Initial water content in mixture replacement

CAT 2024 Slot 3 · QA · Medium · Arithmetic

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A certain amount of water was poured into a 300 litre container and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was

Answer

30

Explanation

Let initial water = ww litres. Then initial milk = (300−w)(300 - w) litres.

Volume of solution removed = 2w2w litres. Fraction of solution remaining = 1−2w300=300−2w3001 - \frac{2w}{300} = \frac{300 - 2w}{300}.

Since no milk is added when refilling with water, the final volume of milk is: Final Milk=(300−w)×300−2w300\text{Final Milk} = (300 - w) \times \frac{300 - 2w}{300}

We are given final milk content is 72%72\% of 300=216300 = 216 litres: (300−w)(300−2w)300=216\frac{(300 - w)(300 - 2w)}{300} = 216 (300−w)(300−2w)=64800(300 - w)(300 - 2w) = 64800 90000−900w+2w2=6480090000 - 900w + 2w^2 = 64800 2w2−900w+25200=0  ⟹  w2−450w+12600=02w^2 - 900w + 25200 = 0 \implies w^2 - 450w + 12600 = 0 (w−30)(w−420)=0(w - 30)(w - 420) = 0

Since capacity is 300300 L, w=30w = 30 litres.

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