Find the speed of the train given the following conditions? A train crosses 2 bridges $430$m and $550$m long in $30$ sec and $36$ sec.Find the speed of the train?
options:
a)$36$kmph  b)$72$kmph c)27kmph d)45kmph
My approach:
I did He crossed 430 m bridge first.Therefore,
(430+L)=30 . speed  1eqn
(120 +L)=6 . speed   2eqn
I am confused about how to solve this problem.Can Anyone give me the Hint?
 A: While it may seem that $\frac{430}{30} \neq \frac{550}{36}$, I think by time it takes the train to cross the bridge, they mean the amount of time from the front of the train reaching the start of the bridge, to the end of the train reaching the end of the bridge.  
Let L be the length of the train, s its speed. Then for any bridge of length B, it will take $t=\frac{B+L}{s}$ seconds for the train to cross the bridge. Rewriting the equation gives us $s=\frac{B+L}{t}$ 
Thus, $s=\frac{430+L}{30}=\frac{550+L}{36}$.  We can solve for L first by cross multiplying: $30 \cdot (550+L)=36 \cdot (430+L)$, I got $L=170$.  Now we easily have $s=\frac{430+170}{30}=\frac{550+170}{36}=20$. 
So the train was moving at $s=20$ m/s (by the way, now the letters like "s" refer to units, not the variables above).  The final step is to convert to km/h: $20 \frac{m}{s} \cdot \frac{3600s}{h} \cdot \frac{km}{1000m}=72$ km/h.
A: Regardless how long the train might be, it tooks him 6 seconds more for 120 meters more, so v = 20.
