Almost every question in this topic is really asking the same thing:
That "times bigger" number is called a multiplier (or a scale factor). Once you can find it, you can answer questions about ratios, fractions, percentages, speed, density, maps, recipes, interest and growth — because they are all the same piece of mathematics wearing different clothes.
• Additive: "8 is 5 more than 3." (subtraction)
• Multiplicative: "12 is 4 times 3." (division)
This whole topic is multiplicative. If you find yourself adding or subtracting to compare two quantities, stop and ask whether you should be dividing instead.
A shop mixes red and blue paint. For every 2 litres of red there are 5 litres of blue.
All four statements say exactly the same thing. Being able to switch between them is the single most useful skill in this topic.
Nearly every question in this topic can be cracked with one of these five methods.
Tool 1 — The unitary method ("find one")
Scale down to 1 of something, then scale up to the amount you want.
5 pens cost £3.50. How much do 8 pens cost?
Tool 2 — The multiplier (scale factor)
Find the number that turns the first quantity into the second, then apply it to everything.
A recipe for 4 people uses 300 g of rice. How much for 10 people?
Tool 3 — The bar model
Draw the ratio as equal-sized boxes. One box = one "part". Brilliant for share-in-a-ratio and "difference" questions.
Tool 4 — The double number line / ratio table
Line the two quantities up in a table and multiply or divide both rows by the same number.
| Litres of petrol | 1 | 5 | 12 |
|---|---|---|---|
| Cost (£) | 1.45 | 7.25 | 17.40 |
Tool 5 — The formula triangle / algebraic rule
For compound measures and proportion, write a rule such as $S = \dfrac{D}{T}$ or $y = kx$ and substitute.
The National Curriculum splits this topic into 16 statements, usually labelled R1 to R16. Each has its own page with explanations, worked examples and ten practice questions.
- R1Units and Compound UnitsChanging between standard units (length, area, volume, mass, time) and compound units (speed, density, rates of pay).
- R2Scale Factors, Scale Diagrams and MapsMap scales such as $1:25\,000$, scale drawings, bearings and real distances.
- R3One Quantity as a Fraction of AnotherWriting $a$ as a fraction of $b$, including improper fractions when $a > b$.
- R4Ratio Notation and SimplifyingWriting, simplifying and comparing ratios, including $1:n$ and $n:1$ form.
- R5Dividing a Quantity in a RatioPart:part and part:whole sharing, "difference" problems, recipes, mixtures and concentrations.
- R6Multiplicative RelationshipsExpressing the link between two quantities as a ratio, a fraction or a multiplier.
- R7Proportion as Equality of RatiosSolving $a:b = c:d$, cross-multiplication and best-buy comparisons.
- R8Ratios, Fractions and Linear FunctionsTurning a ratio into a fraction and into an equation $y = mx$; changing ratios.
- R9Percentages and Percentage ChangeIncrease, decrease, reverse percentages, simple interest and comparing with percentages.
- R10Direct and Inverse Proportion ProblemsRecognising and solving both types, numerically and graphically.
- R11Compound MeasuresSpeed, density, pressure, rates of pay, unit pricing and population density.
- R12Similar Shapes: Length, Area and VolumeScale factors $k$, $k^2$ and $k^3$, and links to similarity and trigonometry.
- R13Equations of Direct and Inverse ProportionUsing $y = kx$, $y = kx^2$, $y = \dfrac{k}{x}$ and finding the constant $k$.
- R14Gradient as a Rate of ChangeReading rates from straight-line graphs; recognising proportion graphs.
- R15Instantaneous Rates of ChangeGradients of chords and tangents; average vs instantaneous rate; areas under graphs.
- R16Growth and DecayCompound interest, depreciation, exponential models and iterative processes.
In the ratio $2:3$ there are $5$ parts altogether. The first quantity is $\frac{2}{3}$ of the second but $\frac{2}{5}$ of the total. Read the question carefully to see which one is wanted.
$50\text{ cm} : 2\text{ m}$ is not $50:2$. Convert first: $50:200 = 1:4$.
A $10\%$ rise followed by a $10\%$ fall is not "no change". The multiplier is $1.10 \times 0.90 = 0.99$, a $1\%$ overall decrease.
If a price after a $20\%$ discount is £48, the original is $48 \div 0.8 = £60$, not $48 \times 1.2 = £57.60$.
