Question

A milk vendor has $2$ cans of milk. The first contains $25\%$ water and the rest milk. The second contains $50\%$ water. How much milk should he mix from each of the containers so as to get $12\mathrm{litres}$ of milk such that the ratio of water to milk is $3:5$?

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Solution

**Step 1: Determine the mean value.**

It is given that the ratio of water to milk in the first can is $25\%:75\%$i.e. $1:3$ and in the second can is $50\%:50\%$ i.e. $1:1$.

Let the cost of $1\mathrm{litre}$ milk be $\mathrm{Re}.1$.

The amount of milk contained in $1\mathrm{litre}$ mixture from the first can is $\frac{3}{4}\mathrm{litres}$.

Therefore, the cost price of $1\mathrm{litre}$ mixture from the first can will be $\frac{3}{4}\times \mathrm{Re}.1=\mathrm{Rs}.\frac{3}{4}$.

Similarly, the amount of milk contained in $1\mathrm{litre}$ mixture from the second can is $\frac{1}{2}\mathrm{litres}$.

Therefore, the cost pricr of $1\mathrm{litre}$ mixture from the second can will be $\frac{1}{2}\times \mathrm{Re}.1=\mathrm{Rs}.\frac{1}{2}$.

The required ratio of water to milk in $12\mathrm{litres}$ of mixture from both the cans is $3:5$.

Hence, the amount of milk contained in $1\mathrm{litre}$ of the required mixture will be $\frac{5}{8}\mathrm{litres}$.

Therefore, the mean price will be $\frac{5}{8}\times \mathrm{Re}.1=\mathrm{Rs}.\frac{5}{8}$.

**Step 2: Apply the rule of alligation to find the required ratio.**

According to the rule of alligation, when different quantities of different ingredients are mixed together to produce a mixture of a mean value, the ratio of their quantities is inversely proportional to the difference in their cost from the mean value.

$\begin{array}{rcl}\therefore \frac{x}{y}& =& \frac{{\displaystyle \frac{3}{4}}-{\displaystyle \frac{5}{8}}}{{\displaystyle \frac{5}{8}}-{\displaystyle \frac{1}{2}}}\\ & =& \frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}\\ & =& \frac{1}{1}\end{array}$

Therefore, the required ratio of the quantities of the mixtures from the two cans is $1:1$.

Hence, the quantity of mixture taken from each can should be $\frac{1}{2}\times 12\mathrm{litres}=6\mathrm{litres}$.

**Therefore, to get **$12\mathrm{litres}$** of milk such that the ratio of water to milk is **$3:5$**, the milk vendor should mix **$6\mathrm{litres}$** of milk from each can.**

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