# Rule of three calculation

## Rule of three, Rule of proportion

A rule of three is used to calculate ratios proportional to each other. Proportional means that when a quantity A is doubled, B is also doubled. For example, e.g. A train with constant speed drives in an hour 80km so it drives in two hours 160km.

### Calculation scheme for Rule of three calculation

$\begin{array}{|cc|}\hline A& B\\ C& x\\ \hline\end{array}$

Example: A train runs in 2 hours 140 km then 2 has to be set for A and 140 for B. The route is asked for 3 hours. So 3 for C and x is the distance searched.

$\begin{array}{|ccccc|}\hline 2& \text{hours}& \stackrel{\wedge}{=}& 140& \text{km}\\ 3& \text{hours}& \stackrel{\wedge}{=}& x& \text{km}\\ \hline\end{array}$

### Calculation

$\begin{array}{|cc|}\hline 1& \frac{B}{A}\\ C& C\frac{B}{A}\\ \hline\end{array}$

$x=\frac{C\cdot B}{A}$

$\begin{array}{|ccccc|}\hline 1& \text{hours}& \stackrel{\wedge}{=}& \frac{140}{2}& \text{km}\\ 3& \text{hours}& \stackrel{\wedge}{=}& 3\frac{140}{2}& \text{km}\\ \hline\end{array}$

$\begin{array}{|ccccc|}\hline 1& \text{hours}& \stackrel{\wedge}{=}& 70& \text{km}\\ 3& \text{hours}& \stackrel{\wedge}{=}& 210& \text{km}\\ \hline\end{array}$

So is x= 210 that means the train runs 210km in 3 hours.

## Rule of three Calculator

## Inverse Rule of three

An inverse rule of three is a calculation scheme for the calculation of ratios which are indirectly proportional to each other. Indirectly proportional, when a quantity A is halved, the quantity B is doubled. If 2 trucks need to transport a certain amount of overburden 10 days so 1 truck needs double time so 20 days.

### Calculation scheme of the inverse rule of three

$\begin{array}{|cc|}\hline A& B\\ C& x\\ \hline\end{array}$

Example: If you need 4 trucks to remove a certain amount of overburden 10 days then 2 is set for A and 10 is for B. Looking for the time that 2 trucks is needed so C is set to 2.

$\begin{array}{|ccccc|}\hline 4& \text{truck}& \stackrel{\wedge}{=}& 10& \text{days}\\ 2& \text{truck}& \stackrel{\wedge}{=}& x& \text{days}\\ \hline\end{array}$

### Calculation

$\begin{array}{|cc|}\hline 1& A\cdot B\\ C& A\frac{B}{C}\\ \hline\end{array}$

$x=\frac{A\cdot B}{C}$

$\begin{array}{|ccccc|}\hline 1& \text{truck}& \stackrel{\wedge}{=}& 4\cdot 10& \text{days}\\ 2& \text{truck}& \stackrel{\wedge}{=}& 4\frac{10}{2}& \text{days}\\ \hline\end{array}$

$\begin{array}{|ccccc|}\hline 1& \text{truck}& \stackrel{\wedge}{=}& 40& \text{days}\\ 2& \text{truck}& \stackrel{\wedge}{=}& 20& \text{days}\\ \hline\end{array}$

So is x= 20 that means a truck needs 20 days.