionality. In everyday terms, \(k\) is the fixed “rate per one”: the price per item, the distance per hour (speed), the mass per litre (density), the marks per question if they’re all worth the same, and so on.

You will meet direct proportion all over GCSE questions: unit pricing and “best buy” comparisons in supermarkets; scaling recipes and construction materials; enlargements in geometry (length scales by a factor, area/volume by related factors); graphs that are straight lines through the origin; physics-style contexts such as constant speed (\(d=vt\)) and constant density (\(m=\rho V\)); and algebraic statements like “\(y\) is directly proportional to \(x\)”. Because so many real problems reduce to a constant “per one”, mastering direct proportion gives you a fast, reliable method for a huge range of tasks.

There are two interchangeable viewpoints you’ll use throughout this tutorial:

• Unitary method: Find the value for one unit, then scale up or down. If 6 notebooks cost £9, then 1 notebook costs \(£9\div6\) and 14 notebooks cost \(14\times(£9\div6)\). This is often the quickest route in word problems.
• Equation method: Write \(y=kx\). From any single data pair \((x,y)\), compute \(k=\dfrac{y}{x}\), then use \(y=kx\) to predict any other value. On a graph of \(y\) against \(x\), the gradient equals \(k\), and the line passes through the origin.

Understanding what isn’t direct proportion is just as important. If a graph doesn’t pass through \((0,0)\), it’s not of the form \(y=kx\); a fixed starting fee (like a delivery charge or setup cost) breaks direct proportion because doubling the number of items doesn’t double the total cost. Likewise, percentage increase or decrease (e.g. “add 20% once”) is a one-off change, not a constant “per one” rate applied across all sizes. Direct proportion is about a steady rate that scales every time you scale the input.

Units make or break proportion questions. The constant \(k\) always carries meaningful units (e.g. £/kg, km/h, g/cm³). Before you calculate, convert everything to consistent units—kilograms rather than grams, hours rather than minutes—to avoid silent errors. Once you’ve pinned down a clean unit rate, most questions become a single multiply or divide.

Graphically, direct proportion produces a straight line through the origin with gradient \(k\). This picture encodes the scaling rule: if you move along the \(x\)-axis by a factor of \(m\), the \(y\)-value also scales by \(m\). That’s why reading or drawing a line that doesn’t pass through the origin is a red flag—you’d be modelling a situation with an extra fixed part (an intercept) rather than pure “per one” scaling.

You’ll also see direct proportion woven into topics that look different on the surface:

• Rates and ratios: “£2.40 per kg” or “60 pages per 8 minutes” are unit rates. Converting to “per 1” (kg, minute, item) is the unitary method in action.
• Similarity and scale: In similar figures, corresponding lengths are in direct proportion to the scale factor \(k\). (Areas scale with \(k^2\), volumes with \(k^3\), which are not direct in the same variable, but still predictable once you know the length scale.)
• Physics-style formulas: With one quantity fixed, many relationships reduce to \(y=kx\): distance–time at constant speed, mass–volume at constant density, amount–time at constant rate of work.

Exam writers like to hide proportion beneath friendly narratives. Clues include phrases such as “at this rate…”, “cost is proportional to quantity…”, “\(y\) varies directly as \(x\)”, “twice as many gives twice the cost/length/mass…”, or “the graph is a straight line through the origin”. When you see any of these, switch on your direct-proportion toolkit: choose unitary or equation, compute \(k\) carefully with consistent units, and scale to the required value. A final sense-check—“does doubling the input double my answer?”—will catch most slips instantly.

In short, direct proportion is the mathematics of fair, even scaling. By the end of this tutorial you’ll move confidently between words, tables, formulas, and graphs; decide when a situation is (or is not) modelled by \(y=kx\); extract and interpret the constant \(k\); and solve real questions quickly using either the unitary method or the equation \(y=kx\).

Key Vocabulary

Before working with direct proportion problems, it’s vital to know the core terms and symbols. These words and notations appear repeatedly in GCSE questions, and understanding them will help you translate everyday situations into mathematical form.

