Ratio, proportion and rates of change — AQA GCSE Mathematics
Test yourself on Ratio, proportion and rates of change with AQA GCSE practice questions.
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Ratio, proportion and rates of change explained
Converting units involves multiplying or dividing.
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First, decide if the answer should be bigger or smaller. Going from a large unit to a small one means more of them, so multiply. For example, 2.5 kg = 2500 g (× 1000) and 3 litres = 3000 ml. For area and volume, the length conversion factor is squared or cubed. Since 1 m = 100 cm, 1 m² = 100² = 10000 cm². For volume, 1 m³ = 100³ = 1,000,000 cm³. Remember that 1 litre = 1000 cm³, so 1 ml = 1 cm³. For compound units like speed or prices, convert each part. To change 90 km/h to m/s, convert 90 km to 90000 m and 1 hour to 3600 s, then divide to get 25 m/s. To compare prices, ensure they are for the same unit of mass or volume.
compound units (eg density, pressure) in numerical and algebraic contexts
Density and pressure each tie three quantities together, and the unit tells you how the formula is built. Density is measured in grams per cubic centimetre, so density = mass ÷ volume. Pressure is measured in newtons per square metre, so pressure = force ÷ area. Rearrange rather than learn three versions of each: from the first relation, mass = density × volume, and volume = mass ÷ density. A typical question gives a block of mass 480 g and volume 60 cm³, and the density is 8 g/cm³. The algebraic version is the same relation with letters in place of numbers, so a force of F newtons spread over an area of x² square metres gives a pressure of F/x² newtons per square metre. Check the units in the question before substituting: a density in g/cm³ next to a volume in m³ means one of them has to be converted first.
use scale factors, scale diagrams and maps
A scale tells you how much real length one drawn length stands for. A map scale of 1 : 25000 means 1 cm on the map is 25000 cm on the ground, which is 250 m. Going from the drawing to real life, you multiply; going back the other way, you divide. So 6 cm on that map is 150000 cm, which is 1.5 km, and a real path of 3 km would be drawn 12 cm long. A scale can also be given as a sentence, such as 1 cm represents 2 m, and that is the same information in different clothes. Many questions expect you to take the length off the diagram yourself with a ruler, so measure to the nearest millimetre and convert afterwards. Scale drawings turn up alongside bearings and constructions, where a length you measure is only as good as the line you drew, so keep your pencil sharp and your lines light.
express one quantity as a fraction of another, where the fraction is less than 1 or greater than 1
Put the amount you are describing on top, the amount you are comparing it with underneath, then cancel. In a class of 30 pupils, 18 walk to school, so the fraction is 18/30, which cancels to 3/5. Both amounts must be in the same unit first: 40 cm out of 2 m is 40/200, not 40/2, and that cancels to 1/5. The answer comes out bigger than 1 whenever the first amount is the larger one, and that is a perfectly good answer. If a plant grows from 20 cm to 35 cm, the new height compared with the old is 35/20, which cancels to 7/4, or 1¾ as a mixed number, and that same value is the multiplier from old height to new. Order is everything here, because swapping the two amounts gives the reciprocal, so decide which one the word 'of' points at before you write anything down.
use ratio notation, including reduction to simplest form
A ratio compares parts without saying how big the whole is, and 3 : 5 is read as three parts to five parts. You reduce it the way you reduce a fraction, by dividing every part by a common factor: 12 : 18 divides by 6 to give 2 : 3, and 20 : 30 : 50 divides by 10 to give 2 : 3 : 5. Units have to match before you start, so 30 cm : 1 m becomes 30 : 100, which is 3 : 10. Decimals and fractions can be cleared by multiplying every part by the same number, so 0.4 : 1.2 becomes 4 : 12 and then 1 : 3. Some questions want the unitary form 1 : n, where you divide both parts by the first, so 4 : 10 becomes 1 : 2.5 and the decimal there is fine. Keep the parts in the order the names appear in the question, because a ratio read backwards describes a different situation.
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)
To share an amount, first find the total number of parts. To split £84 in the ratio 3:4, add 3+4=7 parts. Then find the value of one part: £84 ÷ 7 = £12. Finally, multiply to find each share: 3×£12=£36 and 4×£12=£48. A part:whole ratio is a fraction; if boys to pupils is 2:5, boys are 2/5 of the group. Ratios apply to many contexts. For scaling recipes, if a recipe for 4 people needs 300g flour, a recipe for 6 needs (300/4)×6 = 450g. For conversions, if a map scale is 1:50000, 4 cm on the map is 4×50000 = 200000 cm, which is 2 km in reality. For comparison, you might compare two mixtures to see which is stronger. To express a division like £20 to £30 as a ratio, simplify it to 2:3.
express a multiplicative relationship between two quantities as a ratio or a fraction
A multiplicative relationship compares two quantities by a single multiplier, and this can be written as a fraction or a ratio. The fraction depends on which quantity is the base of the comparison. If Sam has £15 and Tom has £20, then Sam's money as a fraction of Tom's is 15/20 = 3/4, while Tom's as a fraction of Sam's is 20/15 = 4/3. The ratio Sam : Tom is 15 : 20 = 3 : 4. To convert a ratio a : b to a fraction comparing the first quantity to the second, write a/b; to compare a part to the whole, write a/(a+b). For example, 3 : 4 gives 3/4 as first-to-second, but 3/7 as first-to-whole. Wording such as 'twice as many cats as dogs' gives cats : dogs = 2 : 1, and dogs are 1/2 of cats. When two ratios share a quantity, for example a : b and b : c, scale them so the shared quantity matches before combining; 2 : 3 and 4 : 5 become 8 : 12 and 12 : 15, giving a : b : c = 8 : 12 : 15.
