CAT 2024 Slot 3 · Quant · Question 10

Let w litres of water be poured in first, so the milk added is 300-w litres.

The amount removed is twice the water poured, that is 2w litres of the mixture.

Removing a fraction of a uniform mixture removes that same fraction of the milk, and the fraction here is \frac{2w}{300}.

So the milk left is \left( 300-w \right)\left( 1-\frac{2w}{300}\right)=\frac{\left( 300-w \right)\left( 300-2w \right)}{300}.

Refilling with water changes no milk, so the final milk is still that amount.

The final solution is 72% milk of 300 litres, which is 0.72\times 300=216 litres.

Setting these equal, \frac{\left( 300-w \right)\left( 300-2w \right)}{300}=216, so \left( 300-w \right)\left( 300-2w \right)=64800.

Expanding, 90000-900w+2{w}^{2}=64800, which gives 2{w}^{2}-900w+25200=0.

Dividing by 2, {w}^{2}-450w+12600=0.

The discriminant is {450}^{2}-4\times 12600=202500-50400=152100, whose square root is 390.

So w=\frac{450\pm 390}{2}, giving w=30 or w=420.

Only w=30 fits a 300 litre container, so 30 litres of water were poured in initially.

Shortcut. Milk falls from \frac{300-w}{300} to 72%, and the drop factor is exactly \left( 1-\frac{2w}{300}\right). Trying the round value w=30 gives \frac{270}{300}\times \frac{240}{300}=0.9\times 0.8=0.72 — a direct hit, with no quadratic.