A bottle of milk contains $$\frac{3}{4}$$​ of milk and the rest is water. How much of the mixture must be taken away and substituted by equab quantity of water so as to have half milk and half water?

Correct option is A

Given:

1. A bottle contains $$\frac{3}{4}$$​ milk and the rest $$\frac{1}{4}$$​ is water.
2. We need to determine how much of the original mixture must be removed and replaced with water to make the mixture half milk and half water.

Formula Used:

The key principle is that the ratio of milk to water changes proportionally when some of the mixture is removed and replaced with water.

Let:
- x : Fraction of the original mixture to be removed and replaced with water.
- Original amount of milk: $$\frac{3}{4}$$​ .
- Original amount of water: $$\frac{1}{4}$$​ .

When x of the mixture is removed:
- Milk removed: x \times $$\frac{3}{4}$$​ 
- Water removed: x \times $$\frac{1}{4}$$​ 

Substituting water back into the mixture:
- New milk content: $$\frac{3}{4} - x \times \frac{3}{4}$$​ ,
- New water content: $$\frac{1}{4} - x \times \frac{1}{4} + x$$​​

At equilibrium (half milk and half water):
New milk content = New water content

Solution:

1. New milk content:
Milk content = $$\frac{3}{4} - x \times \frac{3}{4} = \frac{3}{4}(1 - x)$$​

2. New water content:
Water content = $$\frac{1}{4} - x \times \frac{1}{4} + x$$​
Simplify:
Water content = $$\frac{1 - x}{4} + x$$​
Take the LCM:
Water content = $$\frac{1 - x}{4} + \frac{4x}{4} = \frac{1 + 3x}{4}$$​

3. At equilibrium (milk = water):
​$$\frac{3}{4}(1 - x) = \frac{1 + 3x}{4}$$​

4. Solve for x :
Multiply through by 4 to eliminate the denominator:
3(1 - x) = 1 + 3x
Expand:
3 - 3x = 1 + 3x
Combine like terms:
3 - 1 = 3x + 3x
2 = 6x
​$$x = \frac{2}{6} = \frac{1}{3}$$​

Final Answer:

The fraction of the original mixture to be removed and replaced is $$\frac{1}{3}$$​ 

Correct Option: (A) $$\frac{1}{3}$$ of original mixture