Two substances of densities `rho_(1)` and `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 `rho_(1)` and `rho_(2)` are:

Options

A. `rho=6 " and " rho_(2)=2`
B. `rho_(3) " and " rho_(2)=5`
C. `rho_(1)=12 " and " rho_(2)=4`
D. None of these

Explanation

To solve the problem, we need to find the densities \( \rho_1 \) and \( \rho_2 \) of two substances based on the information provided about their mixtures.

### Step 1: Understand the first case (equal volumes)
When the two substances are mixed in equal volumes, the relative density of the mixture is given as 4.

The formula for the density of the mixture when mixed in equal volumes is:
\[ \rho_{\text{mixture}} = \frac{\rho_1 V + \rho_2 V}{V + V} = \frac{\rho_1 + \rho_2}{2} \]
Given that \( \rho_{\text{mixture}} = 4 \):
\[ \frac{\rho_1 + \rho_2}{2} = 4 \]
Multiplying both sides by 2:
\[ \rho_1 + \rho_2 = 8 \quad \text{(Equation 1)} \]

### Step 2: Understand the second case (equal masses)
When the substances are mixed in equal masses, the relative density of the mixture is given as 3.

The formula for the density of the mixture when mixed in equal masses is:
\[ \rho_{\text{mixture}} = \frac{m_1 + m_2}{\frac{m_1}{\rho_1} + \frac{m_2}{\rho_2}} \]
Since \( m_1 = m_2 = m \):
\[ \rho_{\text{mixture}} = \frac{m + m}{\frac{m}{\rho_1} + \frac{m}{\rho_2}} = \frac{2m}{\frac{m}{\rho_1} + \frac{m}{\rho_2}} = \frac{2}{\frac{1}{\rho_1} + \frac{1}{\rho_2}} \]
Given that \( \rho_{\text{mixture}} = 3 \):
\[ 3 = \frac{2}{\frac{1}{\rho_1} + \frac{1}{\rho_2}} \]
Cross-multiplying gives:
\[ 3\left(\frac{1}{\rho_1} + \frac{1}{\rho_2}\right) = 2 \]
This simplifies to:
\[ \frac{1}{\rho_1} + \frac{1}{\rho_2} = \frac{2}{3} \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
We have two equations now:
1. \( \rho_1 + \rho_2 = 8 \) (Equation 1)
2. \( \frac{1}{\rho_1} + \frac{1}{\rho_2} = \frac{2}{3} \) (Equation 2)

From Equation 2, we can express it in terms of \( \rho_1 \) and \( \rho_2 \):
\[ \frac{\rho_1 + \rho_2}{\rho_1 \rho_2} = \frac{2}{3} \]
Substituting \( \rho_1 + \rho_2 = 8 \) from Equation 1 into this gives:
\[ \frac{8}{\rho_1 \rho_2} = \frac{2}{3} \]
Cross-multiplying gives:
\[ 8 \cdot 3 = 2 \cdot \rho_1 \rho_2 \]
\[ 24 = 2 \rho_1 \rho_2 \]
Dividing by 2:
\[ \rho_1 \rho_2 = 12 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations
Now we have:
1. \( \rho_1 + \rho_2 = 8 \)
2. \( \rho_1 \rho_2 = 12 \)

Let \( \rho_1 \) and \( \rho_2 \) be the roots of the quadratic equation:
\[ x^2 - (\rho_1 + \rho_2)x + \rho_1 \rho_2 = 0 \]
This becomes:
\[ x^2 - 8x + 12 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2} \]
This gives:
\[ x = \frac{12}{2} = 6 \quad \text{or} \quad x = \frac{4}{2} = 2 \]
Thus, \( \rho_1 = 6 \) and \( \rho_2 = 2 \) (or vice versa).

### Final Answer
The values of \( \rho_1 \) and \( \rho_2 \) are:
\[ \rho_1 = 6, \quad \rho_2 = 2 \]