R Ratio, proportion and rates of change — notes on one page

R1 · Change freely between related standard units (time, length, area, volume/capacity, mass) and compound units (speed, rates of pay, prices, density, pressure) in numerical and algebraic contexts

Definition. Specification R1: change freely between related standard units (time, length, area, volume/capacity, mass) and compound units (speed, rates of pay, prices, density, pressure) in numerical and algebraic contexts.

• A conversion is multiplication or division by a fixed factor, and the direction is checked by size: $8000$ cm$^3$ is a small volume, so it must become $0.008$ m$^3$, not $8\,000\,000$. Lengths use $1$ m $=100$ cm and $1$ km $=1000$ m; areas square the factor and volumes cube it, so $1$ m$^3=100^3=1\,000\,000$ cm$^3$ and $1$ litre $=1000$ cm$^3$.
• Time converts through $60$: $1$ hour $=3600$ seconds. A compound unit converts one part at a time: $180$ km/h becomes m/s by multiplying by $1000$ and dividing by $3600$, giving $50$; each conversion is a method mark.
• Before substituting into a formula, put every quantity into the unit the answer needs, converting millimetres and centimetres to metres for a density in kg/m$^3$.
• In algebraic contexts the same factor multiplies the whole expression.
• Write the conversion line down; the mark is for seeing it.

What the examiner wants. Units that match before the formula is used. The June 2024 Paper 3H report describes a density problem in which the volume of flour had to be found from mass and density before anything else, and the November 2024 Paper 1H report warns that a time for $5$ miles at $30$ mph ranging up to $900$ minutes should have been caught by checking the reasonableness of the answer.

Exam tip. For a multi-mark conversion, write the compatible units before the formula so the method mark is unambiguous.

R2 · Use scale factors, scale diagrams and maps

• A scale written $1:n$ compares a diagram length with the corresponding real length in the same units. Multiply a diagram length by $n$ to obtain the real length, or divide the real length by $n$ to obtain the diagram length.
• Convert units only after establishing which direction the scale factor acts.
• For areas, square the linear scale factor; for volumes, cube it.
• On a map, measure between the stated points accurately and use any given scale bar or numerical scale.
• Examiners award method for applying the scale in the correct direction and then converting units correctly.

Exam tip. Write what $1\,\text{cm}$ represents before scaling, because this makes the direction of the conversion clear.

R3 · Express one quantity as a fraction of another, where the fraction is less than 1 or greater than 1

• To express quantity $A$ as a fraction of quantity $B$, form $\dfrac{A}{B}$ and simplify.
• The order of the wording fixes the order of the fraction: the quantity before “as a fraction of” is the numerator.
• Quantities must first be written in the same units, so that the units cancel.
• The result may be less than $1$, equal to $1$, or greater than $1$; an improper fraction is valid when the first quantity is larger.
• Examiners expect a simplified exact fraction unless the question asks for another form.

Exam tip. Underline the quantity named first and place it in the numerator before simplifying.

R4 · Use ratio notation, including reduction to simplest form

• Ratio notation compares quantities in a fixed order, for example $a:b$. Before simplifying, express every quantity in the same units.
• An equivalent ratio is produced by multiplying or dividing every part by the same non-zero value.
• For an integer ratio, divide all parts by their highest common factor; when decimals occur, first scale them to integers if helpful.
• Ratios can have more than two parts, and a shared part can be matched using a common multiple.
• Examiners require the stated order to be preserved and usually expect the simplest integer form.

Exam tip. For linked ratios, show the common value of the repeated quantity to secure the method mark.

R5 · Divide a quantity in a given part:part or part:whole ratio; express division into two parts as a ratio; apply ratio to real problems (conversion, comparison, scaling, mixing, concentrations)

Definition. Specification R5: divide a given quantity into two parts in a given part:part or part:whole ratio; express the division of a quantity into two parts as a ratio; apply ratio to real contexts and problems, such as those involving conversion, comparison, scaling, mixing and concentrations.

