Modelling Direct and Inverse Proportion Revision Notes | GCSE AQA Higher Maths

1. Why Proportion Matters

Here is a question worth thinking about: if you double the number of workers on a building site, does the job take twice as long, half as long, or the same time? The answer depends entirely on the type of relationship between the two quantities — and that is exactly what proportion is about.

Proportion describes how two quantities change relative to each other. It shows up everywhere: currency exchange, recipe scaling, speed and journey time, fuel consumption, and engineering calculations. Mastering it means you can set up an equation from a real-world situation and solve it efficiently.

There are two flavours: direct proportion, where both quantities grow together, and inverse proportion, where one grows as the other shrinks. The higher-tier specification requires you to handle both algebraically and graphically — so this lesson covers both in full.

2. Direct Proportion — The Core Idea

Think about buying apples. One apple costs 30p, two cost 60p, three cost 90p. Every time you buy one more apple, the cost goes up by exactly the same amount. The cost and the number of apples are in direct proportion.

The key test: if you double one quantity and the other doubles too, they are directly proportional. More precisely, the ratio of the two quantities stays constant. That constant ratio is what we call $k$, the constant of proportionality.

Notice what this means graphically: $y = kx$ is a straight line through the origin. If the graph of two quantities is a straight line passing through $(0, 0)$, they are directly proportional. The gradient of that line is $k$.

The method for every direct proportion problem follows the same three steps: write the proportionality statement using $\propto$, convert it to an equation by introducing $k$, then substitute a known pair to find $k$. After that, the equation is yours to use for any value.

Worked example: Direct Proportion: Finding k and Solving

$y$ is directly proportional to $x$. When $x = 4$, $y = 20$. Find $y$ when $x = 7$. [3 marks]

1. $y \propto x$ — This tells us the relationship type. [Setup — no mark yet, but essential for method marks]
2. $y = kx$ — Replace the $\propto$ symbol with $= k$. [M1 — correct form of equation]
3. $20 = k \times 4 \Rightarrow k = \frac{20}{4} = 5$ — Divide both sides by 4. Always verify: $5 \times 4 = 20$ ✓ [M1 — correct substitution and value of k]
4. $y = 5x$ — This equation works for any value of $x$ in this proportion relationship.
5. $y = 5 \times 7 = 35$ — [A1 — correct final answer]

Answer: $y = 35$

This three-step method — write $\propto$, introduce $k$, substitute to find $k$ — works for every proportion problem on the paper. Commit it to memory now, because it is the same structure whether the relationship is direct or inverse.

Before moving on, check your understanding of the direct proportion setup. This is where students often lose marks by skipping the $k$ step entirely and trying to scale directly — which only works for the simplest cases.

3. Inverse Proportion — When One Goes Up, the Other Goes Down

Back to the building site: if 2 workers take 12 days to finish a job, how long would 4 workers take? Your instinct says 6 days — and you are right. More workers means less time. This is inverse proportion: as one quantity increases, the other decreases at the same rate.

The key test for inverse proportion: if you double one quantity and the other halves, they are inversely proportional. More precisely, the product of the two quantities stays constant. In the example above, $2 \times 12 = 24$ and $4 \times 6 = 24$ — the product is always 24.

The graph of $y = \frac{k}{x}$ is a curve called a reciprocal curve (or hyperbola). It never touches either axis — as $x$ gets very large, $y$ gets very small but never reaches zero, and vice versa. This is completely different from the straight-line graph of direct proportion.

Worked example: Inverse Proportion: Finding k and Solving

$y$ is inversely proportional to $x$. When $x = 3$, $y = 8$. Find $y$ when $x = 6$. [3 marks]

1. $y \propto \frac{1}{x}$ — The $\frac{1}{x}$ signals inverse proportion. [Setup for method marks]
2. $y = \frac{k}{x}$ — [M1 — correct form of inverse proportion equation]
3. $8 = \frac{k}{3} \Rightarrow k = 8 \times 3 = 24$ — Multiply both sides by 3. Verify: $\frac{24}{3} = 8$ ✓ [M1 — correct value of k]
4. $y = \frac{24}{x}$ — This equation now works for any value of $x$.
5. $y = \frac{24}{6} = 4$ — Verify using the product check: $6 \times 4 = 24 = k$ ✓ [A1 — correct final answer]

Answer: $y = 4$

Notice the product check in Step 5: $6 \times 4 = 24$, which equals $k$. This is a powerful sanity check for inverse proportion — if the product of your answer pair does not equal $k$, something has gone wrong.

