0.35 \cdot 160+0.1 x=0.2 \cdot 160+0.2 x, 0.15 \cdot 160=0.1 x, x=240 \text {. }
$$

In total, it results in $160+240=400$ g of solution.
Answer: 400.

## Problem 9

Points $B_{1}$ and $C_{1}$ are the feet of the altitudes of triangle $ABC$, drawn from vertices $B$ and $C$ respectively. It is known that $AB=7, \quad AC=6, \sin \angle BAC=\frac{2 \sqrt{110}}{21}$. Find the length of the segment $B_{1}C_{1}$.

## Solution:

Let the angle $\angle BAC$ be denoted by $\alpha$. There are two possible cases: 1) $00$.

$$
\begin{aligned}
& \frac{\left(x^{2}-4\right)(|2 x+2|+|x+4|)}{(2 x+2)^{2}-(x+4)^{2}}=a x^{2}, \\
& \frac{\left(x^{2}-4\right)(|2 x+2|+|x+4|)}{3\left(x^{2}-4\right)}=a x^{2}, \\
& |2 x+2|+|x+4|=3 a x^{2} .
\end{aligned}
$$

If $a \leq 0$, then the left side $>0$, and the right side $\leq 0$; equation ( ${ }^{*}$ ) has no solutions.

If $a>0$, equation (*) has at least 2 solutions.

The original equation can have fewer solutions only if at least one of the numbers $\pm 2$ is a solution ( ي $\left.^{2}\right)$.

2 is a solution of (*), if $12=12 a, a=1 ;(-2)$ is a solution of (*), if $4=12 a, a=1 / 3$.

When $a=1$, equation (*) becomes $|2 x+2|+|x+4|=3 x^{2}$ and has roots $x_{1}=-1, x_{2}=2$. The original equation has a single root $x=-1$.

When $a=1 / 3$, equation (*) becomes $|2 x+2|+|x+4|=x^{2}$ and has roots $x_{1}=-2, x_{2}=\frac{3+\sqrt{33}}{2}$. The original equation has a single root $x=\frac{3+\sqrt{33}}{2}$.

Answer: the equation has no more than one root when $a \leq 0$, when $a=1 / 3$, and when $a=1$.