Key Stage 3 maths · Ratio, proportion and rates of change

KS3 percentage change, proportion, speed and density: a parent’s guide

This topic is the maths behind everyday decisions: sale prices, savings, journeys and best buys. In Key Stage 3 children work out percentage increases and decreases, find an original price from a changed one, calculate simple interest, solve direct and inverse proportion problems and use compound units such as speed, density and price per kilogram.

Percentage change, proportion, compound units, explained

Percentage change problems are easiest with a multiplier. A 20% increase means the new amount is 120% of the old, so multiply by 1.2. A 15% decrease leaves 85%, so multiply by 0.85. The change can also be measured: the percentage change is the actual change divided by the original, multiplied by 100, so a price rising from £40 to £46 is 6 ÷ 40 × 100 = 15%. The word “original” is the key, because a percentage change is always worked out from the starting amount. Original value problems work backwards: if the sale price is known, divide by the multiplier to find the price before the sale. Simple interest is the same idea applied to savings: the interest is the same amount each year, calculated on the original sum, so £800 at 3% a year for 4 years earns 800 × 0.03 × 4 = £96. Interest that is earned on interest is a later topic.

Two quantities are in direct proportion when they increase in the same ratio: doubling one doubles the other, and their ratio never changes. If 5 notebooks cost £8, then 15 cost £24. On a graph this gives a straight line through the origin, and the algebraic form is y = kx, where k is the constant of proportionality. Inverse proportion is the opposite: as one quantity increases the other decreases so that the product stays constant. If 6 people paint a fence in 8 days, then 4 people take 12 days, because the total work is 6 × 8 = 48 person-days and 48 ÷ 4 = 12. The graph of an inverse relationship is a downward curve that never touches the axes, and the algebra is y = k/x.

Compound units combine two measurements. Speed is distance divided by time, so 150 km in 2.5 hours is 60 km/h, and the same formula can be rearranged to find distance (speed × time) or time (distance ÷ speed). Density is mass divided by volume, such as 54 g in 20 cm³ giving 2.7 g/cm³. Unit pricing is the price for a single unit, such as pence per gram, and lets shoppers compare different pack sizes fairly. The units written after the number, such as km/h or g/cm³, tell the child which division to do.

Throughout, children are expected to reason as well as calculate. They decide which method fits, check whether an answer is sensible, and communicate clearly. A sale price that is higher than the original price is wrong, however neat the working. Time conversions also matter, since 45 minutes is 0.75 of an hour, not 0.45.

Words your child will hear

Multiplier: The number you multiply by to apply a percentage change, such as 1.2 for +20% or 0.85 for −15%.
Percentage change: The change divided by the original amount, written as a percentage.
Simple interest: Interest worked out on the original sum only, so the same amount is added each year.
Direct proportion: Two quantities that increase together in the same ratio, such as the cost of notebooks and the number bought.
Inverse proportion: Two quantities where one halves as the other doubles, such as workers and days to finish a job.
Compound unit: A unit made from two others, such as km/h, g/cm³ or pence per gram.
Density: Mass divided by volume, such as 2.7 g/cm³.

What the National Curriculum says

In Key Stage 3, pupils should be taught to:

Solve problems involving percentage change, including: percentage increase, decrease and original value problems and simple interest in financial mathematics
Solve problems involving direct and inverse proportion, including graphical and algebraic representations
Use compound units such as speed, unit pricing and density to solve problems

Objectives from the National Curriculum in England, © Crown copyright, used under the Open Government Licence v3.0.

A worked example: finding the original price

In a sale, a coat is reduced by 15% and now costs £51. What was the original price?

A 15% reduction means the sale price is 100% − 15% = 85% of the original. So £51 represents 85%.
Find 1% by dividing by 85.
```
85% = £51
 1% = £51 ÷ 85 = £0.60
```
Find 100% by multiplying by 100.
```
100% = £0.60 × 100 = £60
```
Check by applying the 15% reduction to the answer: 15% of £60 is £9, and £60 − £9 = £51, which is the sale price.
The same answer comes from using the multiplier. The original × 0.85 = 51, so the original = 51 ÷ 0.85 = 60.

The coat originally cost £60.

Common mistakes and how to help

Adding the percentage back to the new price: 15% of £51 on top of £51

The reduction was taken off the original price, but the child works out 15% of the sale price and gets £58.65 instead of £60.

What helps: The 15% was of the original price, and here the original price is the unknown. Use 85% = £51 to find 1% first, then 100%.

Thinking an increase and an equal decrease cancel out

A price goes up 20% and then falls 20%, so surely it is back to the start? In fact the 20% decrease is of a larger amount: × 1.2 × 0.8 = × 0.96, a net fall of 4%.

What helps: Try it with £100: it goes to £120, then 20% of £120 is £24, giving £96. Multipliers make this clear because 1.2 × 0.8 is not 1.

Treating minutes as hundredths of an hour in speed questions

A journey of 2 hours 30 minutes is written as 2.30 hours, giving a speed of 150 ÷ 2.30 instead of 150 ÷ 2.5.

What helps: Convert minutes by dividing by 60: 30 minutes is 0.5 hours. Or work in minutes throughout and convert at the end. A sensible check is that 2 h 30 min is between 2 and 3 hours.

Using direct proportion when the situation is inverse

If 6 people take 8 days, a child says 12 people take 16 days because “double” goes with “double”. More workers should make the job faster, not slower.

What helps: Ask whether more of one means more or less of the other. Work out the total (6 × 8 = 48 person-days) and divide by the new number of workers: 48 ÷ 12 = 4 days.

How to help at home

Compare unit prices. Choose two pack sizes of the same product and work out the price per 100 g or per litre, for example 500 g for £1.60 (32p per 100 g) against 750 g for £2.25 (30p per 100 g). Which is better, and is the cheaper pack still worth buying if half of it goes off?
Work out sale prices. When you see 20% off, ask your child to find the new price using a multiplier (× 0.8), not just taking off 20%. Then reverse it: if a jacket is £48 after 20% off, what was the original?
Average speed on a journey. Note the distance and the time of a trip and calculate the average speed. Then ask how long the same journey would take at a different speed, and why it is an average, not the speed all the time.
Try simple interest. If your child saves pocket money, show how much interest £200 would earn at 2% a year over 3 years (£12), and discuss how that compares with what banks actually do, where interest is usually added to interest.

Builds on

Year 6: Ratio, percentages and scale factors

Leads on to

Key Stage 4: Compound units, proportion and growth
Key Stage 4: Proportion equations and rates of change