Maths · 2 Algebra

2.8 Proportion

Before you start

Substituting numbers into a formula such as $y = 3 x^{2}$ .
Squares, cubes, square roots and cube roots, e.g. $\sqrt[3]{27} = 3$ .
Solving a simple equation such as $\frac{24}{x} = 3$ by multiplying both sides by $x$ .
Percentage multipliers: an increase of $20 %$ means multiply by $1.2$ , a decrease of $36 %$ means multiply by $0.64$ .
The graph of $y = m x$ is a straight line through the origin with gradient $m$ .

Syllabus objectives covered (2)

Express direct and inverse proportion in algebraic terms and use this form of expression to find unknown quantities.
Identify the best variation model for given data.

Part 1 of 5

Direct proportion: one quantity is a fixed multiple of the other

Two quantities are directly proportional when one is always the same multiple of the other. A printer that prints $12$ pages each minute prints $24$ pages in $2$ minutes and $60$ pages in $5$ minutes. The number of pages is always $12 \times$ the number of minutes.
The symbol $\propto$ means 'is proportional to'. You write 'pages is directly proportional to time' as $p \propto t$ . The symbol states a fact. You cannot calculate with it, so the first thing you do is replace it with ' $= k \times$ '.
$y \propto x ⟹ y = k x$
The letter $k$ is the constant of proportionality. It has the same value for every pair of values, so you can find it by dividing: $k = \frac{y}{x}$ . For the printer, $k = \frac{60}{5} = 12$ .
Because $y = k x$ has the form $y = m x$ , the graph of $y$ against $x$ is a straight line through the origin, with gradient $k$ . Both conditions matter. The line $y = 2 x + 3$ is straight, but it does not go through the origin, so $y$ is not proportional to $x$ : when $x = 1$ , $y = 5$ , and when $x = 2$ , $y = 7$ , which is not double.
graph diagram — not drawn yet
Axes for x from 0 to 5 and y from 0 to 14. A solid line y = 2x through the origin, labelled 'y = 2x: proportional', and a dashed parallel line y = 2x + 3 crossing the y-axis at (0, 3), labelled 'y = 2x + 3: not proportional'. Mark the points (1, 2) and (2, 4) on the solid line and (1, 5) and (2, 7) on the dashed line.
Both lines are straight, but only the one through the origin shows direct proportion. On it, doubling x doubles y.
Here is the key property: in $y = k x$ , if you multiply $x$ by any number, $y$ is multiplied by the same number. Treble $x$ and you treble $y$ . Halve $x$ and you halve $y$ .

Part 2 of 5

Proportion to a square, a cube or a square root

The syllabus also includes proportion to the square, the cube and the square root. 'The area is directly proportional to the square of the radius' means area $\propto$ radius $^{2}$ . The thing that $k$ multiplies is now $x^{2}$ , not $x$ .
Translate the words exactly. 'The square of $x$ ' is $x^{2}$ . 'The cube of $x$ ' is $x^{3}$ . 'The square root of $x$ ' is $\sqrt{x}$ . These give three new formulas.
$y = k x^{2} y = k x^{3} y = k \sqrt{x}$
The constant is still found by division, but you divide by the power of $x$ in the formula ( $x^{2}$ , $x^{3}$ or $\sqrt{x}$ ), not by $x$ itself. For $y = k x^{2}$ , $k = \frac{y}{x^{2}}$ . If $y = 48$ when $x = 4$ , then $k = \frac{48}{16} = 3$ , not $\frac{48}{4}$ .
The doubling property changes. If $x$ is doubled, $x^{2}$ is multiplied by $2^{2} = 4$ , so $y$ is multiplied by $4$ as well. Put $2 x$ in place of $x$ to see why.
$y = k {(2 x)}^{2} = 4 k x^{2}$
In general, if $x$ is multiplied by $m$ , then for $y \propto x^{2}$ the value of $y$ is multiplied by $m^{2}$ , for $y \propto x^{3}$ by $m^{3}$ , and for $y \propto \sqrt{x}$ by $\sqrt{m}$ . Doubling $x$ multiplies $y$ by $4$ , $8$ or $\sqrt{2} \approx 1.41$ .
The same idea handles percentages. An increase of $20 %$ in $x$ is a multiplier of $1.2$ . For $y \propto x^{2}$ , $y$ is multiplied by ${1.2}^{2} = 1.44$ , which is an increase of $44 %$ . The trap is to answer $40 %$ by doubling the $20$ : you must square the multiplier, not the percentage.

