Percentage increase and decrease using multipliers, reverse percentages, and simple and compound interest including depreciation.

What this dot point is asking

AQA wants you to work out percentage increases and decreases using multipliers, solve reverse-percentage problems (finding the original amount before a change), and calculate simple and compound interest, including depreciation. These are heavily applied to money and growth contexts, and the multiplier method is the efficient route AQA expects, especially for repeated (compound) change.

Percentage change with multipliers

To increase $\pounds 240$ by $15\%$, compute $240 \times 1.15 = \pounds 276$. To decrease it by $8\%$, compute $240 \times 0.92 = \pounds 220.80$. Multipliers also let you chain several changes: a $10\%$ rise followed by a $10\%$ fall gives $\times 1.1 \times 0.9 = \times 0.99$, a net $1\%$ decrease (not back to the start, a frequent surprise).

Reverse percentages

A reverse percentage finds the original amount before a stated change. The new amount equals the original times the multiplier, so dividing the new amount by the multiplier recovers the original.

The key warning: never take $30\%$ off the sale price; the percentage was of the original, so you must divide by the multiplier.

Simple and compound interest

Simple interest pays the same amount each year, based only on the original principal: $\pounds 500$ at $4\%$ simple interest earns $500 \times 0.04 = \pounds 20$ a year, so $\pounds 60$ over three years. Compound interest applies the multiplier each year to the growing balance.

For $\pounds 2000$ at $5\%$ compound interest over $4$ years, the value is $2000 \times 1.05^4 = 2000 \times 1.21551 = \pounds 2431.01$. A car worth $\pounds 12\,000$ depreciating at $15\%$ a year is worth $12\,000 \times 0.85^3 = \pounds 7369.50$ after three years.

Why compound beats simple interest

Over a single year, simple and compound interest at the same rate give the same amount. The difference appears from the second year, because compound interest pays interest on the interest already earned, while simple interest always pays on the original principal only. The gap widens each year, which is why long-term savings and debts are almost always compound. A question may ask you to find how much more compound interest earns than simple over a set period: calculate each total separately and subtract.

Finding the rate or the time

Harder questions reverse the process. Given the start and end values, you can find the annual rate: if $\pounds 5000$ grows to $\pounds 5618$ in two years at compound interest, then $1.0r^2 = \dfrac{5618}{5000} = 1.1236$, so the yearly multiplier is $\sqrt{1.1236} = 1.06$, a rate of $6\%$. Trial and improvement, or systematically testing whole-number rates with a calculator, is the expected GCSE method for finding how many years an investment takes to reach a target, since logarithms are beyond the specification.