• In a part:part ratio $a:b$ the whole is $a+b$ equal shares, so divide the quantity by $a+b$ to find one share and multiply by each part; dividing by $a$ and by $b$ separately is the error every scheme lists. Check that the parts add back to the total.
• A part:whole statement is read differently: if one part is $\dfrac{3}{8}$ of the whole, the other is $\dfrac{5}{8}$ and the ratio is $3:5$. To express a division as a ratio, write the two amounts in order and simplify by their highest common factor.
• Multi-step problems combine a ratio with fractions or percentages of the same total: remove the fraction and percentage shares first, then split the remainder in the ratio.
• For mixing or concentration, track the actual amount of the ingredient, not an average of percentages.
• Label which part is which in the final answer.

What the examiner wants. A correct first step from the ratio and a systematic layout. The November 2024 Paper 3H report says most began well by using the ratio to find the two quantities, but those who divided $240$ by $3$ and by $5$ separately lost the start, and others lost their way through a lack of a systematic approach to the later steps.

Exam tip. Label each share or variable, then check that the parts sum to the stated whole.

R6 · Express a multiplicative relationship between two quantities as a ratio or a fraction

• A multiplicative relationship compares quantities by division rather than subtraction. If $A=kB$, then $\dfrac{A}{B}=k$, so $A:B=k:1$ and $A$ is the fraction $k$ of $B$.
• Convert decimal multipliers to fractions when an integer ratio is required; for example, $1.75=\dfrac{7}{4}$ gives $A:B=7:4$.
• For a chain of relationships, substitute one equation into another or match the shared ratio part.
• Check that a multiplier greater than $1$ corresponds to the larger quantity.
• Examiners expect the direction of the comparison to agree with the wording, because reversing it produces the reciprocal.

Exam tip. Translate “$A$ is $k$ times $B$” into $A=kB$ before converting it to a ratio.

R7 · Understand and use proportion as equality of ratios

• Proportion means that two corresponding ratios are equal. Keep the quantities in the same order on both sides, then solve by scaling, finding a unit value, or cross-multiplying.
• If $\dfrac{a}{b}=\dfrac{c}{d}$ with non-zero denominators, then $ad=bc$.
• A unitary method is often clearest in context: find the amount for one unit and scale to the required number.
• Decimal answers involving money must be interpreted in pounds and pence.
• Examiners award method for a correct proportional relationship even if a later arithmetic slip occurs, so show the equation or unit value.

Exam tip. Show either the equal-ratios equation or the value of one unit before giving the scaled answer.

R8 · Relate ratios to fractions and to linear functions

• For parts in ratio $a:b$, the whole has $a+b$ equal parts, so their fractions of the whole are $\dfrac{a}{a+b}$ and $\dfrac{b}{a+b}$. A constant ratio between variables can also define a linear function.
• If $y:x=m:1$, then $\dfrac{y}{x}=m$ and $y=mx$.
• Its graph is a straight line through the origin, with gradient $m$.
• A line with a non-zero intercept does not represent a constant ratio.
• Examiners may ask you to move between ratio, fraction, equation and graph, so always state the link explicitly.

Exam tip. When linking ratio to a function, verify that the equation gives $y=0$ when $x=0$.

R9 · Define percentage as 'number of parts per hundred'; interpret percentages and percentage changes as fractions/decimals, multiplicatively; percentages > 100%; percentage change and simple interest

Definition. Specification R9: define percentage as “number of parts per hundred”; interpret percentages and percentage changes as a fraction or a decimal, and interpret these multiplicatively; express one quantity as a percentage of another; compare two quantities using percentages; work with percentages greater than $100\%$; solve problems involving percentage change, including percentage increase or decrease and original value problems, and simple interest including in financial mathematics.

• A percentage is a number of parts per hundred, so $p\%=\dfrac{p}{100}$ and the multiplier for a change of $p\%$ is $1\pm\dfrac{p}{100}$. Percentage change is always measured against the original value: $\text{percentage change}=\dfrac{\text{change}}{\text{original}}\times100$, so a rise from $245$ to $883$ is $\dfrac{638}{245}\times100\approx260\%$, above $100\%$ because the value more than doubled.
• Percentage profit uses the cost price as the denominator. For an original value problem, the given amount is the multiplier times the original, so divide by the multiplier: a price of $\pounds46.75$ after a $15\%$ reduction was $46.75\div0.85$.
• Applying the percentage to the reduced price is the standard error and scores nothing.
• Simple interest is the same amount each year, calculated from the original principal only.
• Keep unrounded values until the comparison or final answer.

