Ratio, proportion and rates of change

A ratio compares quantities multiplicatively. A proportion states how two quantities vary together. A rate of change compares a change in one quantity with the corresponding change in another.

These ideas are closely connected. If a car travels at constant speed, distance is directly proportional to time, and the constant of proportionality is the rate of change:

$d=vt,\qquad \frac{d}{t}=v.$

Ratio and proportion appear throughout A-level Mathematics, including similar shapes, trigonometry, probability, exponentials, differentiation, statistics and mechanics.

Prerequisites

You should be able to:

• calculate with fractions and decimals;
• solve linear equations and substitute into formulae;
• convert common units;
• find the gradient of a straight line;
• calculate a percentage of an amount.

Review exact arithmetic, linear equations or straight line graphs where necessary.

Understanding ratio

The ratio $a:b$ means

$a:b=\frac{a}{b},$

provided $b\ne0$. It compares amounts measured in the same units. If there are $12$ red counters and $18$ blue counters, then

$\text{red}:\text{blue}=12:18=2:3.$

This says that there are $2$ red counters for every $3$ blue counters. It does not say that there are $2$ red counters.

Order matters:

$\text{blue}:\text{red}=3:2.$

Simplifying ratios

Divide every part by a common factor. A ratio may have more than two parts:

$24:36:60=2:3:5.$

Fractions or decimals should first be converted to convenient integers.

Worked example 1: unlike units

Simplify $45\text{ minutes}:2\text{ hours}$.

Convert both quantities to minutes:

$45:120.$

The highest common factor is $15$, so

$45:120=3:8.$

The incorrect answer $45:2$ compares different units and has no useful interpretation.

Worked example 2: fractional parts

Simplify

$\frac34:\frac56.$

Multiply both parts by the lowest common multiple of the denominators, $12$:

$\frac34:\frac56=9:10.$

Equivalently,

$\frac{\frac34}{\frac56}=\frac9{10}.$

Part to part and part to whole

If boys:girls is $3:5$, there are $3+5=8$ equal parts altogether. Therefore

$$\frac{\text{boys}}{\text{total}}=\frac38, \qquad \frac{\text{girls}}{\text{total}}=\frac58.$$

Misconception: confusing a ratio with a fraction of the total

If red:blue is $2:7$, the red fraction is $\frac{2}{9}$, not $\frac27$. The ratio compares red with blue; the fraction compares red with all $2+7$ parts.

Self-check 1

1. Simplify $1.5\text{ kg}:600\text{ g}$.
2. If apples:pears is $4:7$, what fraction of the fruit are pears?
3. Simplify $0.4:1.6:2$.

Answers

1. $1500:600=5:2$.
2. There are $11$ parts, so the fraction is $\frac7{11}$.
3. Multiply by $10$ and divide by $4$: $4:16:20=1:4:5$.

Dividing an amount in a ratio

To divide a total $T$ in the ratio $a:b$:

1. find the total number of parts, $a+b$;
2. find one part, $\frac{T}{a+b}$;
3. multiply by $a$ and $b$.

The shares are therefore

$\frac{a}{a+b}T\quad\text{and}\quad\frac{b}{a+b}T.$

Worked example 3: three shares

Divide $£840$ in the ratio $2:3:7$.

There are

$2+3+7=12$

equal parts. One part is

$£840\div12=£70.$

Hence the shares are

$2(£70)=£140,\qquad 3(£70)=£210,\qquad 7(£70)=£490.$

Check that $140+210+490=840$.

Worked example 4: reconstructing a total

Ali and Bea share some money in the ratio $5:3$. Ali receives $£96$ more than Bea. Find the total.

The difference of $5-3=2$ parts corresponds to $£96$. Thus

$1\text{ part}=£96\div2=£48.$

There are $8$ parts altogether, so

$\text{total}=8(£48)=£384.$

This difference method works because the same scale factor multiplies every part.

Direct proportion

Two quantities $y$ and $x$ are directly proportional if their ratio is constant:

$\frac{y}{x}=k,$

so

$y=kx.$

The symbol $\propto$ records the relationship before the constant is known:

$y\propto x\quad\Longrightarrow\quad y=kx.$

The constant $k$ is the constant of proportionality. Doubling $x$ doubles $y$; multiplying $x$ by any factor multiplies $y$ by the same factor.

Worked example 5: finding the constant

Given that $y\propto x$ and $y=18$ when $x=12$, find $y$ when $x=35$.

