Percentage change: percentage increase and decrease using multipliers, percentage profit and loss, reverse percentages (finding the original amount), and simple and compound interest.

What this dot point is asking

Edexcel expects you to find percentage increases and decreases, ideally using multipliers, to calculate percentage profit and loss, to reverse a percentage change to find the original amount, and to work out simple and compound interest. Percentages run through everyday finance, so these questions are common and the multiplier method makes them fast and reliable.

Percentage increase and decrease with multipliers

The multiplier method turns a percentage change into a single multiplication. To increase by a percentage, add it to $100\%$ and convert to a decimal; to decrease, subtract from $100\%$.

So increasing $£250$ by $8\%$ is $250 \times 1.08 = £270$, and decreasing $£90$ by $35\%$ is $90 \times 0.65 = £58.50$. Multipliers also chain neatly: a $10\%$ rise followed by a $10\%$ fall is $\times 1.1 \times 0.9 = \times 0.99$, a net $1\%$ decrease, not back to the start.

Percentage change, profit and loss

To express a change as a percentage, divide the change by the original amount and multiply by $100$. If a shop buys an item for $£40$ and sells it for $£50$, the profit is $£10$, so the percentage profit is $\dfrac{10}{40} \times 100 = 25\%$. The denominator is always the original (cost) value, not the new one, which is the single most common slip.

Reverse percentages

A reverse percentage works backwards from a changed amount to the original. The key insight is that the percentage was applied to the original, so you must divide by the multiplier.

Reverse percentages are easy to spot because they give you the amount after the change and ask for the amount before. Dividing by the multiplier, rather than subtracting the percentage of the new value, is essential.

Simple and compound interest

Simple interest pays the same amount each year, calculated on the original sum. So $£500$ at $4\%$ simple interest earns $£20$ per year, giving $£560$ after three years. Compound interest pays interest on the growing balance, so it is calculated with a repeated multiplier.

So $£500$ at $4\%$ compound interest for three years is $500 \times 1.04^3 = £562.43$, slightly more than simple interest because each year's interest itself earns interest. The same repeated-multiplier idea models depreciation (a falling value) using a multiplier below $1$.

Choosing the right method

The wording of a question signals the method. "Per year" with a fixed cash amount each year is simple interest; "compound" or "the interest is added to the account each year" means a repeated multiplier. "In a sale, reduced to" followed by the new price is a reverse percentage. "Profit" or "loss" as a percentage divides the change by the cost price. Spotting which of these is being asked is half the battle, because each has a clean method once identified. A good habit on the calculator papers is to compute the multiplier first, since the same multiplier is reused whether you are doing one step or raising it to a power.

Try this

Q1. Increase $£80$ by $15\%$ using a multiplier. [2 marks]

• Cue. Multiplier $1.15$: $80 \times 1.15 = £92$.

Q2. A laptop costs $£540$ after a $10\%$ discount. Work out the original price. [3 marks]

• Cue. The discount leaves $90\%$, so original $= £540 \div 0.9 = £600$.