understand and use proportion as equality of ratios
Two ratios match when one is a multiple of the other, so 3 : 5 and 12 : 20 go together because both parts have been multiplied by four. Written as fractions that is 3/5 = 12/20, and it is this equality that lets you fill in a missing value. If four pens cost £2.60, the safest route to the cost of seven is through one pen: £2.60 divided by four is £0.65, and seven of those come to £4.55. You can also scale the whole ratio, multiplying 4 : 2.60 by 1.75 to reach 7 : 4.55, which is the same arithmetic arranged differently. To test whether two pairs of numbers match, simplify both ratios and compare them, or cross-multiply and see whether the two products agree. Recipes scaled up, best-buy comparisons and currency conversions all rest on this one idea.
relate ratios to fractions and to linear functions
A ratio, a fraction and a straight-line graph can all carry the same relationship. If y to x is 3 : 2, then y/x = 3/2, and as a function that is y = 1.5x. Plotted, the pairs lie on a straight line through the origin, and the gradient 1.5 is the ratio expressed as a single number. The line passes through the origin because when x is zero, y is zero as well, which is what marks a graph of direct proportion out from any other straight line. Reading a graph the other way works too, so a line through the origin with gradient 4/5 describes y to x as 4 : 5. Take care with the difference between a part to part ratio and a fraction of the whole: if red counters to blue counters is 2 : 3, the red ones are 2/5 of all the counters but 2/3 of the number of blue ones.
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 per hundred, so 37% is 37/100, or 0.37 as a decimal, and that decimal is what you multiply by to find 37% of an amount. To write one amount as a percentage of another, divide and then multiply by 100, so 18 out of 40 is 0.45, which is 45%. A rise of 15% means multiplying by 1.15 and a fall of 15% means multiplying by 0.85. For percentage change, divide the change by the original amount: a price going from £80 to £92 changes by 12/80, which is 15%. Original value questions give you the new amount and want the old one, so you divide instead of multiplying, and £92 after a 15% rise came from 92 ÷ 1.15, which is £80. Simple interest is worked out on the starting amount every year, so £2000 at 3% for four years earns four lots of £60, which is £240. A percentage above 100% is allowed and means more than the whole.
solve problems involving direct and inverse proportion, including graphical and algebraic representations
Two quantities are in direct proportion when doubling one doubles the other, so their ratio stays fixed and y = kx. They are inversely related when doubling one halves the other, so their product stays fixed and y = k/x. The method is the same either way: use a pair of values you are given to find the constant, then use that constant to answer the question. Example: 6 pens cost £2.10, so one pen costs £0.35 and 11 pens cost £3.85. Inverse example: 4 workers take 9 hours, so the job is 36 worker-hours and 6 workers take 6 hours. On a graph, the first kind is a straight line through the origin; the second is a curve that falls towards the axes without ever touching them.
use compound units such as speed, rates of pay, unit pricing
A compound unit measures one quantity per unit of another, and the unit itself tells you the formula. Miles per hour means miles divided by hours, so speed = distance ÷ time, and rearranging gives distance = speed × time and time = distance ÷ speed. Pounds per hour works the same way for pay: 7.5 hours at £12 an hour is £90. Unit pricing compares value by putting different sizes on the same footing: a 500 g box at £1.80 is £3.60 per kg, while a 750 g box at £2.85 is £3.80 per kg, so the smaller box is better value. Questions often mix units, so convert first: 90 minutes is 1.5 hours, and a journey of 30 km in 20 minutes is 90 km/h.
use compound units such as density and pressure
Density and pressure are compound measures: each divides one quantity by another. Density is mass ÷ volume, in g/cm³ or kg/m³, so mass = density × volume. Pressure is force ÷ area, in newtons per square metre (the pascal), so force = pressure × area. Work in one consistent set of units throughout. Example: a block of mass 480 g and volume 200 cm³ has density 480 ÷ 200 = 2.4 g/cm³; a force of 60 N on 0.5 m² gives pressure 60 ÷ 0.5 = 120 N/m². A common question gives a density and the dimensions of a solid, so find the volume first, then multiply. A pressure question may ask which face an object rests on: a larger contact face means a smaller pressure.
compare lengths, areas and volumes using ratio notation scale factors
When one shape is an enlargement of another, a single scale factor controls everything. If every length is multiplied by k, every area is multiplied by k² and every volume by k³. Written as ratios, lengths in the ratio 2:3 give areas in the ratio 4:9 and volumes in the ratio 8:27. Find the length scale factor first from a pair of corresponding lengths, then square it for an area or cube it for a volume. Going the other way, square root an area ratio or cube root a volume ratio to get back to lengths. Example: two similar cans have heights 10 cm and 15 cm, so k = 1.5, their surface areas are in the ratio 1:2.25 and their volumes in the ratio 1:3.375.
make links to similarity (including trigonometric ratios)
Two shapes are similar when one is an enlargement of the other: their angles match and corresponding sides are in the same ratio. In a right-angled triangle that fact is what makes trigonometry work. Every right-angled triangle containing a 30° angle is similar to every other one, so the opposite side divided by the hypotenuse comes to the same number in all of them; that fixed number is sin 30°, and cosine and tangent are the other two fixed ratios. For a similar-triangles question, match the corresponding sides, write the scale factor as a fraction such as 12/8, and multiply the matching side of the smaller triangle by it. Look for a shared angle or a pair of parallel lines, because that is usually what creates the similarity.
