Study case: calculating compound interest over three years In this Year 10 Compound Interest Problem Tutor, I use one study case to show how repeated percentage change works. The problem is: £500 is invested at 4 percent compound interest per year for 3 years. Find the value of the investment after 3 years. The answer is £562.43 to the nearest penny. I like this example because it shows why compound interest is not the same as adding the same amount each year. Before solving, I ask the student to identify the multiplier. A 4 percent increase means the investment becomes 104 percent of its value each year. As a decimal, 104 percent is 1.04. Since the interest is compound, the multiplier is used repeatedly. For 3 years, we multiply by 1.04 three times, or use 1.04^3. Method 1 is the multiplier formula method. The final amount is 500 x 1.04^3. First, 1.04^3 is approximately 1.124864. Multiplying by 500 gives 562.432. Rounded to the nearest penny, the value is £562.43. I explain that the power of 3 represents the three repeated years of growth. Method 2 is the year-by-year method. After one year, £500 becomes £520. After two years, £520 is multiplied by 1.04 to give £540.80. After three years, £540.80 is multiplied by 1.04 to give £562.432, which rounds to £562.43. This method is useful because students can see the amount of interest increasing each year. Method 3 is the comparison with simple interest. With simple interest at 4 percent, the interest each year would be 4 percent of £500, which is £20. Over three years, that would be £60, giving £560. Compound interest gives £562.43 because the second and third years earn interest on earlier interest as well. This comparison helps students understand the meaning of compound growth. After the three methods, I compare them. The multiplier formula is fastest and prepares students for GCSE exam questions. The year-by-year method builds understanding. The simple interest comparison explains why the compound answer is slightly higher. I want students to know all three so they can choose the best method for the question. Common mistakes include using 0.04 instead of 1.04 as the multiplier, multiplying by 1.04 only once, adding 4 percent of the original amount every year, rounding too early, and forgetting money notation. I use these mistakes as teaching points. If a student uses 0.04, I ask whether the answer should be smaller than the starting amount. Since the investment is growing, the multiplier must be greater than 1. This topic is important for GCSE because compound percentage change appears in interest, depreciation, population growth, inflation and repeated discounts. It also links closely to exponential growth and decay. A student who understands multipliers will find many percentage problems easier and more consistent. A useful extension is to calculate depreciation: a car worth £9000 loses 12 percent of its value each year for 4 years. The multiplier would be 0.88, because the value becomes 88 percent each year. Another extension is to work backwards and find the original amount before compound growth. These questions build flexibility. In one-to-one tutoring, I would finish by asking the student to solve one compound increase and one compound decrease question. I would ask them to write the multiplier first, then the number of repeats. This routine prevents most percentage errors. My aim is for the student to understand that compound interest is repeated multiplication, not repeated addition. 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.