What the examiner wants. The original value as the denominator and no early rounding. The November 2024 Paper 2H report lists the errors that cost marks: choosing the new rather than the original value as the denominator, forgetting to subtract $100$ from the percentage, and rounding $11.84$ to $12$ before comparing; the Paper 3H report adds increasing or decreasing $\pounds46.75$ by $15\%$ in a reverse percentage question.

Exam tip. For reverse percentage, write the multiplier equation and divide by the multiplier.

R10 · Solve problems involving direct and inverse proportion, including graphical and algebraic representations

Definition. Specification R10: solve problems involving direct and inverse proportion, including graphical and algebraic representations. If $y$ is directly proportional to $x$ then $y=kx$; if $y$ is inversely proportional to $x$ then $y=\dfrac{k}{x}$, where $k$ is the constant of proportionality.

• Write the relationship as an equation with a constant before doing anything numerical: $y=kx$, $y=\dfrac{k}{x}$, or at Higher tier $y=kx^2$, $y=k\sqrt{x}$, $y=\dfrac{k}{x^2}$. Substitute the given pair to find $k$, write the formula out, then substitute the new value.
• For inverse proportion the product $xy$ is constant, so the same problem can be solved by $xy=k$; for direct proportion the ratio $\dfrac{y}{x}$ is constant and doubling $x$ doubles $y$, while for $y=kx^3$ doubling $x$ multiplies $y$ by $8$, not $6$.
• The graph of direct proportion is a straight line through the origin; inverse proportion is a decreasing curve approaching both axes.
• In a practical problem such as seed for a lawn, scale by the ratio of areas and compare with what is available, and finish with a supported yes or no.
• Writing $k$ on the wrong side, $y=kx^2$ for $y=\dfrac{k}{x^2}$, loses every mark.

What the examiner wants. The equation with $k$ in the right place. The June 2024 Paper 2H report says that, as is often the case with inverse proportion, a significant number lost all marks by working with $f=kd^2$ rather than $f=\dfrac{k}{d^2}$, and that the most common wrong sketch for a linear relationship was a direct proportion graph.

Exam tip. Write $y=kx$ or $y=\dfrac{k}{x}$ before substituting values; this is the key method step.

R11 · Use compound units such as speed, rates of pay, unit pricing, density and pressure

Definition. Specification R11: use compound units such as speed, rates of pay, unit pricing, density and pressure. A compound unit is a rate formed from two quantities, so its name states the division: metres per second, grams per cubic centimetre, newtons per square metre.

• Write the formula before substituting: $\text{speed}=\dfrac{\text{distance}}{\text{time}}$, $\text{density}=\dfrac{\text{mass}}{\text{volume}}$, $\text{pressure}=\dfrac{\text{force}}{\text{area}}$. Rates of pay and unit prices are a total divided by the hours or the number of items.
• Make the units match the unit you must answer in before you calculate: $180$ km per hour becomes metres per second by multiplying by $1000$ and dividing by $3600$. Rearrange when the unknown is not the subject; mass is density multiplied by volume, and time is distance divided by speed.
• Multi-step problems chain these: mass and density give a volume, subtracting a known volume gives the volume of the remaining material, and dividing its mass by that volume gives its density.
• The unit in the question tells you which quantity is divided by which.
• State the unit with the value; some schemes award it a separate mark.

What the examiner wants. Each stage of the process written down, with the right quantity on top. The June 2024 Paper 3H report says most students divided mass by density to find a volume, but many did not see that the remaining volume was that answer minus $700$, and a surprising number wrote $600/300$ and gave $0.5$ or $0.2$ instead of $2$.

Exam tip. Write the compound-unit formula first, because a correct formula can earn a method mark despite arithmetic error.