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
Saying that one quantity varies inversely with another is the same as saying it varies directly with the reciprocal of that other quantity. So if p varies inversely with q, then p = k/q and the product pq stays constant. Direct proportion instead keeps the quotient constant: p = kq, so p/q is always k. Reading an equation tells you which one you have: a variable on the bottom of a fraction means the inverse kind, a variable multiplied by a constant means the direct kind. Example: in v = 240/t the product of v and t is always 240, so t = 4 gives v = 60, and doubling t to 8 halves v to 30. The constant has a meaning too: in c = 0.85n, the number is the cost of one item.
construct and interpret equations that describe direct and inverse proportion (Higher tier only)
This is Higher tier only. Here the proportion is not always to the variable itself: y may be proportional to the square, the cube or the square root of x, or to the reciprocal of one of those. The method is four steps every time. Write the relationship with a constant, for example y = kx², substitute the pair of values you are given, work out k, then rewrite the equation and use it. Example: y is proportional to x², and y = 45 when x = 3, so 45 = 9k, k = 5 and the formula is y = 5x²; when x = 4, y = 80. If instead y varies inversely with the square of x, the equation is y = k/x², so trebling x divides y by 9.
interpret the gradient of a straight-line graph as a rate of change recognise and interpret graphs that illustrate direct and inverse proportion
On a real-life graph the steepness tells you how fast one quantity changes as the other changes, and its unit is the vertical axis unit per horizontal axis unit. Work it out as the change up divided by the change across, using two points that are easy to read. On a distance-time graph that number is a speed; on a graph of pay against hours worked it is the hourly rate; on a tank filling with water it is litres per minute. Example: a line from (0, 0) to (4, 30) on a distance-time graph measured in km and hours gives 30 ÷ 4, a speed of 7.5 km/h. A straight line through the origin shows the direct kind of proportion. The inverse kind gives a curve falling from the top left, approaching both axes but never 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 (Higher tier only)
On a curve, the rate of change is not constant. The average rate of change between two points is the gradient of the chord connecting them: (y₂ - y₁) / (x₂ - x₁). The instantaneous rate of change at a single point is the gradient of the tangent to the curve at that point. To find this graphically, draw a tangent that just touches the curve at the point. Then, pick two distinct points on your tangent and calculate its gradient (rise/run). In numerical contexts, estimate the instantaneous rate by calculating the average rate over a very small interval around the point. For an algebraic curve like y = x², the average rate between x=2 and x=3 is (3²-2²)/(3-2) = 5. The instantaneous rate at x=2 is the gradient of the tangent there, which is 4. In context, the gradient of a distance-time graph is speed; on a velocity-time graph, it is acceleration.
set up, solve and interpret the answers in growth and decay problems, including compound interest
Repeated percentage change is handled with a multiplier rather than by adding the same amount each time. A rise of 3% is a multiplier of 1.03, a fall of 12% is a multiplier of 0.88, and a number of years means raising the multiplier to that power. So the calculation is the starting amount times the multiplier to the power of the number of periods. Example: £2000 invested at 3% compound interest for five years gives 2000 × 1.03⁵ = 2318.55, so the interest earned is £318.55, which is more than five lots of 3% because each year's interest earns interest of its own. Decay works the same way: a car worth £9000 that loses 15% of its value each year is worth 9000 × 0.85³ after three years.
and work with general iterative processes (Higher tier only)
This is Higher tier only. An iterative method starts from a value and applies the same rule repeatedly, feeding each answer back as the next input. It gives approximate numerical solutions to equations that are awkward to solve exactly, or where an exact method is not available. Given a starting value x₀ and a rule such as xₙ₊₁ = √(2xₙ + 5), put x₀ in to get x₁, put x₁ in to get x₂, and continue; the values settle towards a root. Example: with x₀ = 3, x₁ = 3.3166, x₂ = 3.4108, x₃ = 3.4382, approaching the positive root of x² = 2x + 5. Note that this equation has two roots; iteration from x₀ = 3 tends to the positive one. Keep every digit on the calculator between steps, using the answer key rather than retyping a rounded value. The same idea covers any rule applied repeatedly, such as a balance updated each month.
Your focus
- Describe why a length factor is squared for an area conversion and cubed for a volume conversion.
- Work out a conversion by first deciding whether the answer should be a larger or a smaller number, then multiplying or dividing.
- Calculate a compound unit such as 90 km/h in metres per second by converting both parts before dividing.
Show all 71 objectives
- Compare two prices quoted per different amounts by rewriting both per the same quantity first.
- Write down the relation a compound unit implies, reading grams per cubic centimetre as mass divided by volume.
- Calculate a mass, volume or density by rearranging the relation before substituting, giving the compound unit.
- Explain why a density in g/cm³ beside a volume in m³ needs a conversion before any value is substituted.
- Write down what a scale such as 1 : 25000 means as a usable conversion, with 1 cm standing for 250 m.
- Calculate a real distance by measuring the drawn length to the nearest millimetre and multiplying by the scale.
- Explain why a length taken from a scale drawing is only as accurate as the line drawn and the reading made.
- Explain why order matters here, since swapping the two amounts gives the reciprocal instead.
- Write down one quantity as a fraction of another, converting to a common unit first and cancelling to lowest terms.
- Interpret a fraction above 1 as the multiplier from old value to new, such as 7/4 for a growth from 20 cm to 35 cm.
- Describe what a ratio compares and why 3 : 5 does not tell you how large the whole is.
- Simplify a ratio to lowest whole-number terms, converting to a common unit and clearing decimals first.
- Work out the unitary form 1 : n and explain why a ratio read backwards describes a different situation.
- Calculate the value of one part by dividing the total by the number of parts, then multiply up to give each share.
- Explain the difference between a part to part and a part to whole ratio, reading the second as a fraction of the whole.
- Work out the total from one known share by reversing the steps, as £36 for the smaller part of a 3 : 4 split.
- Calculate the amount of one ingredient in a mixture such as 1 : 6 squash to water, then check the shares add back.
- Describe how a ratio a : b and the fraction a/b carry the same multiplicative link between two quantities.
- Work out a to c from a ratio for a to b and one for b to c by scaling the shared quantity to match.
- Explain which quantity is being described in wording such as twice as many cats as dogs before any ratio is written.
- Explain why two ratios match when every part of one has been multiplied by the same number.
- Calculate a missing value by finding the cost or size of one unit first, as £2.60 for four pens gives £0.65 each.
- Compare two options for value by simplifying both ratios or cross-multiplying, then state which is the better buy.
- Write down the fraction and the equation that match a ratio, turning y to x of 3 : 2 into y = 1.5x.
- Draw the line through the origin for a direct proportion and work out its gradient from a clear pair of points.
