HookThe 2016 Toblerone that got 10% lighter without changing price
In November 2016 the maker of Toblerone, Mondelez, quietly widened the gaps between the triangular chunks of its UK bars, cutting the \(400\) g bar to \(360\) g while the shelf price stayed the same. Shoppers were furious, and the mathematics explains why. Hold the price at \(300\) pence: the price per gram rose from \(\tfrac{300}{400}=0.75\) pence to \(\tfrac{300}{360}\approx 0.833\) pence, so the ratio of new unit price to old is \(\tfrac{400}{360}=\tfrac{10}{9}\approx 1.111\) — an \(11.1\%\) rise, even though the sticker never moved. The weight fell \(10\%\) but the price per gram climbed more than \(11\%\), because reversing a \(10\%\) cut takes an \(11.1\%\) rise, not a \(10\%\) one. That reverse-percentage asymmetry sits at the centre of this section.
R3.1 is the largest section in GCSE maths and the most useful once you leave the exam hall: every 'per', 'for every', 'at this rate', 'best value' and 'how much interest' question lives here. You will handle ratio notation and sharing; proportion as equal ratios; percentages including the reverse-percentage trap that outraged Toblerone buyers; compound units such as speed, density, pressure and price-per-gram; direct and inverse proportion in words, algebra and graphs; rate of change read as the gradient of a graph — an average rate from a chord, an instantaneous rate from a tangent; and growth and decay, from compound interest to depreciation and iterative processes. It threads through both the non-calculator and calculator papers, and it is the section examiners lean on hardest to separate grades.
ModelRatio, its simplest form, and one quantity as a share of another
A ratio compares quantities of the same kind, written \(a:b\) and read 'a to b'. You simplify it exactly as you simplify a fraction — divide both parts by their highest common factor — so \(45:75\) divides by \(15\) to give \(3:5\), and \(0.5:2\) scales up to \(1:4\). A ratio with units must have the same unit on both sides first: \(50\) cm to \(2\) m is \(50:200\), i.e. \(1:4\), not \(50:2\).
You will also express one quantity as a fraction of another, and Higher insists this fraction can be greater than \(1\). If a rope of \(15\) m is compared with one of \(12\) m, the first is \(\tfrac{15}{12}=\tfrac{5}{4}\) of the second — a fraction above \(1\) because the first rope is longer — while the second is \(\tfrac{12}{15}=\tfrac{4}{5}\) of the first. The same relationship written as a ratio is \(15:12=5:4\), and as a multiplier the first is \(1.25\) times the second. Ratio, fraction and multiplier are three views of one multiplicative relationship, and moving fluently between them is the backbone of the whole section.
A recipe uses \(150\) g of flour and \(90\) g of butter. Express the butter as a fraction of the flour, and write the flour-to-butter ratio in simplest form. Butter as a fraction of flour is \(\tfrac{90}{150}\); dividing top and bottom by \(30\) gives \(\tfrac{3}{5}\). The ratio flour\(:\)butter is \(150:90\), and dividing both by their HCF \(30\) gives \(5:3\). Notice the fraction \(\tfrac{3}{5}\) and the ratio \(5:3\) carry the same information from opposite ends — the butter is \(\tfrac{3}{5}\) of the flour, so for every \(5\) parts of flour there are \(3\) of butter.
MechanismDividing a quantity in a ratio — sharing, mixing and concentration
To divide a quantity in a given ratio, add the parts to find the total number of shares, work out the value of one share, then multiply out. Sharing £4,200 in the ratio \(2:3:2\) uses \(2+3+2=7\) shares, so one share is \(4200\div 7=600\), and the three amounts are \(1200,\ 1800\) and \(1200\) pounds. Always check they add back to the original: \(1200+1800+1200=4200\) ✓.
Higher pushes ratio into mixing and concentration problems, where the same idea models real mixtures. Concrete mixed \(1:2:4\) (cement to sand to gravel) by mass, mortar mixed \(1:5\), squash diluted \(1:4\) with water — each is a part-to-part ratio you scale up or down. The subtlety examiners exploit is the difference between part-to-part and part-to-whole: squash diluted '\(1\) part squash to \(4\) parts water' is \(1\) part in \(5\) of the final drink, so the squash is \(\tfrac{1}{5}\) of the mixture, not \(\tfrac{1}{4}\). Reading which ratio you have been given is where the marks are lost or won.
