The Trinomial
The trinomial is used in simple direct and inverse proportion calculations. Usually we know three interdependent figures and have to calculate a fourth. In the trinomial, we have to carefully distinguish between direct and indirect proportion, because it has different calculations.
Motivation
You go to the store to buy a supply of lemonade. You bought 5 bottles of lemonade for 100 crowns. How many lemonades would you buy if you had 200 crowns? This is a typical example that can be solved with a trinomial.
At this point we can do a simple reasoning - in the first case we had 100 crowns, in the second case we had 200 crowns. So we can expect that if we have twice as many crowns available, we will buy twice as many bottles with them. So we would buy 2 · 5 = 10 lemonades with 200 crowns.
The previous "the more... the more" is not entirely obvious, because we can have the following example: 10 masons build a house in four months. How long does it take 20 masons to build a house? If we use the previous procedure - there are twice as many masons, so there are twice as many months - we get that 20 masons would build a house in 2 · 4 = 8 months.
This is obviously not correct, because the more masons, the fewer months it will take them to build the house. So we have to do it the other way around: twice as many masons will build the house in twice as many months. This gives the correct result: 20 masons build a house in 4/2 = 2 months.
The previous two different procedures also have names: direct and inverse proportionality.
Direct proportionality
If "the more... the more" is true, this is direct proportionality. Examples:
- The more money we have, the more lemonade/glasses/scooters we can buy.
- The more articles a journalist writes, the more money he makes.
- The more diggers dig, the more they dig.
- The longer we let the pump pump, the more water we pump out.
Example: A car uses 6 litres of petrol per 100 kilometres. How many litres of petrol will it use after 250 kilometres? This is an example of direct proportionality because the more the car is driven, the more petrol it uses.
Procedure. We use the following table to make it clear what we know and what we want to calculate:
$$\begin{eqnarray} 100\mbox{ km}&\quad…\quad&6 \mbox{ litru }\\ 250\mbox{ km}&\quad…\quad&x\mbox{ litru } \end{eqnarray}$$
In the first row, we have written down the data that we know completely. In the second row, we have on the right-hand side the data that we do not know and on the left-hand side the data for which we want to calculate the unknown. The columns must always contain the same data, in this case the column must always contain either kilometres or litres.
There is a utility to help you calculate this example. First, draw an arrow from bottom to top on the right, like this:
$$\begin{eqnarray} 100\mbox{ km}&\quad…\quad&6 \mbox{ litru }\\ 250\mbox{ km}&\quad…\quad&x\mbox{ litru }\quad\uparrow \end{eqnarray}$$
If this is a direct proportion, then draw the same arrow on the left, from the bottom up:
$$\begin{eqnarray} \quad100\mbox{ km}&\quad…\quad&6 \mbox{ litru }\\ \uparrow\quad250\mbox{ km}&\quad…\quad&x\mbox{ litru }\quad\uparrow \end{eqnarray}$$
In the next step, construct the fractions from the columns in the direction of the arrows. Thus, the numerator corresponds to the number where the arrow starts and the denominator to the number where the arrow ends. We put these fractions into equality. This gives us the fraction $\frac{250}{100}$ from the first column (kilometres) and the fraction $\frac{x}{6}$ from the second column (litres):
$$\frac{250}{100}=\frac{x}{6}$$
This is an ordinary linear equation that we can solve by equivalent modifications. We multiply the whole equation by six to get rid of the fraction on the right hand side:
$$6\cdot\frac{250}{100}=x$$
Now all we have to do is calculate x:
$$x=6\cdot\frac{250}{100}=6\cdot\frac{5}{2}=\frac{30}{2}=15$$
The result is that the car uses 15 litres of petrol after 250 kilometres.
Inverse proportionality
If "the more ... the less" is true, this is an inverse proportion. Example:
- The more pages of a book we read, the fewer pages we have left to finish.
- The faster our internet speed, the faster we download a movie.
- The faster we drive, the sooner we get to our destination.
- The more workers working on a house, the sooner the house will be built.
Example: you have 2 MB per second internet on your computer and you downloaded the recording of Dada Patras' concert in 450 seconds. How long would it take you to download the same concert if you had 6 MB per second internet?
The procedure will be the same as for the direct proportion - we will write the data in a table:
$$\begin{eqnarray} 2\mbox{ MB/s}&\quad…\quad&450 \mbox{ Seconds }\\ 6\mbox{ MB/s}&\quad…\quad&x\mbox{ Seconds } \end{eqnarray}$$
Now the arrows. The right arrow will again be from the bottom up, nothing changes here. But since this is an inverse proportion, the left arrow will go in the opposite direction, from top to bottom.
$$\begin{eqnarray} \quad2\mbox{ MB/s}&\quad…\quad&450 \mbox{ Seconds }\\ \downarrow\quad6\mbox{ MB/s}&\quad…\quad&x\mbox{ Seconds }\quad\uparrow \end{eqnarray}$$
The next step is the same - we'll create fractions from this table, again following the direction of the arrows. This will give us the fractions $\frac26$ and $\frac{x}{450}$, which we will put into equality:
$$\frac26=\frac{x}{450}$$
Multiply the equation by 450:
$$450\cdot\frac26=x$$
And calculate x:
$$x=450\cdot\frac26=450\cdot\frac13=\frac{450}{3}=150$$
We can download Dada Patras with a faster connection in 150 seconds.
