Two liquids of masses 600 g and 500 g and densities $$0.6\text{ g cm}^{-3}$$ and $$1.5\text{ g cm}^{-3}$$ are combined to form a homogenous mixture. A solid block of mass 420 g just floats in this mixture. Find the density of the block material. If the equal volumes of the two liquids were mixed instead, would this block float in that mixture? If it floats find the fraction of the block that is visible above the mixture.

PLACEHOLDER ANSWER - the question has three answers. The volumes are $$\frac{600}{0.6} = 1000\text{ cm}^3$$ and $$\frac{500}{1.5} = \frac{1000}{3}\text{ cm}^3$$, so the mixture's density is $$\frac{1100}{4000/3} = 0.825\text{ g cm}^{-3}$$, and a block that just floats has this density. Mixing equal volumes gives density $$\frac{0.6 + 1.5}{2} = 1.05\text{ g cm}^{-3}$$, which is greater, so the block floats. The fraction submerged is $$\frac{0.825}{1.05} = \frac{11}{14}$$, so $$\frac{3}{14} \approx 0.21$$ of the block is visible.