Task 2 - 130912 Someone wants to create a new mixture with a water content of $98 \%$ from a mixture that consists of $99 \%$ water, by removing water from the original mixture.

Determine how much percent of the water content in the original mixture he must remove for this purpose in total.
The original mixture initially consists of water $\left(w_{1}\right)$ and another substance $(y)$. The total amount is $m_{1}$, so the following equation can be set up: $m_{1}=y+w_{1}$ (1), where the 99 percent water content is given by: $w_{1}=0.99 \cdot m_{1}(2)$.

For the state after water removal, the changed total amount $m_{2}$ with constant substance $y$ is: $m_{2}=y+w_{2}(3)$, and here too, a statement about the water content can be made: $w_{2}=0.98 \cdot m_{2}$ (4).

The fraction of the removed water amount $\left(x_{1}-x_{2}\right)$ relative to the original water amount $\left(x_{1}\right)$ in percent is sought. To do this, equations (1) to (4) are now used. First, (1) and (3) are solved for $y$ and then equated, and subsequently (2) and (4) are used: $y=m_{1}-w_{1}=m_{2}-w_{2} \Rightarrow$ $m_{1}-0.99 \cdot m_{1}=m_{2}-0.98 \cdot m_{2} \Rightarrow 0.01 \cdot m_{1}=0.02 \cdot m_{2}$ or $m_{1}=2 \cdot m_{2}$. Now the sought fraction can be calculated:

$$
\frac{x_{1}-x_{2}}{x_{1}}=\frac{0.99 \cdot m_{1}-0.98 \cdot m_{2}}{0.99 \cdot m_{1}}=\frac{1.98 \cdot m_{2}-0.98 \cdot m_{2}}{1.98 \cdot m_{2}}=\frac{m_{2}}{1.98 \cdot m_{2}}=\frac{1}{1.98} \approx 0.51
$$

The fraction of the removed water relative to the original water amount is thus approximately $51\%$.