• Direct proportion: A relationship where one quantity increases or decreases in the same ratio as another. Formally written as \(y \propto x\) or \(y = kx\). Example: Cost is directly proportional to quantity bought.
• Constant of proportionality (\(k\)): The fixed multiplier that links \(x\) and \(y\). If 5 apples cost £2.50, then \(k=£0.50\) per apple. Equation: \(y = kx\).
• Unitary method: A method that works by finding the value for 1 unit first, then scaling up or down. Example: 8 pencils cost £4 → 1 pencil costs £0.50 → 15 pencils cost £7.50.
• Rate: A measure per unit. Rates are constants of proportionality in disguise. Examples: £/kg (unit price), km/h (speed), g/cm³ (density).
• Proportional symbol (\(\propto\)): Means “is proportional to”. Example: \(y \propto x\) means doubling \(x\) will double \(y\).
• Equation of direct proportion: \(y = kx\). If you know any one pair \((x,y)\), you can calculate \(k\) and then use this equation to find other values.
• Graph of direct proportion: A straight line through the origin \((0,0)\) with gradient equal to \(k\). Steeper line ⇒ larger constant of proportionality.
• Unit conversion: Changing measurements into the same unit before working with proportions. Example: Convert 250 g to 0.25 kg before calculating price per kg.
• Scaling factor: The number you multiply by to enlarge or reduce a quantity. In direct proportion, both variables scale by the same factor.

Core Ideas

Direct proportion questions may look different on the surface (cost, speed, density, recipes, graphs), but they all rest on a few simple principles. Mastering these will allow you to quickly identify proportional situations and apply the right method.

• Definition of direct proportion: Two quantities \(x\) and \(y\) are in direct proportion if there exists a constant \(k\) such that \[ y = kx. \] This means that the ratio \(\tfrac{y}{x}\) is always the same for all values of \(x\) and \(y\).
• Scaling rule: If \(x\) is multiplied by a factor \(m\), then \(y\) is also multiplied by the same factor. Example: If 4 apples cost £2, then 12 apples (triple the quantity) cost £6 (triple the cost).
• Constant of proportionality: The value of \(k\) tells you the “rate per 1 unit”. \[ k = \frac{y}{x}. \] Example: If 10 pencils cost £2.50, then \(k = 2.50 \div 10 = £0.25\) per pencil.
• Graphs of direct proportion: When \(y\) is plotted against \(x\), the graph is a straight line through the origin with gradient equal to \(k\). A larger gradient means a larger unit rate. Example: If cost per kg of apples is £2, the graph of cost vs. mass has gradient 2.
• Unitary method connection: The unitary method is just a practical way of using the constant \(k\). Find the cost for 1 item (\(k\)), then multiply by the number of items to get the total cost.
• Checking if a situation is direct proportion: - Does doubling one double the other? - Is the ratio \(y/x\) constant across different pairs? - Does the graph go through the origin? If the answer is yes, then it is direct proportion.
• What is not direct proportion: Situations with a fixed extra amount (e.g. setup fee, standing charge, delivery cost) are not direct proportion because the relationship is \(y = kx + c\), not \(y=kx\). Example: Taxi fares often have a fixed starting fee + cost per mile.
• Common contexts: - Prices and quantities (shopping, best buy). - Recipes and scaling up/down. - Speed, distance, and time (if speed is constant). - Mass and volume (if density is constant). - Simple straight-line graphs in maths and science.

Step-by-Step Method

Direct proportion problems can be tackled with a clear sequence of steps. Whether you use the unitary method or the equation \(y=kx\), the process is essentially the same. Here’s the method you can apply to any exam-style question.