Part 3 of 5

Inverse proportion: one goes up as the other goes down by the same factor

Suppose $4$ workers, all working at the same rate, take $6$ days to build a wall. Then $8$ workers take $3$ days and $12$ workers take $2$ days. When the number of workers is doubled, the time is halved. This is inverse proportion.
Notice what stays the same: workers $\times$ days is $24$ every time. 'y is inversely proportional to x' means $y$ is proportional to $\frac{1}{x}$ , so $y = \frac{k}{x}$ , which is the same as $x y = k$ .
$y \propto \frac{1}{x} ⟹ y = \frac{k}{x} ⟺ x y = k$
So for inverse proportion you find $k$ by multiplying, not dividing. If $y = 6$ when $x = 4$ , then $k = 4 \times 6 = 24$ .
Powers and roots work in the same way. 'Inversely proportional to the square of $x$ ' is $y = \frac{k}{x^{2}}$ , so $x^{2} y = k$ . 'Inversely proportional to the cube of $x$ ' is $y = \frac{k}{x^{3}}$ . 'Inversely proportional to the square root of $x$ ' is $y = \frac{k}{\sqrt{x}}$ .
$y = \frac{k}{x^{2}} y = \frac{k}{x^{3}} y = \frac{k}{\sqrt{x}}$
The scaling rule turns round. If $x$ is multiplied by $m$ , then for $y = \frac{k}{x}$ the value of $y$ is divided by $m$ , and for $y = \frac{k}{x^{2}}$ it is divided by $m^{2}$ . Doubling $x$ in $y = \frac{k}{x^{2}}$ divides $y$ by $4$ .
The graph of $y = \frac{k}{x}$ for $x > 0$ is a curve, not a line. It gets closer and closer to both axes but never meets them, because $x \times y$ must stay equal to $k$ , which is not zero.
graph diagram — not drawn yet
Axes for x from 0 to 13 and y from 0 to 13. The curve y = 12/x for 0.9 <= x <= 13, with the points (1, 12), (2, 6), (3, 4), (4, 3), (6, 2) and (12, 1) marked and labelled with their coordinates.
At every marked point, x times y equals 12. That constant product is what makes this inverse proportion.
Be careful: not every quantity that decreases is inversely proportional. In $y = 10 - x$ , $y$ falls as $x$ rises, but $1 \times 9 = 9$ and $2 \times 8 = 16$ , so $x y$ is not constant. The test is always the constant product, never just 'goes down'.

