A container contains some amount of milk. A milkman adds $200$ ml of water for each one litre of milk in the container. $6$ litres of the mixture is sold from the container and $10$ litres of milk is added to the remaining mixture. If now the ratio of milk to water in container is $25 : 3$ , find the initial quantity of milk in the container

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Solution

Let the initial quantity of milk $= 10 x$ litres,
For each $1$ litre, $200$ ml of water is added, so after adding water, quantity of mixture become $= 12 x$ litres.
Now $6$ litres of mixture is sold, and $10$ litres of milk is added.
So remaining quantity is $(12 x - 6 + 10) = (12 x + 4)$
In this final quantity, milk $= 10 x - (\frac{10 x}{12 x \times 6}) + 10 = (10 x + 5)$
So, $\frac{(10 x + 5)}{(12 x + 4)} = \frac{25}{(25 + 3)}$ Solving, we get $x = 2$
So initial quantity of milk $= 10 x = 20$ litres

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