Bus and truck traveling problem 
The distance between A and B is $330$ km. At 6 am a truck travels from A to B. At a distance of $150$ km from B the truck met a bus that went out from B to A at $7$ am. Find the velocity of the truck if it's smaller by 15 km/h from the velocity of the bus.

This should be so simple but I got stuck and didn't get a correct answer.
So I tried to build a table of velocity, time and distance.
The bus travels one hour less than the truck. In addition, it travels faster. But they travel the same distance at the end.
$$\begin{array}{c|c|c|} 
 & \text{$v$ velocity} & \text{$t$ time} &\text{$d$ distance} \\ \hline
\text{Truck} & x & t & 330\\ \hline
\text{Bus} & x + 15 & t-1 & 330\\ \hline
\end{array}$$
A equation in this case is:
$$vt = d$$
However, if I resolve this two equations with two variables I don't get a reasonable answer:
$$xt = 350$$
$$(x+15)(t-1) = 350$$
Any help...?
 A: "150 km from B the truck meets a bus...". So in your table the distance that the truck traveled in $t$ is $330 - 150 = 180$ km. Similarly the bus managed to travel $150$km in $t-1$ time. This should give you the desired result $x = 60$.
A: Okay, let's redefine the solution.
Considering truck travel with speed $s$ $km/h$(not velocity as for velocity -ve sign has to be taken through which I was not able to find answer).
Therefore bus will travel with Speed $(s+15) km/h$.
As bus and truck met at $150km$ from $B$.
Therefore truck has travel $180km$ and bus has travel $150 km$.
Time taken by truck to reach at $180 km$ point be $t$ hr ,then time taken by bus will be $(t-1)$ hr.
So for, 
truck ,equation will be
$$st = 180     \,\,   \Rightarrow \,\, t = (180/s)$$
And for ,
bus equation will be 
$$(t-1)(s+15) = 150$$
Substituting $t =(180/s)$ in equation of bus we get two values of s i.e. $60$ & $45$ .
That means one will be of bus and other will be of truck.
since speed of truck is less than speed of bus by $15 km/hr$.
speed of truck is $45 km/h$ & bus = $60 km/hr$.
