A classical mechanics problem by Sabhrant Sachan

Let $\rho_A = \frac{m_{A}}{V_{A}}$ and $\rho_B = \frac{m_{B}}{V_{B}}$ be the respective densities of these two liquids. We require the following formulations to properly model the two described mixtures of A & B:

$4 = \frac{m_{A} + m_{B}}{2V} = \frac{\rho_{A}V + \rho_{B}V}{2V} \Rightarrow \boxed{8 = \rho_{A} + \rho_{B}}$ (i)

$3 = \frac{2m}{V_{A} +V_{B}} = \frac{2m}{\frac{m}{\rho_{A}} + \frac{m}{\rho_{B}}} \Rightarrow 3 = \frac{2}{\frac{1}{\rho_{A}} + \frac{1}{\rho_{B}}} = \boxed{3(\rho_{A} + \rho_{B}) = 2\rho_{A}\rho_{B}}$ (ii)

Equating (i) with (ii) ultimately yields:

$3 \cdot 8 = 2\rho_{A}(8 - \rho_{A}) \Rightarrow 12 = 8\rho_{A} - \rho^{2}_{A} \Rightarrow \rho^{2}_{A} - 8\rho_{A} + 12 = 0 \Rightarrow (\rho_{A} -2)(\rho_{A} - 6) = 0 \Rightarrow \rho_{A} = 2, 6.$

Since we are given $\rho_{A} > \rho_{B}$ , the required densities are $\boxed{\rho_{A} = 6, \rho_{B} = 2}.$