Note that the procedure is consistent with intuition. In the last step, we divide 450 by three, which is expected. Increasing the speed from 2 MB/s to 6 MB/s, we have three times the speed. Therefore, we would expect the time it takes to download a movie to be three times smaller $\rightarrow$ and this is exactly what the fraction $\frac{450}{3}$ expresses.
Examples
First example: a room in a luxury hotel costs ten thousand five hundred crowns for seven nights. How much would the same room cost if we wanted to go there for twelve days?
The first thing to do is to write the figures down in a table:
$$\begin{eqnarray} 7\mbox{ dni}&\quad…\quad&10{,}500 \mbox{ crowns }\\ 12\mbox{ dni}&\quad…\quad&x\mbox{ crowns } \end{eqnarray}$$
Now we have to decide whether this is a direct or an inverse relationship. Is it true that the longer we stay, the less we pay? No, the opposite is true - the longer we stay, the more we pay. So it is a direct proportion. Let's fill in the arrows:
$$\begin{eqnarray} \quad7\mbox{ dni}&\quad…\quad&10{,}500 \mbox{ crowns }\\ \uparrow\quad12\mbox{ dni}&\quad…\quad&x\mbox{ crowns }\qquad\quad\uparrow \end{eqnarray}$$
Use the arrows to make fractions:
$$\frac{12}{7}=\frac{x}{10{,}500}$$
Let's calculate the equation:
$$\begin{eqnarray} \frac{12}{7}&=&\frac{x}{10{,}500}\\ 10{,}500\cdot\frac{12}{7}&=&x\\ x&=&10{,}500\cdot\frac{12}{7}\\ x&=&\frac{10{,}500\cdot12}{7}\\ x&=&18{,}000 \end{eqnarray}$$
A room for 12 days would cost us 18 000 crowns.
The second example: 10 temporary workers pick strawberries and pick 50 kilos of strawberries per day. How many kilograms of strawberries do 7 temporary workers pick in a day?
Let's write the data in a table:
$$\begin{eqnarray} 10\mbox{ Brig }&\quad…\quad&50 \mbox{ kg}\\ 7\mbox{ Brig }&\quad…\quad&x\mbox{ kg} \end{eqnarray}$$
Is this a direct or an inverse relationship? In this example, there is some treachery because, unlike the previous examples, we have fewer workers in the row with the unknown than in the first row. Don't be fooled by this, it is still a direct proportion because the more temps, the more strawberries picked. Or vice versa - the fewer temps, the fewer strawberries. This is also a direct proportion. So both arrows will be from the bottom up.
$$\begin{eqnarray} 10\mbox{ Brig }&\quad…\quad&50 \mbox{ kg}\\ \uparrow\quad7\mbox{ Brig }&\quad…\quad&x\mbox{ kg}\quad\uparrow \end{eqnarray}$$
We'll create fractions:
$$\frac{7}{10}=\frac{x}{50}$$
and calculate x:
$$\begin{eqnarray} \frac{7}{10}&=&\frac{x}{50}\\ 50\cdot\frac{7}{10}&=&x\\ x&=&50\cdot\frac{7}{10}\\ x&=&5\cdot7\\ x&=&35 \end{eqnarray}$$
Seven temporary workers pick 35 kilos of strawberries a day.
Example 3: The average stride length of George is 80 cm. On his way home from school, he counts that he took 1300 steps. How many steps would he have taken if one step had been a metre long?
First, a table. Note that in the problem we have 80 cm and one meter. However, in the table we need to have the data in the same unit, so instead of one metre we use 100 cm - we could also choose the opposite and write 1 metre and 0.8 metres.
$$\begin{eqnarray} 80\mbox{ cm}&\quad…\quad&1300 \mbox{ step }\\ 100\mbox{ cm}&\quad…\quad&x\mbox{ step } \end{eqnarray}$$
And which is the proportion? It is true that the longer the step, the fewer steps we need to cover a certain distance, so it is an inverse proportion. The left arrow will be from top to bottom, the right arrow from bottom to top.
$$\begin{eqnarray} 80\mbox{ cm}&\quad…\quad&1300 \mbox{ step }\\ \downarrow\quad100\mbox{ cm}&\quad…\quad&x\mbox{ step }\qquad\uparrow \end{eqnarray}$$
The fractions will look like this:
$$\frac{80}{100}=\frac{x}{1300}$$
We isolate and calculate x:
$$\begin{eqnarray} \frac{80}{100}&=&\frac{x}{1300}\\ 1300\cdot\frac{80}{100}&=&x\\ x&=&1300\cdot\frac45\\ x&=&\frac{1300\cdot4}{5}\\ x&=&1040 \end{eqnarray}$$
George would only need 1040 steps to walk the same distance.
The formula
As can be seen from the previous calculations, the procedures are always the same and lead to simple formulas. First, let's write the table symbolically:
$$\begin{eqnarray} a&\quad…\quad&b\\ c&\quad…\quad&x \end{eqnarray}$$
Then the following formulas apply. For direct proportionality:
$$x=b\cdot\frac{c}{a}$$
And for inverse proportionality:
$$x=b\cdot\frac{a}{c}$$