Percentage Change and Financial Applications Revision Notes | GCSE AQA Higher Maths

1. Why Percentage Change Matters

A jacket costs £80 in a sale after a 20% reduction. A savings account promises 3% interest per year. A house price rises by 12%. These are all percentage change problems — and they appear constantly in GCSE Higher exams, often disguised in financial contexts.

The good news is that every percentage change question, no matter how it is dressed up, uses the same core toolkit. Master that toolkit and the financial applications follow naturally. In many GCSE exam papers, questions regarding percentage change are posed in diverse ways to assess students' understanding. It is crucial to clearly identify what each quantity represents. When calculating an increase or decrease, one must first determine the base value—the original value—observing that percentage change is computed using this original value. For example, calculating a 20% increase involves multiplying the original value by 1.20, which effectively increases the number by 20%. Conversely, a 25% decrease can be quickly computed by multiplying the original value by 0.75. Furthermore, mastering the multiplier method saves time and ensures greater accuracy during examinations. Students should also be aware that applying percentages directly to the new value is a typical error. It is important to inspect any changes, especially in financial contexts where miscalculations lead to significant errors. In addition to percentage problems involving price reductions and interest calculations, reverse percentage questions are frequently tested in exams. In these problems, students are asked to determine the original price given the reduced price and the percentage reduction, thereby reinforcing the concept of division by the appropriate multiplier. The lesson also covers a variety of financial contexts, including simple interest, which is calculated on the original principal every year. Simple interest problems require the use of the formula I = (P × R × T) / 100, and then the total amount is found by adding the interest to the principal. Paying careful attention to units is vital since omitting symbols such as £ or % can result in loss of marks. To consolidate understanding, multiple step-by-step examples are provided throughout this lesson. They illustrate not only the computational procedures but also effective exam techniques such as writing clear working and verifying answers. Additional practice questions are provided in the math practice section, which reinforces the learned concepts through real-world applications, such as discounting in sales, salary increases, and mortgage calculations. By continuously reviewing common mistakes, such as confusing the base value or misapplying the multiplier, students can build confidence and proficiency in handling percentage change questions. Moreover, integrating these methods into your study practice will improve your speed and accuracy during exams. Finally, understanding the underpinning business applications of percentage changes is beneficial in bridging the gap between mathematical theory and practical financial decision-making.

2. The Core Formula and the Multiplier Method

There are two quantities you always need: the original value and the change. The percentage change formula connects them directly.

For applying a percentage change to a quantity, the multiplier method is faster and less error-prone than working out the percentage separately and then adding or subtracting. The multiplier for an increase of $p\%$ is $\left(1 + \frac{p}{100}\right)$, and for a decrease of $p\%$ it is $\left(1 - \frac{p}{100}\right)$.

3. Percentage Increase and Decrease — Worked Examples

Here is where the multiplier method earns its place. Rather than finding 20% and then adding it on, you multiply by 1.20 in a single step. The worked examples below show both the calculation and the mark-winning annotation examiners look for.

Worked example: Percentage Increase — calculator-allowed exam question Method

A television costs £340. Its price increases by 15%. Calculate the new price. [2 marks]

1. $\text{Multiplier} = 1 + \dfrac{15}{100} = 1.15$ — A 15% increase means we keep 100% and add 15%, giving 115% of the original, which is 1.15 as a decimal. [M1 — correct multiplier]
2. $340 \times 1.15 = 391$ — Verify: $340 \times 1 = 340$ and $340 \times 0.15 = 51$, so $340 + 51 = 391$. [A1 — correct answer with units]
3. $\text{New price} = £391$ — Always include the £ sign in financial questions — missing units can cost a mark.

Answer: £391

Worked example: Percentage Decrease — Without a Calculator

A coat costs £120. It is reduced by 35% in a sale. Work out the sale price. [2 marks]

1. $\text{Multiplier} = 1 - \dfrac{35}{100} = 0.65$ — A 35% decrease means we keep 65% of the original. [M1 — correct multiplier]
2. $120 \times 0.65$ $= 120 \times 0.6 + 120 \times 0.05$ $= 72 + 6$ $= 78$ — In questions without a calculator, split the decimal into parts you can handle mentally. Show all arithmetic working.
3. $\text{Sale price} = £78$ — [A1 — correct answer with £ sign]

Answer: £78

4. Finding the Percentage Change

Sometimes the question gives you both the original and new values and asks you to find the percentage change. This is where the formula at the top of the lesson comes in. The key is always to divide by the original value, not the new one.

