An 80-litre mixture contains milk and water in the ratio 7:1. Some quantity of this mixture is removed and replaced with pure milk. If the final ratio of milk to water becomes 15:1, how many litres of the mixture were replaced?

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Accepted Answer

1. Initial milk = $\frac{7}{8}	imes80=70$ L, water = $10$ L.

2. Let $x$ L be replaced. Milk removed = $\frac{7}{8}x$, water removed = $\frac{1}{8}x$. After refill:

- Milk = $70-\frac{7}{8}x+x=70+\frac{1}{8}x$.
   - Water = $10-\frac{1}{8}x$.

3. Final ratio $\frac{70+\frac{1}{8}x}{10-\frac{1}{8}x}=\frac{15}{1}$. Solve:

$70+\frac{x}{8}=15\left(10-\frac{x}{8}ight)\Rightarrow70+\frac{x}{8}=150-\frac{15x}{8}$

$70+2x=150\Rightarrow2x=80\Rightarrow x=40$ L.

4. So $40$ litres were replaced.

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