Mixing two containers of each water and ethanol to varying degrees of purity
$$\begin{array}{c|c|c}
\ &amp; \text{water} &amp; \text{ethanol}\\ \hline
\text{Volume in concern, in L} &amp; 10.00 &amp; 5.00 \\ \hline
\text{Density in kg/L} &amp; 1.000 &amp; 0.789 \\ \hline
\text{Molar mass in g/mol} &amp; 18.015 &amp; 46.069\\ \hline
\text{mass in kg} &amp; 10.000 &amp; 3.945\\ \hline
\text{moles} &amp; 555 &amp; 85.63\\ \hline
\text{mole fraction} &amp; \frac{555}{555+85.63} = 0.866 &amp; \frac{85.6}{555+85.63} = 0.134
\end{array}$$

From the given graph, it looks like we have an excess volume of $\pu{-0.75 mL/mol}$ (at $\chi_\text{EtOH}=0.134$). Since we have total amount of water and ethanol, $555+85.63=\pu{640.63 mol}$, we can calculate the excess volume of the container:

$$\pu{-0.75 mL/mol} \times \pu{640.63 mol} \times \pu{1 \times 10^{-3} L/mL}= \pu{-0.480 L}$$
Thus the actual volume is not $\pu{15 L}$, but $\pu{(10.00+5.00-0.480) L}=\pu{14.52 L}$

Suppose, we transfer $V_1$ of water/ethanol mixture back to the container-2 containing the remaining of $\pu{5.00 L}$ of ethanol. That $V_1$ contains:
$$\frac{\pu{555 mol}\text{ water}}{\pu{14.52 L}}\times V_1 \ \pu{L} = 38.22V_1 \ \pu{mol}\text{ water}$$ 
and,
$$\frac{\pu{85.63 mol}\text{ EtOH}}{\pu{14.52 L}}\times V_1 \ \pu{L} = 5.90V_1 \ \pu{mol}\text{ EtOH}$$

Thus container-2 has $38.22V_1 \ \pu{mol}$ of water and $\left(5.90V_1 + \pu{85.63 mol}\right)$ of ethanol after the addition of $V_1 \ \pu{L}$ of water/ethanol mixture from container-1.

Suppose we have made both containers have the same purity (by $\%(w/w)$) by this action. Let's calculate the masses of water and ethanol ($m_{w2}$ and $m_{et2}$, respectively) in container-2:

$$m_{w2} = 38.22V_1 \ \pu{mol}\text{ water}\times \frac{\pu{18.015 g}\text{ water}}{\pu{1.0 mol}\text{ water}}= 688.5 V_1 \ \pu{g}\text{ water}$$

Similarly,

$$m_{et2} = \left(5.90V_1 + 85.63\right) \ \pu{mol}\text{ EtOH}\times \frac{\pu{46.069 g}\text{ EtOH}}{\pu{1.0 mol}\text{ EtOH}}= \left(271.8V_1 + 3945\right) \ \pu{g} \text{EtOH}$$

Now, we'll calculate the masses of water and ethanol ($m_{w1}$ and $m_{et1}$, respectively) in container-1. Amounts of water and ethanol in this container is:

$$\frac{\pu{555 mol}\text{ water}}{\pu{14.52 L}}\times (14.52 - V_1) \ \pu{L} = \left(555 - \frac{555V_1}{14.52}\right) \ \pu{mol}\text{ water}$$ 
and,
$$\frac{\pu{85.63 mol}\text{ EtOH}}{\pu{14.52 L}}\times (14.52 - V_1) \ \pu{L} = \left(85.63 - \frac{85.63V_1}{14.52}\right)  \ \pu{mol}\text{ EtOH}$$

Thus,

$$m_{w1} = \left(555 - \frac{555V_1}{14.52}\right) \ \pu{mol}\text{ water}\times \frac{\pu{18.015 g}\text{ water}}{\pu{1.0 mol}\text{ water}}= \left(555 - \frac{555V_1}{14.52}\right)\times   \pu{18.015 g}\text{ water}$$

Similarly,

$$m_{et1} = \left(85.63 - \frac{85.63V_1}{14.52}\right)  \ \pu{mol}\text{ EtOH}\times \frac{\pu{46.069 g}\text{ EtOH}}{\pu{1.0 mol}\text{ EtOH}}= \left(85.63 - \frac{85.63V_1}{14.52}\right)\times   \pu{46.069 g}\text{ EtOH}$$

Theoretically, the condition for same purity by $w/w$ is:

$$\frac{m_{w1}}{m_{w1}+m_{et1}} = \frac{m_{w2}}{m_{w2}+m_{et2}} $$

or

$$\frac{m_{et1}}{m_{w1}+m_{et1}} = \frac{m_{et2}}{m_{w2}+m_{et2}} $$