Four consecutive volumetric transfers

Let the vector $X(k)$ be defined as:

$X(k) = \begin{bmatrix} V_A(k) \\ V_B(k) \\ V_C(k) \\ V_D(k) \end{bmatrix}$

That is, $X(k)$ is the vector of volumes in the four containers after the $k$ -th transfer. Each transfer can be modelled by a matrix $T_k$ multiplying $X(k-1)$ such their product is $X(k )$ , hence,

$X(k) = T_k X(k-1) , \hspace{24pt} k = 1, 2, 3, 4$

It follows by iteration that,

$X(4) = T_4 T_3 T_2 T_1 X(0) = T x(0)$

Note that the entries of $X(0)$ are unknown, only that their sum is $4$ .

Now the matrix $T_1 , T_2, T_3, T_4$ are given by

$T_1 = \begin{bmatrix} 0.8 && 0 && 0 && 0 \\ 0.2 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix}$

$T_2 = \begin{bmatrix} 1 && 0 && 0 && 0 \\ 0 && 0.7 && 0 && 0 \\ 0 && 0.3 && 1 && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix}$

$T_3 = \begin{bmatrix} 1 && 0 && 0 && 0 \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 0.6 && 0 \\ 0 && 0 && 0.4 && 1 \end{bmatrix}$

$T_4 = \begin{bmatrix} 1 && 0 && 0 && 0.5 \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 0.5 \end{bmatrix}$

So that their product $T$ is given by

$T = T_4 T_3 T_2 T_1 = \begin{bmatrix} 0.812&&0.06&&0.2&&0.5\\ 0.14&&0.7&&0&& 0\\ 0.036&& 0.18&& 0.6&& 0\\ 0.012&& 0.06&& 0.2&& 0.5\end{bmatrix}$

Now we have an explicit expression relating $X(4)$ with $X(0)$ . On the other hand, we are given that

$X(4) = G X(0)$

Where $G$ is the following diagonal matrix:

$G = \begin{bmatrix} 1 + R && 0 && 0 && 0 \\ 0&& 1 && 0 &&0 \\ 0 && 0&&1 - R && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix}$

Where $R = \dfrac{r}{100}$ is the fraction corresponding to the percentage $r$ .

Thus we have

$X(4) = T X(0) = G X(0)$

That is,

$(T - G) X(0) = 0$

This equation will have non-trivial (i.e. non-zero) solutions if and only if the matrix $(T - G)$ is singular.

The matrix $T - G$ is given by:

$T - G = \begin{bmatrix} 0.812 - (1 + R) &&0.06&&0.2&&0.5\\ 0.14&&(0.7 - 1)&&0&& 0\\ 0.036&& 0.18&& 0.6 - (1 - R) && 0\\ 0.012&& 0.06&& 0.2&& 0.5 - 1 \end{bmatrix} = \begin{bmatrix} -0.188 - R &&0.06 &&0.2 &&0.5\\ 0.14 &&-0.3 && 0 && 0\\ 0.036 && 0.18 && -0.4 + R && 0\\ 0.012 &&0.06 && 0.2 && -0.5 \end{bmatrix}$

The computation of the determinant of $T - G$ is a tedious job and is best done with the aid of a computer program. The result is

$| T - G | = -0.15 R^2 + 0.042 R$

Setting this to zero, results in $R = 0$ or $R = 0.28$ . Since we're given that $r \gt 0$ , then $R = 0.28$ which means $r = 28$ is the required percentage.

We're not done yet. We need to find $X(0)$ , and this is also a tedious task without the help of a computer program or online app such as the well-known www.wolframalpha.com

Of course, there are infinite solutions, but the normalized one (in the sense that the sum of its entries is $1$ ), is

$X(0) = \begin{bmatrix} 0.339366516 \\ 0.158371041 \\ 0.339366516 \\ 0.162895928 \end{bmatrix}$

Multiplying the first entry (since we're interested in $A$ 's volume ) by $4$ results in $X(0)_1 = 1.357466063$

The corresponding final volume is $(1+\dfrac{28}{100}) (1.357466063) = 1.737556561$

This makes the answer $= 28 + 1.737556561 \approx \boxed{29.737}$

Let $a$ , $b$ , $c$ , and $d$ be the starting amount of water in $A$ , $B$ , $C$ , and $D$ 's containers, respectively.

Since the total volume of water in all the four containers is $4$ liters, $a + b + c + d = 4$ .

After $A$ transfers $20\%$ of his water to $B$ , $A$ has $0.8a$ and $B$ has $b + 0.2a$ .

After $B$ transfers $30\%$ of his water to $C$ , $B$ has $0.7(b + 0.2a)$ and $C$ has $c + 0.3(b + 0.2a)$ .

After $C$ transfers $40\%$ of his water to $D$ , $C$ has $0.6(c + 0.3(b + 0.2a))$ and $D$ has $d + 0.4(c + 0.3(b + 0.2a))$ .

After $D$ transfers $50\%$ of his water to $A$ , $D$ has $0.5(d + 0.4(c + 0.3(b + 0.2a)))$ and $A$ has $0.8a + 0.5(d + 0.4(c + 0.3(b + 0.2a)))$ .

Since $B$ has the same amount that he started with, $b = 0.7(b + 0.2a)$ , which solves to $b = \frac{7}{15}a$ .

Since $D$ has the same amount that he started with, $d = 0.5(d + 0.4(c + 0.3(b + 0.2a)))$ , which after substituting $b = \frac{7}{15}a$ rearranges to $d = 0.4c + 0.08a$ .

Substituting $d = 4 - a - b - c$ and $b = \frac{7}{15}a$ into $d = 0.4c + 0.08a$ gives $c = \frac{20}{7} - \frac{116}{105}a$ .

Since $A$ 's water increased by $r\%$ and $B$ 's water decreased by $r\%$ , $r = \frac{100}{a}(0.8a + 0.5(d + 0.4(c + 0.3(b + 0.2a))) - a) = \frac{100}{c}(c - 0.6(c + 0.3(b + 0.2a)))$ .

Substituting $0.5(d + 0.4(c + 0.3(b + 0.2a))) = d = 0.4c + 0.08a$ , then $b = \frac{7}{15}a$ , and then $c = \frac{20}{7} - \frac{116}{105}a$ into the above equation and then rearranging gives $13039a^2 - 44220a + 36000 = 0$ , which has solutions of $a = \frac{120}{59}$ and $a = \frac{300}{221}$ .

If $a = \frac{120}{59}$ , then $b = \frac{7}{15}a = \frac{56}{59}$ , $c = \frac{20}{7} - \frac{116}{105}a = \frac{36}{59}$ , and $r = \frac{100}{c}(c - 0.6(c + 0.3(b + 0.2a))) = 0$ . However, $r > 0$ , so this cannot be a solution.

If $a = \frac{300}{221}$ , then $b = \frac{7}{15}a = \frac{140}{221}$ , $c = \frac{20}{7} - \frac{116}{105}a = \frac{300}{221}$ , $d = 0.4c + 0.08a = \frac{144}{221}$ , and $r = \frac{100}{c}(c - 0.6(c + 0.3(b + 0.2a))) = 28$ .

Therefore, $V_A = 0.8a + 0.5(d + 0.4(c + 0.3(b + 0.2a))) = \frac{384}{221}$ , so $r + V_A = 28 + \frac{384}{221} = \frac{6572}{221} = \boxed{29.7376}$ .