A Trabant car used a $1: 40$ volume ratio mixture for its operation. What was the density of the mixture? $\left(\varrho_{\text {oil }}=900 \mathrm{~kg} / \mathrm{m}^{3}, \varrho_{\text {gasoline }}=700 \mathrm{~kg} / \mathrm{m}^{3}.\right)$
The volume ratio of $1: 40$ means that to $V$ volume of gasoline, $V/40$ volume of oil is added. Assuming the volume of the resulting mixture is the sum of the volumes of the components, the volume of the resulting mixture is

$$
V^{\prime}=V+\frac{1}{40} V=\frac{41}{40} V
$$

The mass of the mixture is

$$
V \cdot \varrho_{\text {gasoline }}+\frac{1}{40} V \cdot \varrho_{\text{oil}}
$$

The density of the mixture is

$$
\varrho^{\prime}=\frac{V \cdot \varrho_{\text {gasoline }}+V / 40 \cdot \varrho_{\text {oil }}}{41 V / 40}=\frac{40 \varrho_{\text {gasoline }}+\varrho_{\text {oil }}}{41}=\frac{28900 \mathrm{~kg} / \mathrm{m}^{3}}{41}=704.9 \frac{\mathrm{kg}}{\mathrm{m}^{3}}
$$

It is possible that the volume of the mixture is smaller than the sum of the volumes of the components. In such a case, the density will be greater than the calculated value.