R12 · Compare lengths, areas and volumes using ratio notation; make links to similarity (including trigonometric ratios) and scale factors

• Similar shapes have equal corresponding angles and proportional corresponding lengths. If the linear scale factor from one shape to another is $k$, the length ratio is $k$, the area ratio is $k^2$, and the volume ratio is $k^3$.
• Match corresponding measurements and keep the comparison order consistent.
• To recover a length factor from an area ratio take a square root; from a volume ratio take a cube root.
• Trigonometric ratios remain constant in similar right-angled triangles.
• Examiners expect you to identify whether the measurements are lengths, areas or volumes before applying the scale factor.

Exam tip. Annotate the scale factor as $k$, $k^2$ or $k^3$ before calculating.

R13 · Understand that X is inversely proportional to Y is equivalent to X is proportional to 1/Y; construct and interpret equations that describe direct and inverse proportion

• If $X$ is inversely proportional to $Y$, then $X\propto\dfrac{1}{Y}$ and $X=\dfrac{k}{Y}$ for a constant $k$; equivalently, $XY=k$. Foundation questions can require interpreting equations that describe direct and inverse proportion.
• Higher tier: construct equations involving a power, such as $y=kx^2$ or $y=\dfrac{k}{x^n}$.
• Substitute a known pair to find $k$, write the complete equation, then use it to find an unknown.
• Check any stated restrictions, such as a positive length, before choosing a root.
• Examiners usually award separate method marks for the correct proportional form and for finding the constant.

Exam tip. Translate the wording into a formula containing $k$ before substituting any values.

R14 · Interpret the gradient of a straight line graph as a rate of change; recognise and interpret graphs that illustrate direct and inverse proportion

• The gradient of a straight line is $\dfrac{\text{change in }y}{\text{change in }x}$, so it represents a rate of change with units taken from the axes.
• Choose two well-separated points on the line, not merely nearby grid intersections, and calculate rise over run.
• In context, state what the rate means: on a distance-time graph it is speed; on a cost-time graph it is cost per unit time.
• A direct-proportion graph is straight through the origin, while an inverse-proportion graph has constant product $xy$.
• Examiners expect both the gradient and its contextual interpretation.

Exam tip. For “interpret the gradient”, give a value, compound unit and sentence explaining the rate.

R15 · Interpret the gradient at a point on a curve as the instantaneous rate of change; apply average and instantaneous rates of change (gradients of chords and tangents) (not calculus) [Higher only]

• For a curve, the gradient changes. The gradient of the chord joining two curve points gives the average rate of change over that interval.
• The gradient of a tangent at one point estimates the instantaneous rate of change there; GCSE questions use a drawn tangent, not calculus. Choose two well-separated, readable points on the chord or tangent and calculate $\dfrac{\Delta y}{\Delta x}$.
• The chosen tangent points need not lie on the original curve.
• Include compound units and compare rates using their values.
• Examiners allow a sensible range when answers depend on drawing and reading a tangent.

Exam tip. Draw a large tangent triangle and show $\dfrac{\Delta y}{\Delta x}$; a sensible estimate range is normally accepted.

R16 · Set up, solve and interpret the answers in growth and decay problems, including compound interest and work with general iterative processes

Definition. Specification R16: set up, solve and interpret the answers in growth and decay problems, including compound interest, and work with general iterative processes.

• Growth or decay by a fixed percentage each period is repeated multiplication: the multiplier is $1\pm\dfrac{r}{100}$, and after $n$ periods the amount is $\text{initial}\times(\text{multiplier})^n$.
• Compound interest earns interest on the interest, so the multipliers are applied in turn and the rates are never added; simple interest, by contrast, is the same amount every year and scores at most one mark when compound was asked for.
• With different rates in different years, multiply by each year's multiplier, $7000\times1.03\times1.015$.
• To find the rate or the number of years, test powers of the multiplier or use the given data to solve.
• Higher tier: an iterative formula such as $P_{n+1}=1.2P_n-50$ makes each term from the previous one, so evaluate the terms in order, stopping at the term asked for, and keep full precision until the end.

What the examiner wants. Multipliers, not added percentages. The November 2024 Paper 3H report says the most successful solutions used percentage multipliers, while common errors were adding the percentages together and raising that multiplier to the power $3$, working with simple interest for a maximum of one mark, and using $1.006$ instead of $1.06$.

Exam tip. For “first time” questions, show the last value that fails and the next value that meets the condition.