#### Chapter 2 Equations in one Variable

Section 2.1 Simple Equations

# 2.1.3 Proportionality and Rule of Three

A relation between two varying quantities that frequently occurs in practice is the proportionality. Examples include mass and volume, time and travelled distance or weight (quantity) of a product and its price. Often we are given two example values that stand in relation and need to complete another example for which only one value is given. We will illustrate this procedure with an example.
##### Example 2.1.11
$5 \mathrm{kg}$ of apples cost $3$ Euro. How much do $11 \mathrm{kg}$ of apples cost?
In a first step, we convert the information we have about the price of apples in the following traditional notation, which in this example we could read as "costs":

$5 \mathrm{kg} \stackrel{\wedge}{=} 3 \text{Euro} .$

In this example, the symbol in the middle can be read as "costs", but in other examples other readings will be required. The key point is that as the value on one side of the symbol varies, the value on the other will have to vary proportionally. In the second step, we want to scale this proportional relation so as to express it in terms of a unit amount of the quantity for which another value (here: $11 \mathrm{kg}$) is given. So we multiply both sides by $1/5$ and get:

$1 \mathrm{kg} \stackrel{\wedge}{=} \frac{1}{5}·3 \text{Euro}=0.6 \text{Euro} .$

In the third and final step, we multiply both sides by the number of units specified in the problem, which in our example is $11$:

$11 \mathrm{kg} \stackrel{\wedge}{=} 11·0.6 \text{Euro}=6.6 \text{Euro} .$

The required price for $11 \mathrm{kg}$ of apples is therefore $6.60 \text{Euro}$.

We have derived the required relation by deriving a relation for one unit of a quantity from the initial relation. The procedure demonstrated here is called the rule of three and is taught in great detail in the schools of some countries such as Germany and France.
The same problem can also be solved by introducing a proportionality factor. Again, we consider the example above.
##### Example 2.1.12
The price $P$ is proportional to the mass $m$. Hence, there exists a constant $k$ with

$P=km .$

Since this relation also holds for the given values ${m}_{0}=5 \mathrm{kg}$ and ${P}_{0}=3 \text{Euro}$ it follows

hence in this case

$k=\frac{3}{5}=0.6 ,$

taken in the unit of Euro per kg. (As a scientist you would correctly write $k=0.6 \text{Euro}/\mathrm{kg}$, since proportionality factors generally carry a dimensional unit.) Using ${m}_{1}=11 \mathrm{kg}$, one obtains finally

${P}_{1}=k{m}_{1}=0.6·11=6.6 \text{(Euro)}$

which is the same result as for using the rule of three (see previous example).

##### Exercise 2.1.13
A car takes $9$ minutes to travel a distance of $6 \mathrm{km}$.
1. Which distance $s$ the car travels within $15$ minutes?

The solution is ${s}_{15}$ =
$\mathrm{km}$.
2. The proportionality factor between travelled distance $s$ and travelling time $t$ is the velocity $v$ of the car.

The velocity is $v$ =
$\mathrm{km}/ \mathrm{h}$.