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    Ratio, proportion and rates of change — Edexcel GCSE Mathematics

    Test yourself on Ratio, proportion and rates of change with PEARSON EDEXCEL GCSE practice questions.

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    Ratio, proportion and rates of change explained

    A unit conversion is a multiplication, so decide first whether the answer should be a larger or a smaller number, then multiply or divide to match.

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    Length is a single step (1 m = 100 cm), but area squares that factor and volume cubes it: 1 m² = 10 000 cm² and 1 m³ = 1 000 000 cm³. Mass and capacity follow the same pattern: 1 kg = 1000 g and 1 litre = 1000 ml, with 1 cm³ = 1 ml linking volume to capacity. A compound unit is built from two others and its name gives the operation, so km/h is kilometres divided by hours, g/cm³ is grams divided by cubic centimetres, and a price in pence per gram is money divided by mass. To change 15 m/s into km/h, multiply by 3600 to reach metres per hour, then divide by 1000, giving 54 km/h. The same reasoning works with letters, so a speed of x m/s is 3.6x km/h.

    use scale factors, scale diagrams and maps

    A scale tells you how many real units one drawing unit stands for. A map scale of 1 : 25 000 means 1 cm on the map is 25 000 cm on the ground, which is 250 m or 0.25 km. When the scale is written as 1 : n with the map length first, you multiply by n to go from the map to real life and divide by n to come back. So a 6 cm line on that map is 150 000 cm, which is 1.5 km. On a scale drawing you also measure with a ruler and a protractor, so work to the nearest millimetre and the nearest degree. Between two similar shapes the scale factor is a length on the new shape divided by the matching length on the old one, and lengths that are not matching pairs give the wrong factor.

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

    The phrase "A as a fraction of B" means A/B, so the quantity that follows the word "of" goes on the bottom. Put both quantities into the same unit first, then write the fraction and cancel it. For example 45 minutes as a fraction of 2 hours is 45/120, which cancels to 3/8. When the first quantity is the larger one the answer is greater than 1, and that is correct rather than a mistake: 2 hours as a fraction of 45 minutes is 120/45, which cancels to 8/3. Leave it as an improper fraction unless the question asks for a mixed number. The comparison is often hidden in wording such as "what fraction of the class", so decide which quantity is the whole before you write anything.

    use ratio notation, including reduction to simplest form

    A ratio compares parts of the same kind and is written with a colon, so 12 : 18 means twelve of one thing for every eighteen of another. To reduce it, divide every part by their highest common factor: the highest common factor of 12 and 18 is 6, so 12 : 18 becomes 2 : 3. All parts must be in the same unit first, so 40 cm : 2 m becomes 40 : 200 and then 1 : 5. Three-part ratios work the same way, using a number that divides all three parts. You can also write a two-part ratio in the form 1 : n by dividing both parts by the first, and n need not be a whole number, so 4 : 10 becomes 1 : 2.5. That form makes two ratios quick to compare.

    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, concentrations)

    Ratio problems involve dividing quantities or expressing a division as a ratio. To divide £60 in the ratio 3:2, find the total number of parts (3 + 2 = 5), the value of one part (£60 ÷ 5 = £12), then each share (3 × £12 = £36 and 2 × £12 = £24). For a part:whole ratio such as 3:5, the whole is 5 parts and the other part is 5 − 3 = 2, so £60 divides as 3/5 × £60 = £36 and 2/5 × £60 = £24. To express a division as a ratio, write the quantities and simplify: £36 and £24 give 36:24, which simplifies to 3:2 by dividing both by 12. Distinguish part:part (3:2) from part:whole (3:5, meaning 3 parts out of 5). Problems may give one part's value or the difference between parts, from which you can find one part's value. Ratios apply to scaling recipes, currency conversion, comparing value, mixing and concentrations, such as diluting squash in the ratio 1:4.

    express a multiplicative relationship between two quantities as a ratio or a fraction

    Two quantities are linked multiplicatively when one is always a fixed multiple of the other, and there are three ways to record that one fact. If y is always 3 times x, you can write the ratio x : y as 1 : 3, say that x is 1/3 of y, or write the equation y = 3x. To find the multiplier, divide one quantity by the other. With 8 red counters and 20 blue ones, red : blue is 8 : 20, which simplifies to 2 : 5, so blue is 2.5 times red and red is 2/5 of blue. The phrase "for every" signals a ratio, and a phrase such as "two fifths of" signals the same relationship written as a fraction, so a question can ask for either form.

    understand and use proportion as equality of ratios

    Two ratios are in proportion when they reduce to the same simplified ratio, so 3 : 4 and 9 : 12 are in proportion because both reduce to 3 : 4. That lets you find a missing value: if 3 : 4 and 9 : x are in proportion, the first part has been multiplied by 3, so x = 12. Recipe, best-value and scaling questions all run on this. If 4 pens cost £3, then one pen costs £0.75 and 10 pens cost £7.50; you could also scale straight from 4 to 10 by multiplying by 2.5. Writing the two ratios as equal fractions gives you a check, since 4/3 and 10/7.5 both come to the same value. This kind of proportion is direct, so doubling one quantity doubles the other.

    relate ratios to fractions and to linear functions

    A ratio a : b carries the same information as a fraction and as a straight line through the origin. If two quantities are in the ratio 1 : 3, then y is 3 times x, so the equation is y = 3x, and every matching pair of values lies on a line through the origin of gradient 3. The gradient of such a line is the multiplier between the two quantities, so a conversion graph with pounds across and euros up has a gradient equal to the number of euros to the pound. Starting from the ratio 2 : 5, the multiplier is 5/2 = 2.5, giving y = 2.5x, and the line rises 5 units for every 2 units across. The fraction 2/5 compares one part with the other part, while 2/7 compares one part with the whole. A ratio question and a graph question can be the same question wearing different clothes.

