Ratio, Proportion and Rate of Change

A ratio compares two or more quantities of the same type. It is written as $a : b$.

Simplifying ratios: divide all parts by their HCF. Ensure all parts are in the same unit before simplifying.

$18 : 24 = 3 : 4$ (divide by 6)

$1.5 \text{ m} : 60 \text{ cm} = 150 \text{ cm} : 60 \text{ cm} = 5 : 2$

Dividing a quantity in a ratio:

Share £40 in the ratio $3 : 5$.

Total parts $= 3 + 5 = 8$. One part $= £40 \div 8 = £5$.

Shares: $3 \times £5 = £15$ and $5 \times £5 = £25$.

Sharing in a three-part ratio:

Divide 120 in the ratio $1 : 3 : 2$.

Total $= 6$ parts; one part $= 20$. Shares: $20 : 60 : 40$.

Finding a missing quantity from a ratio:

If $a : b = 3 : 7$ and $a = 12$, find $b$:

$\frac{b}{a} = \frac{7}{3} \implies b = \frac{7 \times 12}{3} = 28$

Combining ratios: if $A : B = 2 : 5$ and $B : C = 3 : 4$, find $A : B : C$.

Scale $B$ to the LCM of 5 and 3 (=15): $A : B = 6 : 15$, $B : C = 15 : 20$. So $A : B : C = 6 : 15 : 20$.

Ratio and fractions:

In ratio $3 : 7$, the first quantity is $\frac{3}{10}$ of the total and the second is $\frac{7}{10}$ of the total.

Direct proportion: $y$ is directly proportional to $x$ means $y = kx$ for some constant $k$.

• Written: $y \propto x$
• Graph: straight line through the origin
• If $x$ doubles, $y$ doubles

Other direct proportions:

Relationship	Equation	Graph
$y \propto x^2$	$y = kx^2$	Parabola through origin
$y \propto x^3$	$y = kx^3$	Cubic through origin
$y \propto \sqrt{x}$	$y = k\sqrt{x}$	Square-root curve

Inverse proportion: $y$ is inversely proportional to $x$ means $y = \frac{k}{x}$.

• Written: $y \propto \frac{1}{x}$
• Graph: reciprocal curve (hyperbola); as $x$ increases, $y$ decreases
• If $x$ doubles, $y$ halves

Finding the constant $k$:

1. Write the proportionality equation ($y = kx$ or $y = \frac{k}{x}$, etc.)
2. Substitute a known pair of values to find $k$
3. Use the complete equation to answer the question

Example: $y \propto x^2$; when $x = 3$, $y = 36$. Find $y$ when $x = 5$.

$36 = k(3)^2 \implies k = 4$. So $y = 4x^2$. When $x = 5$: $y = 4(25) = 100$.

Graphical recognition:

• Direct proportion graphs pass through the origin; the gradient equals $k$
• Inverse proportion ($y \propto \frac{1}{x}$) gives a hyperbola — both axes are asymptotes

Percentage of an amount: multiply by the decimal equivalent ($23\% = 0.23$).

Percentage change:

$\% \text{ change} = \frac{\text{new value} - \text{original value}}{\text{original value}} \times 100\%$

Multiplier method (most efficient):

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

Example: increase £240 by 15%: $£240 \times 1.15 = £276$

Reverse percentage (finding the original value):

A price after a 20% increase is £144. Find the original price.

The multiplier was $1.20$, so: original $= £144 \div 1.20 = £120$.

Never subtract the percentage from the final value — always divide by the multiplier.

Compound interest:

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

where $P$ is the principal, $r$ is the annual rate, $n$ is the number of years, and $A$ is the final amount.

Example: £5000 invested for 3 years at 4% per annum:

$A = 5000 \times 1.04^3 = 5000 \times 1.124864 = £5624.32$

Compound decay (depreciation):

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

Example: car worth £12 000 depreciates at 18% per year. Value after 4 years:

$A = 12000 \times 0.82^4 \approx £5 464$

A compound measure is formed by combining two other measures. The three most common are:

$\text{Speed} = \frac{\text{Distance}}{\text{Time}} \qquad (S = \frac{D}{T})$

$\text{Density} = \frac{\text{Mass}}{\text{Volume}} \qquad (D = \frac{M}{V})$

$\text{Pressure} = \frac{\text{Force}}{\text{Area}} \qquad (P = \frac{F}{A})$

Use the formula triangle to rearrange: cover the quantity you want — the remaining two show whether to multiply or divide.

Units:

Measure	Common units
Speed	m/s, km/h, mph
Density	g/cm³, kg/m³
Pressure	N/m², Pa (pascal)

Unit conversion for compound units:

To convert speed from m/s to km/h:

$1 \text{ m/s} = \frac{1}{1000} \text{ km per } \frac{1}{3600} \text{ h} = \frac{3600}{1000} \text{ km/h} = 3.6 \text{ km/h}$

So multiply m/s by 3.6 to get km/h.

Distance–time and velocity–time graphs:

• Distance–time: gradient = speed
• Velocity–time: gradient = acceleration; area under graph = distance

Average speed:

$\text{Average speed} = \frac{\text{total distance}}{\text{total time}}$

This is not necessarily the mean of speeds for individual stages.

The gradient of a straight-line graph equals the rate of change of the quantity on the $y$-axis with respect to the quantity on the $x$-axis.

$\text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{\text{change in } y}{\text{change in } x}$

Units of gradient: the units of $y$ divided by the units of $x$. For a distance–time graph the gradient has units of m/s (speed); for a cost–weight graph the gradient has units of £ per kg.

Interpreting graphs in context:

• A steeper section means a faster rate of change
• A horizontal section ($\text{gradient} = 0$) means no change — e.g. stationary in a distance–time graph
• A negative gradient means the quantity is decreasing

Estimating gradient of a curve: draw a tangent at the point of interest, then calculate the gradient of the tangent using two points far apart on it.

Area under a graph: the area under a speed–time (velocity–time) graph equals the distance travelled. Estimate the area by counting squares or treating the region as trapezoids.

Interpreting real-life graphs:

AQA questions often show graphs in unfamiliar contexts (e.g. litres of water in a tank vs time, temperature vs time). The approach is always the same:

1. Identify what the $x$- and $y$-axes represent and their units
2. Calculate the gradient (its units are $y$-units per $x$-unit)
3. State what the gradient represents in context

Exponential growth and decay occur when a quantity increases or decreases by a fixed percentage each time period.

General formula:

$A = A_0 \times m^t$

where $A_0$ is the initial amount, $m$ is the multiplier per time period, $t$ is the number of time periods, and $A$ is the final amount.

• Growth: $m > 1$ (e.g. population increasing by 3% per year: $m = 1.03$)
• Decay: $0 < m < 1$ (e.g. radioactive material losing 10% per year: $m = 0.90$)

Compound interest (growth):

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

Depreciation (decay):

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

Finding the number of years: set up the equation and use trial and improvement (or logarithms at A-level):

A car is worth £10 000. It depreciates by 15% per year. After how many complete years is it worth less than £5 000?

$10000 \times 0.85^n < 5000 \implies 0.85^n < 0.5$

Try: $0.85^4 \approx 0.522$, $0.85^5 \approx 0.444$. After 5 complete years.

Recognising exponential graphs:

• Growth: $y = A_0 m^t$ with $m > 1$ produces a curve that increases rapidly
• Decay: $m < 1$ produces a curve that approaches zero but never reaches it (asymptote at $y = 0$)