T(n) = T(n/4) + T(n/2) + c n^2
I think what U can do is, U can write subproblems of size (n/4) is nearly equivalent (n/2).
Now The Recurrence Relation looks something like this: T(n) <= T(n/2) + T(n/2) + c n^2
T(n) <= 2 T(n/2) + cn^2
Now u can apply masters theorm
Ans will be O(n^2) ie; B