    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/decrease and original value problems, and simple interest including in financial mathematics

    Per cent means out of 100, so 23% is 23/100, which is 0.23 as a decimal. Using a single multiplier is faster and safer than finding the change and adding it on: a 15% increase means multiplying by 1.15, a 15% decrease means multiplying by 0.85, and 130% of an amount means multiplying by 1.3. To write one quantity as a percentage of another, divide, then multiply by 100, so 18 out of 40 is 18 ÷ 40 = 0.45, which is 45%. A reverse percentage undoes the multiplier: if a price is £42 after a 20% rise, the original was £42 ÷ 1.2 = £35. Simple interest adds the same amount each year, so £500 at 4% for 3 years earns £500 × 0.04 × 3 = £60.

    solve problems involving direct and inverse proportion, including graphical and algebraic representations

    Two quantities are in direct proportion when one is a fixed multiple of the other, so y = kx and the graph is a straight line through the origin. They are in inverse proportion when their product stays fixed, so y = k/x, and with k positive the graph is a curve that falls as x grows for positive values of x. Either way the method is the same: use the pair of values you are given to find k, write the rule down, then use it on the value the question asks about. If five pens cost £3.50, one pen costs £0.70 and a dozen pens cost £8.40. If four people working at the same rate take six hours for a job, that job is worth twenty-four person-hours, so eight people take three hours. Decide first whether doubling one quantity doubles the other or halves it.

    use compound units such as speed, rates of pay, unit pricing, density and pressure

    A compound unit is built from two other measures, and the unit itself tells you which division to do. Speed is distance divided by time, in units such as km/h. Density is mass divided by volume, in units such as g/cm³. Pressure is force divided by area, in units such as N/m². A unit price follows the same shape, total cost divided by the number of items, and a rate of pay is total pay divided by the time worked. Rearrange the relationship when the missing quantity is on top, so distance = speed × time. Make the units agree before you start. A metal block of mass 240 g and volume 30 cm³ has a density of 8 g/cm³. A journey of 90 km taking one and a half hours gives an average speed of 60 km/h, which is the same as one kilometre every minute.

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

    When two shapes or solids are similar, every length is multiplied by the same scale factor k. Areas are then multiplied by k² and volumes by k³, because an area is built from two lengths and a volume from three. So lengths in the ratio 2 : 3 give areas in the ratio 4 : 9 and volumes in the ratio 8 : 27. Going the other way, take the square root of an area ratio or the cube root of a volume ratio to get back to lengths: volumes in the ratio 8 : 125 come from lengths in the ratio 2 : 5, so the areas are in the ratio 4 : 25. Similar right-angled triangles are also where sine, cosine and tangent come from, since the ratio of two sides depends only on the angle, not on how big the triangle is.

    understand that X is inversely proportional to Y is equivalent to X is proportional to 1/Y; interpret equations that describe direct and inverse proportion (Foundation tier)

    The equation tells you which kind of proportion you are looking at. If y = 4x then y is a fixed multiple of x: double x and y doubles, and where x is zero y is zero too. If y = 24/x the two quantities multiply to a fixed number, so doubling x halves y. Saying y is inversely proportional to x is the same as saying y is proportional to 1/x, the reciprocal of x. To read an equation, substitute a value and watch what happens: in y = 24/x, x = 3 gives y = 8, and x = 6 gives y = 4. In a table of values, a constant product points to an inverse relationship and a constant ratio between the two rows points to a direct one.

    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 (Higher tier)

    Higher tier only. The key equivalence is that X inversely proportional to Y means X is proportional to 1/Y, so X = k/Y, equivalently XY = k. Direct proportion gives X = kY; inverse proportion gives X = k/Y. For example, if y is inversely proportional to x and y = 5 when x = 2, then k = 5 × 2 = 10, so y = 10/x; when x = 5, y = 2. Powers and roots extend this: y = kx², y = k√x, y = k/x². For instance, if y is inversely proportional to x² and y = 5 when x = 2, then k = 5 × 2² = 20, so y = 20/x²; when x = 5, y = 0.8. Always write the finished formula down, as later parts often need it, sometimes rearranged to find x from a given y.

    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 graph measures how fast one quantity changes as the other changes, and its units are the unit on the vertical axis for each unit on the horizontal axis. Work it out as the change up divided by the change across, using two points you can read exactly. On a distance-time graph a rise of 120 km across two hours gives a gradient of 60, meaning a speed of 60 km/h; a steeper section means a faster speed and a flat section means no movement. A graph of direct proportion is a straight line through the origin, and its gradient is the constant multiplier. With both quantities positive, a graph of inverse proportion is a curve that drops steeply and then flattens, moving closer to the axes without ever meeting them.

    interpret the gradient at a point on a curve as the instantaneous rate of change; apply the concepts of average and instantaneous rate of change (gradients of chords and tangents) in numerical, algebraic and graphical contexts (this does not include calculus) (Higher tier only)

    This topic covers rates of change for curves, which are constantly varying. The average rate of change between two points is the gradient of the chord connecting them. For a function like y = x² + 2, the average rate between x=1 and x=4 is found using the points (1, 3) and (4, 18). The gradient is (18−3)/(4−1) = 5. This links algebraic, numerical and graphical contexts. The instantaneous rate of change at a single point is the gradient of the tangent to the curve at that point. To estimate this, draw a tangent line that just touches the curve and follows its direction. Then, pick two points far apart on your tangent and calculate the gradient (change in y / change in x). On a distance-time graph, this gradient represents instantaneous speed. As this involves drawing, a range of answers is accepted.

    set up, solve and interpret the answers in growth and decay problems, including compound interest (Foundation tier)

    This is a core Foundation tier topic and essential knowledge for Higher tier. Repeated percentage change is handled with a decimal multiplier. An increase of 5% uses a multiplier of 1.05; a decrease of 12% uses 0.88. This multiplier is applied for each period, so for n periods, it is raised to the power n. For example, £2000 invested at 3% compound interest for four years becomes 2000 × 1.03⁴ = £2251.02. The interest earned is £251.02. Decay problems use the same method with a multiplier between 0 and 1. A car worth £8000 that loses a fifth (20%) of its value each year is worth 8000 × 0.8³ = £4096 after three years. Always check if the question asks for the final amount or the total change, and round money to two decimal places.

