Subject: Mathematics | Level: GCSE | Exam Board: Edexcel
Master the essential principles of Ratio, Proportion, and Rates of Change. This topic is heavily tested in GCSE Mathematics, offering significant marks for students who can accurately apply multipliers, interpret compound measures, and distinguish between direct and inverse proportion.
Revision Notes & Key Concepts
Revision Podcast Transcript
GCSE Mathematics Podcast — Ratio, Proportion and Rates of Change Running time: approximately 10 minutes Voice: Female, warm, conversational, enthusiastic tutor tone --- INTRO (approximately 1 minute) --- Hello and welcome! I'm so glad you're here, because today we're diving into one of the most practical and frequently-examined topics in your GCSE Maths course — Ratio, Proportion, and Rates of Change. Now, I know what some of you might be thinking — ratios? That sounds dry. But here's the thing: this topic is absolutely everywhere in real life. When a chef scales up a recipe, when an engineer calculates speed, when a bank applies compound interest to your savings — all of that is ratio and proportion in action. And examiners absolutely love testing it, because it rewards students who can think logically and apply methods carefully. In today's episode, we're going to cover the core concepts you need to know, walk through some exam-style examples together, highlight the most common mistakes candidates make — and I promise you, once you hear them, you'll never make them yourself — and then we'll finish with a quick-fire quiz to test your recall. By the end of this podcast, you'll feel genuinely confident walking into any question on this topic. So let's get started. --- CORE CONCEPTS (approximately 5 minutes) --- Let's begin with ratios. A ratio is simply a way of comparing two or more quantities. If I say the ratio of boys to girls in a class is 3 to 2, I'm saying that for every 3 boys, there are 2 girls. We write that as 3 colon 2. The first thing examiners always check is whether you've simplified your ratio correctly. To simplify, you find the highest common factor of all the parts and divide through. So if you have 12 to 8, the highest common factor is 4, giving you 3 to 2. Simple. Now here's a really important distinction that trips up so many candidates: the difference between a part-to-part ratio and a part-to-whole ratio. If the ratio of boys to girls is 3 to 2, that's part to part — boys compared to girls. But the fraction of the class that are boys is 3 over 5 — that's part to whole, because the total number of parts is 3 plus 2, which equals 5. Mixing these up is one of the most common errors in exam papers, so always ask yourself: am I comparing part to part, or part to the total? When you need to divide a quantity into a given ratio, use the unitary method. Here's how it works. Suppose you need to share £240 in the ratio 3 to 5. First, add the parts: 3 plus 5 equals 8 parts in total. Then find one part: 240 divided by 8 equals 30. Finally, multiply each part by 30: 3 times 30 is £90, and 5 times 30 is £150. Always check by adding your answers: 90 plus 150 equals 240. Correct. Now let's move on to percentage change, which is tested constantly. The most efficient method is the multiplier method. For a percentage increase, you add the percentage to 100 and divide by 100. So a 15% increase gives a multiplier of 1.15. For a decrease, you subtract from 100. So a 20% decrease gives a multiplier of 0.80. To apply it: if a jacket costs £80 and is reduced by 20%, you multiply 80 by 0.80 to get £64. Quick, clean, and far less prone to error than working out 20% and subtracting separately. The reverse percentage question is where many candidates lose marks. If you're told the price after a 20% reduction is £64, and you need to find the original price — do NOT add 20% back on. That's wrong. Instead, divide by the multiplier: 64 divided by 0.80 equals £80. The original value is always new value divided by the multiplier. Now, compound interest and growth and decay. These use repeated multiplication. The formula is: Final Amount equals Original Amount multiplied by the multiplier raised to the power of n, where n is the number of time periods. So if you invest £1000 at 3% compound interest for 4 years, you calculate 1000 multiplied by 1.03 to the power of 4. That gives you approximately £1125.51. This is Higher tier content, but it's well worth knowing. Let's talk about direct and inverse proportion. In direct proportion, as one variable increases, the other increases at the same rate. The equation is y equals k times x, where k is the constant of proportionality. The graph is a straight line through the origin. In inverse proportion, as one variable increases, the other decreases. The product of the two variables is always constant. The equation is y equals k divided by x. The graph is a curved hyperbola — it never touches the axes. To find k in either case, substitute a known pair of values. For example, if y is directly proportional to x and y equals 15 when x equals 5, then k equals 15 divided by 5, which equals 3. So the equation is y equals 3x. Finally, compound measures. These are measures that involve two different units combined. The three you must know are speed, density, and pressure. Speed equals distance divided by time. Density equals mass divided by volume. Pressure equals force divided by area. Think of the formula triangles — cover the quantity you want to find, and the triangle shows you whether to multiply or divide the other two. But critically — and I cannot stress this enough — always check your units are consistent before you start. If distance is in kilometres and time is in minutes, you cannot just divide and get kilometres per hour. You must convert first. Rates of change on graphs are read as gradients. The gradient of a distance-time graph gives speed. The gradient of a velocity-time graph gives acceleration. A steeper gradient means a faster rate of change. A horizontal line means no change — zero rate of change. --- EXAM TIPS AND COMMON MISTAKES (approximately 2 minutes) --- Right, let's talk exam technique. These are the mistakes I see most often in mark schemes, and they cost candidates marks every single year. Mistake number one: confusing part-to-part and part-to-whole ratios. If the ratio is 3 to 2 and the question asks for the fraction of the total, the answer is 3 over 5, not 3 over 2. Always add the parts to find the total. Mistake number two: adding a percentage back on to reverse a decrease. If something decreased by 20% and you want