- Explain why red to blue of 2 : 3 makes the red counters 2/5 of all of them but 2/3 of the number of blue ones.
- Write down the decimal multiplier for a percentage change, 1.15 for a rise of 15% and 0.85 for a fall of the same size.
- Calculate one quantity as a percentage of another, and a percentage change by dividing by the original amount.
- Work out an original value by dividing the new amount by the multiplier instead of multiplying by it.
- Calculate simple interest as the same amount earned on the starting sum in each year of the term.
- Explain why a percentage above 100% is allowed and compare two quantities fairly by putting both as percentages.
- Describe the difference between direct and inverse proportion by what stays fixed, the ratio or the product.
- Work out the constant from a given pair of values and use it to answer the question asked.
- Calculate a time for a different number of workers by finding the total worker-hours first.
- Explain why direct proportion gives a straight line through the origin and inverse proportion a curve that never touches the axes.
- Write down the formula a compound unit implies, reading miles per hour as distance divided by time.
- Calculate a speed after converting the time first, so 30 km in 20 minutes gives 90 km/h.
- Compare two package sizes by working out the price per kilogram for each and stating which is the better value.
- Calculate density as mass divided by volume and explain that the answer gives the mass of one cubic centimetre of the material when the density is in g/cm³, or one cubic metre when it is in kg/m³.
- Work out a force or a mass by rearranging the relationship, converting every measurement into one set of units first.
- Explain why an object resting on its largest face exerts the smallest pressure on the ground.
- Describe how a length scale factor k multiplies every area by k squared and every volume by k cubed.
- Calculate the area or volume of a similar shape by squaring or cubing the scale factor found from a matching pair of lengths.
- Explain why an area ratio must be square rooted, and a volume ratio cube rooted, to get back to the ratio of lengths.
- Explain why every right-angled triangle containing a 30 degree angle gives the same value for opposite divided by hypotenuse.
- Work out a missing length in a pair of similar triangles by matching corresponding sides and multiplying by the scale factor.
- Show that two triangles are similar by naming two pairs of equal angles created by a shared angle or by parallel lines.
- Compare direct and inverse proportion by whether the quotient or the product of the two quantities stays constant.
- Interpret an equation such as v = 240/t to say which kind of proportion it shows and work out a missing value.
- Explain what the constant in an equation such as c = 0.85n stands for in the situation the question describes.
- Write down the relationship as an equation containing a constant, such as y = kx squared or y = k over root x.
- Calculate k by substituting the given pair of values, then use the completed formula to find y for a new x.
- Explain why trebling x divides y by nine when y varies inversely with the square of x.
- Work out the gradient of a straight-line graph from two points read off the axis scales rather than off the squares.
- Interpret that gradient in context, saying whether it is a speed, an hourly rate of pay or a filling rate, with its unit.
- Describe how a graph showing direct proportion differs in shape from one showing inverse proportion.
- Describe the difference between the gradient of a chord as an average rate and the gradient of a tangent as an instantaneous rate.
- Estimate an instantaneous rate of change by drawing a tangent at the named point and dividing the rise by the run.
- Interpret a chord or tangent gradient on a distance-time or speed-time graph as a speed or an acceleration.
- Apply chord and tangent gradients in numerical and algebraic contexts as well as graphical ones.
- Explain why a hand-drawn tangent means answers are accepted over a range rather than as one exact value.
- Write down the multiplier for a percentage change, such as 1.03 for a rise of 3% and 0.88 for a fall of 12%.
- Calculate a compound total as the starting amount times the multiplier to the power of the periods, then subtract to give the interest.
- Explain why five years of 3% compound interest earns more than five separate lots of 3% simple interest.
- Describe how an iterative process feeds each answer back into the same rule so the values close in on a solution.
- Work out the first three iterates from a given rule and starting value, keeping full calculator accuracy between steps.
- Show that a rearrangement of a given equation produces the iterative rule quoted in the question.
- Recognise that an iterative process gives an approximate root and that the starting value can affect which root is approached.
Ratio, proportion and rates of change exam tips
Quick Revision Summary (Key Takeaway)
Ratio, proportion and rates of change covers comparing quantities using multiplicative relationships, sharing amounts in given ratios, solving direct and inverse proportion problems, and calculating with compound units such as speed, density and pressure. It is a major strand of AQA GCSE Mathematics, appearing in both calculator and non-calculator papers and underpinning topics like percentages, similarity and graphs.
Topic Overview
This topic develops your ability to compare quantities multiplicatively using ratio notation, to share amounts in given ratios, and to solve problems involving direct and inverse proportion. You will also work with rates of change, including compound measures like speed, density and pressure, and interpret graphs representing proportional relationships.
Ratio, proportion and rates of change is fundamental to many areas of mathematics and real-life contexts, from scaling recipes and mixing concrete to calculating fuel consumption and currency exchange. It appears across all GCSE papers and provides essential foundations for algebra, geometry and statistics, making it one of the most heavily weighted topics in the AQA GCSE Mathematics specification.
Key Concepts
- →Ratio compares quantities of the same kind using division; it can be simplified like fractions and used to share amounts by finding the value of one part.
- →Direct proportion means two quantities increase or decrease at the same rate (y = kx), while inverse proportion means one increases as the other decreases (y = k/x).
- →Rates of change describe how one quantity changes with respect to another, often involving compound units such as m/s, g/cm^3 or £/kg.
- →Compound measures combine two different units, and problems often require rearranging formulas such as speed = distance / time or density = mass / volume.
- →Proportional reasoning can be represented graphically: direct proportion gives a straight line through the origin, while inverse proportion gives a reciprocal curve.
Marking Points
- A correct conversion fact written down, such as 1 kg = 1000 g or 1 hour = 3600 seconds, even if the arithmetic that follows is incorrect.
- Multiplying or dividing by the correct conversion factor in the right direction.
- Using the squared or cubed factor in an area or volume conversion, e.g., using 100² for m² to cm².
- A rate formed as one quantity divided by the other, for example distance divided by time for a speed.