    set up, solve and interpret the answers in growth and decay problems, including compound interest and work with general iterative processes (Higher tier)

    Higher tier only. For compound interest, the amount after n periods is A = P(1 + r)ⁿ, where r is the decimal rate per period. For example, £2000 at 5% per year for 3 years gives 2000 × 1.05³ = £2315.25. Decay uses a multiplier below 1, such as 0.85 for a 15% fall. An iteration turns one value into the next, written xₙ₊₁ in terms of xₙ. Start from x₀, work out x₁, feed that answer back in to get x₂, and carry on; the calculator answer key does the feeding back. With xₙ₊₁ = √(3xₙ + 4) and x₀ = 2, x₁ ≈ 3.162, x₂ ≈ 3.672 and x₃ ≈ 3.875, closing in on 4, the positive solution of x² = 3x + 4. Growth questions can run backwards, asking for the rate or the number of periods, so try whole-number powers until the target is passed.

    Your focus

    1. Describe why an area factor is the square of a length factor and a volume factor the cube of it.
    2. Calculate a converted value by multiplying or dividing by the conversion factor, deciding first whether the answer grows or shrinks.
    3. Calculate a two-stage compound conversion such as 15 m/s into 54 km/h and state the answer with its unit.
    Show all 72 objectives
    1. Apply the same reasoning in algebra, writing a speed of x m/s as a speed of 3.6x km/h.
    2. Write down what a scale such as 1 : 25 000 means as a multiplier, with 1 cm representing 250 m.
    3. Calculate a real distance from a map length by multiplying by the scale, giving the unit the question asks for.
    4. Measure a length on a scale drawing to the nearest millimetre and an angle to the nearest degree.
    5. Work out the scale factor between two similar shapes by dividing matching lengths, and explain why unmatched pairs fail.
    6. Explain which quantity goes on the denominator, namely the one named after the word of.
    7. Calculate one quantity as a fraction of another after putting both into the same unit, then cancel it down.
    8. Justify an answer greater than 1 as correct whenever the first quantity is the larger of the two.
    9. Describe what a ratio compares and why all parts must be in the same unit before anything is cancelled.
    10. Simplify a ratio by dividing every part by their highest common factor, as 12 : 18 becomes 2 : 3.
    11. Work out the 1 : n form by dividing both parts by the first, keeping a decimal such as 1 : 2.5.
    12. Compare two ratios by putting both into 1 : n form and giving a reason for which is the larger.
    13. Divide a quantity in a given part:part ratio by finding the total number of parts and the value of one part.
    14. Divide a quantity in a given part:whole ratio by treating the whole as the total and finding the other part by subtraction.
    15. Express the division of a quantity into two parts as a ratio and simplify it to its lowest terms.
    16. Apply ratio to real contexts such as scaling, conversion, mixing and concentrations.
    17. Calculate the multiplier between two quantities by dividing one of them by the other.
    18. Write one relationship in all three forms, as a simplified ratio, as a fraction and as an equation such as y = 3x.
    19. Explain which form the wording asks for, with for every signalling a ratio and two fifths of signalling a fraction.
    20. Explain that two ratios are in proportion when they reduce to the same simplified ratio.
    21. Calculate a missing value using the scale factor between two equal ratios, or by finding the cost of one item first.
    22. Show that a pair of ratios is in proportion by writing them as fractions and checking both come to the same value.
    23. Explain how a ratio a : b, the fraction a/b and a line y = kx through the origin all carry the same relationship.
    24. Write down the equation y = kx from a given ratio, taking k as the multiplier between the two quantities, for example the ratio 2 : 5 gives y = 2.5x.
    25. Work out the gradient of a conversion graph as a change in y over the matching change in x.
    26. Interpret that gradient in context, such as the number of euros to the pound on a conversion graph.
    27. Explain percentage as parts per hundred and write a given percentage as both a fraction and a decimal.
    28. Calculate a percentage increase or decrease with a single multiplier, such as 1.15 for a rise of 15%.
    29. Calculate one quantity as a percentage of another by dividing and then multiplying by 100.
    30. Work out an original value by dividing by the multiplier, so £42 after a 20% rise came from £35.
    31. Calculate simple interest as amount times rate times years, and state whether the answer is interest or total.
    32. Compare two quantities using percentages and justify which has changed by more.
    33. Explain the difference between direct and inverse proportion by what happens when one quantity is doubled.
    34. Calculate the constant k from a given pair of values and write the rule as y = kx or y = k/x.
    35. Apply that rule to the value the question asks about, giving the answer with its unit and any rounding.
    36. Interpret a straight line through the origin as direct proportion and a falling curve as inverse proportion.
    37. Describe how a compound unit names its own calculation, so that g/cm³ means mass divided by volume.
    38. Calculate a speed, density, pressure, unit price or rate of pay and state the answer with its compound unit.
    39. Apply a rearranged relationship when the unknown is on top, such as distance = speed × time.
    40. Compare two rates fairly by converting first so that the two measures agree.
    41. Explain why a length scale factor k gives an area factor of k² and a volume factor of k³.
    42. Calculate an area or volume ratio from a length ratio, and work back with a square root or a cube root.
    43. Apply a scale factor in the right direction, multiplying to enlarge and dividing to reduce, and state the unit.
    44. Explain why sine, cosine and tangent depend only on the angle, since similar right-angled triangles share side ratios.
    45. Explain that X inversely proportional to Y says the same as X proportional to 1/Y.
    46. Interpret an equation such as y = 24/x by substituting a value of x and working out y.
    47. Explain which kind of proportion a table shows, giving a constant product or a constant ratio as the reason.
    48. Explain the effect of a change, such as trebling one quantity dividing the other by three in an inverse relationship.
    49. Explain that X inversely proportional to Y is equivalent to X proportional to 1/Y, giving X = k/Y.
    50. Write a proportion statement containing k that matches the power or root named, such as y = k/x².
    51. Calculate k by substituting the given pair of values, then write the finished formula down.
    52. Apply the completed formula to the part asked, rearranging it when the other variable is the unknown.
    53. Work out the gradient of a straight line graph as the change up over the matching change across.
    54. Interpret that gradient in context with its unit, so a gradient of 60 is a speed of 60 km/h.
    55. Describe direct proportion as a straight line through the origin and inverse proportion as a curve nearing both axes.
    56. Explain what a steeper section and a flat section of a distance-time graph each show.
    57. Explain the difference between an average rate of change over an interval and an instantaneous rate of change at a point.
    58. Calculate an average rate of change from an algebraic function or graph as the gradient of the chord joining two points.
    59. Estimate an instantaneous rate of change by drawing a tangent to a curve and calculating its gradient.
    60. Interpret a tangent's gradient in context with its compound unit, such as speed or acceleration.
    61. Explain why repeated percentage change uses a multiplier raised to a power.
    62. Calculate a value after compound growth using the formula: Initial Value × (multiplier)ⁿ.
    63. Calculate a value after decay or depreciation using a multiplier between 0 and 1.
    64. Interpret the question to determine whether to provide the final amount or the total change.
    65. Set up and solve compound interest and decay problems using a multiplier raised to the number of periods.
    66. Work out successive terms of an iteration from the given starting value, feeding each answer back into the rule.
    67. Find the value the terms settle on, given to the number of decimal places the question asks for.
    68. Work out a rate or a number of periods backwards, trying whole-number powers until the target is passed.
    69. Explain what the settled value represents, such as the positive solution of x² = 3x + 4.