If lengths are doubled, areas are $\times 4$ and volumes are $\times 8$.
Driving 60 km at 60 km/h and 60 km at 30 km/h does not average 45 km/h. Use total distance $\div$ total time: $120 \div 3 = 40$ km/h.
Share in a ratio
Add the parts → divide the total by the number of parts → multiply each share.
Percentage multiplier
Increase by $p\%$: $\times\left(1 + \frac{p}{100}\right)$. Decrease: $\times\left(1 - \frac{p}{100}\right)$.
Direct proportion
$y = kx$. Doubling $x$ doubles $y$. Graph: straight line through the origin.
Inverse proportion
$y = \dfrac{k}{x}$, so $xy = k$. Doubling $x$ halves $y$. Graph: a hyperbola.
Speed / Density / Pressure
$S = \dfrac{D}{T}$, $\rho = \dfrac{M}{V}$, $P = \dfrac{F}{A}$.
Similar shapes
Length $\times k$, Area $\times k^2$, Volume $\times k^3$.
Compound growth
$A = P(1 + r)^n$ for growth; $A = P(1 - r)^n$ for decay.
Rate from a graph
Gradient $=\dfrac{\text{change in } y}{\text{change in } x}$ — always give the units.
These ten questions sample the whole topic. If one type catches you out, follow the link in Section 3 to the page that covers it.
Simplify the ratio $45\text{ minutes} : 2\text{ hours}$.
▶ Show solution
Convert to the same unit first: $2$ hours $= 120$ minutes.
$45 : 120$
Divide both by the HCF, which is $15$: $\;45 \div 15 = 3$, $\;120 \div 15 = 8$.
Answer: $3 : 8$
£420 is shared between Amir and Beth in the ratio $4 : 3$. How much does each receive?
▶ Show solution
Total parts $= 4 + 3 = 7$.
One part $= 420 \div 7 = £60$.
Amir $= 4 \times 60 = £240$. Beth $= 3 \times 60 = £180$.
Check: $240 + 180 = 420$ ✓
A jacket costs £85. In a sale the price is reduced by 24%. Find the sale price.
▶ Show solution
Multiplier for a 24% decrease $= 1 - 0.24 = 0.76$.
$85 \times 0.76 = £64.60$
Sale price = £64.60
After a 15% increase, a train fare is £69. What was the fare before the increase?
▶ Show solution
This is a reverse percentage: divide by the multiplier.
Multiplier $= 1.15$.
Original $= 69 \div 1.15 = £60$
Check: $60 \times 1.15 = 69$ ✓
A car travels 189 km in 2 hours 15 minutes. Find its average speed in km/h.
▶ Show solution
Time in hours: $2\text{ h }15\text{ min} = 2.25$ h.
$$S = \frac{D}{T} = \frac{189}{2.25} = 84$$
Average speed = 84 km/h
$y$ is directly proportional to $x$. When $x = 6$, $y = 15$. Find $y$ when $x = 10$.
▶ Show solution
$y = kx$, so $15 = k \times 6 \Rightarrow k = 2.5$.
Formula: $y = 2.5x$.
When $x = 10$: $y = 2.5 \times 10 = 25$.
It takes 4 painters 9 days to paint a building. Assuming they all work at the same rate, how long would 6 painters take?
▶ Show solution
More painters → less time, so this is inverse proportion.
Total work $= 4 \times 9 = 36$ painter-days.
Time for 6 painters $= 36 \div 6 = 6$ days.
Two similar cylinders have heights 5 cm and 15 cm. The smaller has volume 40 cm³. Find the volume of the larger.
▶ Show solution
Length scale factor $k = 15 \div 5 = 3$.
Volume scale factor $= k^3 = 27$.
Volume $= 40 \times 27 = 1080\text{ cm}^3$.
£2500 is invested at 3% compound interest per year. Find the value after 4 years, to the nearest penny.
▶ Show solution
$$A = P(1 + r)^n = 2500 \times 1.03^{4}$$
$1.03^4 = 1.12550881$
$A = 2500 \times 1.12550881 = £2813.77$ (nearest penny).
On a map with scale $1 : 50\,000$, two towns are 7.4 cm apart. Find the real distance in kilometres.
▶ Show solution
Real distance $= 7.4 \times 50\,000 = 370\,000$ cm.
$370\,000 \text{ cm} \div 100 = 3700$ m $\div 1000 = 3.7$ km.
Answer: 3.7 km