Compound rule of three

A compound rule of three is formed by several simple rules of three and it is used when relating more than two magnitudes.

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:

$$\begin{eqnarray} 5\ \mbox{trucks} & \rightarrow & 120 \ \mbox{tm} & \rightarrow & 2 \ \mbox{days} \\\\ 7\ \mbox{trucks} & \rightarrow & x \ \mbox{tm} & \rightarrow & 3 \ \mbox{days} \end{eqnarray}$$

In fractions this would be:

$$\dfrac{\frac{5}{120}}{2}=\dfrac{\frac{7}{x}}{3} \Rightarrow \dfrac{5\cdot2}{120}=\dfrac{7\cdot3}{x} \Rightarrow \dfrac{10}{120}=\dfrac{21}{x}$$

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:


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

$$\dfrac{120}{x}=\dfrac{10}{21} \Rightarrow 10x=120\cdot21 \Rightarrow 10x=2520 \Rightarrow x=\dfrac{2520}{10}=252$$ tm

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

Note that the relation of the unknown with the other two magnitudes is direct in this case: the more trucks, the more tons transported; and the more days, more tons will be also transported. The relation can be expressed as follows:

$$\begin{eqnarray} & d & & d & \\\\ 5\ \mbox{trucks} & \rightarrow & 120 \ \mbox{tm} & \rightarrow & 2 \ \mbox{days} \\\\ 7\ \mbox{trucks} & \rightarrow & x \ \mbox{tm} & \rightarrow & 3 \ \mbox{days} \\\\ & d & & d & \end{eqnarray}$$

where $$d$$ expresses a direct relation.

But it can be the case that the relation between the magnitudes is not direct, but inverse.In these cases, as with the simple rules of three, we will invert those fractions that have an inverse relation with the unknown.

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:

$$\begin{eqnarray} & i & & i & \\\\ 10\ \mbox{workers} & \rightarrow & 8 \ \mbox{hours} & \rightarrow & 15 \ \mbox{days} \\\\ x\ \mbox{workers} & \rightarrow & 4 \ \mbox{hours} & \rightarrow & 10 \ \mbox{days} \\\\ & i & & i & \end{eqnarray}$$

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:


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


And now it can be solved:

$$\dfrac{10}{x}=\dfrac{4}{8}\cdot\dfrac{10}{15} \Rightarrow \dfrac{10}{x}=\dfrac{40}{120} \Rightarrow 40x=120\cdot10 \Rightarrow 40x=1200 \Rightarrow$$ $$x=\dfrac{1200}{40}=30$$ people.

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

Finally, in the same problem there can be magnitudes directly proportional to the unknown and inversely proportional magnitudes to the unknown.

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:

$$\begin{eqnarray} & d & & i & \\\\ 2\ \mbox{machines} & \rightarrow & 1600 \ \mbox{packages} & \rightarrow & 8 \ \mbox{hours} \\\\ x\ \mbox{machines} & \rightarrow & 2400 \ \mbox{packages} & \rightarrow & 6 \ \mbox{hours} \\\\ & d & & i & \end{eqnarray}$$

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:

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

$$\Rightarrow 9600x=2\cdot19200 \Rightarrow 9600x=38400 \Rightarrow x=\dfrac{38400}{9600}=4$$ machines.

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

At the time of raising these problems the best thing for obtaining an integer result is to start from a known result and then to raise the relation between magnitudes across the statement. There are many daily relations that are proportional, directly or inversely: from €/kg of a food product up to the work that a certain number of persons can do, even the length of a document and the time that it takes in reading it.

Nevertheless, it is not important that the result is an integer. In the previous example, with slightly different information, it might have been given $$4,3$$ machines as a result, and it would nothave been serious. With this result, the post office workers should know that with $$4$$ machines they will get close to accomplishing not only the volume of work but also the term, while if they put on another machine the work will be done before time.