    Ratio, proportion and rates of change exam tips

    Quick Revision Summary (Key Takeaway)

    Ratio, proportion and rates of change covers comparing quantities multiplicatively, scaling recipes and maps, sharing amounts in a given ratio, direct and inverse proportion, compound measures such as speed and density, and percentage change including compound interest. It is a major strand of Pearson Edexcel GCSE Mathematics, assessed across all three papers and heavily applied in real-world problem-solving questions.

    Topic Overview

    Ratio, proportion and rates of change is a fundamental GCSE Mathematics topic that develops your ability to compare quantities multiplicatively and solve problems involving scaling, sharing, and comparing different amounts. It covers ratio notation, dividing quantities in a given ratio, direct and inverse proportion, compound measures such as speed, density and pressure, and percentage change including simple and compound interest. These ideas appear in everyday contexts like recipes, maps, currency exchange, and financial calculations.

    This topic is assessed across all three Pearson Edexcel GCSE Mathematics papers (non-calculator, calculator, and both tiers) and often forms the basis of multi-step problem-solving questions. Mastery of ratio and proportion is essential for success in algebra, geometry, and statistics, as well as for everyday life beyond the classroom. A strong understanding here will boost your confidence in tackling unfamiliar problems and improve your overall grade.

    Key Concepts
    • →Ratio compares quantities of the same kind using division; it can be simplified like fractions and used to share a total into parts.
    • →Direct proportion means as one quantity increases, the other increases at the same rate (e.g., cost of petrol vs litres); inverse proportion means as one increases, the other decreases (e.g., number of workers vs time taken).
    • →A rate of change compares two different quantities with different units, such as speed (km/h), density (g/cm³), or pressure (N/m²).
    • →Percentage change is calculated using (change / original) x 100, and compound interest uses repeated percentage multipliers over time.
    • →Proportion problems can be solved using the unitary method (find the value of one unit first) or by setting up equations.
    Marking Points
    • a correct conversion factor, such as 1 m³ = 1 000 000 cm³ or 1 kg = 1000 g, wherever it appears in the working
    • the method of multiplying or dividing by that factor in the right direction, even if the arithmetic afterwards is wrong
    • carrying a two-stage conversion through both stages, for example seconds into hours and metres into kilometres
    • the final value with its unit stated, because a rate without its unit is not an answer to a compound-unit question
    • for a compound unit such as a price or pressure, converting both the numerator and denominator units consistently before combining them
    • stating the scale as a multiplier, for example that 1 cm represents 25 000 cm
    • the method of multiplying or dividing a measured length by that scale, even if the final unit is wrong
    • a length measured within the tolerance allowed on the drawing, a small amount either way
    • the real distance given in the unit the question asks for
    • converting both quantities into the same unit before the fraction is formed
    • the fraction written the right way round, with the quantity named after "of" as the denominator
    • dividing both numbers by a common factor, even where the cancelling is not taken all the way
    • the fraction in its simplest form, which is greater than 1 whenever the first quantity is the larger of the two
    • converting the parts into a common unit before any cancelling
    • dividing every part by the same number, even where the number chosen is not the highest common factor
    • the fully simplified ratio, written with a colon rather than as a fraction
    • a correct 1 : n form with the decimal kept, where the question asks for that form
    • For a part:part ratio, calculating the total number of shares by adding the parts.
    • For a part:whole ratio, treating the whole as the total shares and finding the other part by subtracting the given part from the whole.
    • Finding the value of one share by dividing the total quantity by the number of shares.
    • Correctly calculating both final amounts after finding the value of one share.
    • Expressing two quantities as a ratio and simplifying it to its lowest terms.
    • Applying ratio to a real context, such as scaling a recipe or a concentration, and interpreting the result.
    • the multiplier, found by dividing one quantity by the other
    • the relationship written as a ratio with the parts in the order the question names them
    • the equivalent fraction, or for an equation linking the two quantities
    • the answer in the form the question asks for, a ratio in its simplest form or a single fraction
    • setting up two ratios that are equal, or an equation between two fractions, from the figures given
    • the scale factor between the ratios, even where it is afterwards applied to the wrong part
    • a unitary step such as the cost of one item, where that is the route taken
    • the missing value, to the accuracy the question demands
    • turning a given ratio into a fraction or a multiplier, distinguishing part-to-part (2/5) from part-to-whole (2/7)
    • writing the equation of the line in the form y = kx, with k taken from the ratio as the multiplier between the two quantities
    • reading a gradient from a graph as a change in y divided by the matching change in x
    • putting an answer back into the context, such as an amount in euros rather than a bare number
    • a correct multiplier, such as 1.15 for a 15% increase or 0.85 for a 15% decrease
    • dividing rather than multiplying in an original value question, even if the figure that follows is wrong
    • a percentage found by dividing one quantity by the other and multiplying by 100
    • interest for one year before the total over several years in a simple interest question
    • the final amount or percentage, followed through from a correct method even where an earlier figure was wrong
    • a correct method that finds the multiplier or the constant, such as dividing a given output by its matching input
    • a proportion equation such as y = kx or y = k/x with the value of k written into it
    • substituting the value the question asks about into that equation, even if the final answer is wrong
    • the final answer with its unit and any rounding the question demands
    • choosing the correct relationship, for example writing density = mass ÷ volume before any numbers go in
    • a correct conversion so that the two measures agree, such as changing minutes into hours
    • the calculation carried out correctly, even if the unit is then left off or written wrongly
    • the final answer stated with its compound unit
