Correct Option (1)

Let the initial volume of the milk sample be 100 units.

Given that the sample contains 50% water:

Amount of water = 50 units
Amount of pure milk = 50 units

When 1/3rd of this milk sample is taken:

Volume of portion taken = $\frac{1}{3} \times 100 = \frac{100}{3}$ units
Water in this portion = $\frac{1}{3} \times 50 = \frac{50}{3}$ units
Pure milk in this portion = $\frac{1}{3} \times 50 = \frac{50}{3}$ units

To this portion, an equal amount of pure milk is added. The 'equal amount' refers to the volume of the portion taken, which is $\frac{100}{3}$ units.

Composition of the new mixture:

Total water = $\frac{50}{3}$ units (from the initial portion)
Total pure milk = $\frac{50}{3}$ (from the initial portion) + $\frac{100}{3}$ (added pure milk) = $\frac{150}{3} = 50$ units

Total volume of the new mixture = Total water + Total pure milk = $\frac{50}{3} + 50 = \frac{50}{3} + \frac{150}{3} = \frac{200}{3}$ units.

Percentage of water in the new mixture:

$Percentage of water = \frac{Total water}{Total volume of new mixture} \times 100$ $= \frac{\frac{50}{3}}{\frac{200}{3}} \times 100$ $= \frac{50}{200} \times 100$ \[ = \frac{1}{4} \times 100 = 25\% \]

Incorrect Options:

Options 2, 3, and 4 are incorrect. The accurate calculation, based on the given conditions and mixture proportions, demonstrates that the water content in the new mixture is 25%, not 30%, 35%, or 40%.