What is the final pressure of helium and neon gas mixture in this case?

[WMDhamnekar]

The answer given to the above question is Final pressure$=\frac{(10.0mol\times 5.00bar) + (5.00 mol \times 20.0 bar)}{(10.0 mol +5.00 mol)}= 10.0 bar$ Is this answer correct? if yes, How and why?

My question is while computing this answer, the volumes of each gas is not considered.

Note:- While computing the volumes of each gas, temperature is not considered as it is a constant.The computed volume of helium gas $=\frac{(10.0mol \times 8.314472 J/mol\cdot K)}{5.00 bar}=0.16628944 Litres$

The computed volume of neon gas $= \frac{(5.00 mols \times 8.314472 J/mol\cdot K)}{20.0 bar}= 0.02078618 Litres$

How to use this additional information to compute the final pressure of the gas mixture?

 

[I like Serena]

We can treat noble gasses as ideal gasses, which means we can use the formula $p V = n R T$, which is independent of the actual type of gas.
Let the first container have $n_1$ moles, pressure $p_1$, and volume $V_1$.
Let the second container have $n_2$, $p_2$, and $V_2$.
Let the temperature be $T$, and let the final pressure be $p$.

Then we have:
\begin{cases}p_1V_1 = n_1 R T \\ p_2 V_2 = n_2 R T \\ p(V_1+V_2) = (n_1+n_2) R T \end{cases}
Since we do not know the volumes, we will eliminate them from the equations:
$$\begin{cases}V_1 = \frac{n_1 R T}{p_1} \\ V_2 = \frac{n_2 R T}{p_2} \\ p(V_1+V_2) = (n_1+n_2) R T \end{cases}
\implies p = \frac{(n_1+n_2) R T}{V_1+V_2} = \frac{(n_1+n_2) R T}{\frac{n_1 R T}{p_1}+\frac{n_2 R T}{p_2}}
=\frac{n_1+n_2}{\frac{n_1}{p_1}+\frac{n_2}{p_2}}= \frac{10.0+5.00}{\frac{10.0}{5.00}+\frac{5.00}{20.0}}=6.67\,\text{bar}$$

So it looks as if the given answer is incorrect.
The final pressure should be $p=6.67\,\text{bar}$.