Compound rule of three

Example

If $5$ trucks transport $120$ tons of goods in $2$ days: How many goods will $7$ trucks transport in $3$ days?

The compound rule of three can be expressed as follows:

$\begin{array}{rclrc}5\text{trucks}& \to & 120\text{tm}& \to & 2\text{days}\\ \\ 7\text{trucks}& \to & x\text{tm}& \to & 3\text{days}\end{array}$

In fractions this would be:

${\displaystyle \frac{\frac{5}{120}}{2}}={\displaystyle \frac{\frac{7}{x}}{3}}\Rightarrow {\displaystyle \frac{5\cdot 2}{120}}={\displaystyle \frac{7\cdot 3}{x}}\Rightarrow {\displaystyle \frac{10}{120}}={\displaystyle \frac{21}{x}}$

To make the operations easier, generally the fraction that contains the unknown quantity will remain on one side of the equality, while the other two will remain multiplied on the opposite side:

${\displaystyle \frac{120}{x}}={\displaystyle \frac{5}{7}}\cdot {\displaystyle \frac{2}{3}}$

If we operate the side without the unknown, we obtain a simple rule of three:

${\displaystyle \frac{120}{x}}={\displaystyle \frac{10}{21}}\Rightarrow 10x=120\cdot 21\Rightarrow 10x=2520\Rightarrow x={\displaystyle \frac{2520}{10}}=252$ tm

So, $7$ trucks will transport $252$ tons in $3$ days.

Example

It takes $15$ days for a team of $10$ workers working $8$ hours a day to finish an order. How many persons with part time jobs will be needed to make the same work in $10$ days?

The first thing to do is to make the scheme of the rule of three analyzing the relations between the magnitudes:

$\begin{array}{rclrc}& i& & i& \\ \\ 10\text{workers}& \to & 8\text{hours}& \to & 15\text{days}\\ \\ x\text{workers}& \to & 4\text{hours}& \to & 10\text{days}\\ \\ & i& & i& \end{array}$

So, the relation of the unknown with the rest of magnitudes is inverse (i): the more persons, the less hours will have to be worked to finish the order; and less days will require more persons to finish the work.

If we convert it into fractions:

${\displaystyle \frac{10}{x}}={\displaystyle \frac{8}{4}}\cdot {\displaystyle \frac{15}{10}}$

We have to invert the fractions on the right side of the equality:

${\displaystyle \frac{10}{x}}={\displaystyle \frac{4}{8}}\cdot {\displaystyle \frac{10}{15}}$

And now it can be solved:

${\displaystyle \frac{10}{x}}={\displaystyle \frac{4}{8}}\cdot {\displaystyle \frac{10}{15}}\Rightarrow {\displaystyle \frac{10}{x}}={\displaystyle \frac{40}{120}}\Rightarrow 40x=120\cdot 10\Rightarrow 40x=1200\Rightarrow$ $x={\displaystyle \frac{1200}{40}}=30$ people.

So, to realize the work in half-day shifts in $10$ days, a team of $30$ persons will be needed.

Example

In a central post office, $2$ machines categorize $1.600$ packages in $8$ hours. How many machines are needed to categorize $2.400$ packages in $6$ hours?

By means of a scheme, the relations between the magnitudes and the unknown are analyzed:

$\begin{array}{rclrc}& d& & i& \\ \\ 2\text{machines}& \to & 1600\text{packages}& \to & 8\text{hours}\\ \\ x\text{machines}& \to & 2400\text{packages}& \to & 6\text{hours}\\ \\ & d& & i& \end{array}$

The relation between the machines and the processed packages is direct since the more machines functioning the more packages will be classified. On the other hand, the relation between the machines and the hours of working is inverse since the more machines the less hours to finish the same volume of work.

Now, we translate the relation into fractions and we solve it, bearing in mind that it will be necessary to invert the fraction regarding the working time:

${\displaystyle \frac{2}{x}}={\displaystyle \frac{1600}{2400}}\cdot {\displaystyle \frac{8}{6}}\Rightarrow {\displaystyle \frac{2}{x}}={\displaystyle \frac{1600}{2400}}\cdot {\displaystyle \frac{6}{8}}\Rightarrow {\displaystyle \frac{2}{x}}={\displaystyle \frac{9600}{19200}}\Rightarrow$

$\Rightarrow 9600x=2\cdot 19200\Rightarrow 9600x=38400\Rightarrow x={\displaystyle \frac{38400}{9600}}=4$ machines.

So , $4$ machines will be needed to finish the work assigned in the time planned.