2. There are two solutions of wine spirit in water ${ }^{1}$ ): The first solution weighs 4002 and contains $30 \%$ spirit, the second solution weighs 6002 and contains $80 \%$ spirit.

From these two solutions, one solution is made. What percentage of spirit will it contain? (O. 117, U. 126, P. 215.$)$
2. a) Let's represent the solutions using material points. For this, we will take a segment $AB$ of one unit length (Fig. 172). Let point $A$ represent pure water

![](https://cdn.mathpix.com/cropped/2024_05_21_27f86caa754616304e40g-216.jpg?height=89&width=979&top_left_y=1446&top_left_x=119)
[Boda] (cimupm.)

Fig. 172.

$(100\%$ water), and $B$—pure alcohol ( $100\%$ alcohol). The first solution will then be represented by the material point $K_{1} \equiv\left(K_{1}, 400\right)$, where $A K_{1}=0.3$ (the mark "30" next to point $K_{1}$ indicates that $A K_{1}$ contains 30 parts out of 100). The second solution will be represented by the material point $K_{2} \equiv\left(K_{2}, 600\right)$. The combination of these two solutions will be the solution $K \equiv(K, 1000)$, where $K \equiv Z\left[K_{1}, K_{2}\right]$. According to the lever rule,

$$
\frac{K_{1} K}{K K_{2}}=\frac{600}{400}=\frac{3}{2}
$$

Let $A K=x$. Then we get: $\frac{x-0.3}{0.8-x}=\frac{3}{2}$, from which $x=0.6$. Therefore, the concentration of alcohol in the solution $(1000)$ is $60\%$. The same result could have been obtained from the relation

$$
\operatorname{Mom}_{A} K \equiv \operatorname{Mom}_{A} K_{1}+\text{Mom}_{A} \boldsymbol{K}_{2},
$$

i.e., $1000 \cdot x=400 \cdot 0.3+600 \cdot 0.8$. The problem can also be solved purely graphically.