the original, divide by 0.80. Do not multiply by 1.20. Those are not inverse operations of each other. Mistake number three: not converting units before using compound measure formulas. If speed is in metres per second and you need kilometres per hour, you must convert. One metre per second equals 3.6 kilometres per hour. Mistake number four: treating inverse proportion like direct proportion. In inverse proportion, the product is constant. In direct proportion, the ratio is constant. They are fundamentally different relationships. Mistake number five: on graph questions, candidates often read the gradient incorrectly by not using two clearly separated points, or by forgetting to include units in their answer. The gradient of a distance-time graph is not just a number — it is a speed, and it needs units. For command words: when a question says "Calculate", you must show every step of working. Even if you get the final answer wrong, you can still earn method marks. When it says "Show that", you must demonstrate the full working — you cannot just write the answer. When it says "Estimate", round values to one significant figure first, then calculate. Time management: allow roughly one minute per mark. A 4-mark question should take around 4 to 5 minutes. If you're spending much longer, move on and come back. --- QUICK-FIRE RECALL QUIZ (approximately 1 minute) --- Okay, quiz time! I'll ask the question, give you a few seconds to think, then give the answer. Ready? Question one: What is the multiplier for a 35% increase? ... The answer is 1.35. Question two: If y is inversely proportional to x, and y equals 6 when x equals 4, what is the value of the constant k? ... k equals y times x, so 6 times 4 equals 24. k equals 24. Question three: A car travels 150 kilometres in 2.5 hours. What is its average speed? ... Speed equals distance divided by time: 150 divided by 2.5 equals 60 kilometres per hour. Question four: A price is reduced by 30% to £56. What was the original price? ... Divide by the multiplier: 56 divided by 0.70 equals £80. Question five: What does the gradient of a distance-time graph represent? ... Speed, or rate of change of distance. How did you do? If you got all five, brilliant — you've really got this. If you missed a couple, go back and review those sections. No stress — that's exactly what this podcast is for. --- SUMMARY AND SIGN-OFF (approximately 1 minute) --- Let's wrap up with the key things to take away from today. One: Always simplify ratios by dividing by the highest common factor, and know the difference between part-to-part and part-to-whole. Two: Use the multiplier method for percentage change. For increases, the multiplier is greater than 1. For decreases, it's less than 1. To reverse, divide by the multiplier. Three: For compound interest and growth or decay, use the formula: Final Amount equals Original Amount times the multiplier to the power of n. Four: Direct proportion gives a straight-line graph through the origin. Inverse proportion gives a curved hyperbola. Know your equations: y equals kx and y equals k over x. Five: For compound measures — speed, density, pressure — always check units before calculating. Use formula triangles to rearrange. Six: On graph questions, the gradient is the rate of change. Always include units in your answer. You've got this. Ratio, proportion and rates of change is a topic that rewards careful, methodical working — and that is absolutely something you can practise and get right every time. Good luck with your revision, and I'll see you in the next episode. Take care!
Key Terms & Definitions
- Ratio
- A mathematical statement showing the relative sizes of two or more values.
- Proportion
- An equation stating that two ratios are equivalent.
- Multiplier
- A decimal used to calculate percentage changes in a single step.
- Compound Measure
- A measure made up of two or more other measurements.
- Direct Proportion
- A relationship where two quantities increase or decrease at the same rate; their ratio is constant.
- Inverse Proportion
- A relationship where one quantity increases as the other decreases; their product is constant.
Worked Examples
Worked Example
Question: A shop has a sale with 20% off all prices. The sale price of a coat is £52. Calculate the normal price of the coat. (3 marks)
Solution: Step 1: Identify the multiplier for a 20% decrease. $100\% - 20\% = 80\%$. The multiplier is $0.80$. Step 2: Set up the equation. $\text{Original Price} \times 0.80 = £52$. Step 3: Rearrange to solve for the original price. $\text{Original Price} = £52 \div 0.80$. Final answer: £65
Worked Example
Question: y is inversely proportional to x. When x = 4, y = 7.5. Find the value of y when x = 5. (3 marks)
Solution: Step 1: Write the relationship as an equation: $y = \frac{k}{x}$ Step 2: Substitute the known values to find k: $7.5 = \frac{k}{4}$ Step 3: Calculate k: $k = 7.5 \times 4 = 30$. So the formula is $y = \frac{30}{x}$ Step 4: Substitute x = 5 into the formula: $y = \frac{30}{5}$ Final answer: $y = 6$
Worked Example
Question: A solid metal block has a mass of 4.5 kg and a volume of 500 cm³. Calculate the density of the metal in g/cm³. (3 marks)
Solution: Step 1: Note the required units for the answer (g/cm³). The mass is given in kg, so it must be converted to grams. Step 2: Convert mass: $4.5 \text{ kg} = 4500 \text{ g}$. Step 3: Use the density formula: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$. Step 4: Substitute values: $\text{Density} = \frac{4500}{500}$. Final answer: 9 g/cm³
Practice Questions
Question: A map has a scale of 1:50000. The distance between two towns on the map is 8 cm. Calculate the actual distance between the towns in kilometres. (3 marks)
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Question: In a school, the ratio of teachers to students is 1:15. There are 1200 students. How many teachers are there? (2 marks)
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Question: A car depreciates in value by 15% each year. It was bought for £18,000. Calculate its value after 3 years. (3 marks)
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Question: The time taken (t) to build a wall is inversely proportional to the number of workers (w). It takes 4 workers 15 days to build the wall. How long would it take 6 workers? (3 marks)
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Question: A cyclist travels at an average speed of 18 km/h for 40 minutes. Calculate the distance travelled in kilometres. (3 marks)
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