Mixture and Alligation::: > A solution contains alcohol and water in the ratio 3:1, 16 litres of the Mixture is drawn off and 12 litres of water is added.11 litres of Mixture is replaced by 11 litres of water.The alcohol and water ratio is 9:13.Find the Intial amount of solution? **How to solve this sum and please guide the steps? I am unable to proceed sum further?**

Say initial solution is $4x \, \mathrm{L}$.

Then quantity of alcohol $=3x \, \mathrm{L}$
And quantity of water $=x \, \mathrm{L}$.

Now out of the $16 \, \mathrm{L}$ of mixture taken away, alcohol $=12 \, \mathrm{L}$ and water $=4 \, \mathrm{L}$. And again $12 \, \mathrm{L}$ of water is added.

So, now quantity of alcohol $=(3x-12) \, \mathrm{L}$
And quantity of water $=(x-4+12) \, \mathrm{L}$.

Again, $11 \, \mathrm{L}$ of mixture is replaced by $11 \, \mathrm{L}$ of water.

So, now quantity of alcohol $=(3x-12-\frac{33}{4}) \, \mathrm{L}$
And quantity of water $=(x-4+12-\frac{11}{4}+11) \, \mathrm{L}$.

By the problem, $$\frac{3x-12-\frac{33}{4}}{x-4+12-\frac{11}{4}+11}=\frac{9}{13}$$

Solve for $x$ and your answer will be $4x$.

Hope this helps you.