A motorboat going downstream overcame a raft at a point A (Kinematics question) 
A motorboat going downstream overcame a raft at a point A. $T$ = $60$ min later it turned back and after some time passed the raft at a distance $l$ = $6$ km from the point $A$. Find the flow velocity.
  Assuming duty of motorboat is constant.

This is a question from IE irodov (mechanics).
Now i tried doing this problem without using relative motion/ velocity:

Let $$\begin{align}v_r &= \text{ velocity of raft}\\
v_f &= \text{ velocity of flow}\\
v_b &= \text{ velocity of boat}\end{align}$$
Let raft reach $C$ when boat reaches $B$.
$$\therefore AB=(v_b+v_f)60\\
AC = (v_r+v_f)60\\
\implies CB = 60(v_b - v_r)$$
Now boat turns back. Let boat and raft meet at $D$ after time $t$ after the bot reaches B. Therefore:
$$(v_r + v_f)t + (v_b - v_f)t = 60(v_b - v_r)\\
\implies t = \dfrac{60(v_b - v_r)}{v_r + v_b}$$
Now, $AC + CD = 6$
$$(v_r + v_f)60 + (v_r + v_f)60\left(\dfrac{v_b - v_r}{v_b + v_r}\right) = 6$$
Simplifying for $v_f$, we get:
$$v_f = \dfrac{v_b +v_r - 10 v_b^2}{10v_b} {km\over min}$$
I cant simplify further. The answer given is 3km/hr
EDIT
I just realised that $v_r = 0$. Therefore last equation gives:
$$v_f = \frac{6(1-10v_b) km}{hr}$$
 A: All that is quite unnecessary, really, just change the frame of reference to the river, and visualize it as a moving conveyor belt on which   the raft is merely a marked  point A.
The motorboat moves away from A for for $1$ hr on the "conveyor belt" , so it will take exactly the same time to get back to it.
In the meantime, A has moved downstream $6$ km w.r.t. the bank,
thus flow velocity $= \dfrac62 = 3\;$km/hr.  
A: I have already given a creative solution, but if your faculty is intolerant of unorthodoxy, here is a simplified algebraic one.
I take the velocity of flow as $v$, and that of the boat as $kv$, where $k$ is a multiplier (often a very useful ploy  in simplifying), and count time in hours to further simplify.
Following your diagram,
Going with the flow and using relative velocities,
$AB = (k+1)v, \quad AC = v, \quad CB = kv$
For return meeting at $D$ against the flow, 
$[v+ v(k-1)]t = kv \Rightarrow t = 1\quad$ (in fact we already knew this !)
Finally, as $(1+t)= 2$ hrs have elapsed from start,
$v + v  = 6, \Rightarrow v = 3$
