T(n) = T(pn) + T(qn) + f(n)	(*)
	vs.
T(n) = T((p+q)n) + f(n)		(**)

Assume 0 < p,q < 1.  Let's also assume p+q < 1, 
since otherwise only (*) makes sense.

For f(n) = n, both are Theta(n) (same as the exam question).
For f(n) = n^c (c>0), they have the same solution, Theta(n^c), 
	iff (p^c + q^c) < 1:
We have
(*):	T(n) = n^c [ 1 + (p^c+q^c) + (p^c+q^c)^2  + (p^c+q^c)^3 + ... ]
	by examining the recursion tree as in the case on the exam, and
(**):	T(n) = n^c [ 1 + (p+q)^c + (p+q)^2c  + (p+q)^3c + ... ]	
Since p+q < 1 by assumption, so is (p+q)^c and (**) is therefore Theta(n^c).
If (p^c+q^c) < 1 then (*) is also Theta(n^c).

Of course, it's possible that (p^c + q^c) > 1 even though p+q < 1.
So, for example, the following recurrences have different solutions, 
	T(n) = T(2n/3) + T(n/4) + \sqrt{n}	T(n) = Theta(n^1.2)
	T(n) = T(11n/12) + \sqrt{n}		T(n) = Theta(n^0.5)
since \sqrt{2/3} + \sqrt{1/4} > 1.
	[This would be a good final exam question!]

This generalizes to  T(n) = aT(n/b) + n^c  (*) in the natural way:
Again, T(n) = T(an/b) + n^c makes sense only for a/b < 1, in which case
the solution is Theta(n^c).  (*) has the same solution iff a/b^c < 1.