- The final value given with its correct unit, since an unlabelled number in a compound-unit question is not a complete answer.
- a correct rearrangement, such as mass = density × volume, before any numbers are substituted
- any unit conversion needed first, such as changing cubic metres into cubic centimetres
- substituting the given values into the correct relation, even if the final answer is wrong
- the answer written with its compound unit rather than as a bare number
- a measured length that falls inside the tolerance allowed when the length has to be read off the diagram
- the scale stated as a usable conversion, for example 1 cm standing for 250 m
- multiplying or dividing by the scale in the correct direction, even if the final answer is wrong
- the real distance given in a sensible unit, with centimetres converted to metres or kilometres as asked
- the two quantities written the right way round, with the one being described on top
- converting both quantities to a common unit before the fraction is formed
- cancelling to lowest terms, or for an equivalent decimal or percentage where the question allows one
- a value greater than 1 left as an improper fraction or a mixed number, either being acceptable unless a form is named
- a common factor used to divide every part of the ratio, not only one of them
- converting to a common unit before simplifying, even if the simplification that follows is wrong
- the ratio in simplest whole-number form, with the parts in the order the question names them
- a correct unitary form when the question asks for 1 : n, where a decimal or fraction part is acceptable
- Calculating the total number of parts by adding the numbers in the ratio.
- Finding the value of one part by dividing the total amount by the total number of parts.
- Correctly multiplying the one-part value by each number in the ratio to find the final shares.
- Interpreting a part:whole ratio as a fraction of the total.
- Expressing a division as a ratio in its simplest form by finding a common factor.
- Applying a ratio correctly for scaling, such as for a recipe or map scale.
- dividing the two quantities in the correct order to obtain the multiplier, for example Sam's money ÷ Tom's money = 15 ÷ 20 = 3/4
- converting correctly between ratio and fraction forms, stating clearly whether the fraction compares first-to-second (a/b) or part-to-whole (a/(a+b))
- scaling one or both ratios so that the shared quantity matches before they are combined, for example 2 : 3 and 4 : 5 become 8 : 12 and 12 : 15
- stating the relationship with the quantities named, so it is clear which quantity is being described as a fraction or ratio of which
- when combining ratios, writing the final three-part ratio in the correct order, for example a : b : c = 8 : 12 : 15
- the two ratios or fractions set equal in matching order, with like quantities in like positions
- the value of one unit, or for the scale factor between the two ratios
- the missing value, even if it is later rounded wrongly
- a money answer written to two decimal places
- a stated conclusion when the question asks which option is better value or whether the pairs match
- turning the ratio into a fraction or a multiplier in the correct order
- an equation such as y = 1.5x with the multiplier taken from the ratio
- a straight line drawn through the origin where the relationship is one of direct proportion
- a gradient read off using a clear pair of points, with the vertical change divided by the horizontal change
- the correct multiplier, such as 1.15 for a rise of 15% or 0.85 for a fall of the same size
- dividing by the original amount rather than the new amount when finding a percentage change
- a division by the multiplier in an original value question, even if the final answer is wrong
- interest calculated on the starting amount for each year of a simple interest question
- a money answer written to two decimal places, and for a percentage rounded as the question asks
- setting up the correct relationship, for example y = kx for the direct case or xy = k for the inverse case
- substituting a given pair of values to find the constant, even if it is then used wrongly
- a correct unitary step, such as the cost of one item or the time for one worker, even if the final answer is wrong
- the final value, with money written to two decimal places and units given
- a correct rearrangement, such as writing distance = speed × time before any numbers go in
- converting a time or a mass into the units the answer needs, for example changing 45 minutes into 0.75 hours
- the correct division or multiplication, credited on the method even if the value is wrong
- the answer written with its compound unit, such as km/h or pounds per hour
- stating the relationship, for example density = mass ÷ volume, before substituting
- an intermediate quantity such as the volume of the solid, even if the final answer is wrong
- a correct unit conversion, such as grams into kilograms or cm³ into m³
- the final value with the matching compound unit
- for a pressure question, identifying the area of the face in contact with the ground
- the length scale factor found from a pair of corresponding lengths
- squaring it for an area or cubing it for a volume, even if the arithmetic after that is wrong
- taking a square root or cube root when the question starts from an area or a volume
- the final length, area or volume with the correct unit, such as cm² or cm³
- showing the triangles are similar, for example by naming two pairs of equal angles
- a correct pair of corresponding sides written as a ratio or a fraction
- the scale factor, or for the trigonometric ratio chosen, even if the final length is wrong
- the length or angle given to the accuracy the question asks for
- choosing the right form, p = kq for the direct case or p = k/q for the inverse case
- substituting a given pair of values to find k, even if k is then used wrongly
- writing the full equation with k replaced by its value
- substituting into that equation to answer the part asked, and for interpreting the constant when asked what it represents
- writing the relationship as an equation containing a constant, such as y = kx² or y = k/√x
- substituting the given pair correctly to form an equation in k
- the value of k, and one for the completed formula written out in full
- substituting the new value into that formula, credited even if the constant found earlier was wrong
- reading two points off the graph correctly, using the axis scales rather than the squares
- the division that gives the gradient, credited even if a point was misread
- the value of the gradient, and one for saying what it means in context with its unit
- identifying the shape correctly, such as a straight line through the origin for the direct kind
- Accurately drawing a tangent to the curve at the specified point.
- Identifying the coordinates of two distinct points on the drawn tangent (or the given chord).
- Correctly calculating the gradient using the formula (change in y) / (change in x).
- Stating a final answer that falls within the acceptable range of values for a hand-drawn tangent.
- Interpreting the gradient in the context of the problem, including the correct units (e.g., m/s for speed).
