Two substances of densities $\rho_1$ and $\rho_2$ are mixed in equal v

Two substances of densities $\rho_1$ and $\rho_2$ are mixed in equal volume and their relative density is 4. When they are mixed in equal masses, relative density is 3. The values of $\rho_1$ and $\rho_2$ respectively are

6, 2
3, 5
12, 4
9, 3
This question was previously asked in
UPSC NDA-2 – 2019
Let the densities be $\rho_1$ and $\rho_2$. Relative density is numerically equal to density assuming water density is 1.
Case 1: Mixed in equal volume ($V$). Total mass $m = m_1 + m_2 = \rho_1 V + \rho_2 V = (\rho_1 + \rho_2)V$. Total volume $V_{total} = V + V = 2V$. Density $\rho_{mix, V} = \frac{(\rho_1 + \rho_2)V}{2V} = \frac{\rho_1 + \rho_2}{2}$. Given relative density is 4, so $\frac{\rho_1 + \rho_2}{2} = 4 \implies \rho_1 + \rho_2 = 8$.
Case 2: Mixed in equal masses ($m$). Volume $V_1 = \frac{m}{\rho_1}$, $V_2 = \frac{m}{\rho_2}$. Total mass $m_{total} = m + m = 2m$. Total volume $V_{total} = \frac{m}{\rho_1} + \frac{m}{\rho_2} = m\left(\frac{\rho_1 + \rho_2}{\rho_1 \rho_2}\right)$. Density $\rho_{mix, m} = \frac{2m}{m\left(\frac{\rho_1 + \rho_2}{\rho_1 \rho_2}\right)} = \frac{2\rho_1 \rho_2}{\rho_1 + \rho_2}$. Given relative density is 3, so $\frac{2\rho_1 \rho_2}{\rho_1 + \rho_2} = 3$.
Substitute $\rho_1 + \rho_2 = 8$ into the second equation: $\frac{2\rho_1 \rho_2}{8} = 3 \implies \frac{\rho_1 \rho_2}{4} = 3 \implies \rho_1 \rho_2 = 12$.
We need two numbers whose sum is 8 and product is 12. These are the roots of the quadratic equation $x^2 – 8x + 12 = 0$, which factors as $(x-2)(x-6) = 0$. The roots are 2 and 6. So, the densities are 2 and 6. Option A is (6, 2), which satisfies both conditions: (6+2)/2 = 4 and (2*6*2)/(6+2) = 24/8 = 3.
– Density of a mixture depends on the densities of the components and the proportions in which they are mixed (by mass or by volume).
– Average density when mixed by volume is the arithmetic mean of the densities.
– Average density when mixed by mass is the harmonic mean of the densities, weighted by mass proportion (or a form related to it). The formula derived $\frac{2 \rho_1 \rho_2}{\rho_1 + \rho_2}$ is 2 times the harmonic mean of $\rho_1$ and $\rho_2$.
– Relative density (or specific gravity) is the ratio of the density of a substance to the density of a reference substance (usually water).
– The problem effectively provides the arithmetic mean and a form of the harmonic mean of the two densities, allowing us to solve for the densities.
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