Solving Direct and Inverse Proportion Problems Revision Notes | GCSE AQA Higher Maths

1. Why Proportion Matters

Two variables that change together in a predictable, locked ratio appear everywhere — from currency exchange to speed and time. Recognising the type of relationship immediately tells you which equation to write, which is exactly what examiners reward.

There are two flavours of proportion at GCSE: direct (both go up together) and inverse (one goes up as the other comes down). The entire topic hinges on telling them apart and then finding the constant $k$ that locks the relationship in place.

2. Direct Proportion

Think about buying apples. If one apple costs 30p, two cost 60p, three cost 90p. The cost and the number of apples are always in the same ratio — double the apples, double the cost. That is direct proportion.

The symbol $\propto$ is shorthand for 'is proportional to', so $y \propto x$ means $y = kx$. You will see this notation in exam questions — it is a signal to write the equation immediately.

The method is always the same three-step routine: write the proportion equation, substitute the known pair to find $k$, then substitute $k$ back to answer the question. Let's walk through a clean example.

Worked example: Direct Proportion: Finding an Unknown Value

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

1. $y = kx$ — The phrase 'directly proportional' tells us to write $y = kx$. [M1 — correct equation structure]
2. $20 = k \times 4 \Rightarrow k = \frac{20}{4} = 5$ — Divide both sides by 4. Always find $k$ before moving on. [M1 — correct substitution and rearrangement]
3. $y = 5x$ — Now $k$ is known, the relationship is fully defined.
4. $y = 5 \times 7 = 35$ — Substitute $x = 7$ into the equation. [A1 — correct final answer]

Answer: $y = 35$

Notice that $k = 5$ throughout — it never changes. That fixed ratio is the whole point of direct proportion. If you recalculate $\frac{y}{x}$ for any pair in this relationship, you will always get 5.

3. The Graph of Direct Proportion

A direct proportion relationship always produces a straight line through the origin. The gradient of that line is exactly $k$. If the line does not pass through $(0, 0)$, the variables are not in direct proportion — they may have a linear relationship, but that is different.

The steeper line ($y = 5x$, from our worked example) has the larger constant of proportionality. The shallower line ($y = 2x$) represents a smaller $k$. Both pass through the origin — that is the visual signature of direct proportion.

4. Inverse Proportion Enhanced Coverage

Now think about sharing a pizza. If one person gets the whole pizza, they have a lot. If eight people share it, each person gets much less. More people means less pizza per person — as one quantity goes up, the other comes down. That is inverse proportion. This relationship is pivotal for numerous real-world applications, illustrating how resources are distributed among increasing groups. Understanding this concept is essential for answering exam questions effectively.

The key insight is that the product $x \times y$ stays constant. Multiply any pair of values together and you always get the same number — that number is $k$.

Here is a crucial equivalence that The higher-tier specification explicitly requires you to know: saying '$x$ is inversely proportional to $y$' is exactly the same as saying '$x$ is directly proportional to $\frac{1}{y}$'. In symbols: $x \propto \frac{1}{y}$ means $x = \frac{k}{y}$, which rearranges to $xy = k$.

This equivalence is not just a curiosity — it explains why the graph of inverse proportion is a curve (a reciprocal curve), not a straight line. If you plotted $x$ against $\frac{1}{y}$ instead, you would get a straight line through the origin with gradient $k$.

Worked example: Inverse Proportion: Finding an Unknown Value

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

1. $y = \frac{k}{x}$ — 'Inversely proportional' means $y = \frac{k}{x}$. Write this immediately. [M1 — correct equation structure]
2. $8 = \frac{k}{3} \Rightarrow k = 8 \times 3 = 24$ — Multiply both sides by 3. For inverse proportion, $k = xy$, so $k = 8 \times 3 = 24$. [M1 — correct substitution]
3. $y = \frac{24}{x}$ — The relationship is now fully defined.
4. $y = \frac{24}{6} = 4$ — Substitute $x = 6$. Check: $6 \times 4 = 24 = k$. The product is constant, confirming inverse proportion. [A1 — correct final answer]

Answer: $y = 4$

Notice the sanity check in Step 4: multiply the new pair together and confirm you get $k$. This takes five seconds and catches arithmetic slips before they cost you marks.

5. The Graph of Inverse Proportion

The graph of $y = \frac{k}{x}$ is a reciprocal curve — it has two branches, one in the first quadrant (both $x$ and $y$ positive) and one in the third quadrant (both negative). The curve approaches but never touches the axes, which act as asymptotes.

This curve ($y = \frac{24}{x}$, from our worked example) is the visual signature of inverse proportion. As $x$ increases, $y$ decreases, and the curve approaches — but never reaches — the $x$-axis. The curve never crosses either axis because $x = 0$ and $y = 0$ are both undefined in this relationship.

6. Comparing Direct and Inverse Proportion

Before tackling harder problems, it helps to see both types side by side. The table below summarises the key differences — this is exactly the kind of comparison an examiner might ask you to make.

The 'what stays constant' row is the fastest way to identify which type you are dealing with. Calculate $\frac{y}{x}$ for two pairs — if it is the same, direct. Calculate $xy$ for two pairs — if it is the same, inverse.

8. Deliberate Practice

To aid understanding of the foundational content for this topic, it is important to recognize how both direct and inverse proportions are applied in various contexts. This includes how relationships between quantities can be defined using equations and how these equations relate to graphical representations.

9. Exam Technique and Common Mistakes

The proportion topic is one of the most reliably mark-able on the paper — the method is always the same. The marks students drop are almost always due to one of the errors below, not a lack of understanding.

One more examiner habit worth knowing: questions sometimes give you three or four pairs of values in a table and ask you to 'show that' the variables are in inverse proportion. The expected method is to calculate $xy$ for every pair and demonstrate they are all equal — state the constant explicitly.

10. A Harder Exam-Style Question

Some exam questions combine proportion with a multi-step context. The method does not change — but you need to read carefully to extract the known pair and identify which variable is which.

Worked example: Exam-Style: Inverse Proportion in Context

The time $T$ (in hours) taken to complete a journey is inversely proportional to the average speed $v$ (in km/h). When $v = 60$ km/h, $T = 2.5$ hours. Calculate the time taken when $v = 75$ km/h. [3 marks]

1. $T = \frac{k}{v}$ — 'Inversely proportional' gives $T = \frac{k}{v}$. [M1 — correct equation]
2. $k = T \times v = 2.5 \times 60 = 150$ — For inverse proportion, $k = T \times v$. Verify: $2.5 \times 60 = 150$. [M1 — correct substitution and value of $k$]
3. $T = \frac{150}{v}$ — The relationship is now fully defined with $k = 150$.
4. $T = \frac{150}{75} = 2$ — Substitute $v = 75$. Check: $75 \times 2 = 150 = k$. [A1 — correct answer with implicit units check]

Answer: $T = 2$ hours

Notice that increasing the speed from 60 to 75 km/h reduced the time from 2.5 to 2 hours — the time went down as the speed went up. That is the hallmark of inverse proportion, and it is a useful sanity check: if both quantities move in the same direction, something has gone wrong.