8. A cube consists of eight smaller identical cubes. Three of the smaller cubes were replaced with ones of the same size but with a density three times greater. Determine the ratio of the final to the initial density of the large cube. (10 points)
Answer: 1.75

Solution. The relationship between mass and volume: $m=\rho V$ (2 points), that is, the new cubes, with the same volume, are three times heavier. Initial density: $\rho_{\text {initial }}=\frac{8 m_{0}}{8 V_{0}}$

(3 points). Final density: $\rho_{\text {final }}=\frac{5 m_{0}+3 m_{1}}{8 V_{0}}=\frac{5 m_{0}+3 \cdot 3 m_{0}}{8 V_{0}}=\frac{14 m_{0}}{8 V_{0}}$ (3 points).

Final result: $\frac{\rho_{\text {final }}}{\rho_{\text {initial }}}=\frac{14}{8}=1.75$ (2 points).