There is a milk sample with 50% water in it. If 1/3rd of this milk is added to an equal amount of pure milk, then water in the new mixture will fall down to

The correct option is 25%.

Let us solve this problem step-by-step by analyzing the composition of each component in the mixture.

Step 1: Understand the initial milk sample
Suppose we have a sample of milk with a volume of $V$ units.
Since the sample contains 50% water, the remaining 50% must be pure milk.
Therefore, in 1 unit of this initial milk sample, we have:
• Pure milk = 0.5 units
• Water = 0.5 units

Step 2: Determine the composition of the portion added
We take $\frac{1}{3}$rd of this milk sample.
Let the volume of this $\frac{1}{3}$rd portion be $x$ units.
Since the concentration of water remains 50% in any portion of the uniform mixture:
• Water in this portion = $0.5\times x=\frac{x}{2}$ units
• Pure milk in this portion = $0.5\times x=\frac{x}{2}$ units

Step 3: Add an equal amount of pure milk
We add an equal amount of pure milk to this portion, which means we add $x$ units of 100% pure milk.
• Amount of pure milk added = $x$ units
• Amount of water added = $0$ units

Step 4: Calculate the total quantities in the new mixture
• Total volume of the new mixture = $x+x=2x$ units
• Total water in the new mixture = $\frac{x}{2}+0=\frac{x}{2}$ units

Step 5: Calculate the percentage of water in the new mixture
The percentage of water in the new mixture is given by:

$\text{Percentage of water}=\left(\frac{\text{Total Water}}{\text{Total Volume}}\right)\times 100$

$\text{Percentage of water}=\left(\frac{\left(\frac{x}{2}\right)}{2x}\right)\times 100$

$\text{Percentage of water}=\left(\frac{1}{4}\right)\times 100=25\%$

Thus, the percentage of water in the new mixture falls down to 25%.