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(A) Let the relative densities of the two metals be $\rho_1$ and $\rho_2$.When equal volumes $V$ are mixed,the density of the mixture is $\rho_{mix} =

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$\rho_1 = 6$ and $\rho_2 = 2$

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(A) Let the relative densities of the two metals be $\rho_1$ and $\rho_2$.When equal volumes $V$ are mixed,the density of the mixture is $\rho_{mix} = \frac{V\rho_1 + V\rho_2}{2V} = \frac{\rho_1 + \rho_2}{2}$.Given $\frac{\rho_1 + \rho_2}{2} = 4$,so $\rho_1 + \rho_2 = 8$ (Equation $i$).When equal masses $M$ are mixed,the density of the mixture is $\rho_{mix} = \frac{2M}{V_1 + V_2} = \frac{2M}{\frac{M}{\rho_1} + \frac{M}{\rho_2}} = \frac{2\rho_1\rho_2}{\rho_1 + \rho_2}$.Given $\frac{2\rho_1\rho_2}{\rho_1 + \rho_2} = 3$,so $2\rho_1\rho_2 = 3(\rho_1 + \rho_2)$ (Equation $ii$).Substituting $\rho_1 + \rho_2 = 8$ into Equation $ii$:$2\rho_1\rho_2 = 3(8) = 24$,so $\rho_1\rho_2 = 12$.We have $\rho_1 + \rho_2 = 8$ and $\rho_1\rho_2 = 12$. Solving the quadratic equation $x^2 - 8x + 12 = 0$:$(x - 6)(x - 2) = 0$.Thus,the relative densities are $6$ and $2$.

When two metals of equal volume are mixed,the relative density of the mixture is $4$. When equal masses of these two metals are mixed,the relative density of the mixture is $3$. What are the relative densities of the two metals?

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