1. Identify that the situation is direct proportion. Look for key phrases like “at this rate”, “directly proportional”, “twice as many costs twice as much”, or graphs that pass through the origin. If doubling one value doubles the other, it’s a direct proportion problem.
2. Choose your approach: unitary or formula. - Unitary method: Find the value for 1 unit, then scale up or down. - Equation method: Write \(y=kx\), then calculate \(k\) using the given data.
3. Calculate the constant of proportionality \(k\). Use the known pair \((x,y)\) to compute: \[ k = \frac{y}{x}. \] This tells you the “per 1” rate, such as £/kg, km/h, or g/cm³.
4. Use the constant or unit rate to find the required value. - If using unitary: multiply the unit value by the desired number. - If using the equation: substitute the new \(x\) into \(y=kx\).
5. Check units carefully. Make sure quantities are in consistent units (e.g. convert grams to kilograms, minutes to hours). An incorrect unit conversion is one of the most common sources of mistakes.
6. Sense-check your answer. Does the result scale logically? If you doubled the input, did the output also double? Answers that don’t follow this rule indicate a calculation error.
7. Show working clearly. Exams award method marks even if the final answer is wrong. Write down your ratio, unit rate, or \(y=kx\) step before substituting numbers.

Worked Examples

Let’s see how direct proportion works in practice. We’ll start with simple foundation-level problems using the unitary method, then move on to higher-level applications involving graphs, constants of proportionality, and compound units.

Example 1 (Foundation — Unitary Method)

6 notebooks cost £9. How much do 14 notebooks cost at the same price per notebook?

Step 1: Find the cost of 1 notebook: \(£9 \div 6 = £1.50\). Step 2: Multiply by 14: \(14 \times 1.50 = £21\). Answer: £21.

Example 2 (Foundation — Recipe Scaling)

A recipe needs 300 g of flour to make 12 cookies. How much flour is needed to make 18 cookies?

Step 1: Flour per cookie = \(300 \div 12 = 25\) g. Step 2: For 18 cookies: \(18 \times 25 = 450\) g. Answer: 450 g.

Example 3 (Foundation — Graph)

A graph of \(y\) against \(x\) is a straight line through the origin with gradient 2.5. Find \(y\) when \(x=7\).

Since \(y = kx\) and \(k = 2.5\), \(y = 2.5 \times 7 = 17.5\). Answer: \(y=17.5\).

Example 4 (Higher — Find Constant and Predict)

If \(y\) is directly proportional to \(x\), and \(y=13.2\) when \(x=4\), find the constant \(k\) and then \(y\) when \(x=9\).

Step 1: \(k = \dfrac{y}{x} = 13.2 \div 4 = 3.3\). Step 2: For \(x=9\), \(y = 3.3 \times 9 = 29.7\). Answer: \(k=3.3\), \(y=29.7\).

Example 5 (Higher — Best Buy)

Bag A: 750 g of rice for £1.80. Bag B: 1.2 kg for £2.76. Which is cheaper per kilogram?

Bag A: £1.80 ÷ 0.75 kg = £2.40 per kg. Bag B: £2.76 ÷ 1.2 kg = £2.30 per kg. Cheaper option: Bag B.

Example 6 (Higher — Physics Context)

Mass \(m\) is directly proportional to volume \(V\) for a fixed material: \(m = \rho V\). If \(m=4.5\) kg when \(V=3\) L, find the mass when \(V=8\) L.

Density \(\rho = 4.5 \div 3 = 1.5\) kg/L. Then \(m = 1.5 \times 8 = 12\) kg. Answer: 12 kg.

Example 7 (Higher — Graph Gradient)

A straight line through the origin passes through \((10,25)\). Write the equation linking \(y\) and \(x\).

Gradient \(k = 25 \div 10 = 2.5\). Equation: \(y = 2.5x\). Answer: \(y=2.5x\).

Common Mistakes & Fixes

Direct proportion is one of the most accessible GCSE topics, but it’s also a frequent source of avoidable errors. Many exam questions are designed to trap students who rush or overlook key details. Here are the main mistakes — and how to avoid them.