A gardener mixes fertiliser and water in the ratio \(3:250\) by volume and needs \(5\) litres of solution. How much fertiliser is that, in millilitres? The total is \(3+250=253\) parts for \(5000\) mL, so one part is \(5000\div 253\approx 19.76\) mL, and the fertiliser is \(3\) parts, \(3\times 19.76\approx 59.3\) mL. A quicker route uses the fraction directly: fertiliser is \(\tfrac{3}{253}\) of the mixture, so \(\tfrac{3}{253}\times 5000\approx 59.3\) mL. Both routes agree, and the fraction method is the one that scales to any total.
MechanismProportion as equal ratios — and the link to fractions and lines
Proportion is the statement that two ratios are equal, \(a:b=c:d\), and it is the engine behind every 'scale the recipe' or 'convert the currency' problem. If \(a:b=c:d\) then \(\tfrac{a}{b}=\tfrac{c}{d}\), so you can cross-multiply to find a missing value: from \(3:4=x:20\) you get \(\tfrac{3}{4}=\tfrac{x}{20}\), hence \(x=\tfrac{3}{4}\times 20=15\). This equality of ratios is what proportion means, and it is the same machinery whether you are scaling a drawing, a recipe or a map.
A ratio also connects straight to a fraction and to a linear function. If two quantities are in a fixed ratio — say height to base is \(y:x=3:2\) — then \(\tfrac{y}{x}=\tfrac{3}{2}\), a constant, which rearranges to the straight line \(y=\tfrac{3}{2}x\) through the origin. The ratio has become a gradient. That is why every direct-proportion relationship plots as a straight line through the origin whose gradient is the ratio in fraction form: ratios, fractions and linear graphs are three languages for the same idea, and Higher expects you to translate between them.
A map has a scale of \(1:25\,000\). Two churches are \(8\) cm apart on the map. How far apart are they in real life, in kilometres? The scale is a proportion, map to real \(=1:25\,000\), so the real distance is \(8\times 25\,000=200\,000\) cm. Convert: \(200\,000\) cm \(=2000\) m \(=2\) km. The single ratio \(1:25\,000\) did all the work, because a scale is nothing more than a fixed proportion between map length and true length.
ModelPercentages, multipliers and the reverse-percentage trap
A percentage is a fraction out of \(100\), and the fastest, most reliable way to work with one is the multiplier. To increase by \(20\%\) multiply by \(1.20\); to decrease by \(15\%\) multiply by \(0.85\); to find \(37\%\) of an amount multiply by \(0.37\). Percentage change is \(\dfrac{\text{change}}{\text{original}}\times 100\), and expressing one quantity as a percentage of another is the same fraction times \(100\). Percentages can exceed \(100\%\): a price that triples has risen by \(200\%\), a multiplier of \(3\). Simple interest adds the same percentage of the original each year, so \(4\%\) simple interest on \(500\) for \(3\) years adds \(3\times(0.04\times 500)=60\).
The examiner's favourite trap is the reverse percentage: you are given the amount after a change and must find the original. The rule is to divide by the multiplier, never to subtract the percentage. A jacket costs \(48\) in a '\(20\%\) off' sale; the sale price is \(80\%\) of the original, so the original is \(48\div 0.8=60\), not \(48+20\%=57.60\). The same logic explains the Toblerone asymmetry: a \(10\%\) cut is a multiplier of \(0.9\), and undoing it needs \(\div 0.9=\times 1.111\), an \(11.1\%\) rise. Whenever the phrase 'before the increase or decrease' appears, reach for division by the multiplier.
A laptop is reduced by \(15\%\) and now sells for £544. What was the price before the reduction? The sale price is \(85\%\) of the original, a multiplier of \(0.85\), so the original is \(544\div 0.85=640\), i.e. £640. Check forwards: \(640\times 0.85=544\) ✓. The classic wrong answer adds \(15\%\) to \(544\) to reach £625.60 — but that takes \(15\%\) of the wrong number (the sale price, not the original), which is exactly the mistake reverse percentages are set to catch.
DataCompound units: speed, density, pressure and best-buy
A compound unit is built from two others by division, and each encodes a rate. Speed is distance per time (m/s, km/h); density is mass per volume (g/cm\(^3\)); pressure is force per area (N/m\(^2\)); a rate of pay is money per hour; a unit price is money per gram or per litre. Each rearranges like any formula: from speed \(=\dfrac{\text{distance}}{\text{time}}\) you get distance \(=\) speed \(\times\) time and time \(=\dfrac{\text{distance}}{\text{speed}}\).