Consider this worked example of inverse proportion. If $y$ is inversely proportional to $x$ and we know $y = 8$ when $x = 2$, we can determine $k$ by using the equation $y = \frac{k}{x}$. Here, $8 = \frac{k}{2}$, so multiplying both sides by 2 gives us $k = 16$. Now, if we want to find $y$ when $x = 4$, we substitute into the equation: $y = \frac{16}{4} = 4$.

4. Graphs of Direct and Inverse Proportion

The graph is one of the most reliable ways to identify the type of proportion — and Examiners regularly ask you to sketch or interpret these graphs. Knowing the shape instantly tells you which equation to use.

For direct proportion ($y = kx$): the graph is a straight line through the origin. The gradient equals $k$. A steeper line means a larger value of $k$, meaning $y$ grows faster relative to $x$.

For inverse proportion ($y = \frac{k}{x}$): the graph is a reciprocal curve. It sits in the first quadrant (when $k > 0$), curving steeply near the $y$-axis and flattening out near the $x$-axis. Crucially, it never crosses either axis — both axes are asymptotes.

In the graph above, the straight line through the origin represents $y = 2x$ (direct proportion with $k = 2$). The curve represents $y = \frac{8}{x}$ (inverse proportion with $k = 8$). Notice how the curve approaches but never reaches the axes.

When a question gives you a table of values and asks you to determine the type of proportion, apply both tests systematically. Calculate $\frac{y}{x}$ for each pair — if it is constant, direct proportion. Calculate $xy$ for each pair — if it is constant, inverse proportion. If neither is constant, it is neither type.

5. Harder Problems and Exam Variations

Some exam questions make proportion problems harder by using a table of values rather than giving you a pair directly, or by asking you to find $x$ given $y$ (working backwards). The method is identical — find $k$ first, write the equation, then solve.

Another common variation: the question gives you a table and asks you to show that the quantities are in direct (or inverse) proportion. Here, you must demonstrate the constant ratio or constant product for all pairs in the table — not just one. Show all your checks explicitly to earn full marks.

Worked example: Working Backwards: Finding x Given y

$y$ is inversely proportional to $x$. When $x = 4$, $y = 9$. Find $x$ when $y = 6$. [3 marks]

1. $k = xy = 4 \times 9 = 36$ — For inverse proportion, $k$ is always the product $xy$. [M1]
2. $y = \frac{36}{x}$ — This is now a usable equation for any pair of values.
3. $6 = \frac{36}{x}$ — We are working backwards — $y$ is given, $x$ is unknown. [M1]
4. $x = \frac{36}{6} = 6$ — Multiply both sides by $x$, then divide both sides by 6. Verify: $6 \times 6 = 36 = k$ ✓ [A1]

Answer: $x = 6$

Here is a table-based problem — the type that trips students up most often. The question does not tell you which type of proportion it is; you have to figure that out from the data.

The ratio $\frac{y}{x}$ is not constant (it changes from 9 to 4 to 1), so this is not direct proportion. But the product $xy = 36$ for every pair — so this is inverse proportion with $k = 36$. The equation is $y = \frac{36}{x}$.

Consider this worked example of inverse proportion. If $y$ is inversely proportional to $x$ and we know $y = 8$ when $x = 2$, we can determine $k$ by using the equation $y = \frac{k}{x}$. Here, $8 = \frac{k}{2}$, so multiplying both sides by 2 gives us $k = 16$. Now, if we want to find $y$ when $x = 4$, we substitute into the equation: $y = \frac{16}{4} = 4$.

6. Deliberate Practice

The following problems cover the full range of proportion question types you will meet on the Higher-tier paper. Work through each one using the three-step method: write $\propto$, introduce $k$, substitute to find $k$. Then use the equation to answer the question.

7. Exam Technique and Common Mistakes

Proportion questions are reliable mark-earners if you follow the method consistently. The most common reason students lose marks is not misunderstanding the concept — it is skipping the algebraic setup and trying to scale mentally, which breaks down the moment the numbers are not straightforward.

Always write the proportionality statement and the equation with $k$ before substituting. These steps earn method marks even if you make an arithmetic error later. In questions without a calculator, keep $k$ as a fraction if it does not simplify to a whole number — never round mid-calculation.

When a question asks you to 'show' that two quantities are in proportion, you must check every pair in the table, not just one. Checking a single pair and concluding proportion is a common error that earns zero marks for that part.

One final check before you move on to the quiz: make sure you can distinguish the two graph shapes instantly. A straight line through the origin means direct proportion. A curve that approaches but never touches the axes means inverse proportion. Examiners sometimes show a graph and ask you to identify the type — this is a free mark if you know the shapes.

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