There are three kinds of liquids X, Y , Z,. Three jars J1, J2, J3 contain 100 ml of liquids X, Y , Z, respectively. By an operation we mean... Problem: 
There are three kinds of liquids X, Y , Z,. Three jars J1, J2, J3 contain 100 ml of liquids X, Y , Z, respectively. By an operation we mean three steps in the following order:   


*

*stir the liquid in J1 and transfer 10 ml from J1 into J2 ;   

*stir the liquid in J2 and transfer 10 ml from J2 into J3 ;   

*stir the liquid in J3 and transfer 10 ml from J3 into J1 ;    


After performing the operation four times, let x, y, z be the amounts of X, Y , Z, respectively, in J1. Then
(A) x > y > z
(B) x > z > y
(C) y > x > z
(D) z > x > y 
My approach:
I can't think of any way except performing all the operations manually and computing the result each time. Is there an easier and shorter way to find the answer to this question?
 A: Let us call $x_0$ the amount of X in J1 at the beginning and $x_n$  the amount of X in J1 after the $n^{th}$ operation. If we analyze the changes from $x_0$ to $x_1$ in the first operation we can note that the amount of X in J1:


*

*is reduced by a factor $9/10\,$ in the first step; 

*is not directly affected by the second step;

*increases by a small quantity given by $ \frac{x_0}{10 \cdot 11 \cdot 11}\,\,\,\,$ in the third step (this is because, in the first three steps, $1/10\,$ of $x_0$ is moved to J2, then $1/11\,$ of this is moved to J3, and finally $1/11\,$ of the remaining quantity is moved back to J1). 


So, after the first operation, we have 
$$x_1=\frac{9}{10} x_0 + \frac{x_0}{1210}$$
Similar considerations can be made for the successive operations, where the only difference is that some additional amount of X arrives to J1 in the third step (although it is not necessary to exactly determine this to answer the problem, we can note that, for the $n^{th}$ operation, this additional quantity $K_n$ is $1/11\,$ of the amount of X in J3 at the beginning of the operation). This leads to
$$x_n=\frac{9}{10} x_{n-1} + \frac{x_{n-1}}{1210} + K_n$$
Since this implies $x_n>\frac{9}{10} x_{n-1}\,\,$, we get that after $4$ operations
$$x_4>\left(\frac{9}{10} \right)^4 x_{0}\,\, \approx 0.656 \, x_{0}$$
and then
$$ x_4>65.6\, \text{ml}$$
This shows that, after $4$ operations, X still accounts for the largest proportion of the liquid in J1, i.e. $x>y\,$ and $x>z\,$. 
Note that a similar procedure can be used to show that, after $4$ operations, Y still accounts for the largest proportion of the liquid in J2 and Z still accounts for the largest proportion of the liquid in J3. In fact, the amount of Y in J2 during any operation does not change or increases in the first step (10 ml of some mixture moved from J1 to J2) and is reduced by a factor $10/11\,$ in the second step. In the same manner, the amount of Z in J3 during any operation does not change or increases in the second step (10 ml of some mixture moved from J2 to J3) and is reduced by a factor $10/11\,$ in the third step. This  means that, for each operation, the final amount of Y and Z in the second and third jar, respectively, is larger than $10/11$ of the initial one. So, after $4$ operations, the amount of Y and Z in the second and third jar, respectively, is $>(10/11)^4 \cdot 100\, \text{ml}\,\,\,\,\,$, that is to say $>68.3\, \text{ml}\,\,\,$. 
Taking into account these considerations, to determine which is the largest between $y$ and $z\,$, it is sufficient to note that, in each of the first $4$ operations, the third step moves from J3 to J1 $10$ ml of mixture that is necessarily composed by a predominant proportion of Z (i.e., it moves to J1 a larger amount of Z than Y). Also, the first step of each operation removes from J1 an equal proportion  ($1/10 \,$) of all liquids. Since J1 starts with a zero amount of Y and Z, the final result is that $z>y\,\,$. Therefore, we conclude that the correct answer is B, i.e. 
$$x>z>y$$
I tested this result by a brief algorithm written in QB and got the following values after $4$ operations: 
$$x\approx 66.834894... \text{ml}$$
$$y\approx 6.159210...\text{ ml}$$
$$z\approx 27.005895... \text{ml}$$