Write the equation first:

$y=kx.$

Use the known pair:

$18=12k\quad\Longrightarrow\quad k=\frac32.$

Therefore

$y=\frac32x.$

When $x=35$,

$y=\frac32(35)=\frac{105}{2}=52.5.$

Direct proportion to a power

The phrase “directly proportional to the square of $x$” means

$y\propto x^2\quad\Longrightarrow\quad y=kx^2,$

not $y=kx$. More generally,

$y\propto x^n\quad\Longrightarrow\quad y=kx^n.$

Worked example 6: a square law

The energy $E$ of an object is proportional to the square of its speed $v$. When $v=4$, $E=120$. Find $E$ when $v=10$.

$E=kv^2.$

Substitute the known values:

$120=k(4^2)=16k,$

so $k=7.5$. Hence

$E=7.5v^2.$

At $v=10$,

$E=7.5(10^2)=750.$

Increasing the speed by a factor of $\frac{10}{4}=2.5$ increases the energy by a factor of $2.5^2=6.25$. Indeed, $120(6.25)=750$.

Graphs of direct proportion

The graph of $y=kx$ is a straight line through the origin with gradient $k$. Both conditions matter. A straight line such as

$y=3x+2$

does not show direct proportion because it does not pass through $(0,0)$.

For $y=kx^2$, the graph is not a straight line, but plotting $y$ against $x^2$ gives a straight line through the origin with gradient $k$. This idea is developed in linearising data.

Misconception: every increasing relationship is proportional

A quantity can increase with $x$ without being directly proportional to $x$. Test whether $y/x$ is constant, or whether the graph is a straight line through the origin.

Inverse proportion

Two quantities are inversely proportional if their product is constant:

$xy=k,$

so

$y=\frac{k}{x},\qquad x\ne0.$

Thus

$y\propto\frac1x.$

Doubling $x$ halves $y$; multiplying $x$ by a factor $c$ divides $y$ by $c$.

Worked example 7: inverse proportion

The time $t$ required to complete a fixed job is inversely proportional to the number $n$ of identical machines. Six machines take $15$ hours. How long would ten machines take?

$t=\frac{k}{n}.$

Using $t=15$ and $n=6$:

$15=\frac{k}{6}\quad\Longrightarrow\quad k=90.$

Therefore

$t=\frac{90}{n}.$

For ten machines,

$t=\frac{90}{10}=9\text{ hours}.$

The model assumes identical machines working at constant rates with no interference or setup delay. Mathematical models should always be checked against their assumptions.

Inverse proportion to a power

If $y$ is inversely proportional to $x^2$, then

$$y\propto\frac1{x^2} \quad\Longrightarrow\quad y=\frac{k}{x^2}.$$

Worked example 8: inverse square scaling

The intensity $I$ of radiation from a point source is inversely proportional to the square of distance $r$. At $r=3$ m, $I=80$ units. Find the intensity at $r=12$ m.

Since the distance is multiplied by $4$, intensity is divided by $4^2=16$:

$I=\frac{80}{16}=5\text{ units}.$

Using the constant gives the same result:

$I=\frac{k}{r^2},\qquad 80=\frac{k}{9},\qquad k=720,$

then

$I=\frac{720}{12^2}=5.$

Self-check 2

1. $y\propto x$ and $y=14$ when $x=8$. Find $y$ when $x=20$.
2. $p\propto q^3$ and $p=54$ when $q=3$. Find a formula for $p$.
3. $a\propto \frac1b$ and $a=12$ when $b=5$. Find $b$ when $a=8$.

Answers

1. $k=\frac{14}{8}=\frac74$, so $y=\frac74(20)=35$.
2. $54=k(27)$, so $k=2$ and $p=2q^3$.
3. $ab=k=60$. Therefore $8b=60$, so $b=7.5$.

Recipes and scale factors

A unitary method finds the amount for one unit, then scales to the required number. It is direct proportion in numerical form.

Worked example 9: scaling a recipe

A recipe for $6$ portions uses $450$ g of flour. Find the flour required for $14$ portions.

Flour per portion is

$450\div6=75\text{ g}.$

For $14$ portions:

$14(75)=1050\text{ g}=1.05\text{ kg}.$

Equivalently, the scale factor is $\frac{14}{6}=\frac73$, so

$450\times\frac73=1050.$

Keep units attached to values. See units and compound measures for systematic unit conversion.

Percentage change and multipliers

A percentage is a ratio with denominator $100$. If an original value $V$ changes by $r\%$, use a multiplier:

$\text{increase:}\quad V_{\text{new}}=V\left(1+\frac{r}{100}\right),$

$\text{decrease:}\quad V_{\text{new}}=V\left(1-\frac{r}{100}\right).$

Percentage change is measured relative to the original value:

$$\text{percentage change} =\frac{\text{new}-\text{original}}{\text{original}}\times100\%.$$

Worked example 10: repeated change

An investment of $£2400$ grows by $3.5\%$ each year. Find its value after $4$ years.

The annual multiplier is $1.035$. Repeated multiplication gives

$V=2400(1.035)^4=2754.056\ldots$

so the value is $£2754.06$ to the nearest penny.

