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Consider three mixtures - the first having water and liquid $A$ in the ratio $1:2$, the second having water and liquid $B$ in the ratio $1:3$, and the third having water and liquid $C$ in the ratio $1:4$. These three mixtures of $A$, $B$, and $C$ respectively, are further mixed in the proportion $4:3:2$. Then the resulting mixture has

1. The same amount of water and liquid $B$
2. The same amount of liquids $B$ and $C$
3. More water than liquid $B$
4. More water than liquid $A$

Given that,

• In mixture $\text{A:}$
• $\text{Water : Liquid} =1:2$
• In mixture $\text{B:}$
• $\text{Water : Liquid} =1:3$
• In mixture $C:$
• $\text{Water : Liquid} =1:4$

These three mixtures of $\text{A, B}$, and $\text{C}$ respectively, are further mixed in the proportion $4:3:2.$

Let the volumes of $\text{A, B}$, and $\text{C}$ be $4k, 3k$, and $2k$ respectively.

$$\begin{array}{cccc} \text{Mixture} & \text{Water : Liquid} & \text{Volume of water} & \text{Volume of liquid} \\ A & 1:2 & \frac{1}{3} \times 4k & \frac{2}{3}\times 4k \\ B & 1:3 & \frac{1}{4} \times 3k & \frac{3}{4} \times 3k \\ C & 1:4 & \frac{1}{5} \times 2k & \frac{4}{5} \times 2k \end{array}$$

Let,  $k= \text{LCM}(3,4,5) = 60$

$$\begin{array}{cccc} \text{Mixture} & \text{Water : Liquid} & \text{Volume of water} & \text{Volume of liquid} \\ \text{A} & 1:2 & \frac{1}{3} \times 4(60) = 80 & \frac{2}{3} \times 4(60) = 160 \\ \text{B} & 1:3 & \frac{1}{4} \times 3(60) = 45 & \frac{3}{4} \times 3(60) = 135 \\ \text{C} & 1:4 & \frac{1}{5} \times 2(60) = 24 & \frac{4}{5} \times 2(60) = 96 \end{array}$$
The volume of water $= 80 + 45 + 24 = 149 > \underbrace{135}_{\text{Liquid B}} > \underbrace{96}_{\text{Liquid C}}.$

$\therefore$ The quantity of water is greater than liquid $\text{B}$.

Correct Answer $:\text{C}$

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