Converting compound units trips people because you must convert both parts. To turn \(20\) m/s into km/h, multiply by \(3600\) (seconds in an hour) and divide by \(1000\) (metres in a km): \(20\times\dfrac{3600}{1000}=72\) km/h — the reason \(20\) m/s and \(72\) km/h are the same speed. The other Higher demand is comparison: to decide which pack is better value, reduce both to the same unit rate — price per \(100\) g, or grams per penny — and compare like with like. The bigger pack is not automatically cheaper per gram, which is exactly what supermarket shelf-edge unit prices exist to reveal.
A \(750\) g jar of coffee costs £5.40 and a \(300\) g jar costs £2.34. Which is better value? Compare price per gram in pence. Large: \(540\div 750=0.72\) pence per gram. Small: \(234\div 300=0.78\) pence per gram. The large jar is cheaper per gram (\(0.72<0.78\)), so it is better value — by \(0.06\) pence per gram, about \(8\%\). Working in pence per gram keeps the numbers clean; comparing grams per pound instead would reach the same verdict from the other direction.
MechanismDirect and inverse proportion — words, algebra and graphs
Two quantities are in direct proportion when one is a constant multiple of the other: \(y\propto x\) means \(y=kx\) for a fixed constant of proportionality \(k\). Its graph is a straight line through the origin with gradient \(k\), and doubling \(x\) doubles \(y\). To solve a direct-proportion problem, find \(k\) from a given pair, then use \(y=kx\) for any other value.
Two quantities are in inverse proportion when one is a constant multiple of the reciprocal of the other: \(y\propto\dfrac{1}{x}\) means \(y=\dfrac{k}{x}\), equivalently \(xy=k\). Here the product stays constant, so doubling \(x\) halves \(y\) — more workers, less time; more speed, less journey time. Its graph is the reciprocal curve, falling and levelling towards the axes, never a straight line. The commonest error is to treat 'as one goes up the other goes down' as inverse proportion when it is really linear: inverse proportion specifically keeps \(xy\) constant, not the difference. Higher also uses squared versions, \(y\propto x^2\) or \(y\propto\dfrac{1}{x^2}\), solved the same way once you find \(k\).
The time \(t\) to fill a tank is inversely proportional to the number of pumps \(p\). With \(4\) pumps it takes \(9\) hours. How long with \(6\) pumps? Inverse proportion means \(t=\dfrac{k}{p}\), so \(k=t\times p=9\times 4=36\). With \(6\) pumps, \(t=\dfrac{36}{6}=6\) hours. The sense-check is that the product stays constant: \(4\times 9=36=6\times 6\). Had the problem instead said \(t\propto\dfrac{1}{p^2}\), then \(k=9\times 4^2=144\) and \(6\) pumps would give \(t=\dfrac{144}{36}=4\) hours — the same method, one squared step more.
ModelScale factors for length, area and volume; maps and similarity
When one shape is an enlargement of another by scale factor \(k\), the three measures scale by different powers of \(k\): lengths by \(k\), areas by \(k^2\), and volumes by \(k^3\). This single fact answers a whole family of Higher questions and catches a whole family of students, who scale area or volume by \(k\) instead of \(k^2\) or \(k^3\). If a model car is built to \(1:43\) scale, its bonnet area is \(43^2\) times smaller and its fuel-tank volume \(43^3\) times smaller than the real car's.
The length ratios themselves come from similarity: two shapes are similar when one is an enlargement of the other, so corresponding sides are in a fixed ratio and corresponding angles are equal. That fixed side ratio is exactly what the trigonometric ratios \(\sin\), \(\cos\) and \(\tan\) capture — they are the side ratios shared by all right-angled triangles of a given angle, which is why a single value of \(\tan 30^\circ\) works for every such triangle regardless of size. Scale drawings and maps are the everyday face of this: a scale of \(1:50\,000\) means every length on the map is \(50\,000\) times longer in reality, and areas on the ground are \(50\,000^2\) times the map area.
Two similar bottles have heights \(20\) cm and \(30\) cm. The smaller holds \(500\) mL. What does the larger hold? The length scale factor from small to large is \(k=\dfrac{30}{20}=1.5\). Volume scales by \(k^3\), so the volume factor is \(1.5^3=3.375\). The larger bottle holds \(500\times 3.375=1687.5\) mL. The trap is to scale volume by \(1.5\) and answer \(750\) mL; capacity is a volume, so it must scale by the cube of the length factor, not the length factor itself.
