## Task 5 - 180835

A 30% salt solution is needed for experimentation. However, only 2 liters of 10% salt solution and a bottle of 42% salt solution are available.

Determine how many liters of 42% salt solution need to be added to the 2 liters of 10% salt solution to create a 30% salt solution!
Assuming that adding $x$ liters of 42% salt solution to the 2 liters of 10% salt solution results in $(x+2)$ liters of a 30% salt solution, it follows that:

2 liters of 10% salt solution contain $0.20 \mathrm{L}$ of salt, $x$ liters of 42% salt solution contain $0.42x \mathrm{L}$ of salt, and $(x+2)$ liters of 30% salt solution contain $0.30(x+2) \mathrm{L}$ of salt.

Since the amount of salt in the mixed solution is always equal to the sum of the amounts of salt in the mixed solutions, the following must hold:

$$
0.20 + 0.42x = 0.30(x + 2) \Rightarrow 20 + 42x = 30x + 60 \Rightarrow x = \frac{10}{3}
$$

Therefore, $\frac{10}{3}$ liters of 42% salt solution must be added to obtain the required 30% salt solution.

Verification: The mixture results in 2 liters + $\frac{10}{3}$ liters = $\frac{16}{3}$ liters of solution with the following salt content: 2 liters of 10% salt solution contain 0.20 liters of salt, $\frac{10}{3}$ liters of 42% salt solution contain 1.40 liters of salt. Together, this is 1.60 liters of salt.

Of the $\frac{16}{3}$ liters of total liquid, these 1.60 liters of salt are $1.60 : \frac{16}{3} \cdot 100\% = 30\%$, as required.