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

When we have mixture problems, we need to track each component separately. Let's break this down systematically.

---

First mixture (Water : Liquid A = 1 : 2)

Total parts = 1 + 2 = 3 parts

Water makes up $\dfrac{1}{3}$ of this mixture

Liquid A makes up $\dfrac{2}{3}$ of this mixture

Second mixture (Water : Liquid B = 1 : 3)

Total parts = 1 + 3 = 4 parts

Water makes up $\dfrac{1}{4}$ of this mixture

Liquid B makes up $\dfrac{3}{4}$ of this mixture

Third mixture (Water : Liquid C = 1 : 4)

Total parts = 1 + 4 = 5 parts

Water makes up $\dfrac{1}{5}$ of this mixture

Liquid C makes up $\dfrac{4}{5}$ of this mixture

---

These three mixtures are combined in the ratio 4 : 3 : 2.

This means:

4 parts of the first mixture (A-mixture)

3 parts of the second mixture (B-mixture)

2 parts of the third mixture (C-mixture)

Total = 4 + 3 + 2 = 9 parts

---

Amount of Liquid A:

Comes only from the first mixture

First mixture contribution = $\dfrac{4}{9}$ of total

A content in first mixture = $\dfrac{2}{3}$

Amount of A = $\dfrac{4}{9} \times \dfrac{2}{3} = \dfrac{8}{27}$

Amount of Liquid B:

Comes only from the second mixture

Second mixture contribution = $\dfrac{3}{9}$ of total

B content in second mixture = $\dfrac{3}{4}$

Amount of B = $\dfrac{3}{9} \times \dfrac{3}{4} = \dfrac{9}{36} = \dfrac{1}{4}$

Amount of Liquid C:

Comes only from the third mixture

Third mixture contribution = $\dfrac{2}{9}$ of total

C content in third mixture = $\dfrac{4}{5}$

Amount of C = $\dfrac{2}{9} \times \dfrac{4}{5} = \dfrac{8}{45}$

---

Water comes from all three mixtures:

Water from first mixture = $\dfrac{4}{9} \times \dfrac{1}{3} = \dfrac{4}{27}$

Water from second mixture = $\dfrac{3}{9} \times \dfrac{1}{4} = \dfrac{3}{36} = \dfrac{1}{12}$

Water from third mixture = $\dfrac{2}{9} \times \dfrac{1}{5} = \dfrac{2}{45}$

Total water = $\dfrac{4}{27} + \dfrac{1}{12} + \dfrac{2}{45}$

To add these fractions, we need a common denominator. The LCM of 27, 12, and 45 is 540.

$\dfrac{4}{27} = \dfrac{80}{540}$

$\dfrac{1}{12} = \dfrac{45}{540}$

$\dfrac{2}{45} = \dfrac{24}{540}$

Total water = $\dfrac{80}{540} + \dfrac{45}{540} + \dfrac{24}{540} = \dfrac{149}{540}$

---

Let's convert everything to the same denominator (540):

Liquid A = $\dfrac{8}{27} = \dfrac{160}{540}$

Water = $\dfrac{149}{540}$

Liquid B = $\dfrac{1}{4} = \dfrac{135}{540}$

Liquid C = $\dfrac{8}{45} = \dfrac{96}{540}$

Order from highest to lowest: A > Water > B > C

---

In mixture problems, always:

Find the fraction of each component in individual mixtures

Calculate the contribution of each mixture to the final blend

Multiply to get the final amount of each component

Convert to common denominators for easy comparison

The resulting mixture has more liquid A than any other component, followed by water, then liquid B, then liquid C.