USSR Exam problem I obtained this problem from here.

A car starts from point $A$ towards $B$ at the same time as a motorcycle starts from $B$ to $A$ (but with a lesser speed). At the moment they meet, a second motorcycle starts from $B$ and also meets the car. The distance between the two meeting places is $\frac{2}{9}$ the distance between $A$ and $B$. Had the car's speed been less by $20$kmh, the distance between the two points would have been $72$km and the first encounter would have taken place $3$ hours after the car started from $A$. Assuming the two motorcycles have the same speed, find the distance between $A$ and $B$.

I started off using this method.  First I know that the car and the first motorcycle will travel a distance $i$ in $3$ hours.  Let $v_m$ denote the speed of the motorcycle and $v_c$ denote the velocity of the car.  Then, $i=(v_c-20)(3)$ and $i=(v_m)(3)$.  We know that the distance between the two meeting points is $72$ and after the first meeting, we have two more equations:  $i+72=(v_c-20)(3+t)$ and $d-i-72=(v_m)(3+t)$ where $d$ is the distance between $A$ and $B$.
I'm wondering if I'm even on the right track with this problem.  Or, is the answer as obvious as noting that $72$ is $2/9$ of the distance.

$S$ - path length
$v_m$ - motorcycle speed
$v_c$ - car speed (in the 1st case)   
Case 1: Car speed is $v_c$
$t_1$ - time to 1st meeting
$t_2$ - time between 1st and 2nd meeting    
Case 2: Car speed is $v_c-20$
$t_3$ - time to 1st meeting
$t_4$ - time between 1st and 2nd meeting      
Just express all statements with equations
(and verify their logic 3-4 times):   
$$v_c.t_1 + v_m.t_1 = S$$
$$v_c.t_2 + v_m.t_2 = S - v_c.t_1$$
$$v_c.t_2 = \frac{2}{9} S$$
$$(v_c-20).t_3 + v_m.t_3 = S$$
$$(v_c-20).t_4 + v_m.t_4 = S - (v_c-20).t_3$$
$$(v_c-20).t_4 = 72$$
$$t_3 = 3$$
Now we have 7 equations and 7 variables, right?
So the rest should be more or less routine.   
EDIT:
The rest is not routine at all. One may easily get swamped in calculations (if one is solving it manually, as I did) and thus lose too much time to say the least.   
OK, I got 4 different solutions/paths of which 3 turned out impossible/invalid due to facts are which are implicitly implied or explicitly given in the problem statement (the statement implicitly implies that $v_c-20 \geq0$, and also it says explicitly "the motorcycle has lesser speed").
The only valid solution turns out to be the following:
$v_c = 80, v_m=40, t_4 = 6/5, t_3 = 3, S = 300$.      
Btw, this problem doesn't seem too recreational :) which is expected taking into account its source. Based on my modest experience the idea behind such problems is that all but the best candidates would get swamped in calculations, and would waste too much time on this problem, and thus will be left with too little time for the other problems. So the idea is to test the candidates' math techniques too (any 6th grader can write the above equations but the rest is not so easy at all). But on the other hand, from all 5 problems this one seems the easiest maybe :) It's not an easy exam to say the least.  