    • a correct length scale factor, found by dividing a length on one shape by the length it corresponds to on the other
    • squaring or cubing that scale factor as the question needs, even if the arithmetic afterwards goes wrong
    • applying the factor in the right direction, multiplying to enlarge and dividing to reduce
    • the final length, area or volume with its unit
    • identifying the type of proportion from the equation or the table, supported by a reason such as a constant product
    • a correct substitution into the given equation, even if the arithmetic that follows is wrong
    • the value found, given to the accuracy the question asks for
    • a correct statement of the effect of a change, such as saying that in an inverse relationship trebling one quantity divides the other by three
    • a correct proportion statement or equation containing k, matching the power or root named, such as y = k/x for simple inverse proportion
    • using the equivalence X inversely proportional to Y ⇔ X proportional to 1/Y to write X = k/Y
    • substituting the given pair of values to form an equation in k
    • the value of k, even if the answer that follows is wrong
    • using the completed formula on the part asked, including rearranging it when the other variable is the unknown
    • a correct gradient method, such as reading a change in the vertical value with the matching change in the horizontal value
    • the value of the gradient, even if it is left without a unit
    • interpreting that number in context, naming the quantity and the unit, for example a speed in kilometres per hour
    • a correct description of the graph, such as identifying direct proportion from a straight line through the origin
    • a tangent drawn accurately at the correct point, touching the curve rather than cutting across it
    • a correct method for the gradient, using two points on the drawn tangent (or chord) to find the change in y and change in x
    • a final gradient value inside the accepted tolerance, even if it is not interpreted in context
    • the correct interpretation of the gradient in the given context, including the correct compound unit (e.g., metres per second, m/s)
    • a correct multiplier for the percentage change, e.g., 1.03 for a 3% increase or 0.8 for a 20% decrease
    • raising the multiplier to the correct power, matching the number of time periods in the question
    • a correct final value after the full period, before any final subtraction if only the change is required
    • the final answer correctly rounded as requested, typically to the nearest penny for monetary contexts
    • a correct compound interest multiplier, such as 1.05 for 5% growth or 0.85 for 15% decay, used with the correct power
    • the correct final amount after n periods, with the calculation shown rather than just an answer
    • a correct first iteration, with the starting value substituted into the rule
    • two further terms that follow correctly from your own previous answer, even where the first term was wrong
    • the value the terms settle on, given to the number of decimal places asked for
    • a correct multiplier raised to a trial power in a growth or decay problem, and one for naming the period at which the target is first passed
    Examiner Tips
    • 💡Write the conversion as an equation between the two units before you use it, so the direction is on the page rather than in your head.
    • 💡Check the size of your answer: the same distance in centimetres must be a bigger number than in metres.
    • 💡Convert both rates into the same units before you compare them, not after you have compared them.
    • 💡Use a sharp pencil and measure to the nearest millimetre, because scale drawing marks are awarded inside a narrow tolerance
    • 💡Write the unit beside every line of a map calculation; most lost marks here are centimetres reported as metres
    • 💡A scale written as a ratio with no units has both sides in the same unit, so convert only once at the end
    • 💡Underline the words that come after "of"; that quantity is the denominator every time
    • 💡Convert the units before you write the fraction, so the one you cancel is already the right one
    • 💡Leave the answer as a fraction unless the question asks for a decimal or a percentage
    • 💡Look for one number that divides every part at once; if you cannot see one, halve or divide by 3 repeatedly until nothing divides again
    • 💡Check a simplified ratio by multiplying it back up to the numbers you started with
    • 💡Keep the parts in the order the question names them, because 3 : 5 and 5 : 3 are different answers
    • 💡When sharing a quantity, check your answers by adding them together. They should sum to the original total.
    • 💡When forming a ratio, always check if it can be simplified. Look for common factors of both numbers.
    • 💡For a part:whole ratio, remember the whole is the total number of parts, so the other part is found by subtraction.
    • 💡Test your multiplier on the pair of numbers you were given before you use it anywhere else
    • 💡If the answer is a fraction, decide whether it should be smaller or larger than 1 before you write it down
    • 💡Write the ratio in the order the words appear in the question, then simplify it
    • 💡Write the two ratios one above the other with matching parts lined up, then look across for the multiplier
    • 💡Keep a unit value as an exact fraction or a full decimal until the last line of working
    • 💡Check the finished pair by simplifying both ratios; they must reduce to the same thing
    • 💡Read a gradient using two points that sit exactly on grid intersections, not two points you estimated
    • 💡Check the direction of a conversion: multiplying by the gradient goes one way and dividing goes back
    • 💡If a ratio is given as part to whole, change it to part to part before you use it as a multiplier
    • 💡Remember that the multiplier is the second part divided by the first part, so the ratio 2 : 5 gives 5/2 = 2.5, not 2/5
    • 💡Turn every percentage change into one multiplier; it makes a two-stage question a single calculation
    • 💡For an original value question, ask what the amount you were given is a percentage of, then divide by that multiplier
    • 💡Write money answers to two decimal places and put the pound sign in once, at the front
    • 💡Estimate before you calculate: 15% of £80 is about £12, so an answer of £120 has a decimal point in the wrong place
    • 💡Check the direction before you calculate: if one quantity rising sends the other down so that their product stays the same, the relationship is inverse rather than direct.
    • 💡Write the equation with k in it before any numbers go in, so the two-step method is visible to the examiner.
    • 💡Keep the units consistent all the way through, and give money answers to two decimal places.
    • 💡Write the relationship out in words first, then put the numbers in; the division you need becomes obvious and the method is on the page.
    • 💡Convert everything into the units named on the answer line before you calculate, not afterwards.
