Ratio, Proportion & Percentages

Ratio and the bar model

A ratio compares quantities of the same kind. We write $3:5$ to mean "3 parts to 5 parts". Ratios behave like fractions, so we simplify by dividing every part by a common factor.

For example, $18:24$ both divide by $6$, giving $3:4$. To compare $1.5:2$, multiply through to clear decimals: $3:4$. To compare different units, convert first: $40\text{ cm}:1\text{ m} = 40:100 = 2:5$.

Key terms

Ratio — a comparison of parts, e.g. $2:3$.

Unitary ratio — written as $1:n$ (or $n:1$), useful for scales and recipes.

Parts — the building blocks; total parts $=$ sum of the ratio numbers.

To get a ratio into the form $1:n$, divide both sides by the first number. For $5:12$, divide by $5$ to get $1:2.4$.

Sharing in a given ratio

To split an amount in a ratio, find the value of one part by dividing the total by the number of parts.

Worked example

Share £240 in the ratio $3:5$.

Total parts $= 3 + 5 = 8$. One part $= 240 \div 8 = £30$.

Worked example

The ratio of red to blue counters is $4:7$. There are $35$ blue counters. How many red?

$7$ parts $= 35$, so $1$ part $= 5$. Red $= 4 \times 5 = 20$ counters.

Exam tip

When a question gives you the value of one part of the ratio (not the total), find the value of a single part first — then scale up the other quantities.

Direct and inverse proportion

Two quantities are in direct proportion if one is a constant multiple of the other: as one doubles, so does the other. We write $y \propto x$, meaning $y = kx$ for some constant $k$.

In inverse proportion, one increases as the other decreases so their product is constant: $y \propto \dfrac{1}{x}$, i.e. $y = \dfrac{k}{x}$.

The method is always the same: find $k$ from the given pair, then use it.

Worked example

$y$ is directly proportional to $x$. When $x = 6$, $y = 15$. Find $y$ when $x = 10$.

Worked example

$y$ is inversely proportional to $x$. When $x = 4$, $y = 9$. Find $x$ when $y = 12$.

Higher tier also uses powers, e.g. $y \propto x^2$ gives $y = kx^2$, and $y \propto \dfrac{1}{x^2}$ gives $y = \dfrac{k}{x^2}$.

Watch out

"$y$ increases as $x$ increases" is not enough to mean direct proportion — the graph must pass through the origin and be a straight line. Inverse proportion graphs are curves (hyperbolae), never straight lines.

Percentages of an amount

A percentage is a fraction out of $100$. Convert to a decimal and multiply:

$$17.5\% \text{ of } 240 = 0.175 \times 240 = 42.$$

To write one quantity as a percentage of another, divide and multiply by $100$: $\dfrac{18}{40} \times 100 = 45\%$.

Percentage change and multipliers

The multiplier method is the fast, exam-friendly way to handle increase and decrease.

Increase by $r\%$: multiply by $\left(1 + \dfrac{r}{100}\right)$.
Decrease by $r\%$: multiply by $\left(1 - \dfrac{r}{100}\right)$.

So a $15\%$ increase uses $\times 1.15$; a $20\%$ decrease uses $\times 0.8$.

For percentage change:

$$\text{percentage change} = \frac{\text{change}}{\text{original}} \times 100.$$

Worked example

A coat costs £85 and is reduced by $30\%$. Find the sale price.

Multiplier $= 1 - 0.30 = 0.70$. Sale price $= 85 \times 0.70 = £59.50$.

Worked example

A share rises from £250 to £290. Find the percentage increase.

Change $= 290 - 250 = 40$. Percentage $= \dfrac{40}{250} \times 100 = 16\%$.

Reverse percentages

Here you are given the value after a change and must find the original. Identify the multiplier used, then divide by it.

Worked example

After a $20\%$ increase, a salary is £33 600. What was the original?

The multiplier was $1.20$, so original $= 33600 \div 1.20 = £28\,000$.

Worked example

A sofa is reduced by $15\%$ in a sale to £680. Find the pre-sale price.

Multiplier $= 0.85$. Original $= 680 \div 0.85 = £800$.

Watch out

Reverse percentages are not found by adding the same percentage back. Taking $20\%$ off £33 600 gives £26 880 — wrong. You must divide by the multiplier, never add or subtract a flat percentage.

Simple vs compound interest

Simple interest is paid only on the original amount each year. Interest of $r\%$ on principal $P$ for $n$ years is $\dfrac{P r n}{100}$, so the total is $P + \dfrac{Prn}{100}$.

Compound interest is paid on the running total, so each year's interest earns interest too. Using the multiplier method:

$$A = P\left(1 + \frac{r}{100}\right)^{n},$$

where $A$ is the final amount, $P$ the principal, $r$ the rate per period and $n$ the number of periods.

Worked example

£2000 is invested at $4\%$ compound interest per year for $3$ years. Find the value and the interest earned.

$A = 2000 \times 1.04^{3} = 2000 \times 1.124864 = £2249.73$ (to the nearest penny).

Worked example

Compare £5000 at $3\%$ simple interest with $3\%$ compound, both over $4$ years.

Simple: $\dfrac{5000 \times 3 \times 4}{100} = £600$ interest, total £5600.

Depreciation

Depreciation is a percentage decrease repeated each year — common for cars and equipment. Use the same formula with a decrease multiplier:

$$A = P\left(1 - \frac{r}{100}\right)^{n}.$$

Worked example

A car bought for £18 000 depreciates by $12\%$ each year. Find its value after $3$ years.

$A = 18000 \times 0.88^{3} = 18000 \times 0.681472 = £12\,266.50$ (to the nearest penny).

Exam tip

Always raise the multiplier to the power $n$ in one go on your calculator — don't round between years, or your final answer will drift. Round only at the very end.

Compound measures

A compound measure combines two different units. The three you must know:

Rearrange as needed, e.g. $\text{distance} = \text{speed} \times \text{time}$ and $\text{mass} = \text{density} \times \text{volume}$.

Worked example

A block has mass $480\text{ g}$ and volume $60\text{ cm}^3$. Find its density.

$\text{density} = \dfrac{480}{60} = 8\text{ g/cm}^3$.

Worked example

A train travels $150\text{ km}$ in $1$ hour $15$ minutes. Find its average speed in km/h.

Time $= 1.25$ hours. Speed $= \dfrac{150}{1.25} = 120\text{ km/h}$.

Unit conversions for compound measures

Convert speeds carefully. To change km/h to m/s, multiply by $\dfrac{1000}{3600}$ (i.e. divide by $3.6$); to reverse, multiply by $3.6$.

So $72\text{ km/h} = 72 \div 3.6 = 20\text{ m/s}$.

Watch out

Times like "$2$ hours $30$ minutes" must become $2.5$ hours, not $2.30$. And for density, $1\text{ g/cm}^3 = 1000\text{ kg/m}^3$, so check your units match before substituting.

Real world

Banks quote savings as "AER", which is exactly the compound interest multiplier in action. Car insurers and accountants use depreciation formulae to value vehicles and equipment each year — the same $A = P(1 \pm r/100)^n$ you use in the exam.