Connecting Ratios, Fractions and Linear Functions Revision Notes | GCSE AQA Higher Maths

1. Why Ratios, Fractions and Linear Functions Connect

You already know ratios compare quantities and fractions show parts of a whole. What's surprising is how these connect to straight-line graphs in ways that make exam questions much easier.

Think about mixing paint. If you mix red and blue in the ratio 3:2, you're actually using 3/5 red paint and 2/5 blue paint. The ratio tells you the relationship, the fractions tell you the actual proportions, and if you graph how much red paint you need against total paint, you get a straight line.

This connection appears everywhere in GCSE maths. Currency exchange rates are ratios that create linear graphs. Recipe scaling uses ratios that behave like fractions. Speed calculations involve ratios of distance to time that form linear relationships.

Understanding these connections means you can tackle ratio problems using fraction methods, solve fraction problems using ratio techniques, and recognise when a linear function is really just a ratio in disguise. This flexibility is exactly what examiners test. In the following section, we will explore practical applications through worked examples to solidify these ideas.

2. Converting Between Ratios and Fractions

Every ratio can be written as fractions, and every fraction comparison can be written as a ratio. The key is understanding what the denominator represents in each case.

When you have a ratio like 4:5, you're comparing two parts. To convert to fractions, you need to find what fraction each part is of the total. The total is 4 + 5 = 9 parts, so the first quantity is 4/9 and the second is 5/9.

Going the other way works too. If you have fractions 2/7 and 5/7, these represent parts of the same whole. The ratio is 2:5, because you're comparing the numerators when the denominators are equal.

Worked example: Converting Ratio 3:7 to Fractions

Express the ratio 3:7 as fractions of the total amount

1. Total parts = 3 + 7 = 10 — Add all parts in the ratio together
2. First part = $\frac{3}{10}$ of the total — 3 parts out of 10 total parts
3. Second part = $\frac{7}{10}$ of the total — 7 parts out of 10 total parts
4. $\frac{3}{10} + \frac{7}{10} = \frac{10}{10} = 1$ — The fractions must add to 1 (the whole amount)

Answer: 3:7 becomes 3/10 and 7/10 of the total

3. Ratios and Linear Functions

Here's where it gets interesting. The gradient of a straight line is actually a ratio in disguise. When you calculate gradient as 'rise over run', you're finding the ratio of vertical change to horizontal change.

Consider the line y = 2x + 3. The gradient is 2, which means for every 1 unit you move right, you move 2 units up. This creates the ratio 2:1 for vertical change to horizontal change. You can also think of this as the fraction 2/1 = 2.

This connection works both ways. If you know two quantities are in a constant ratio, their relationship will be linear. Currency exchange is a perfect example - if £1 = $1.30, then the ratio of pounds to dollars is always 1:1.30, creating the linear function D = 1.30P.

Worked example: Finding Gradient Using Ratios

Find the gradient of the line passing through points (2, 5) and (6, 13)

1. Horizontal change = 6 - 2 = 4 — Change in x-coordinates (right - left)
2. Vertical change = 13 - 5 = 8 — Change in y-coordinates (top - bottom)
3. Vertical : Horizontal = 8 : 4 — This is the ratio of changes
4. 8 : 4 = 2 : 1 — Divide both parts by their HCF (4)
5. Gradient = $\frac{8}{4} = 2$ — Gradient is vertical change ÷ horizontal change

Answer: Gradient = 2 (or ratio 2:1 for rise:run)

4. Proportional Relationships

When two quantities are in direct proportion, their ratio stays constant. This creates a special type of linear function that passes through the origin (0,0). The equation has the form y = kx, where k is the constant of proportionality.

Think about buying apples at £2 per kilogram. The cost is directly proportional to the weight, with constant of proportionality k = 2. The ratio of cost to weight is always 2:1, and if you graph cost against weight, you get a straight line through the origin.

This is different from other linear functions like y = 2x + 3. Here, the ratio of y to x changes as x increases because of the constant term. Only when the line passes through the origin do we have true proportionality.

Worked example: Identifying Proportional Relationships

Determine if y is directly proportional to x from this table: (1,3), (2,6), (3,9), (4,12)

1. Point (1,3): ratio = 3:1 = 3 Point (2,6): ratio = 6:2 = 3 Point (3,9): ratio = 9:3 = 3 Point (4,12): ratio = 12:4 = 3 — Calculate y÷x for each point
2. All ratios equal 3 — Constant ratio means direct proportion
3. When x = 0, y = 3 × 0 = 0 — Proportional relationships pass through (0,0)
4. y = 3x — Direct proportion: y = kx where k = 3

Answer: Yes, y is directly proportional to x with constant of proportionality 3

The graph above shows the difference between proportional (y = 2x) and non-proportional (y = 2x + 1) linear relationships. Both have the same gradient ratio of 2:1, but only the first represents true proportion because it passes through the origin.

5. Worked Examples and Applications

Let's tackle some exam-style problems that combine ratios, fractions, and linear functions. These questions often appear in the higher-tier papers and test your ability to move flexibly between different representations.

Worked example: Currency Exchange Problem

The exchange rate is £1 = €1.20. Sarah exchanges £150. Express the ratio of pounds to euros and find the linear function relating euros to pounds.

1. Euros = 150 × 1.20 = 180 — Multiply pounds by exchange rate
2. Pounds : Euros = 150 : 180 — Compare the two amounts
3. 150 : 180 = 5 : 6 — Divide both by HCF of 30
4. E = 1.20P — Euros = 1.20 × Pounds (direct proportion)
5. Gradient = $\frac{1.20}{1} = 1.20$ — This matches our exchange rate

Answer: Ratio is 5:6 (pounds:euros), Linear function is E = 1.20P

Worked example: Recipe Scaling Problem

A recipe for 8 people uses flour and butter in ratio 5:2. For 12 people, how much flour is needed if 300g butter is used?

1. Flour : Butter = 5 : 2 — Given ratio from recipe
2. If butter = 300g, then flour = $\frac{5}{2} \times 300 = 750$g — Use ratio as fraction: flour = (5/2) × butter
3. $\frac{\text{flour}}{\text{butter}} = \frac{750}{300} = \frac{5}{2}$ — Verify the ratio is maintained
4. F = 2.5B — Flour = 2.5 × Butter (since 5/2 = 2.5)

Answer: 750g flour needed (using linear relationship F = 2.5B)

6. Practice Problems

Now you understand the connections, let's practise with some targeted problems. These will help you recognise when to use ratio methods, fraction techniques, or linear function approaches.

These practice problems show the key skill: recognising which representation (ratio, fraction, or function) makes each problem easiest to solve. Sometimes converting between forms reveals the quickest path to the answer.

7. Exam Techniques and Common Mistakes

Exam questions love to test whether you can spot the connections between ratios, fractions, and linear functions. The key is recognising which form makes the problem easiest to solve.

Many students treat these as completely separate topics, missing opportunities to use the most efficient method. For example, a ratio problem might be quicker to solve using fraction methods, or a linear function question might be clearer when you think about it as a ratio.

Watch out for questions that give you information in one form but ask for the answer in another. These test your ability to convert flexibly between representations while maintaining mathematical accuracy.

Remember that understanding these connections gives you multiple ways to approach problems. If one method seems complicated, try converting to a different representation - often this reveals a much simpler solution path.