It is incorrect to add $4\times3.5\%=14\%$, because each year’s increase is calculated from a new value. This is exponential growth, explored in growth and decay.

Worked example 11: reverse percentage

After a $20\%$ reduction, a coat costs $£72$. Find its original price.

The sale price is $80\%=0.8$ of the original price $P$:

$0.8P=72.$

Therefore

$P=\frac{72}{0.8}=90.$

The original price was $£90$. Adding $20\%$ of $72$ would not reverse the reduction because it uses the wrong reference amount.

Rates of change

An average rate of change is

$\frac{\text{change in output}}{\text{change in input}}.$

For a function $y=f(x)$ between $x=a$ and $x=b$,

$$\text{average rate of change} =\frac{f(b)-f(a)}{b-a}.$$

Geometrically, this is the gradient of the chord joining $(a,f(a))$ and $(b,f(b))$.

Worked example 12: average speed

A runner’s distance from the start changes from $120$ m at $t=20$ s to $390$ m at $t=50$ s. Find the average velocity over this interval.

$$\begin{aligned} \text{average velocity} &=\frac{390-120}{50-20}\\ &=\frac{270}{30}\\ &=9\text{ m s}^{-1}. \end{aligned}$$

This does not imply that the runner travelled at exactly $9\text{ m s}^{-1}$ throughout.

Worked example 13: average rate for a curve

For $f(x)=x^2+1$, find the average rate of change from $x=2$ to $x=5$.

$f(2)=5,\qquad f(5)=26.$

Therefore

$\frac{f(5)-f(2)}{5-2}=\frac{26-5}{3}=7.$

The rate varies along the curve. The value $7$ is the gradient across the whole interval, not the gradient at every point.

Instantaneous rate of change

An instantaneous rate of change is the rate at one particular input value. It is the gradient of the tangent to a curve at that point.

For a small non-zero change $h$, the average rate from $x$ to $x+h$ is

$\frac{f(x+h)-f(x)}{h}.$

As $h$ approaches zero, the chord approaches the tangent. The limiting value is the derivative:

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$

For $f(x)=x^2$,

$$\begin{aligned} \frac{f(x+h)-f(x)}{h} &=\frac{(x+h)^2-x^2}{h}\\ &=\frac{2xh+h^2}{h}\\ &=2x+h. \end{aligned}$$

As $h\to0$, this approaches $2x$, so $f'(x)=2x$. At $x=3$, the instantaneous rate is $6$.

This preview becomes rigorous and systematic in differentiation basics.

Units of a rate

Rate units are output units divided by input units:

$$\frac{\text{metres}}{\text{seconds}}=\text{m s}^{-1}, \qquad \frac{\text{litres}}{\text{minute}}=\text{L min}^{-1}.$$

Checking units often reveals an inverted fraction. Speed is distance divided by time, not time divided by distance.

Mixed self-check

1. Divide $£495$ in the ratio $4:5:6$.
2. $y$ is directly proportional to $\sqrt{x}$. Given $y=15$ when $x=25$, find $y$ when $x=81$.
3. $T$ is inversely proportional to $r^2$. If $T=18$ when $r=2$, find $T$ when $r=6$.
4. A value rises from $160$ to $188$. Find the percentage increase.
5. Find the average rate of change of $g(x)=3x^2-2$ from $x=1$ to $x=4$.
6. Explain why the graph $y=5x-1$ does not represent direct proportion.

Answers

1. There are $15$ parts, each worth $£33$. The shares are $£132$, $£165$ and $£198$.
2. $y=k\sqrt{x}$. Since $15=5k$, $k=3$. Thus $y=3\sqrt{81}=27$.
3. $T=\frac{k}{r^2}$ and $18=\frac{k}{4}$, so $k=72$. Hence $T=\frac{72}{36}=2$.
4. The increase is $28$, so $\frac{28}{160}\times100\%=17.5\%$.
5. $g(1)=1$ and $g(4)=46$, so the average rate is $\frac{46-1}{4-1}=15$.
6. Its graph has intercept $-1$, so it does not pass through the origin. Also $y/x=5-1/x$ is not constant.

Key facts

• Simplify ratios only after expressing quantities in the same units.
• To share in a ratio, add the parts before finding the value of one part.
• $y\propto x^n$ means $y=kx^n$.
• $y\propto \frac1{x^n}$ means $y=\frac{k}{x^n}$.
• A graph of $y=kx$ is a straight line through the origin with gradient $k$.
• Percentage changes use the original value as the denominator.
• Average rate of change is $\frac{\Delta y}{\Delta x}$; instantaneous rate of change is a tangent gradient.

Next, strengthen these ideas through similarity and transformations, units and compound measures, and graphs of functions.