• Forgetting to check if it’s direct proportion. Mistake: Assuming everything scales directly, even when there’s a fixed starting amount (e.g. taxi fare with a base charge). Fix: Look for evidence: does the graph go through the origin? Does doubling one double the other? If not, it’s not direct proportion.
• Confusing ratio with difference. Mistake: Thinking “if 4 apples cost £2, then 8 apples cost £4 more”. That’s correct here, but some students mistakenly treat it as +2 apples = +£2 (additive), instead of doubling both. Fix: Always think in multiples, not just “add-ons”. Direct proportion is multiplicative.
• Mixing up the constant of proportionality. Mistake: Writing \(k = \dfrac{x}{y}\) instead of \(k = \dfrac{y}{x}\). Fix: Remember the formula is \(y = kx\). Rearranging gives \(k = \dfrac{y}{x}\).
• Skipping unit conversions. Mistake: Comparing “£ per 100 g” with “£ per kg” without converting. This leads to wrong comparisons in best-buy problems. Fix: Always standardise units (e.g. to kilograms, hours, litres) before comparing rates.
• Misreading graphs. Mistake: Using a line that doesn’t pass through the origin as if it were direct proportion. Fix: Check: all direct proportion graphs pass through \((0,0)\).
• Forgetting proportional scaling in recipes. Mistake: Scaling one ingredient but not the others. Fix: If it’s direct proportion, every ingredient must scale by the same factor.
• Not sense-checking the answer. Mistake: Getting an answer smaller than the starting value when the question clearly asked for “more of the same”. Fix: Always ask: if I increased the input, did my output also increase? If not, re-check the calculation.

Practice Questions — Foundation

These practice problems will help you master the basics of direct proportion. Use either the unitary method or the formula \(y = kx\). Show all working clearly — this is where method marks are earned in exams.

1. 5 cans of drink cost £3.25. How much will 12 cans cost at the same rate?
2. A car travels 180 km in 3 hours at constant speed. How far will it travel in 5 hours?
3. A photocopier produces 120 pages in 8 minutes at a steady rate. How many pages will it produce in 15 minutes?
4. On a graph of \(y\) against \(x\), the line goes through the points \((0,0)\) and \((6,9)\). Find the value of \(y\) when \(x=14\).
5. Sugar costs £0.84 for 600 g. Work out the price per kilogram.
6. A recipe uses 250 mL of milk to make 8 pancakes. How much milk is needed for 20 pancakes?
7. A bus covers 96 km in 2 hours at constant speed. How long will it take to travel 180 km at the same speed?
8. 7 pencils cost £2.45. Find the cost of 20 pencils.

Practice Questions — Higher

These problems extend direct proportion into more complex contexts: graphs, constants, compound units, and best-buy reasoning. Work systematically and check that your answers scale correctly.

1. \(y \propto x\). When \(x=7\), \(y=17.5\). (a) Find the constant \(k\). (b) Find \(y\) when \(x=12\).
2. The fuel cost of a car journey is directly proportional to the distance travelled. A 240 km trip costs £43.20. (a) Find the cost per 100 km. (b) Predict the cost of a 390 km trip.
3. The mass of a cable is directly proportional to its length. An 18 m cable weighs 7.2 kg. (a) Find the constant of proportionality (kg per metre). (b) Find the mass of a 42 m cable.
4. Two offers for batteries: - Pack A: 9 batteries for £6.57 - Pack B: 14 batteries for £10.36 Work out the cost per battery in each pack. Which is the better buy?
5. \(m\) is directly proportional to \(r\). If \(m=14\) when \(r=3.5\): (a) Express \(m\) in terms of \(r\). (b) Find the value of \(m\) when \(r=11\).
6. The cost of electricity is directly proportional to the number of units used. 320 units cost £52.80. (a) Find the cost per unit. (b) Work out the cost of 500 units.
7. A liquid has constant density. A volume of 250 mL has a mass of 200 g. (a) Find the density in g/mL. (b) Find the mass of 1.5 L of the liquid.
8. A graph of \(y\) against \(x\) passes through \((0,0)\) and \((15, 42)\). (a) Find the equation of the line. (b) Use it to calculate \(y\) when \(x=24\).

