What will be quantity of wine at the end in the vessel? There are two vessels containing 12 liter of water the other contains 6 liters of wine respectively.One liter of mixture was taken out from each vessel and poured into the other and if this process is to be done four times. Find how much liters of wine will be at the end in second vessel?
 A: Hint:
We know that the two vessels will always have respectively 12L and 6L of mixture. Let $v_i(n)$ the percentage of wine in vessel $i$ after $n$ times.
We have $v_1(0)=0$ and $v_2(0)=1$.
And the to get the percentage of wine we will have after the next time, here is how I'd go (here is for the first vessel, it goes the same for the second):
Since I remove one liter and replace it with one liter from the second vessel, the concentration of wine will be 11/12 what I had before in the first vessel, plus 1/12 what I have in the second one.
This is indeed how concentration works when you mix two liquids that don't react with each other (we suppose they don't of course).
Finally you can come up with a formula:
$$
\begin{cases}
v_1(n+1) = \frac{11}{12}v_1(n)+\frac1{12}v_2(n)\\
v_2(n+1) = \frac56v_2(n)+\frac16v_1(n)\\
\end{cases}
$$
You compute the values one after the other from here, or try and calculate an explicit forumla. This is not so easy, but this answer could set you on the way.

Solution:
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Diagonalizing the matrix describing our sequences as in the linked answer, I get:

Eigenvalues $1$ and $\frac34$, eigenvectors $\begin{bmatrix}1 \\1\end{bmatrix}$ and $\begin{bmatrix}1 \\-2\end{bmatrix}$

Thus the following equality:

$$\begin{bmatrix}\frac{11}{12} & \frac1{12}\\\frac16 & \frac56\end{bmatrix} = \frac13 \begin{bmatrix}1 & 1\\1 & -2\end{bmatrix} \begin{bmatrix}1 & 0\\0 & \frac34\end{bmatrix} \begin{bmatrix}2 & 1\\1 & -1\end{bmatrix}$$

Putting that to the $n$th, we finally have:

$$\forall n,\begin{cases}v_1(n) = \frac13 \left(1-\left(\frac34\right)^n\right)\\v_2(n) = \frac13 \left(1+2\left(\frac34\right)^n\right)\end{cases}$$

And for your special case,

$v_1(4) = \frac{175}{768} \approx 22.8\%$ and $v_2(4) = \frac{209}{384} \approx 55.4\%$

This means that the answer to your very problem is :

There are $3.27$ liters of wine ($209/384$ of $6$ liters) contained in the mixture of the second vessel after the fourth trade.

To be sure that your answer is correct, compute the amount of wine in each vessel, and add them: it should always be $6$, since we didn't add or remove wine.
