8. (15 points) Density is defined as the ratio of the mass of a body to its volume. There are two cubes. The second cube is made of a material with half the density of the first, and the side length of the second cube is 100% greater than the side length of the first. By what percentage is the mass of the second cube greater than the mass of the first?
Answer: by $300 \%$

Solution. From the condition, it follows that $\rho_{2}=\frac{1}{2} \rho_{1}$, i.e., $\frac{m_{2}}{V_{2}}=\frac{1}{2} \cdot \frac{m_{1}}{V_{1}}$.

It also follows from the condition that the side length of the second cube $a_{2}=2 a_{1}$.

(2 points)

Therefore, their volumes are related by the ratio $V_{2}=a_{2}^{3}=\left(2 a_{1}\right)^{3}=8 V_{1}$.

(3 points)

We obtain $\frac{m_{2}}{8 V_{1}}=\frac{1}{2} \cdot \frac{m_{1}}{V_{1}}$. (2 points) From which it follows that $m_{2}=4 m_{1}$.

That is, the mass of the second cube is greater by $300 \%$.