Challenge Questions

These problems push direct proportion into trickier, exam-style contexts. They mix unitary reasoning, algebra, and real-world applications. Work carefully, show your steps, and always check that your results scale logically.

1. A printer charges a fixed setup fee of £5 plus £0.12 per page. (a) Is the total cost in direct proportion to the number of pages? Explain why or why not. (b) Suggest a formula that would represent direct proportion in this context.
2. The graph of \(y\) against \(x\) passes through \((0,0)\) and \((a,b)\), where \(a\) and \(b\) are positive constants. (a) Show that \(y = \dfrac{b}{a}x\). (b) Interpret \(\dfrac{b}{a}\) in a cost vs. quantity context.
3. A fabric roll weighs 2.8 kg for 4.0 m. (a) Find the mass per metre. (b) If 12% waste must be allowed, calculate the mass needed for 9.5 m of usable fabric.
4. Two shops sell fruit: - Shop A: 3.5 kg for £7.00 - Shop B: 2.8 kg for £5.46 (a) Work out the price per kilogram in each shop. (b) If you must buy exactly 10 kg, what is the minimum total cost by combining the two offers?
5. A straight line through the origin has gradient \(k\). Another line through the origin has gradient \(1.6k\). (a) Which line gives the larger \(y\) for the same positive \(x\)? (b) In terms of “cost per item”, what does increasing \(k\) represent?
6. A company prints 120 flyers for £9.60. (a) Find the cost per flyer. (b) Work out the cost of 850 flyers. (c) If another company offers 1000 flyers for £80, which company is cheaper per flyer?
7. The mass of a liquid is directly proportional to its volume. A 0.5 L sample has a mass of 0.44 kg. (a) Find the mass of 1 L. (b) Find the volume if the mass is 1.76 kg. (c) Write the equation linking mass and volume.
8. The table shows how much it costs to hire bikes for different hours.

   Hours	Cost (£)
   2	£18
   5	£45

   (a) Check whether cost is in direct proportion to time. (b) If so, find the cost per hour and the cost of 8 hours.

Quick Revision Sheet

Use this page as a last-minute checklist for direct proportion. It condenses the whole topic into formulas, spotting cues, rapid methods, unit reminders, and quick self-tests.

Key Formulas

• Model: \(y \propto x \;\Longleftrightarrow\; y = kx\) where \(k\) is constant.
• Constant of proportionality: \(k=\dfrac{y}{x}\) (units matter: £/kg, km/h, g/cm³, …).
• Scaling rule: If \(x \mapsto m x\), then \(y \mapsto m y\).
• Graph fact: Line through the origin; gradient \(=\;k\).
• Unitary method: “Per 1” first, then multiply/divide to the target.

How to Spot Direct Proportion (Fast)

• Keywords: “directly proportional”, “at this rate”, “cost is proportional to quantity”.
• Table ratios \(y/x\) are equal for different rows.
• Graph passes through \((0,0)\) and is a straight line.
• Word cue: “Double the input → double the output” (same with triple/half, etc.).

5-Step Method (Exam Speed)

1. Confirm it’s direct proportion (\(y=kx\)).
2. Convert units to be consistent (kg, L, h, etc.).
3. Find \(k\) from one data pair or via unitary (“per 1”).
4. Use \(y=kx\) (or multiply the unit rate) to answer the question.
5. Sence-check: did scaling behave as expected? (Bigger in → bigger out.)

Unit Conversions (Common Pitfalls)

• Mass: 1000 g = 1 kg
• Volume: 1000 mL = 1 L
• Time: 60 min = 1 h
• Distance: 1000 m = 1 km

Tip: Standardise rates to “per kg”, “per L”, or “per hour” before comparing.

