Mathematical Analysis of Mixture Proportions

Let $Q_X$ and $Q_Y$ represent the quantities of solution $X$ and solution $Y$ being mixed, respectively. The ingredients $A, B, C$ are present in solution $X$ in the proportion $a:b:c$, and in solution $Y$ in the proportion $c:b:a$. The total proportion units for both solutions is $a+b+c$.

The amount of each ingredient from each solution can be expressed as follows:

• In $Q_X$ quantity of solution $X$: Amount of $A = Q_X \frac{a}{a+b+c}$, Amount of $B = Q_X \frac{b}{a+b+c}$, Amount of $C = Q_X \frac{c}{a+b+c}$.
• In $Q_Y$ quantity of solution $Y$: Amount of $A = Q_Y \frac{c}{a+b+c}$, Amount of $B = Q_Y \frac{b}{a+b+c}$, Amount of $C = Q_Y \frac{a}{a+b+c}$.

Deriving Necessary Conditions

The total amount of each ingredient in the resultant mixture is the sum of the amounts from solution $X$ and solution $Y$. For the resultant mixture to have $A, B, C$ in a $1:1:1$ proportion, the total amounts must be equal:

Total $A$ = Total $B$ = Total $C$

This implies:

1. Equating Total $A$ and Total $B$: $Q_X \frac{a}{a+b+c} + Q_Y \frac{c}{a+b+c} = Q_X \frac{b}{a+b+c} + Q_Y \frac{b}{a+b+c}$ Multiplying by $(a+b+c)$ gives: $Q_X a + Q_Y c = Q_X b + Q_Y b$ Rearranging terms: $Q_X (a-b) = Q_Y (b-c)$ \quad (Equation 1)
2. Equating Total $B$ and Total $C$: $Q_X \frac{b}{a+b+c} + Q_Y \frac{b}{a+b+c} = Q_X \frac{c}{a+b+c} + Q_Y \frac{a}{a+b+c}$ Multiplying by $(a+b+c)$ gives: $Q_X b + Q_Y b = Q_X c + Q_Y a$ Rearranging terms: $Q_X (b-c) = Q_Y (a-b)$ \quad (Equation 2)

Solving the System of Equations

Let $\Delta_{ab} = a-b$ and $\Delta_{bc} = b-c$. The equations become:

1. $Q_X \Delta_{ab} = Q_Y \Delta_{bc}$

2. $Q_X \Delta_{bc} = Q_Y \Delta_{ab}$

For a non-trivial solution where $Q_X > 0$ and $Q_Y > 0$, we must have $\Delta_{ab} \neq 0$ and $\Delta_{bc} \neq 0$. Dividing Equation 1 by Equation 2 (assuming $Q_X, Q_Y, \Delta_{ab}, \Delta_{bc}$ are non-zero):

This implies $\Delta_{ab}^2 = \Delta_{bc}^2$, which means $(a-b)^2 = (b-c)^2$. This leads to two possibilities:

1. $a-b = b-c \implies a+c = 2b$. This indicates that $a, b, c$ must form an arithmetic progression.
2. $a-b = -(b-c) \implies a-b = -b+c \implies a=c$. If $a=c$, the proportions are $a:b:a$ for both solutions, making them identical. The mixture would always maintain this $a:b:a$ ratio. For this to be $1:1:1$, $a$ must equal $b$, resulting in the trivial case $1:1:1$. Thus, $a=c$ is not a necessary condition for a non-trivial mixture.

Therefore, the necessary condition is that $a,b,c$ must form an arithmetic progression ($a+c=2b$).

Checking the Options

We examine which of the given options satisfy the condition $a+c = 2b$:

• Option 1: $1:2:3$. Here $a=1, b=2, c=3$. $1+3 = 4$ and $2(2) = 4$. Satisfies $a+c=2b$.
• Option 2: $2:1:3$. Here $a=2, b=1, c=3$. $2+3 = 5$ and $2(1) = 2$. Does not satisfy $a+c=2b$.
• Option 3: $1:3:2$. Here $a=1, b=3, c=2$. $1+2 = 3$ and $2(3) = 6$. Does not satisfy $a+c=2b$.
• Option 4: $3:2:1$. Here $a=3, b=2, c=1$. $3+1 = 4$ and $2(2) = 4$. Satisfies $a+c=2b$.

Both options $1:2:3$ and $3:2:1$ satisfy the derived necessary condition. However, the question asks for *the* necessary ratio, implying a unique answer among the options. Based on the provided correct answer being Option D, the necessary proportion is $3:2:1$.