Part 4 of 5

The four steps to find an unknown

Every proportion calculation follows the same four steps. (1) Write the proportion statement with $\propto$ and the correct power. (2) Replace $\propto$ with $= k$ . (3) Put in the pair of values you are given and solve for $k$ . (4) Write the formula with the number in place of $k$ , then use it.
Example: $p$ is inversely proportional to the square root of $q$ , and $p = 6$ when $q = 16$ . Steps (1) and (2) give $p = \frac{k}{\sqrt{q}}$ . Step (3): $6 = \frac{k}{\sqrt{16}} = \frac{k}{4}$ , so $k = 24$ .
$p = \frac{k}{\sqrt{q}}, 6 = \frac{k}{4}, k = 24$
Step (4): write $p = \frac{24}{\sqrt{q}}$ on a line of its own. When a question says 'find a formula for $p$ in terms of $q$ ', this line is the answer. A formula that still contains $k$ is not finished.
Now use it forwards. When $q = 9$ , $p = \frac{24}{\sqrt{9}} = \frac{24}{3} = 8$ .
Or use it backwards. To find $q$ when $p = 3$ , write $3 = \frac{24}{\sqrt{q}}$ . Multiply by $\sqrt{q}$ and divide by $3$ to get $\sqrt{q} = 8$ . The last thing done to $q$ was a square root, so undo it by squaring: $q = 64$ .
When the backwards step is a square root, check the question for a condition. If $x > 0$ is given, or $x$ is a length, a time or a number of objects, give the positive root only. If there is no condition, $3 x^{2} = 108$ has two solutions, $x = 6$ and $x = - 6$ .

Part 5 of 5

Choosing the best variation model for a table of data

Sometimes you are not told the type of proportion. You get a table of values and must decide which model fits. Each model has its own quantity that stays constant, so you test which one does.
For direct proportion the test quantity is a quotient. For inverse proportion it is a product.
$y \propto x : \frac{y}{x} y \propto x^{2} : \frac{y}{x^{2}} y \propto x^{3} : \frac{y}{x^{3}} y \propto \sqrt{x} : \frac{y}{\sqrt{x}} y \propto \frac{1}{x} : x y y \propto \frac{1}{x^{2}} : x^{2} y$
Sort first, then test. If $y$ increases as $x$ increases, only the direct models can fit. If $y$ decreases as $x$ increases, only the inverse models can fit. This halves the work.
With exact data, one test quantity is exactly the same for every pair. For $x = 1, 2, 3, 4$ and $y = 2, 16, 54, 128$ , the values of $\frac{y}{x}$ are $2, 8, 18, 32$ , which are not constant. The values of $\frac{y}{x^{3}}$ are $2, 2, 2, 2$ . So $y = 2 x^{3}$ .
With measured data, no test quantity is exactly constant, because measurements contain small errors. The best model is the one whose values stay closest together with no steady trend. A test quantity that keeps rising across the table means $y$ grows faster than your model: try a higher power. One that keeps falling means $y$ grows more slowly: try a lower power or a root. These two hints are for the direct models. For an inverse model, a product that keeps rising means $y$ falls more slowly than your model: try a smaller power underneath, such as $\frac{1}{\sqrt{x}}$ instead of $\frac{1}{x}$ .
A quick cross-check uses the scaling rule. Find two $x$ values where one is $4$ times the other. If $y$ is multiplied by about $4$ , try $y \propto x$ . By about $16$ , try $x^{2}$ . By about $64$ , try $x^{3}$ . By about $2$ , try $\sqrt{x}$ . If $y$ is divided by about $4$ , try $\frac{1}{x}$ , and by about $16$ , try $\frac{1}{x^{2}}$ .
When you have chosen the model, take $k$ as the value the test quantities cluster around (their mean is a sensible choice). Then state the model in full, for example ' $y \propto x^{2}$ , so $y = 0.8 x^{2}$ '. The model with its value of $k$ is what the question asks for, and you can then use it to estimate new values.
The graph gives the same test. If $y \propto x^{2}$ , then a graph of $y$ against $x^{2}$ (not against $x$ ) is a straight line through the origin with gradient $k$ .

Worked examples

Example 1. $y$ is directly proportional to the square of $x$ . When $x = 4$ , $y = 48$ . (a) Find a formula for $y$ in terms of $x$ . (b) Find $y$ when $x = 5$ . (c) Find $x$ when $y = 108$ , given that $x > 0$ .