Graph & Interpretation

• Straight line through origin; gradient is the unit rate \(k\).
• Steeper line ⇒ larger rate (e.g. higher £/kg, faster km/h).
• If there’s a non-zero intercept (\(y=kx+c\)), it’s not direct proportion.

Common Traps (One-liners)

• Forgetting unit conversion → wrong rate.
• Using \(k=\dfrac{x}{y}\) instead of \(k=\dfrac{y}{x}\).
• Assuming direct proportion when there’s a fixed fee (intercept).
• Scaling one ingredient in a recipe but not the others.

Mini Checklist

• Did I verify “through the origin” or constant ratio?
• Are all units consistent?
• Have I written \(y=kx\) or found the “per 1” first?
• Does my answer scale logically?

Micro Practice (2 quick wins)

1. A line passes through \((0,0)\) and \((8, 18.4)\). Find \(k\) and the equation. Answer: \(k=18.4/8=2.3\); \(y=2.3x\).
2. £1.92 buys 600 g of pasta. What’s the price per kg and for 1.25 kg? Answer: £1.92/0.6 = £3.20 per kg; \(1.25\times£3.20=£4.00\).

Answers

Foundation

1. 5 cans = £3.25 → £3.25 ÷ 5 = £0.65 each. 12 cans = 12 × £0.65 = £7.80.
2. Speed = 180 ÷ 3 = 60 km/h. In 5 h: 5 × 60 = 300 km.
3. 120 pages ÷ 8 min = 15 pages/min. In 15 min: 15 × 15 = 225 pages.
4. Gradient = 9 ÷ 6 = 1.5. Equation: \(y = 1.5x\). At \(x=14\): 1.5 × 14 = 21.
5. £0.84 ÷ 0.6 = £1.40 per kg.
6. 250 ÷ 8 = 31.25 mL per pancake. 20 × 31.25 = 625 mL.
7. Speed = 96 ÷ 2 = 48 km/h. Time for 180 km = 180 ÷ 48 = 3.75 h (3 h 45 min).
8. £2.45 ÷ 7 = £0.35 each. 20 × 0.35 = £7.00.

Higher

1. (a) \(k = 17.5 ÷ 7 = 2.5\). (b) \(y = 2.5 × 12 = \mathbf{30}\).
2. (a) £43.20 ÷ 240 = £0.18/km = £18/100 km. (b) 390 × 0.18 = £70.20.
3. (a) 7.2 ÷ 18 = 0.4 kg/m. (b) 42 × 0.4 = 16.8 kg.
4. Pack A: £6.57 ÷ 9 = £0.73 each. Pack B: £10.36 ÷ 14 ≈ £0.74 each. Better buy: Pack A.
5. (a) \(k = 14 ÷ 3.5 = 4\). Equation: \(m = 4r\). (b) For \(r=11\): 4 × 11 = 44.
6. (a) £52.80 ÷ 320 = £0.165 per unit. (b) 500 × 0.165 = £82.50.
7. (a) 200 ÷ 250 = 0.8 g/mL. (b) 1.5 L = 1500 mL. Mass = 1500 × 0.8 = 1200 g (1.2 kg).
8. (a) Gradient = 42 ÷ 15 = 2.8. Equation: \(y = 2.8x\). (b) For \(x=24\): 2.8 × 24 = 67.2.

