4. There are three alloys of nickel, copper, and manganese. In the first, $30\%$ is nickel and $70\%$ is copper, in the second - $10\%$ is copper and $90\%$ is manganese, and in the third - $15\%$ is nickel, $25\%$ is copper, and $60\%$ is manganese. A new alloy of these three metals is needed with $40\%$ manganese. What values can the percentage content of copper in the new alloy take?
Answer: from $40 \%$ to $43 \frac{1}{3} \%$.

Solution. Let the masses of the original alloys, from which the new alloy is obtained, be denoted by $m_{1}, m_{2}, m_{3}$, and the mass of the new alloy by $m$. The equalities $m_{1}+m_{2}+m_{3}=m$ and $0.9 m_{2}+0.6 m_{3}=0.4 m$ hold. The amount of copper in the new alloy is $0.7 m_{1}+0.1 m_{2}+0.25 m_{3}$. Let $x_{i}=m_{i} / m$ be the fraction of the $i$-th alloy in the new alloy. Then, for the fraction of copper, we have the expression $y=0.7 x_{1}+0.1 x_{2}+0.25 x_{3}$.

Thus, the problem is formalized as follows. Under the conditions

$$
\left\{\begin{array}{r}
x_{1}+x_{2}+x_{3}=1 \\
0.9 x_{2}+0.6 x_{3}=0.4 \\
0 \leqslant x_{1}, x_{2}, x_{3} \leqslant 1
\end{array}\right.
$$

find the set of values of $y=0.7 x_{1}+0.1 x_{2}+0.25 x_{3}$.

From the system of linear equations, express $x_{3}, x_{1}$, and $y$ in terms of $x_{2}$:

$$
x_{3}=\frac{2}{3}-\frac{3}{2} x_{2} ; \quad x_{1}=\frac{1}{3}+\frac{1}{2} x_{2} ; \quad y=0.4+0.075 x_{2}
$$

Now, solving the system of inequalities, we find the values that $x_{2}$ can take. We get $0 \leqslant x_{2} \leqslant \frac{4}{9}$. Since $y$ is a linear function of the variable $x_{2}$, it is easy to find the set of values of $y$. Next, we need to convert the fraction to a percentage.

Evaluation. 14 points for a correct solution.

## Final Stage

2016-2017 academic year
## Problems, answers, and evaluation criteria

## 11th grade

Variant II Ћวff Һว+Gra