Hello,
         
        Here is the question.

We can treat noble gasses as ideal gasses, which means we can use the formula pV=nRT, which is independent of the actual type of gas.

Let the first container have n₁ moles, pressure p₁, and volume V₁.

Let the second container have n₂, p₂, and V₂.

Let the temperature be T, and let the final pressure be p.

Then we have:

p₁V₁=n₁RT

p₂V₂=n₂RT,

p(V₁+V₂)=(n₁+n₂)RT

Since we do not know the volumes, we will eliminate them from the equations:

V₁=n₁RT/p1

V₂=n₂RT/p2

p(V₁+V₂)=(n₁+n₂)RT

⟹p=(n₁+n₂)RT/(V₁+V₂)=(n₁+n₂)RT/(n₁RT/p₁+n₂RT/p₂)

=n₁+n₂/(n₁/p₁+n₂p₂)

=10.0+5.00/(10.0/5.00+5.00/20.0)=6.67bar

The final pressure should be p=6.67bar.

Your method is right -If I use PV =  nRT
However, we don't require that big formula to solve this question.
The volume of helium gas = 10 RT/5 and volume of neon gas = 5RT/20
The total volume of the container after mixing of the gases is 10 RT/ 5 +5 RT/20.=45RT/20

P(total) =15RT/(45 RT/20) = 6.67bar