- the correct multiplier, such as 1.045 for a rise of 4.5% a year
- raising the multiplier to the right power, matching the number of periods
- a complete calculation such as 5000 × 1.045⁶, credited even if the arithmetic is wrong
- the answer rounded to the nearest penny, and for subtracting the starting amount when the interest rather than the total is asked for
- substituting the starting value into the rule correctly to get the first iterate
- two or more further iterates, each written to enough decimal places, even if the last one is wrong
- the solution given to the accuracy asked for, with the iterates written down as evidence
- showing that a rearrangement of the equation leads to the given rule, when that is what is asked
- recognising that an iterative process gives an approximate root, and that a different starting value may tend to a different root
Examiner Tips
- 💡Write the conversion factor down as a simple statement (e.g., 1 m = 100 cm) before you start calculating. Then check if your answer should be a bigger or smaller number.
- 💡Keep the units attached to every number in your working to avoid confusion and ensure the final answer has the correct units.
- 💡For compound units, write the conversion for each part separately before combining them. This breaks the problem down and reduces errors.
- 💡Read the unit you are asked for and build the formula from it: newtons per square metre means force divided by area
- 💡Sense-check against something familiar; water has a density of about 1 g/cm³, so a metal coming out at 0.02 g/cm³ signals a flipped division
- 💡In an algebraic question keep the letters in the same order as the words, so the expression can be read back against the question
- 💡Turn the ratio into a sentence in words, such as one centimetre stands for 250 metres, before you calculate anything
- 💡Check the answer is a believable real distance; a garden path coming out at 40 km means the conversion went the wrong way
- 💡Take a ruler in with you, because a question that says measure expects one and an answer judged by eye loses accuracy marks
- 💡Underline the quantity that follows the word 'of', because that is the one on the bottom
- 💡Cancel with a factor you are sure about rather than hunting for the largest one in a single step
- 💡If the answer is bigger than 1, reread the wording once and then leave it alone, since that is allowed here
- 💡Ask whether the parts still share a factor; if all of them are even, you can go further
- 💡Write the labels above the numbers the first time you form the ratio, so the order cannot drift as you work
- 💡For a unitary answer, divide by the first part even when the result is not a whole number
- 💡Add your final shares together at the end. If they do not add up to the original total, you have made a mistake.
- 💡When a question gives you the value of one share instead of the total, find the value of one part first. All other values can be found from there.
- 💡Label every number in your working with what it represents (e.g., 'parts', '£', 'ml') to avoid getting confused.
- 💡Decide which quantity the words say the other is measured against; that one goes on the bottom of the fraction. For example, 'Sam's money is what fraction of Tom's?' puts Tom's amount on the bottom.
- 💡Test the multiplier you found on the numbers given, and see whether it really lands on the other quantity. For example, if you claim Sam's money is 3/4 of Tom's, check that 3/4 × £20 = £15.
- 💡When two ratios share a quantity, write them one above the other and scale until the shared column matches. For example, 2 : 3 and 4 : 5 become 8 : 12 and 12 : 15, so a : b : c = 8 : 12 : 15.
- 💡Set the working out in two labelled columns, so the numbers you divide stay paired with the right quantity
- 💡Finding the value of one item is usually the safest route and earns method marks even when the arithmetic slips
- 💡Sense-check direction: more items must cost more, so an answer that falls as the amount rises is wrong
- 💡On a straight line through the origin the ratio of the two quantities is the same everywhere, so read it from whichever point has easy coordinates
- 💡Turn the ratio into a decimal multiplier and test it on one pair of values before using it on the rest
- 💡If a graph is meant to show direct proportion, check it starts at the origin before you trust anything read from it
- 💡Decide first whether the percentage is taken from the original amount or from the new one; the word original is your signal to divide
- 💡Write the multiplier down before you calculate, so an increase cannot get applied as a decrease
- 💡Keep full accuracy in the calculator between steps and round only the final money answer
- 💡For a comparison question, turn both quantities into percentages of their own totals before saying which is larger
- 💡Write down what one unit costs, or how long one worker takes, before you scale up; showing this unitary step makes your method clear to the examiner.
- 💡Check the direction of your answer before you write it: more workers must mean less time, more items must mean more cost.
- 💡If a graph is given, read a pair of values off it and test them, rather than guessing which kind of proportion it shows.
- 💡Write the formula down before you touch the numbers, so the rearrangement is on the page even if the arithmetic slips.
- 💡Convert every time into hours, or every mass into kilograms, at the start and write the converted value down.
- 💡Read the unit the answer must be in from the question, then check your last line matches it.
- 💡Write the relationship out in full and rearrange it before substituting; a clear rearranged form makes your method easy to follow.
- 💡Decide your units at the start: if the answer is wanted in kg/m³, convert every measurement before you divide.
- 💡For a solid, work out the volume as a separate labelled step, so a slip later does not cost you that mark.
- 💡Before you start, write the three scale factors on your page: length k, area k² and volume k³.
- 💡If the question gives an area and asks for a length, take the square root first and write that step down.
- 💡Check the direction: going from the larger shape to the smaller one, the scale factor is less than 1.
- 💡Redraw the two triangles separately in the same orientation and mark the equal angles before you write any ratio.
- 💡Write the fraction with the unknown side on top, so the last step is a multiplication rather than a rearrangement.
- 💡When you are asked to prove similarity, give a reason for each pair of equal angles, such as alternate angles or a common angle.
- 💡Turn the words into an equation on your first line; that equation is where the method marks sit.
- 💡Test your equation by putting the given values back in and checking both sides agree.
- 💡For the inverse kind, check that each pair of values has the same product; for the direct kind, check the quotient is the same.
- 💡Set the work out as four lines: the proportion statement, the equation with k, the value of k, and the finished formula.
- 💡Read whether the proportion is to the square, the cube or the root before you substitute anything.
- 💡When you are asked what happens to y as x is multiplied by a number, the powers alone answer it: if y is proportional to x³, trebling x multiplies y by 27.
- 💡Pick two points where the line crosses grid intersections and mark them on the graph so your reading can be seen.
- 💡Say what the number means in the words of the question, for example pounds per hour, not only its value.
- 💡Check the origin before claiming proportion: a straight line that misses the origin is linear but not proportional.
