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

When the substance are mixed in equal volumes, then

    ${\mathrm{V\rho }}_1+{\mathrm{V\rho }}_2= 2\mathrm{V} \times 4 ------\left(\mathrm{i}\right)$

When the two substances are mixed in equal masses, then

    $\frac{\mathrm{m}}{{\mathrm{\rho }}_1}+\frac{\mathrm{m}}{{\mathrm{\rho }}_2} = \frac{2\mathrm{m}}{3}----\left(\mathrm{ii}\right)$

From eq.(i): ${\mathrm{\rho }}_1 + {\mathrm{\rho }}_2 = 8$

From eq. (ii): $\frac{1}{{\mathrm{\rho }}_1}+\frac{1}{{\mathrm{\rho }}_2} = \frac{2}{3}$

or $\frac{{\mathrm{\rho }}_1+{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1{\mathrm{\rho }}_2} = \frac{2}{3}$

or $\frac{9}{{\mathrm{\rho }}_1{\mathrm{\rho }}_2} = \frac{2}{3} \mathrm{or} {\mathrm{\rho }}_1{\mathrm{\rho }}_2 = 12$------(iv)

Now, ${\mathrm{\rho }}_1 - {\mathrm{\rho }}_2 = {[{({\mathrm{\rho }}_1+{\mathrm{\rho }}_2)}^2-4{\mathrm{\rho }}_1{\mathrm{\rho }}_2]}^{\frac{1}{2}}$------(v)

Solving eqs. (iii) and (v), we get ${\text{ρ}}_1 = 6 \mathrm{and} {\mathrm{\rho }}_2 = 2$