    • 💡When a rate is per hour and the time is in minutes, turn the minutes into a fraction of an hour, so 24 minutes becomes 0.4 of an hour.
    • 💡Set the three factors out before you calculate: length k, area k², volume k³, then decide which line the question sits on.
    • 💡Mark the pair of corresponding lengths on the diagram before dividing, so the scale factor is the right way up.
    • 💡If an enlarged shape comes out smaller than the original, you divided where you should have multiplied; check that before moving on.
    • 💡Test the equation with an easy value such as one, then double it and see what happens to the other quantity; that identifies the relationship quickly.
    • 💡For an inverse relationship, multiply each pair in the table together; if every product agrees, you have found the constant.
    • 💡Answer using the names of the quantities in the question, not only the letters.
    • 💡Read the wording closely: the square of x means x², while the square root of x means √x, and the two give different formulas.
    • 💡Keep k exact, as a fraction if it will not come out neatly, and round only the final answer.
    • 💡Check the formula by putting the original pair of values back in; it should return the value you started with.
    • 💡Pick two points where the line crosses grid corners exactly, so the values you read carry no estimating error.
    • 💡Give the gradient as a rate with its unit, for example 60 kilometres per hour, rather than the number on its own.
    • 💡A negative gradient means the quantity is falling; say so in words when the question asks what the graph shows.
    • 💡Use a ruler and turn the page until the tangent looks balanced on either side of the point before you draw it.
    • 💡Take the two points on your tangent as far apart as the grid allows; a bigger triangle gives a more reliable gradient.
    • 💡State what the gradient means in the context given, with units, as this interpretation usually carries its own mark.
    • 💡Write down the multiplier before using the calculator, then check it makes sense: above 1 for growth, below 1 for decay.
    • 💡Use the power key (e.g., xʸ) on your calculator instead of multiplying repeatedly; it is faster and more accurate.
    • 💡Keep the full calculator value on screen until the very end of the calculation, then round just once as the final step.
    • 💡For compound interest, write the multiplier and the power before reaching for the calculator, so the method is visible even if the arithmetic slips.
    • 💡Enter the starting value, press equals, then key the iteration rule using the answer button; each further press gives the next term.
    • 💡Keep every digit on the calculator between steps and round only when you write a term down.
    • 💡For growth with an unknown number of periods, work out the total for a few whole numbers of periods and pick the first that passes the target.
    • 💡Always show your working clearly, especially in ratio sharing and proportion problems. Method marks are awarded even if the final answer is wrong due to a small arithmetic slip.
    • 💡For compound measures, write down the formula you are using (e.g., speed = distance / time) and substitute carefully. This helps you gain method marks and reduces errors.
    • 💡In percentage change and compound interest questions, use multipliers (e.g., 1.05 for 5% increase) rather than calculating step-by-step, as this is more efficient and less error-prone.
    Common Mistakes
    • converting m² to cm² by multiplying by 100 instead of by 100², so an area comes out 100 times too small; correct by squaring the length factor
    • dividing where you should multiply, which usually shows as an answer far too small for the context; correct by checking the expected size first
    • writing 1 hour 45 minutes as 1.45 hours when working in hours, rather than 1.75 hours; correct by converting the 45 minutes as 45 ÷ 60
    • reporting a density as grams per cubic metre after working in cubic centimetres, so the unit does not match the figure; correct by keeping numerator and denominator units consistent
    • reversing a scale of 1 : 50 000, so a map length is divided by the scale instead of multiplied by it
    • turning 250 000 cm into kilometres by dividing by 1000 only, giving 250 km rather than 2.5 km
    • measuring from the end of the ruler instead of the zero mark, so every length is out by the same amount
    • taking a scale factor from a pair of sides that do not correspond, such as the base of one shape against the height of the other
    • writing the fraction upside down, so 20 g as a fraction of 50 g is given as 50/20
    • comparing 30 cm with 2 m without converting, giving 30/2 rather than 30/200
    • assuming an answer above 1 must be wrong and inverting it to force a proper fraction
    • cancelling part way and stopping at a fraction that still has a common factor
    • dividing only one part, so 12 : 18 is written as 2 : 18
    • stopping at 6 : 9 because the numbers look smaller, when 3 still divides both parts
    • simplifying 30 minutes : 2 hours as 30 : 2 instead of 30 : 120
    • reversing the order of the parts, so an answer of 3 : 5 is given for a question that named the quantities the other way round
    • Dividing the total quantity by one of the parts of the ratio instead of the sum of the parts.
    • Confusing a part:part ratio with a part:whole ratio (e.g., treating 3:5 as 3 parts to 5 parts, not 3 out of 5 total).
    • When forming a ratio from two quantities, failing to simplify it to its lowest terms or writing the parts in the wrong order.
    • In a part:whole ratio, adding the two numbers to find the total shares instead of using the whole as the total.
    • inverting the multiplier, so saying the second quantity is 2/5 of the first when it is 5/2 of it
    • reading "three times as many" as 3 : 1 when the question names the quantities in the other order
    • using a difference instead of a multiplier, so 8 and 20 are described as twelve apart when the question asks for a ratio
    • comparing a part with the whole when the question compares one part with the other part
    • adding the same number to both parts instead of multiplying, so 3 : 4 is scaled to 6 : 7
    • pairing the parts the wrong way round when the two ratios are written in different orders
    • rounding a unit cost to the nearest penny early, so the scaled-up total is out by several pence
    • scaling a quantity that does not scale, such as a fixed delivery charge added to a per-item price
    • turning the ratio 2 : 5 into the fraction 2/7 when the question compares one part with the other part rather than with the whole
    • using the fraction 2/5 as the multiplier when the ratio 2 : 5 gives the multiplier 5/2, so the equation is y = 2.5x
    • reading the gradient as a change in x over a change in y, so the multiplier comes out inverted
    • drawing a line that misses the origin for a relationship where zero of one quantity means zero of the other
    • reading a value off the graph at a point that is not on a grid intersection and rounding it silently