- 💡Use a transparent ruler to draw your tangent. Align it carefully so that it just touches the curve at the required point and the angle with the curve appears equal on both sides.
- 💡To improve accuracy, extend your tangent line across the entire grid. This allows you to choose two points for your gradient calculation that are far apart, which minimises reading errors.
- 💡Don't just write a number. Always interpret your calculated gradient in the context of the question, including the correct units (e.g., acceleration in m/s²).
- 💡Write the multiplier as a separate step so the percentage method is clear before evaluating.
- 💡Count the periods carefully: interest paid at the end of each of three years is a power of three, not four.
- 💡Keep the full value in the calculator and round to the nearest penny only on the last line.
- 💡Type the rule once with the answer key standing in for the previous value, then press equals repeatedly to generate the sequence.
- 💡Write each iterate to at least four decimal places and round only at the very end.
- 💡If you are asked to show a rearrangement gives the rule, start from the equation and work towards the rule, showing each move.
- 💡Always show your working clearly, especially when setting up equations for proportion problems, as method marks are awarded even if the final answer is wrong.
- 💡For ratio sharing, write down the total number of parts and the value of one part before calculating individual shares; this makes your reasoning clear and helps you check your answer.
- 💡When using compound measures, write down the formula you are using and substitute values carefully, paying close attention to units; examiners look for correct unit conversions and final units in answers.
Common Mistakes
- Multiplying by 100 instead of dividing when converting centimetres to metres; for example, 250 cm ÷ 100 = 2.5 m, whereas 250 × 100 = 25,000 and is not the conversion.
- Using a linear conversion factor for area or volume, such as 1 m² = 100 cm² instead of 10000 cm².
- Converting only one part of a compound unit, for example changing kilometres to metres but leaving hours as hours.
- Writing a decimal time incorrectly, for example 2 hours 30 minutes as 2.3 hours instead of 2.5 hours.
- Turning the formula over, dividing volume by mass, which gives the reciprocal of the density rather than the density, so 60 cm³ ÷ 480 g comes out as 0.125 instead of 8
- Substituting a mass in kilograms into a formula that expects grams, so the density is wrong by a factor of 1000
- Giving a pressure with no unit, or with the unit of the area attached instead
- Rearranging by moving a letter to the other side without changing whether it multiplies or divides
- Leaving a real distance in centimetres, writing 150000 cm where 1.5 km was wanted
- Dividing by the scale when going from map to ground, which shrinks a real distance instead of enlarging it
- Reading the ruler from the 1 cm mark rather than from zero, or measuring from the wrong end of the line
- Treating a scale of 1 : 25000 as though it were a ratio of areas rather than of lengths
- Writing the fraction upside down, giving 30/18 when 18 as a fraction of 30 was asked for
- Comparing amounts in different units, so 40 cm out of 2 m becomes 40/2
- Flipping an improper fraction such as 7/4 to 4/7 because a top-heavy fraction looks wrong
- Cancelling only part of the way and stopping, leaving 18/30 as 9/15
- Dividing one part only, turning 12 : 18 into 2 : 18
- Ignoring the units, so 30 cm : 1 m is simplified to 30 : 1
- Stopping too early, leaving 12 : 18 as 6 : 9 when the simplest form was wanted
- Reversing the order, answering a question about red to blue with the blue to red ratio
- Dividing the total amount by one of the numbers in the ratio instead of by the sum of the parts.
- Confusing a part:part ratio with a part:whole ratio, for example treating a 2:5 ratio of boys to pupils as 2 boys for every 5 girls.
- Incorrectly setting up a scaling problem, for example by adding the difference instead of multiplying by a scale factor.
- Forgetting to simplify a ratio to its lowest terms when asked to express a division as a ratio.
- Giving the reciprocal: saying Tom's money is 3/4 of Sam's when it is Sam's that is 3/4 of Tom's. Correct by identifying the base quantity from the wording.
- Reading 'A is three times B' as B being three times A, which reverses the multiplier. Correct by checking which quantity is described relative to which.
- Joining a ratio for a to b with one for b to c without matching the shared part, so 2 : 3 and 4 : 5 are written as 2 : 5 rather than 8 : 12 : 15. Correct by scaling both ratios so the shared quantity is the same number.
- Treating the ratio 3 : 4 as the fraction 3/7 when the comparison asked for is between the two quantities themselves. Correct by using a/b for first-to-second and a/(a+b) for part-to-whole.
