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 shape of a typical GCSE Higher power-proportion question.

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.

Ready to fix this? The Proportion lesson works through this misconception and the others in Proportion, one altitude at a time.

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

An exam question that triggers it

Here is a typical AQA Higher question of this shape:

$A$ 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

Of this shape: $A \propto B^4$ , $B$ 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).

Of this shape: $d \propto t^2$ , 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 $300 \div 10 = 30 \text{ s}$ .

Of this shape: $G \propto \sqrt{H}$ , $H$ 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).

Find out if this is costing you marks

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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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