Worked example: Calculating Percentage Change

Mohammed bought a car for £8,500. He sold it for £6,460. Calculate the percentage change in the value of the car. [3 marks]

1. $\text{Change} = 6460 - 8500 = -2040$ — The negative sign tells us this is a decrease. [M1 — correct subtraction]
2. $\text{Percentage change} = \dfrac{-2040}{8500} \times 100$ — Always divide by the original value (£8,500), not the new value. [M1 — dividing by original]
3. $\dfrac{2040}{8500} \times 100 = 0.24 \times 100 = 24$ — Verify: $8500 \times 0.24 = 2040$. Correct.
4. $\text{Percentage decrease} = 24\%$ — [A1 — correct value and direction stated] The question asks for percentage change, so state whether it is an increase or decrease.

Answer: 24% decrease

5. Reverse Percentages — Finding the Original Value

This is the concept that trips up the most students. The question gives you the value after a percentage change and asks for the value before. The trap is to apply the percentage to the number you are given — but that number is already the changed value, not the original.

The method is to identify the multiplier that was applied, then divide by it to reverse the change. Think of it as undoing the multiplication.

Worked example: Reverse Percentage — Finding the Original Price

A pair of trainers costs £63 after a 30% reduction. Work out the original price. [3 marks]

1. $100\% - 30\% = 70\%$ So £63 represents 70% of the original price. — After a 30% reduction, the customer pays 70% of the original. [M1 — correct identification]
2. $\text{Original} \times 0.70 = 63$ — The multiplier for a 30% decrease is 0.70.
3. $\text{Original} = \dfrac{63}{0.70} = 90$ — Dividing by 0.70 undoes the 30% reduction. [M1 — dividing by correct multiplier]
4. $90 \times 0.70 = 63$ ✓ — Always check: multiply your original by the multiplier and confirm you get the sale price. [A1 — correct answer]

Answer: £90

6. Simple Interest in Financial Mathematics

Simple interest is a direct application of percentage change to money over time. Unlike compound interest (which is a separate topic), simple interest is calculated only on the original amount — the principal — every year. The interest earned each year stays the same.

The formula $I = \frac{P \times R \times T}{100}$ gives the total interest earned. To find the total amount in the account, add the interest to the principal: $\text{Total} = P + I$.

Worked example: Simple Interest Calculation

Ben invests £4,500 in a savings account that pays simple interest at a rate of 3.5% per year. Calculate the total amount in the account after 4 years. [3 marks]

1. $P = 4500$, $R = 3.5$, $T = 4$ — P is the principal (starting amount), R is the annual interest rate as a percentage, T is the time in years.
2. $I = \dfrac{P \times R \times T}{100} = \dfrac{4500 \times 3.5 \times 4}{100}$ — [M1 — correct substitution into formula]
3. $4500 \times 3.5 = 15750$ $15750 \times 4 = 63000$ — Work step by step: $4500 \times 3.5 = 4500 \times 3 + 4500 \times 0.5 = 13500 + 2250 = 15750$. Then $15750 \times 4 = 63000$.
4. $I = \dfrac{63000}{100} = £630$ — [M1 — correct interest value]
5. $\text{Total} = P + I = 4500 + 630 = £5130$ — [A1 — correct total with £ sign] The question asks for the total amount, not just the interest.

Answer: £5,130

7. Bringing It Together — Multi-Step Financial Problems

GCSE Higher exam questions often combine percentage change with a financial context across multiple steps. The method stays the same — identify the multiplier, apply it, and check your answer makes sense in context.

8. Exam Technique and Common Mistakes

Percentage change questions are reliable mark-earners when you follow a consistent method. The mistakes below are the ones that appear most often in mark scheme reports — knowing them in advance puts you ahead.