Crossmultiplication
The crossmultiplication is a solution method for tasks with (inversely) proportional sizes.
For crossmultiplication exercises, one looks for a certain size with given values. You then try to increase or decrease the values to the size you are looking for.
Method
 Is it a proportional or inversely proportional relationship?
 Calculate auxiliary value (usually 1)
 Convert auxiliary value to searched size
Proportional relationship
For proportional relationships applies:
and vice versa
We imagine "more" as multiplication and "less" as division. The same arithmetic operation must be carried out on both sides.
Example
10 breads cost 30 euros. How much are 4 bread?
$\begin{array}{cccc} \text{Number} & & \text{Price} & \\ \hline 10 & \color{red}{:10} & 30 & \color{red}{:10} \\ 1 & {\color{green}{\cdot4}} & 3 & { \color{green}{\cdot4}} \\ 4 & & 12 & \end{array}$
A little more detailed explanation:
Proportional or inversely proportional relationship?
It's a proportional relationship, because the more breads you buy, the more you have to pay.

Enter values and determine the auxiliary value
We enter the given sizes.
$\begin{array}{cc} \text{Number} & \text{Price} \\ \hline 10 & 30 \\ 4 & ? \end{array}$
We calculate the price per piece, i.e. we calculate the value to 1:
$\frac{\text{Price}}{\text{Number}}$ $=\frac{30}{10}$ $=\color{blue}{3}$

Calculate the size you are looking for
The unit price will now be multiplied by the number.
$\text{Number}\cdot$ $\text{Unit price}$ $=4\cdot\color{blue}{3}$ $=12$
4 breads cost 12 euros.
Inversely proportional relationship
For inversely proportional relationships applies:
and vice versa
We imagine "more" as multiplication and "less" as division. On both sides, therefore, the reverse arithmetic operation must be performed.
Example
3 workers need 15 hours. How many hours do 5 workers need?
$\begin{array}{cccc} \text{Workers} & & \text{Hours} & \\ \hline 3 & \color{red}{:3} & 15 & \color{green}{\cdot3} \\ 1 & {\color{green}{\cdot5}} & 45 & { \color{red}{:5}} \\ 5 & & 9 & \end{array}$