Challenge

1. (a) Not direct proportion (fixed fee breaks it). (b) A direct model would be “Cost = 0.12 × pages”.
2. (a) Gradient = \(b ÷ a\). Equation: \(y = \tfrac{b}{a}x\). (b) \(\tfrac{b}{a}\) = cost per unit.
3. Mass per metre = 2.8 ÷ 4 = 0.7 kg/m. Required length = 9.5 × 1.12 = 10.64 m. Mass = 10.64 × 0.7 = 7.448 kg ≈ 7.45 kg.
4. Shop A: £7 ÷ 3.5 = £2.00/kg. Shop B: £5.46 ÷ 2.8 = £1.95/kg. Best option: Shop B. For 10 kg = 10 × 1.95 = £19.50.
5. (a) The line with gradient \(1.6k\). (b) A larger \(k\) means higher cost per unit (steeper rate).
6. (a) £9.60 ÷ 120 = £0.08 per flyer. (b) 850 × 0.08 = £68. (c) Other company: £80 ÷ 1000 = £0.08 per flyer (same cost). Both are equal.
7. (a) Mass per litre = 0.44 ÷ 0.5 = 0.88 kg/L. (b) Volume = 1.76 ÷ 0.88 = 2 L. (c) Equation: \(m = 0.88V\).
8. (a) 18 ÷ 2 = £9 per hour; 45 ÷ 5 = £9 per hour. Ratio constant ⇒ direct proportion confirmed. (b) Cost = 9 × 8 = £72.

Conclusion & Next Steps

Direct proportion is the mathematics of fair, even scaling: double one quantity and the other doubles, halve one and the other halves. Across shopping problems, recipe scaling, physics-style formulas, and straight-line graphs through the origin, the same core idea appears: a constant “per one” rate links two variables via \(y=kx\). Once you can identify that constant (either by the unitary method or from a data pair using \(k=\tfrac{y}{x}\)), most questions reduce to a single multiply or divide.

Here’s what you should now be confident with:

• Spotting direct proportion: Constant ratio \(y/x\); graph passes through \((0,0)\); “at this rate”, “directly proportional”, or similar phrasing in word problems.
• Two interchangeable toolkits: (1) Unitary — go to “per 1” then scale; (2) Equation — compute \(k\) and use \(y=kx\).
• Interpreting \(k\): It is the gradient on a \(y\)–\(x\) graph and has meaningful units (e.g. £/kg, km/h, g/cm³). A larger \(k\) means a steeper line and a higher rate.
• Consistency of units: Convert carefully (g ↔ kg, mL ↔ L, min ↔ h) before comparing or calculating. Many marks are lost here.
• Sense-checking: If the input increases, the output should increase in the same ratio. Use “double → double” as a quick logic test.

To deepen and connect this knowledge, follow these next learning steps:

1. Inverse Proportion (\(y \propto \tfrac{1}{x}\)): Learn how fixed products lead to hyperbolic graphs and “double one, halve the other” behaviour. You’ll practise rearranging \(y=\tfrac{k}{x}\), interpreting tables, and solving word problems where increased input reduces output (e.g. speed vs time for a fixed distance).
2. Proportion in Context (Rates & Best Buy Mastery): Consolidate real-world reasoning: unit pricing, density–mass–volume, speed–distance–time, recipe scaling, and comparing mixed offers. You’ll build fluency in converting to a clean “per 1” and evaluating deals fairly.
3. Graphs & Gradients (Revision Link): Revisit straight-line graphs with and without intercepts. Distinguish \(y=kx\) (through the origin) from \(y=mx+c\) (with a fixed part). This prevents modelling errors in exam questions that include setup fees or standing charges.
4. Unit Conversions & Compound Units (Skill Drill): Practise quick conversions (e.g. p per 100 g → £/kg, m/s ↔ km/h). Build a mini “conversion map” to avoid last-minute slips under time pressure.
5. Mixed Practice Sets: Interleave direct proportion with ratio, percentages, and equations of lines. This mirrors exam conditions where topics are blended and improves long-term retention.

Exam strategy recap: (1) Confirm it’s direct proportion. (2) Standardise units. (3) Find \(k\) (or the unit rate). (4) Apply \(y=kx\) or scale the unit rate. (5) Sense-check for proportional scaling. Write the model first — it earns method marks and keeps your reasoning clean.

When you’re ready, move on to the next tutorial in this series:

• Inverse Proportion: \(y=\dfrac{k}{x}\); recognising “increase–decrease” patterns; solving tables, graphs, and context problems.
• Proportion in Context: multi-step rate problems, best-buy comparisons, density/speed/pressure families, and interpreting gradients in real scenarios.