## Task 6 - 140836

Given three numbers $p, p_{1}, p_{2}$ with $0<p_{1}<p<p_{2}<100$.

From a suitable amount $x$ kg of a $p_{1}$-percent solution of a substance (i.e., a solution that contains $p_{1}$ % of this substance and the rest water) and a suitable amount $y$ kg of a $p_{2}$-percent solution of the same substance, a $p$-percent solution is to be produced by mixing.

a) Determine the required mixing ratio, i.e., the number $x: y$, first specifically for the values $p_{1}=25, p_{2}=60$ and $p=35$!

b) Then derive a formula for the mixing ratio that is valid for arbitrary values of $p_{1}, p_{2}$ and $p$!

Note: The given percentages refer to the mass, and are not to be interpreted as volume percentages.
a) In $x \text{ kg}$ of the 25% solution, there are $\frac{25 x}{100} \text{ kg}$ of the dissolved substance, and in $y \text{ kg}$ of the 60% solution, there are $\frac{60 y}{100} \text{ kg}$ of the dissolved substance. Therefore, $x: y$ has the desired value if and only if in the $(x+y) \text{ kg}$ obtained by mixing, there are exactly $\frac{35(x+y)}{100} \text{ kg}$ of the dissolved substance, i.e., if and only if

$$
\frac{25 x}{100}+\frac{60 y}{100}=\frac{35(x+y)}{100}
$$

holds. This is successively equivalent to $25 x + 60 y = 35 x + 35 y$, $25 y = 10 x$, $\frac{5}{2} = \frac{x}{y}$. The desired mixing ratio is thus $5: 2$.

b) With analogous reasoning as in a), $x: y$ has the desired value if and only if

$$
\frac{p_{1}}{100} x + \frac{p_{2}}{100} y = \frac{p}{100}(x+y) \quad \Rightarrow \quad \frac{x}{y} = \frac{p_{2} - p}{p - p_{1}}
$$

The desired mixing ratio is: $\frac{p_{2} - p}{p - p_{1}}$

Solutions of the III. Round 1974 adopted from [5]

### 5.17 XV. Olympiad 1975

### 5.17.1 I. Round 1975, Class 8