DataRate of change: gradients, chords and tangents
The gradient of a graph is a rate of change, and reading it is a Higher staple. On a straight-line distance–time graph the gradient is speed; on a straight-line graph of cost against quantity it is the price per item. When the graph is a straight line you read the gradient exactly with \(\dfrac{\text{rise}}{\text{run}}\).
When the graph is a curve, the rate is changing from instant to instant, and Higher distinguishes two readings. The average rate of change between two points is the gradient of the chord joining them — total change in \(y\) over total change in \(x\), such as average speed over a whole journey. The instantaneous rate of change at a single point is the gradient of the tangent there — the straight line just grazing the curve — which gives, for example, the exact speed at one moment. Because a tangent is drawn by eye, its gradient is an estimate, and saying so is part of the mark scheme. A curve steepening over time means the rate is increasing (accelerating); a curve flattening means the rate is falling.
On a distance–time graph a sprinter's distance is \(10\) m at \(t=2\) s and \(90\) m at \(t=6\) s. Estimate her average speed over that interval, and describe how to find her speed exactly at \(t=6\) s. Average speed is the chord gradient: \(\dfrac{90-10}{6-2}=\dfrac{80}{4}=20\) m/s. That is her mean speed across the four seconds, not her speed at any single instant. For the instantaneous speed at \(t=6\) s you draw a tangent to the curve at that point and measure its gradient with \(\dfrac{\text{rise}}{\text{run}}\); if the curve is steeper there than the chord, the tangent gradient — and so her speed at \(t=6\) — exceeds \(20\) m/s. Always label such a reading an estimate.
CaseGrowth and decay: compound interest, depreciation and iteration
Growth and decay apply the same percentage change repeatedly, so the quantity is multiplied by a fixed multiplier again and again — a geometric process. Compound interest is the headline case: money earning \(3\%\) a year is multiplied by \(1.03\) each year, so after \(n\) years an amount \(P\) becomes \(P\times 1.03^{\,n}\). This beats simple interest because each year's interest itself earns interest the next year. Depreciation is decay: a car losing \(18\%\) of its value a year is multiplied by \(0.82\) each year, so after \(n\) years it is worth \(P\times 0.82^{\,n}\).
The power of the multiplier method is that any repeated-change problem — population growth, radioactive decay, a savings account, a shrinking rainforest — is the same calculation with a different multiplier and a different number of steps. This is also the model behind iterative processes, where you apply a rule over and over and watch the output settle or run away, exactly as the doubling in the sequences section did. Compound growth turns a modest yearly rate into a dramatic long-run change, which is why examiners lean on it to test whether you truly understand percentages rather than just 'find \(3\%\) once'.
£2,000 is invested at \(3\%\) compound interest per year. What is it worth after \(4\) years, and how much of that is interest? Each year multiplies by \(1.03\), so the value is \(2000\times 1.03^{4}\). Now \(1.03^{4}=1.12550881\), so the amount is \(2000\times 1.12550881=2251.02\) to the nearest penny, i.e. £2,251.02. The interest earned is \(2251.02-2000=251.02\), i.e. £251.02. Simple interest would have given only \(4\times(0.03\times 2000)=240\), i.e. £240, so compounding added about £11.02 — small over four years, but the gap widens sharply over decades, which is the whole point of the model.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This section rewards the multiplier method above all — reach for \(\times 1.2\), \(\times 0.85\) or \(\div 0.9\) rather than finding a percentage and adding, because the multiplier survives multi-step problems and is exactly what reverse-percentage and compound-interest questions demand. Whenever a question gives you the amount after a change and asks for the original, divide by the multiplier; whenever it repeats a change over several years, raise the multiplier to a power.
Show the structure the mark scheme is looking for: for sharing, write the total number of parts and the value of one part; for proportion, write \(y=kx\) or \(y=\tfrac{k}{x}\), find \(k\) explicitly, then substitute; for compound units, carry the units through every line and convert both parts of a compound unit. Scale-factor questions turn on the power: lengths \(k\), areas \(k^2\), volumes \(k^3\) — state which you are using. On rate-of-change questions from a curve, distinguish the average rate (a chord) from the instantaneous rate (a tangent), draw the line on the graph, label the answer an estimate, and attach the unit. A missing unit or an unstated 'estimate' is a routinely dropped mark across the whole section.