Show the working

Steps (1) and (2): write the statement with the square, then replace the proportion sign with $= k$ .
$y \propto x^{2} ⟹ y = k x^{2}$
Step (3): substitute $x = 4$ , $y = 48$ . Square the $4$ before you divide.
$48 = k \times 4^{2} = 16 k ⟹ k = 3$
Step (4): write the formula with $k = 3$ . This line answers part (a).
$y = 3 x^{2}$
(b) Substitute $x = 5$ .
$y = 3 \times 5^{2} = 3 \times 25 = 75$
(c) Substitute $y = 108$ and undo the operations in reverse order: divide by $3$ , then take the square root.
$108 = 3 x^{2} ⟹ x^{2} = 36 ⟹ x = 6$
The question says $x > 0$ , so $x = - 6$ is rejected. Check: $3 \times 6^{2} = 108$ . Correct.

Answer: (a) $y = 3 x^{2}$ (b) $y = 75$ (c) $x = 6$

Example 2. $y$ is inversely proportional to the cube of $x$ . When $x = 2$ , $y = 5$ . Find $y$ when $x = 4$ , and find $x$ when $y = 320$ .

Show the working

Write the equation. Inverse means $k$ goes on top, and the cube goes underneath.
$y = \frac{k}{x^{3}}$
Substitute $x = 2$ , $y = 5$ . Multiply both sides by $2^{3} = 8$ to find $k$ .
$5 = \frac{k}{8} ⟹ k = 40$
The formula is $y = \frac{40}{x^{3}}$ . Now $x = 4$ .
$y = \frac{40}{4^{3}} = \frac{40}{64} = 0.625$
Check with the scaling rule: $x$ was doubled from $2$ to $4$ , so $y$ should be divided by $2^{3} = 8$ . $5 \div 8 = 0.625$ . It agrees.
For $y = 320$ , multiply both sides by $x^{3}$ , then divide by $320$ .
$320 = \frac{40}{x^{3}} ⟹ x^{3} = \frac{40}{320} = 0.125$
Undo the cube with a cube root. A cube root has only one real value, so there is no sign to choose.
$x = \sqrt[3]{0.125} = 0.5$

Answer: $y = 0.625$ when $x = 4$ ; $x = 0.5$ when $y = 320$ .

Example 3. $y$ is inversely proportional to the square of $x$ . The value of $x$ is increased by $25 %$ . Work out the percentage change in $y$ .

Show the working

You do not need $k$ . You need the multiplier. An increase of $25 %$ means $x$ is multiplied by $1.25$ .
$y$ depends on $x^{2}$ , so square the multiplier.
${1.25}^{2} = 1.5625$
The proportion is inverse, so $y$ is divided by this number.
$\frac{1}{1.5625} = 0.64$
A multiplier of $0.64$ means $y$ becomes $64 %$ of its old value. That is a decrease of $100 % - 64 % = 36 %$ .
Check with numbers: take $y = \frac{100}{x^{2}}$ and $x = 4$ , so $y = 6.25$ . Then $x = 5$ gives $y = 4$ . $\frac{4}{6.25} = 0.64$ . It agrees.

Answer: $y$ decreases by $36 %$ .

Example 4. A ball rolls from rest down a slope. A student measures the distance, $d$ metres, that it has travelled after $t$ seconds. $\begin{matrix} t & 1 & 2 & 3 & 4 \\ d & 0.8 & 3.1 & 7.3 & 12.7 \end{matrix}$ Find the variation model that best fits these data, and use it to estimate $d$ when $t = 5$ .