- Comparing ratios written in opposite orders, so 3 : 5 is tested against 20 : 12 and judged not to match
- Adding rather than multiplying, turning 3 : 5 into 6 : 8 by adding three to each part
- Rounding the value of one unit too early, so three items for £2.00 are called 67p each and ten of them then come to £6.70 rather than £6.67
- Comparing two prices without putting them on the same footing, such as cost per pack against cost per item
- Reading a part to part ratio as a fraction of the whole, so 2 : 3 is written as 2/3 of the total instead of 2/5
- Writing the gradient upside down, giving x/y where y/x was needed
- Drawing a direct proportion graph that misses the origin, or joining the plotted points with a curve
- Using the coordinates of one point on their own to get the gradient of a line that does not pass through the origin, where the gradient has to come from the change between two points
- Dividing by the new amount in a percentage change, so a rise from £80 to £92 is given as 12/92
- Reversing an increase by taking the same percentage off again, subtracting 15% of £92 instead of dividing by 1.15
- Using simple interest when the question says compound, so later years are charged on the starting amount only
- Reading 'of' the wrong way round, working out 40 as a percentage of 18 when the reverse was asked for
- Treating an answer above 100% as an error and shrinking it to fit
- treating an inverse situation as a direct one, so scaling both quantities up together when one of them should go down
- working out the constant and then leaving it as the answer instead of using it
- adding or subtracting the difference rather than multiplying by a scale factor, for example saying that if 4 workers take 9 hours then 6 workers take 7 hours
- reading minutes as decimal hours, so treating 2 hours 30 minutes as 2.30 hours rather than 2.5 hours
- dividing the wrong way round, so working out a time divided by a distance when the question asks for a speed
- comparing two sizes without putting them in the same unit, so a price per 100 g is set against a price per kg
- mixing units inside one calculation, such as a mass in grams with a volume in cubic metres, which leaves the density out by a factor of a thousand
- dividing the volume by the mass instead of the mass by the volume
- using the whole surface area of a solid when only the face in contact with the ground carries the force
- multiplying an area by the length scale factor instead of by its square, so doubling the lengths is treated as doubling the area
- cube rooting an area ratio or square rooting a volume ratio, so the wrong root is taken
- pairing a length in one shape with a length that does not correspond to it, such as a height against a slant height
- pairing sides that do not correspond, usually because the two triangles were not redrawn the same way up
- calling the side opposite the marked angle the hypotenuse, when the hypotenuse is always opposite the right angle
- using the scale factor upside down, so a length that should grow shrinks instead
- reading an inverse relationship as a direct one, so writing p = kq when it is the product of the two quantities that stays fixed
- recalculating the constant from the new values, as though it changed between parts of the question
- confusing the reciprocal with the negative, so writing p = −kq instead of p = k/q
- squaring the constant as well as the variable, so writing y = (kx)² when the relationship is y = kx²
- using a proportion to x when the question says the square of x, which turns a factor of 4 into a factor of 2
- leaving out the reciprocal in an inverse relationship, so writing y = kx² instead of y = k/x²
- counting squares instead of reading the scales, so the answer is out by whatever the scale is
- dividing the change across by the change up, which gives the reciprocal of the gradient
- giving the gradient as a bare number, so a speed is never identified as a speed
- calling any straight line direct proportion when it does not pass through the origin
- Calculating the gradient of a chord by using two points on the curve when asked for an instantaneous rate. Correction: An instantaneous rate requires the gradient of a tangent, a straight line that touches the curve at that single point.
- Misreading the scales on the axes, for example by counting squares instead of using the marked units. Correction: Always check the value each square represents on both the vertical and horizontal axes before reading coordinates for your gradient calculation.
- Drawing an inaccurate tangent that crosses the curve (a secant) instead of just touching it. Correction: Use a clear ruler and ensure the line only makes contact at the required point, with the curve moving away from the line on both sides.
- adding the same interest every year, which is simple interest, when the question says compound
- using the percentage itself as the multiplier, so multiplying by 0.05 rather than 1.05 for a rise of 5%
- using a multiplier of 0.15 rather than 0.85 for a fall of 15%
- giving the total when only the interest earned was asked for, or the other way round
- rounding each iterate to two decimal places and feeding the rounded value back in, so the later values drift
- stopping at x₁ when the question asks for x₃, or stopping before the values have settled
- putting the previous answer into the wrong place in the rule, for example squaring it when the rule takes a root
- assuming an iterative method is needed only when an equation cannot be solved exactly; it is also used when an exact method is unavailable or inconvenient
- Students often think that adding the same amount to both parts of a ratio preserves the ratio, but only multiplying or dividing both parts by the same non-zero number keeps the ratio equivalent.
- When solving inverse proportion, many students incorrectly set up a direct proportion equation; they should remember that the product of the two quantities remains constant.
- In rate problems, students sometimes mix units (e.g., minutes and hours) without converting, leading to incorrect answers; always convert to consistent units before calculating.
Revision Plan
- 1Start by reviewing the basics of ratio notation and simplifying ratios; practice sharing amounts in one-part, two-part and three-part ratios with and without a calculator.
- 2Move on to direct and inverse proportion, learning to identify each type from word problems and setting up the correct equation; use the constant of proportionality method.
- 3Study compound measures, focusing on speed, density and pressure; memorise the formulas and practice rearranging them to find any missing quantity.
- 4Work through past AQA GCSE exam questions on ratio, proportion and rates of change, timing yourself to build exam technique and familiarity with question styles.
- 5Create a summary sheet of key formulas and methods, and use active recall to test yourself on definitions and problem types regularly.
Exam Question Types
- 📋Ratio sharing problems: often ask to share an amount in a given ratio, sometimes with a twist such as one share is known and you must find the total. Advice: always find the value of one part first.
- 📋Direct and inverse proportion word problems: typically involve real-life contexts like recipes, workers, or speed. Advice: identify the type of proportion, set up an equation using k, and solve.
- 📋Compound measures calculations: questions on speed, density or pressure, sometimes requiring unit conversion. Advice: write down the formula, convert units if necessary, and show substitution clearly.
- 📋Graph interpretation: questions may ask you to read values from a conversion graph or identify whether a graph shows direct or inverse proportion. Advice: check if the graph is a straight line through the origin (direct) or a curve (inverse).
Command Word Expectations (AQA)
You must work out a numerical answer, showing sufficient working to justify your result. Method marks are available even if the final answer is incorrect, so always write down intermediate steps.
You must provide a clear, logical chain of reasoning that leads to the given result. All steps must be shown, and the final statement should match the given value exactly.
You must give reasons for your answer, often referring to mathematical concepts such as proportionality or ratio equivalence. Use precise mathematical language and link your explanation to the context.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A recipe uses flour and sugar in the ratio 5:2. If a baker uses 750 g of flour, how much sugar is needed?
- 1.Step 1: Identify the ratio of flour to sugar is 5:2, and the known amount of flour is 750 g.
- 2.Step 2: Find the value of one part by dividing the flour amount by its ratio part: 750 / 5 = 150 g per part.
- 3.Step 3: Multiply the value of one part by the sugar ratio part: 150 x 2 = 300 g.
Question: The density of a metal block is 8.5 g/cm^3 and its volume is 20 cm^3. Calculate its mass.
- 1.Step 1: Recall the formula: density = mass / volume, so mass = density x volume.
- 2.Step 2: Substitute the given values: mass = 8.5 g/cm^3 x 20 cm^3.
- 3.Step 3: Calculate: 8.5 x 20 = 170 g.