Study case: finding the original price after a discount In this GCSE Reverse Percentages Price Problem Tutor, I use one practical study case to show how reverse percentage questions work. The problem is: a coat is reduced by 20 percent in a sale and now costs £64. Find the original price. The answer is £80. I like this example because many students incorrectly take 20 percent off 64, but the £64 is already the reduced price, not the original price. Before solving, I ask the student to identify what £64 represents. If the coat has been reduced by 20 percent, then the sale price is 80 percent of the original price. That means £64 corresponds to 80 percent. We need 100 percent. This interpretation is the key to reverse percentages. Method 1 is the multiplier method. A 20 percent reduction means multiply the original price by 0.8. So original price times 0.8 = 64. To reverse this, divide 64 by 0.8. This gives 80. Therefore the original price was £80. I explain that dividing by the multiplier reverses the percentage change. Method 2 is the unitary method. If 80 percent is £64, then 10 percent is £8 because 64 divided by 8 is 8. Therefore 100 percent is 10 times £8, which is £80. Some students prefer this method because it feels more concrete. It also works well when percentages break down into simple chunks. Method 3 is the checking method. If the original price was £80, then 20 percent of £80 is £16. Subtracting £16 from £80 gives £64. This matches the sale price, so the answer is correct. I always include this check because reverse percentage answers can feel counterintuitive at first. After the methods, I compare them. The multiplier method is fastest and prepares students for compound percentage questions. The unitary method builds understanding. The checking method confirms the context. I want students to understand why 20 percent of £64 is not the right route. The discount was taken from the original price, not from the sale price. Common mistakes include subtracting 20 percent from the reduced price, multiplying by 1.2 instead of dividing by 0.8, confusing increase and decrease, and forgetting money units. I use these mistakes as teaching points. If a student uses the wrong base value, we identify what amount represents 100 percent and what amount represents the changed value. This topic is important for GCSE and IGCSE because reverse percentages appear in finance, tax, discounts, profit, interest and depreciation. They also prepare students for compound percentage change, where the multiplier may need to be applied several times or reversed over several years. A strong foundation in reverse percentages makes those questions much easier. A useful extension is: a price increases by 15 percent and becomes £92. Find the original price. In that case, the multiplier is 1.15, so the original price is 92 divided by 1.15. Another extension is to combine two changes, such as a discount followed by VAT. These questions help students become flexible with multipliers. In one-to-one tutoring, I would finish by asking the student to write a sentence before calculating: the final value is what percent of the original? This habit prevents many mistakes. Once the student can identify the correct percentage value, the calculation becomes straightforward. For support with this topic, students can book one-to-one tutor services: https://www.mastermathstutoring.co.uk/services. For the wider course page, visit GCSE and IGCSE Maths tutoring: https://www.mastermathstutoring.co.uk/gcse-igcse. To ask about lessons or availability, use contact MasterMaths Tutoring: https://www.mastermathstutoring.co.uk/contact-me.