GCSE Algebraic proportion Practice Questions

Turn the proportional statement into an equation

"Directly proportional to the square of $x$" becomes $y = k x^{2}$ once the constant $k$ is put in.

Substitute the pair of values given

Replace $x$ with $2$ and $y$ with $12$.

Solve for the constant

So the constant of proportionality is $k = 3$, and the model is $y = 3 x^{2}$.

Turn the words into an equation

The shape of the model is fixed by the words: $y = \frac{k}{x^{2}}$.

Work out the constant

The constant of proportionality is $36$.

Write the model and check it

Substituting $x = 2$ gives $y = 9$ — the option $y = \frac{36}{x^{2}}$ is the correct one.

Turn the words into an equation

The shape of the model is fixed by the words: $y = k \sqrt{x}$.

Substitute the pair of values

Only the constant is unknown, so one pair of values is enough.

Work out the constant

The constant of proportionality is $5$.

Write the model and check it

Substituting $x = 4$ gives $y = 10$ — the option $y = 5 \sqrt{x}$ is the correct one.

Test the first wrong option

At $x = 4$ it gives $20$, not $10$.

State the answer

So the correct formula is $y = 5 \sqrt{x}$.

Write the relationship as an equation

Keep the constant $k$ in — it will cancel out later, which is exactly why the answer does not depend on it.

Replace the variable with its new value

$x$ is multiplied by 4, so every $x$ in the model becomes $4x$.

Compare the new value with the old one

The constant $k$ and the $x$ cancel, leaving a pure number: $\frac{1}{2}$. $y$ is divided by 2.

Try it with numbers to be sure

Taking $x = 1$ and then $x = 4$ gives $1k$ and $\frac{1}{2}k$ — a factor of $\frac{1}{2}$, confirming the algebra.

Note the trap

The scale factor on $y$ is not the same as the scale factor on $x$ unless $y \propto x$. The power in the model changes it.

Write the general rule for this model

Multiplying $x$ by $c$ multiplies $y$ by $\frac{1}{\sqrt{c}}$ — that is the whole of the scale-factor idea.

Check one of the wrong statements

$y$ is divided by 4. would need a factor of $\frac{1}{4}$, not $\frac{1}{2}$, so it is false.

Check another wrong statement

$y$ is multiplied by 2. would need a factor of $2$, not $\frac{1}{2}$, so it is false too.

Try a second set of numbers

A different starting value gives the same factor $\frac{1}{2}$ — the result is independent of where you start and of $k$.

State the conclusion

So $y$ is divided by 2.

Write the relationship using the proportional sign

"Directly proportional to the square of $v$" is written $E \propto v^{2}$.

Replace the proportional sign with a constant of proportionality

Every proportional statement becomes an equation once the constant $k$ is included. Finding $k$ is always the first job.

Substitute the pair of values you are given

Putting $v = 12$ and $E = 36$ into the equation gives one equation in one unknown.

Work out the power or root of the given value

The relevant power of $12$ is $144$, so the equation now has $k$ as its only unknown.

Rearrange to make k the subject

Divide (or multiply) to isolate $k$ — keep everything exact, never round.

Work out the constant

The constant of proportionality is $\frac{1}{4}$.

Write the complete model

This single equation now links $E$ and $v$ for every pair of values.

Check the model reproduces the given pair

Substituting $v = 12$ back gives $E = 36$, which matches the question — the model is right.

Say why that check matters

A round trip through the original data is the fastest way to catch an arithmetic slip in $k$.

Substitute the value you are asked about

This time $E$ is known, so the equation must be solved for $v$.

Rearrange to isolate the power of the unknown

Undo the multiplication by $\frac{1}{4}$ first.

Undo the power or root

Take the positive root: a length or size in a proportion question is positive, so $v = 20$.

State the final answer

So $v = 20$.

Sense check the size of the answer

$E$ grows as $v$ grows, so a larger $v$ must give a larger $E$ — check the answer moves the right way.

Reflect on the structure

Once $k$ is known the model works in both directions: forwards to find $E$, backwards to find $v$.