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Higher - Percentages
infoWhy this? We are teaching this unit so that pupils can work confidently with percentages in a range of contexts, including percentage change, reverse percentages, compound interest and depreciation, enabling them to solve financial and growth/decay problems and interpret percentage information accurately
scheduleWhy now? We are teaching it now in Year 10 so that pupils can apply secure percentage and growth/decay skills across their current GCSE topics and real-life contexts, particularly in questions involving money, proportion and exponential change
neurologyYou need to know
- A percentage is a proportion out of 100, so `p\% = \frac{p}{100}` of a quantity.
- To find `p\%` of an amount `A`, you can calculate `\frac{p}{100} \times A`.
- A decimal percentage can be converted to a decimal multiplier by dividing by 100 (for example, `12.5\% = 0.125`).
- To express one quantity `x` as a percentage of another `y` (with `y \ne 0`), the percentage is `\frac{x}{y} \times 100\%`.
- Increasing an amount by `p\%` is equivalent to multiplying by the growth multiplier `1 + \frac{p}{100}`.
- Decreasing an amount by `p\%` is equivalent to multiplying by the decay multiplier `1 - \frac{p}{100}`.
- Percentage change from an original value `O` to a new value `N` is `\frac{N-O}{O} \times 100\%` (positive for an increase, negative for a decrease).
- Simple interest adds the same amount each time period because it is calculated only on the original principal.
- Simple interest can be calculated using `I = P \times r \times t`, where `P` is principal, `r` is the interest rate per time period as a decimal, and `t` is the number of time periods.
- The amount after simple interest is `A = P + I`.
- Compound interest (and repeated depreciation) changes by a constant percentage each time period because it is calculated on the current amount.
- Compound growth can be modelled by `A = P(1+r)^t`, where `r` is the rate per time period as a decimal and `t` is the number of time periods.
- Compound depreciation can be modelled by `A = P(1-r)^t`, where `r` is the depreciation rate per time period as a decimal.
- Reverse percentages work by dividing by the correct multiplier (for example, if the final amount is after a `p\%` increase, then `\text{original} = \frac{\text{final}}{1+\frac{p}{100}}`).
- In growth and decay problems, the multiplier must match the context and time unit (for example, a monthly rate must be applied monthly, and the power `t` counts the number of months).
rocket_launchYou must be able to
- Calculate `p\%` of an amount by converting `p\%` to a decimal and multiplying, selecting an efficient method (fraction, decimal, or multiplier) where appropriate.
- Express one quantity as a percentage of another by forming the fraction `\frac{\text{part}}{\text{whole}}`, multiplying by 100, and stating the answer with the `%` sign.
- Increase or decrease a value by a given percentage by applying the correct multiplier `1 \pm \frac{p}{100}` and giving the result to an appropriate level of accuracy for the context (e.g. money to 2 d.p.).
- Compute simple interest over time by identifying `P`, `r`, and `t`, calculating `I = P \times r \times t`, and then finding the final amount.
- Calculate percentage change by substituting into `\frac{N-O}{O} \times 100\%` and interpreting the sign to decide increase or decrease.
- Solve reverse percentage problems by writing an equation with the appropriate multiplier and rearranging to isolate the original value (usually by dividing by the multiplier).
- Calculate compound interest or depreciation by setting up `A = P(1 \pm r)^t`, substituting correctly, and evaluating the power accurately.
- Set up, solve, and interpret growth/decay problems by defining variables, forming an exponential equation from the context, solving for the unknown (amount, rate, or time), and checking the answer is realistic (e.g. decay gives a smaller value).