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

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}$31​×100=3100​ units
• Water in this portion = $\frac{1}{3} \times 50 = \frac{50}{3}$31​×50=350​ units
• Pure milk in this portion = $\frac{1}{3} \times 50 = \frac{50}{3}$31​×50=350​ 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}$3100​ units.

Composition of the new mixture:

• Total water = $\frac{50}{3}$350​ units (from the initial portion)
• Total pure milk = $\frac{50}{3}$350​ (from the initial portion) + $\frac{100}{3}$3100​ (added pure milk) = $\frac{150}{3} = 50$3150​=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}$350​+50=350​+3150​=3200​ units.

Percentage of water in the new mixture:

$$\text{Percentage of water} = \frac{\text{Total water}}{\text{Total volume of new mixture}} \times 100$$Percentage of water=Total volume of new mixtureTotal water​×100 $$= \frac{\frac{50}{3}}{\frac{200}{3}} \times 100$$=3200​350​​×100 $$= \frac{50}{200} \times 100$$=20050​×100 $$= \frac{1}{4} \times 100 = 25\%$$=41​×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%.