Show the working

Sort first: $d$ increases as $t$ increases, so test the direct models only.
Test $d \propto t$ . The values of $\frac{d}{t}$ are $0.8, 1.55, 2.43, 3.18$ (3 s.f.). They rise steadily, so $d$ grows faster than $t$ . Try a higher power.
Test $d \propto t^{2}$ . The values of $\frac{d}{t^{2}}$ are $\frac{0.8}{1} = 0.8$ , $\frac{3.1}{4} = 0.775$ , $\frac{7.3}{9} = 0.811$ and $\frac{12.7}{16} = 0.794$ (3 s.f.). They stay close to $0.8$ with no steady trend.
For completeness, test $d \propto t^{3}$ . The values of $\frac{d}{t^{3}}$ are $0.8, 0.388, 0.270, 0.198$ (3 s.f.). They fall steadily, so the cube is too high a power.
Cross-check with scaling: from $t = 1$ to $t = 2$ the time is doubled, and $d$ goes from $0.8$ to $3.1$ , which is about $4$ times. That fits a square.
Take $k$ as the mean of the four values of $\frac{d}{t^{2}}$ .
$k = \frac{0.8 + 0.775 + 0.8111 + 0.79375}{4} = 0.795 (3 s.f.)$
Use the model at $t = 5$ .
$d = 0.795 \times 5^{2} = 19.9 (3 s.f.)$
This is an estimate from measured data, so a value such as $20$ m from $k = 0.8$ is just as reasonable.

Answer: $d \propto t^{2}$ , with $d \approx 0.795 t^{2}$ . When $t = 5$ , $d \approx 19.9$ m (about $20$ m).

Where people go wrong

Mistake: Writing 'y is proportional to $x^{2}$ ' as $y = x^{2} + k$ .

Instead: Proportion always multiplies: $y = k x^{2}$ . For inverse proportion, $y = \frac{k}{x^{2}}$ .

Proportional means 'a fixed multiple of'. Adding a constant gives a graph that does not pass through the origin, and doubling $x$ then does not scale $y$ by a fixed factor.

Mistake: Finding $k$ with the wrong power, e.g. $k = \frac{48}{4} = 12$ for $y \propto x^{2}$ with $y = 48$ when $x = 4$ .

Instead: Substitute into the full formula first: $48 = k \times 4^{2}$ , so $k = 3$ .

The constant is $\frac{y}{x^{2}}$ , not $\frac{y}{x}$ . Writing the substitution line out in full makes the power visible before you divide.

Mistake: Saying that doubling $x$ doubles $y$ in every direct proportion.

Instead: Doubling $x$ multiplies $y$ by $2^{n}$ when $y \propto x^{n}$ : by $4$ for a square, $8$ for a cube, $\sqrt{2}$ for a square root.

Only $y \propto x$ (the first power) scales $y$ by the same factor as $x$ . Putting $2 x$ into the formula shows the true factor.

Mistake: Calling any decreasing relationship 'inverse proportion'.

Instead: Check that the product $x y$ (or $x^{2} y$ , and so on) is the same for every pair.

$y = 10 - x$ decreases, but $1 \times 9 \neq 2 \times 8$ . Inverse proportion is a constant product, not just a falling value.

Mistake: Treating a percentage change as if the percentage itself were squared or added, e.g. $x$ up $20 %$ in $y \propto x^{2}$ gives ' $y$ up $40 %$ '.

Instead: Convert to a multiplier ( $1.2$ ), apply the power ( ${1.2}^{2} = 1.44$ ), then convert back: up $44 %$ .

Proportion acts on multipliers. A percentage is only a way of writing a multiplier, so the power must be applied to $1.2$ , not to $20$ .