    • treating an original value question as a straightforward decrease, so £42 after a 20% rise is reduced by 20% to £33.60 instead of £35
    • expecting a 10% rise followed by a 10% fall to return the starting amount, when the two multipliers together give 0.99
    • writing 0.4 as 4% instead of 40% when changing a decimal into a percentage
    • using simple interest where the question describes interest paid on the balance at the end of each year
    • finding the second percentage of the original amount when the question means a percentage of the new amount
    • treating every pair of quantities as directly proportional, so multiplying when the total amount of work is fixed and a division was needed
    • finding the constant and then offering that constant as the final answer instead of using it
    • dividing the wrong way round when finding the constant, writing k = x/y for a direct relationship when it should be k = y/x
    • reading an inverse proportion curve as though it had one fixed gradient
    • dividing the wrong way round, for example working out volume ÷ mass and calling the result a density
    • mixing units inside one calculation, such as a distance in metres with a time in hours, without converting first
    • reading a decimal time wrongly, treating 1.5 hours as one hour and fifty minutes
    • averaging two speeds by adding them and halving, instead of using total distance ÷ total time
    • using the length scale factor on an area or a volume, so multiplying by three rather than by nine when lengths are in the ratio 1 : 3
    • taking a square root of a volume ratio, or a cube root of an area ratio, when converting back to lengths
    • pairing sides that do not correspond, usually by not matching the shapes up by their equal angles first
    • applying the scale factor the wrong way round, so the larger solid comes out smaller than the one you started from
    • reading y = 24/x as direct proportion because both quantities are changing, then multiplying where a division was needed
    • assuming any equation with the variable underneath is inverse proportion, including ones such as y = 2/x + 5 where no product stays constant
    • swapping the roles of the two quantities, so working out x when the question asked for y
    • calling a graph direct proportion because it passes through the origin, even when it is a curve
    • writing y = kx for inverse proportion instead of y = k/x, so the variables move in the same direction rather than opposite directions
    • squaring the whole product instead of the variable, writing y = (kx)² where the question says y is proportional to the square of x
    • finding k and stopping there, without going back for the value the question actually wanted
    • writing y = k/x where the relationship is inverse to the square of x, so dividing by x rather than by x²
    • rounding k partway through, which pushes the final answer outside the accepted range
    • counting squares instead of reading the axis scales, which fails whenever one square is not one unit
    • dividing the horizontal change by the vertical change, which gives the reciprocal of the gradient
    • calling a straight line direct proportion when it does not pass through the origin
    • leaving the gradient as a bare number when the question asked what it represents
    • drawing a chord between two points on the curve when asked for the rate at a single instant (which requires a tangent)
    • reading the two points for the gradient calculation from the curve itself, rather than from the tangent line that was drawn
    • using a very small triangle on the tangent, which makes the calculation very sensitive to small reading inaccuracies
    • giving the y-coordinate of the point on the curve instead of the gradient of the tangent at that point
    • calculating one period of interest and multiplying it by the number of periods (this is simple interest, not compound)
    • using a multiplier of 0.2 for a decrease of 20% when it should be 1 - 0.2 = 0.8
    • multiplying the initial value by the number of years instead of raising the multiplier to the power of the number of years
    • giving the total final amount when the question asked only for the interest earned, or vice versa
    • adding simple interest each year instead of multiplying by the same factor each period, so the total drifts below the compound answer
    • rounding each term to two decimal places and feeding the rounded value back in, so later terms drift away from the true value
    • starting again from the first value each time instead of using the term just found
    • naming the last period before a target is passed when the question asked for the first period at which the amount is above it
    • Students often think that a ratio 2:3 means 2/3 and 3/2 as fractions of the total, but it actually means 2/5 and 3/5 of the total. Correct by adding the parts to find the denominator.
    • In inverse proportion, students sometimes multiply instead of dividing, or assume the relationship is linear. For example, if 4 workers take 6 days, 8 workers take 3 days (not 12 days).
    • When converting between units in rates, students forget to convert both quantities consistently, e.g., using minutes with kilometres per hour. Always convert to matching units before calculating.
    Revision Plan
    1. 1Day 1-2: Revise ratio basics: simplifying ratios, dividing amounts in a given ratio, and solving word problems. Practice with textbook exercises and online quizzes.
    2. 2Day 3-4: Move on to direct and inverse proportion. Learn to recognise each type from context and solve using the unitary method. Complete past paper questions on these topics.
    3. 3Day 5-6: Study compound measures (speed, density, pressure) and rates of change. Memorise the formulas and practice unit conversions thoroughly.
    4. 4Day 7-8: Cover percentage change, including simple and compound interest, depreciation, and reverse percentages. Use multipliers and practice exam-style questions.
    5. 5Day 9-10: Mixed practice: attempt a full past paper section on ratio, proportion and rates of change under timed conditions. Review mistakes and revisit weak areas.
    Exam Question Types
    • 📋Ratio sharing problems: Often ask to divide a total into a given ratio, sometimes with a twist such as one share being given. Advice: always find the value of one part first and check your shares add up.
    • 📋Direct and inverse proportion: Presented as word problems or in a table. Advice: identify the type of proportion, then use the unitary method or set up an equation with a constant of proportionality.
    • 📋Compound measures: Calculate speed, density, or pressure, sometimes requiring unit conversion. Advice: write the formula, convert units if needed, and include the correct unit in your answer.
    • 📋Percentage change and compound interest: Calculate percentage increase/decrease, simple interest, or compound interest over multiple years. Advice: use multipliers for efficiency and show your working.
    Command Word Expectations (PEARSON EDEXCEL)
    Calculate

