Two substances of densities {rho _1} and {rho | vedclass

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(A) When substances are mixed in equal volumes,the density of the mixture is given by $\rho_{mix} = \frac{\rho_1 + \rho_2}{2}$.Given $\rho_{mix} = 4$,

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${\rho _1} = 6$ and ${\rho _2} = 2$

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(A) When substances are mixed in equal volumes,the density of the mixture is given by $\rho_{mix} = \frac{\rho_1 + \rho_2}{2}$.Given $\rho_{mix} = 4$,so $\frac{\rho_1 + \rho_2}{2} = 4$,which implies $\rho_1 + \rho_2 = 8$ .......$(i)$When substances are mixed in equal masses,the density of the mixture is given by $\rho_{mix} = \frac{2\rho_1 \rho_2}{\rho_1 + \rho_2}$.Given $\rho_{mix} = 3$,so $\frac{2\rho_1 \rho_2}{\rho_1 + \rho_2} = 3$,which implies $2\rho_1 \rho_2 = 3(\rho_1 + \rho_2)$ .......$(ii)$Substituting equation $(i)$ 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 $\rho_1 \rho_2 = 12$. These are the roots of the quadratic equation $x^2 - 8x + 12 = 0$.$(x - 6)(x - 2) = 0$,so $x = 6$ or $x = 2$.Thus,the values are ${\rho _1} = 6$ and ${\rho _2} = 2$ (or vice versa).

Two substances of densities ${\rho _1}$ and ${\rho _2}$ are mixed in equal volumes and the relative density of the mixture is $4$. When they are mixed in equal masses,the relative density of the mixture is $3$. The values of ${\rho _1}$ and ${\rho _2}$ are

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