Check yourself

Write each statement as an equation with a constant $k$ . (a) $A$ is directly proportional to the square of $r$ . (b) $T$ is inversely proportional to the square root of $m$ .
Answer
(a) $A = k r^{2}$ (b) $T = \frac{k}{\sqrt{m}}$
$p$ is directly proportional to $q$ . When $q = 4$ , $p = 14$ . Find $p$ when $q = 10$ .
Answer
$k = \frac{14}{4} = 3.5$ , so $p = 3.5 q$ . When $q = 10$ , $p = 35$ .
$y$ is inversely proportional to $x$ . When $x = 3$ , $y = 8$ . Find $y$ when $x = 6$ , and find $x$ when $y = 2$ .
Answer
$k = 3 \times 8 = 24$ , so $y = \frac{24}{x}$ . When $x = 6$ , $y = 4$ . When $y = 2$ , $x = 12$ .
$y$ is directly proportional to the square root of $x$ . When $x = 9$ , $y = 12$ . Find $x$ when $y = 20$ .
Answer
$12 = k \sqrt{9} = 3 k$ , so $k = 4$ and $y = 4 \sqrt{x}$ . Then $20 = 4 \sqrt{x}$ gives $\sqrt{x} = 5$ , so $x = 25$ .
$V$ is directly proportional to the cube of $r$ . Describe what happens to $V$ when $r$ is multiplied by $3$ .
Answer
$V$ is multiplied by $3^{3} = 27$ .
$y$ is directly proportional to the cube of $x$ . The value of $x$ is increased by $10 %$ . Work out the percentage increase in $y$ .
Answer
Multiplier ${1.1}^{3} = 1.331$ , so $y$ increases by $33.1 %$ .
Which variation model fits this data? Give the formula. $\begin{matrix} x & 2 & 3 & 6 & 8 \\ y & 18 & 12 & 6 & 4.5 \end{matrix}$
Answer
$y$ decreases as $x$ increases, so test the inverse models. $x y = 36$ for every pair, so $y \propto \frac{1}{x}$ and $y = \frac{36}{x}$ .
Which variation model fits this data? Give the formula. $\begin{matrix} x & 1 & 4 & 9 & 16 \\ y & 5 & 10 & 15 & 20 \end{matrix}$
Answer
$\frac{y}{x}$ is $5, 2.5, 1.67, 1.25$ , which is not constant. $\frac{y}{\sqrt{x}}$ is $5$ for every pair, so $y \propto \sqrt{x}$ and $y = 5 \sqrt{x}$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The mass, $m$ grams, of a solid sphere made from a certain material is directly proportional to the cube of its radius, $r$ cm. A sphere of radius $2$ cm has a mass of $60$ g.

(a) Find a formula for $m$ in terms of $r$ . 3 marks No calculator

Mark scheme

M1 m = kr³ oe
M1 60 = k × 2³ oe, or k = 60 ÷ 8 soi
A1 m = 7.5r³ oe, e.g. m = (15/2)r³

(b) Work out the mass of a sphere of radius $4$ cm. 1 mark No calculator

Mark scheme

B1 480 (g); FT 64 × their k

(c) Calculate the radius of a sphere that has a mass of $202.5$ g. 2 marks No calculator

Mark scheme

M1 7.5r³ = 202.5 oe, or r³ = 27; FT their k
A1 3 (cm) cao

Question 2. A floor is covered with square tiles that fit exactly with no gaps. The number of tiles needed, $N$ , is inversely proportional to the square of the side length, $s$ cm, of one tile. When $s = 20$ , $N = 600$ .

(a) Find a formula for $N$ in terms of $s$ . 2 marks

Mark scheme

M1 600 = k ÷ 20² oe, or k = 600 × 400 soi
A1 N = 240000 ÷ s² oe

(b) Work out the number of tiles needed when $s = 25$ . 1 mark

Mark scheme

B1 384; FT their k ÷ 625

(c) Calculate the side length of a tile when $2400$ tiles are needed. 2 marks

Mark scheme

M1 2400 = 240000 ÷ s² oe, or s² = 100; FT their k
A1 10 (cm) cao; −10 not given as the final answer

(d) Describe what happens to $N$ when the side length of the tile is halved. 1 mark

Mark scheme

B1 N is multiplied by 4 oe, e.g. 'four times as many tiles'

Question 3. A student measures a quantity $y$ for five values of $x$ . The results are in the first two rows of the table. $\begin{matrix} x & 4 & 9 & 16 & 25 & 36 \\ y & 5.9 & 9.1 & 12.0 & 14.8 & 18.1 \\ \frac{y}{\sqrt{x}} & \dots & \dots & \dots & \dots & \dots \end{matrix}$

