Two liquids of densities $$1\text{ g/cm}^3$$ and $$3\text{ g/cm}^3$$ are mixed in equal volumes. A cube of external dimension 20 cm has an inner cubical portion of side 10 cm whose density is twice that of the outer portion. If this cube is just floating in the liquid mixture, find the density of the inner portion. If the density of the inner portion is half of this value, what fraction of the volume of the cube will be submerged?

PLACEHOLDER ANSWER - the question has two answers. The mixture has density $$\frac{1 + 3}{2} = 2\text{ g/cm}^3$$. With outer density $$\rho$$, the cube's mass is $$7000\rho + 1000 \times 2\rho = 9000\rho$$, and just floating needs this to equal $$8000 \times 2 = 16000\text{ g}$$, so $$\rho = \frac{16}{9}$$ and the inner density is $$\frac{32}{9} \approx 3.56\text{ g/cm}^3$$. If the inner density is halved to $$\frac{16}{9}$$, the whole cube has density $$\frac{16}{9}$$, so the fraction submerged is $$\frac{16/9}{2} = \frac{8}{9}$$.