Differentiating milk from water

Question

A milk seller has a milk of Rs.100/- per litre. In what ratio should water be mixed in that milk, so that after selling the mixture at Rs.80/- per litre, he may get a profit of 50%? 7:8/7:9/9:7/7:5

My attempt: To get 50% profit, selling price should be Rs.150/- per litre (only milk). But we are selling at Rs.80/- per litre (milk+water). So, maybe water is 70% (150-80), and milk is 80%. Even answer is 7:8. But I am not feeling confident of this solution. It's appearing fuzzy. (and even non-sensical)

Answer 1

The seller needs to turn every litre of pure milk into $\frac{150}{80}$ litres of diluted milk.

A litre bought for $100$ RS would give back $\frac{150}{80}\cdot80=150$ RS, hence $50\%$ profit.

Ratio of $7:8$ means that you take $1$ litre of pure milk, and add $\frac78$ litres of water.

The result is $1+\frac78=\frac{15}{8}$ litres of diluted milk.

Answer 2

Let's say he mixes milk to water in the ratio $x:(1-x)$. For $1L$ of mixture, his cost would be $Rs 100x$ and his selling price would be $Rs 80$. This would give a profit of $Rs (80 - 100x)$. His percentage profit would be $\frac{(80 - 100x)}{100x}$. Equating that to $50 \%$ (or half), you get:

$\frac{(80 - 100x)}{100x} = \frac 12$

$160 - 200x = 100x$

$300x = 160$

$x = \frac{160}{300} = \frac{8}{15}$

That's the proportion of the mixture that was milk, so the proportion that was water was $1 - \frac{8}{15} = \frac {7}{15}$ giving a ratio of water to milk of $7:8$