(a) Complete the third row of the table. Give each value correct to 3 significant figures. 2 marks

Mark scheme

B2 2.95, 3.03, 3.00, 2.96, 3.02 (accept 3 for 3.00); B1 for at least three of the five correct

(b) Explain why the results show that $y$ is not directly proportional to $x$ . 1 mark

Mark scheme

B1 y ÷ x is not constant, with evidence, e.g. 1.475 when x = 4 and 0.503 when x = 36; or x is multiplied by 4 (4 to 16) but y is only about doubled (5.9 to 12.0) oe

(c) Write the variation model that best fits the data as a formula for $y$ in terms of $x$ . 2 marks

Mark scheme

B1 y ∝ √x, or y = k√x, soi (implied by a correct final formula)
B1 y = 3√x (accept k from 2.95 to 3.03)

(d) Calculate an estimate of the value of $y$ when $x = 64$ , using your model. 1 mark

Mark scheme

B1 24; FT 8 × their k

How this is examined

Turning a proportion sentence into a formula with a number for k: A 2- or 3-mark Find: a sentence such as 'm is directly proportional to the cube of r' and one pair of values, then 'find a formula for m in terms of r' The equation with $k$ and the right power, e.g. $m = k r^{3}$ (method mark). The given pair substituted with the power applied, e.g. $60 = k \times 2^{3}$ (method mark). The final formula with the number in, e.g. $m = 7.5 r^{3}$ (accuracy mark). An answer that still contains $k$ , or one written with $\propto$ , does not earn the last mark.
Choosing the variation model that fits a table of values: A table of x and y values, followed by a Complete (work out a ratio or product for each pair), an Explain or Write down (which model fits), and a Calculate (use the model) The test quantity worked out for every pair, e.g. $\frac{y}{\sqrt{x}} = 2.95, 3.03, 3.00, \dots$ . Then a statement that names the model and the constant: ' $y \propto \sqrt{x}$ , $y = 3 \sqrt{x}$ '. To explain why a model does not fit, quote two values of its test quantity that differ, e.g. $\frac{y}{x} = 1.475$ and $0.503$ .
Using the formula forwards and backwards: A 1-mark Calculate of y from x, then a 2-mark Calculate of x from y, straight after the formula has been found Forwards: the correct value, which is follow-through from your own $k$ . Backwards: the equation with the given value in, e.g. $7.5 r^{3} = 202.5$ , or the rearranged form $r^{3} = 27$ (method mark), then the answer after the correct inverse operation, $r = 3$ (accuracy mark). For a length, give the positive root only.
The effect on y when x is multiplied by a number or changed by a percentage: A 1- or 2-mark Describe or Work out: 'x is halved' or 'x is increased by 25%', with no pair of values given The multiplier with the power applied, e.g. ${1.25}^{2} = 1.5625$ and $\frac{1}{1.5625} = 0.64$ , then the effect in words: ' $y$ is multiplied by $4$ ' or ' $y$ decreases by $36 %$ '. 'y gets bigger' or 'y gets smaller' alone gives no number, so it does not answer the question.

Words to know

Directly proportional: $y$ is directly proportional to $x$ when $y = k x$ for a fixed $k$ : multiplying $x$ by a number multiplies $y$ by the same number.
Inversely proportional: $y$ is inversely proportional to $x$ when $y = \frac{k}{x}$ for a fixed $k$ , so the product $x y$ is constant.
Proportional symbol $\propto$: Means 'is proportional to'. $y \propto x^{2}$ is read 'y is directly proportional to the square of x'. Replace it with $= k$ before you calculate.
Constant of proportionality: The fixed number $k$ in a proportion formula, found by substituting one known pair of values.
Variation model: A proportion formula, such as $y = k x^{2}$ or $y = \frac{k}{x}$ , chosen to describe how one quantity varies with another.

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