IAS 2017 — CSAT Maths

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%.