Arithmetic

Mixture & Alligation

Medium

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

Entered answer:

✅ Correct Answer: 30

We need to track what happens to the milk at each step.

Let $x$ = amount of water initially poured into the container (in liters)

The container holds 300 liters total, so initially we have:

Water: $x$ liters

Milk: $(300 - x)$ liters

Instead of tracking both water and milk separately, we'll focus on tracking the milk since that's what determines our final percentage.

Initially, we have $(300 - x)$ liters of pure milk in the container.

When we remove $2x$ liters of the mixture, we're not just removing water - we're removing some of the milk too!

The mixture at this point has:

Total volume: 300 liters

Milk volume: $(300 - x)$ liters

Milk concentration = $\frac{300-x}{300}$

When we remove $2x$ liters of mixture, the amount of milk lost is:

Milk lost = Volume removed × Milk concentration

Milk lost = $2x \times \frac{300-x}{300} = \frac{2x(300-x)}{300}$ liters

After both operations, total milk remaining:

$300 - x - \frac{2x(300-x)}{300}$

The final solution is 72% milk, which means 216 liters of milk:

$300 - x - \frac{2x(300-x)}{300} = 216$

Rearrange the equation:

$300 - x - \frac{2x(300-x)}{300} = 216$

$-x - \frac{2x(300-x)}{300} = -84$

$-300x - 2x(300-x) = -25200$

$-300x - 600x + 2x^2 = -25200$

$2x^2 - 900x = -25200$

$2x^2 - 900x + 25200 = 0$

$x^2 - 450x + 12600 = 0$

In this case, it is actually easier to find $x$ using the quadratic formula & using the calculator instead of trying to figure out the factorisation.

For $ax^2 + bx + c = 0$ , the solution is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Here: $a = 1$ , $b = -450$ , $c = 12600$

$x = \frac{450 \pm \sqrt{(-450)^2 - 4(1)(12600)}}{2(1)}$

$x = \frac{450 \pm \sqrt{202500 - 50400}}{2}$

$x = \frac{450 \pm \sqrt{152100}}{2}$

$x = \frac{450 \pm 390}{2}$

This gives us: $x = \frac{450 + 390}{2} = 420$ or $x = \frac{450 - 390}{2} = 30$

Since we can't pour 420 liters into a 300-liter container, $x = 30$ liters.

Therefore, the amount of water initially poured was 30 liters.

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