Determining change in entropy and enthalpy during mixing of gases Question taken from KVPY 2011. Answer key is A.

Logically it makes sense that $\Delta S_{\text{system}}&gt;0$, because the process is spontaneous and as per second law, this means $\Delta S_{\text{universe}}&gt;0)$. System is isolated ($\Delta S_{\text{surrounding}}=0$), hence system's entropy must be increasing.

However, it is also given that $$\Delta S_{\text{system}}=\frac{\Delta q}T$$ So does this mean that heat is transferred in some way? How is that possible for an isolated system?

Also, if there is no heat transfer, we can conclude that $\Delta H = 0$. But is that true?
Your equation for $\Delta S_{system}$ is incorrect.  The correct equation is $\Delta S_{system}=\int{\frac{dq_{rev}}{T}}$.  I hope you understand the difference, and the fact that the subscript "rev" refers exclusively to a reversible path between the initial and final states of the system.  To get that, you need to forget entirely about the actual irreversible path, and instead focus on the two end states.  You then have to devise a reversible path between these same two states so that you can determine the value of the integral.  For your problem, this recipe will yield a positive value for $\Delta S_{system}.$

Regarding the enthalpy change, you are aware that the enthalpy of mixing of ideal gases is zero, correct?