    Work out a numerical answer using the given information. You must show sufficient working to demonstrate the method, and give the answer with correct units where appropriate. Marks are awarded for correct method and accuracy.

    Show that

    Prove a given result by showing each step of your working clearly. You must not just state the answer; you need to demonstrate the logical progression that leads to the given result. All steps must be mathematically correct.

    Explain

    Give a reason or justification for a statement or result, often in words. You need to refer to the mathematical concept or rule used, not just describe the calculation. For example, 'because the ratio of workers to time is inversely proportional'.

    How Students Lose Marks (Examiner Pitfalls)
    Pitfall: Students often add the parts of a ratio and divide the total by that sum, but then forget to multiply back by each individual share, or they mix up which share belongs to which person. Another common error is treating a ratio such as 3:5 as fractions 3/5 and 5/3 rather than 3/8 and 5/8 of the total.
    ❌ Weak Answer (Loses Marks):Total = 48 sweets, ratio 3:5, so 48 / 8 = 6. Answer: 6 sweets each.
    Example improved answer:Total parts = 3 + 5 = 8. One part = 48 / 8 = 6 sweets. Ali gets 3 x 6 = 18 sweets. Ben gets 5 x 6 = 30 sweets. Check: 18 + 30 = 48.
    Examiner Tip: Always write the number of parts, find the value of one part, then multiply by each share. Finish with a quick addition check to confirm the shares sum to the original total.
    Pitfall: In compound measures and rates of change, students frequently use the wrong formula triangle or fail to convert units before calculating. For example, mixing minutes with hours in a speed calculation, or using grams with kilograms in a density problem.
    ❌ Weak Answer (Loses Marks):Speed = distance / time = 150 / 2.5 = 60 km/h. (Correct here, but many write 150 / 2.5 = 60 and then state 60 m/s without checking units.)
    Example improved answer:Distance = 150 km, time = 2 hours 30 minutes = 2.5 hours. Speed = distance / time = 150 / 2.5 = 60 km/h. Units are consistent because distance is in km and time is in hours.
    Examiner Tip: Before substituting into any formula, convert all quantities to consistent units. Write the unit in your final answer and check it matches what the question asks for.
    Step-by-Step Worked Solutions

    Question: A recipe for 12 cupcakes uses 180 g of flour and 120 g of sugar. How much flour and sugar are needed for 30 cupcakes?

    1. 1.Step 1: Identify the given ratio: 12 cupcakes need 180 g flour and 120 g sugar.
    2. 2.Step 2: Find the amount per cupcake: flour = 180 / 12 = 15 g per cupcake; sugar = 120 / 12 = 10 g per cupcake.
    3. 3.Step 3: Multiply by the new number of cupcakes: flour = 15 x 30 = 450 g; sugar = 10 x 30 = 300 g.
    4. 4.Step 4: State final answer with units: 450 g flour and 300 g sugar.
    Final Answer: 450 g of flour and 300 g of sugar are needed for 30 cupcakes.

    Question: A car travels 210 km in 3 hours 30 minutes. Calculate its average speed in km/h.

    1. 1.Step 1: Convert time to hours: 3 hours 30 minutes = 3.5 hours.
    2. 2.Step 2: Use the formula speed = distance / time.
    3. 3.Step 3: Substitute: speed = 210 / 3.5 = 60 km/h.
    4. 4.Step 4: State final answer with correct units: 60 km/h.
    Final Answer: The average speed is 60 km/h.
    Active Recall Memory Test
    What is the formula for calculating speed?
    Key Fact: Speed = distance / time. The units must be consistent, e.g., km/h or m/s.
    How do you share an amount in a given ratio?
    Key Fact: Add the parts of the ratio to find the total number of parts, divide the total amount by this sum to find the value of one part, then multiply by each share.
    What is the difference between direct and inverse proportion?
    Key Fact: Direct proportion: as one quantity increases, the other increases at the same rate (e.g., cost vs quantity). Inverse proportion: as one quantity increases, the other decreases (e.g., number of workers vs time taken).
    How do you calculate compound interest?
    Key Fact: Use the formula: Final amount = Principal x (1 + r/100)^n, where r is the annual interest rate and n is the number of years. Alternatively, use a multiplier for each year.
    Frequently Asked Questions
    What is the difference between ratio and proportion?
    A ratio compares two or more quantities of the same kind, such as 3:2. A proportion is an equation stating that two ratios are equal, such as 3:2 = 6:4. In GCSE Mathematics, you will use both to solve problems involving scaling and sharing.
    How do I solve inverse proportion problems?
    In inverse proportion, the product of the two quantities is constant. For example, if 4 workers take 6 days, the total work is 24 worker-days. So 8 workers will take 24 / 8 = 3 days. Always identify that as one quantity increases, the other decreases.
    What is a compound measure?
    A compound measure combines two or more different units, such as speed (distance per unit time), density (mass per unit volume), or pressure (force per unit area). You need to remember the formulas and ensure units are consistent before calculating.
    How do I calculate percentage change?
    Percentage change = (change / original value) x 100. If the value increases, it is a percentage increase; if it decreases, it is a percentage decrease. For example, a price rising from £40 to £50 is a change of £10, so (10/40) x 100 = 25% increase.
    What is the unitary method in proportion?
    The unitary method involves finding the value of one unit first, then multiplying to find the required value. For example, if 5 pens cost £2, then 1 pen costs £0.40, so 8 pens cost £3.20. It is a reliable method for both direct and inverse proportion.
    How do I revise ratio, proportion and rates of change effectively?
    Start by revising the basics of ratio and proportion, then move on to compound measures and percentage change. Practice a variety of exam-style questions, especially multi-step problems, and always check your units. Use past papers to familiarise yourself with the question formats and timing.