A is directly proportional to B to the power 4. B is doubled. By how many times does A increase?

A increases by a factor of 16, not 2. The scale factor on B is 2, and the power is 4, so the scale factor on A is 2 to the power 4 = 16. The trap is multiplying by 2 (linear) or by 8 (using 4 times 2 instead of 2 to the power 4). This is the exact question from AQA Higher November 2023 Paper 2 Q19.

A stone falls d metres in t seconds where d is proportional to t squared. It falls 20 m in 2 s. How long does it take to fall 300 m?

7.75 seconds, not 30 seconds. Step 1: find k using k = d divided by t squared = 20 divided by 4 = 5. Step 2: set up 300 = 5t squared, so t squared = 60. Step 3: t = root 60 = 7.746 seconds, which rounds to 7.75 s to 3 significant figures. The trap is treating the relationship as linear: 20 m in 2 s gives 10 m/s, so 300 divided by 10 = 30 s. But the relationship is quadratic, not linear.

G is proportional to the square root of H. H increases from 16 to 100. What is the scale factor on G?

The scale factor on G is 2.5, not 6.25. The scale factor on H is 100 divided by 16 = 6.25. Because G is proportional to root H, the scale factor on G is the square root of 6.25, which is 2.5. The trap is applying the H scale factor of 6.25 directly to G, ignoring the square root.

GCSE Maths Higher · AQA · Proportion

Proportion to a power scales linearly: the scale factor carries the power

The most common error on AQA Higher power-proportion questions: double $x$ in a $y \propto x^2$ relationship and the student writes that $y$ also doubles. But the power applies to the scale factor, not to the raw change. Double $x$ and $y$ multiplies by $2^2 = 4$ . Triple $B$ in a $A \propto B^4$ relationship and $A$ multiplies by $3^4 = 81$ .

The fix is one step: take the scale factor on $x$ , raise it to the same power as in the relationship, and that is the scale factor on $y$ . For root proportion, take the square root of the scale factor instead.

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How to spot it in your own work

You doubled $A$ when $B$ doubled in a $A \propto B^4$ relationship, writing $\times 2$ instead of $2^4 = 16$ (NOV23 Paper 2 Q19).
You wrote $k = d/t = 10$ instead of $k = d/t^2 = 5$ for a $d \propto t^2$ relationship (NOV24 Paper 3 Q17).
You multiplied $G$ by the full scale factor of $H$ (6.25) instead of its square root (2.5) when $G \propto \sqrt{H}$ (JUN23 Paper 3 Q20b).
You wrote $\times 2$ for $3x^2$ when $x$ doubled, rather than $\times 4$ (NOV24 Paper 3 Q15).

An exam question that triggers it

Here is NOV23 Paper 2 Question 19, reproduced in structure:

$A$ is directly proportional to $B^4$ .

The value of $B$ is doubled.

By how many times does the value of $A$ increase?

The misconception produces $\times 2$ or $\times 8$ (using $4 \times 2$ instead of $2^4$ ). The correct answer is $2^4 = 16$ . The scale factor on $B$ is 2, and the power is 4, so the scale factor on $A$ is $2^4 = 16$ .

Why students fall for this

The default model for proportion is linear: more of one means proportionally more of the other. That reflex is correct for $y = kx$ but fires incorrectly for $y = kx^2$ or $y = k\sqrt{x}$ . Students see “proportional” and apply the direct-proportion scaling rule without noticing the power.

A second version of the error arises specifically with power 4: the student treats the exponent as a multiplier, writing $4 \times 2 = 8$ instead of $2^4 = 16$ . This confuses repeated multiplication with scalar multiplication.

For root proportion the error runs the other way: the student sees a large scale factor on $H$ (6.25) and applies it directly to $G$ , forgetting that the root compresses the effect.

The fix: Raise the scale factor to the power of the relationship

Step 1: identify the power. Read the proportionality statement. Is it $y \propto x^2$ , $A \propto B^4$ , or $G \propto \sqrt{H}$ ?

Step 2: find the scale factor on x. What is $x$ multiplied by? (Often 2, 3, or a ratio like 100/16 = 6.25.)

Step 3: raise to the power. For $y \propto x^n$ , the scale factor on $y$ is the scale factor on $x$ raised to the power $n$ . For $y \propto \sqrt{x}$ , take the square root of the scale factor on $x$ .

Step 4: verify by substitution. Replace $x$ with the scaled value and compute $y$ directly to confirm.

Worked example

NOV23·2·19: $A \propto B^4$ , $B$ doubled.

Power: 4.
Scale factor on B: 2.
Scale factor on A:
$2^4 = 16$
Trap: $\times 2$ (linear) or $4 \times 2 = 8$ (exponent as multiplier).

NOV24·3·17: $d \propto t^2$ , falls 20 m in 2 s. How long to fall 300 m?

Find k.
$k = \frac{d}{t^2} = \frac{20}{4} = 5$
Set up equation.
$300 = 5t^2 \implies t^2 = 60$
Solve.
$t = \sqrt{60} \approx 7.75 \text{ s (3 s.f.)}$
Trap: treating as linear, $10 \text{ m/s}$ gives $300 \div 10 = 30 \text{ s}$ .

JUN23·3·20b: $G \propto \sqrt{H}$ , $H$ from 16 to 100.

Scale factor on H: $100 \div 16 = 6.25$ .
Scale factor on G: $\sqrt{6.25} = 2.5$ .
Trap: $\times 6.25$ (ignores the root).

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Common questions

$A \propto B^4$ . $B$ is doubled. By how many times does $A$ increase?: 16 times. The scale factor on $B$ is 2, and the power is 4, so the scale factor on $A$ is $2^4 = 16$ . The trap $\times 8$ comes from treating the exponent as a multiplier ( $4 \times 2 = 8$ ), not as a power. The trap $\times 2$ applies the linear rule.
How do you find $k$ in $d = kt^2$ ?: Divide $d$ by $t^2$ , not by $t$ . From (t = 2, d = 20): $k = 20 \div 4 = 5$ . The trap is $k = 20 \div 2 = 10$ , which uses the linear formula. With $k = 5$ and $d = 300$ : $t = \sqrt{300/5} = \sqrt{60} \approx 7.75 \text{ s}$ .
$G \propto \sqrt{H}$ . $H$ goes from 16 to 100. What is the scale factor on $G$ ?: The scale factor on $G$ is $\sqrt{100/16} = \sqrt{6.25} = 2.5$ , not 6.25. The square root in the proportionality compresses the effect: a 6.25-fold increase in $H$ gives only a 2.5-fold increase in $G$ . Check: if $G = 2\sqrt{H}$ , then $G(16) = 8$ and $G(100) = 20$ , ratio $20/8 = 2.5$ .

Related misconceptions

Inverse proportion treated as directWriting y = kx (or scaling y up as x rises) for a y proportional to 1/x relationship. The product xy is fixed, not the ratio.
Constant of proportionality and graph shapeFinding k = y/x for an inverse law, or expecting a reciprocal curve to be a straight line.

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