A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

Given: Distance traveled by the train with average speed $=$54 km and distance traveled by the train with 6 km/h more speed than the average$=$63 km/h. Time taken to complete the journey$=$3 hours.

To do: To find the first speed of the train.

Solution: Let us consider s to be the first speed of the train,

As known $time=\frac{distance}{speed}$

Time taken to travel 54 km with the first speed, $t_{1}=\frac{distance}{speed}=\frac{54}{s}$                         .................i

Similarly time taken to travel next 63 km with 6km/h more speed than the first $t_{1}=\frac{distance}{speed}=\frac{63}{s+6}$     ................ii

As given, time taken to travel the total journey=$t_{1}+t_{2}$

$\Rightarrow$$\frac{54}{s}+\frac{63}{s+6}=3$  

$\Rightarrow$$54( s+6) +63s=3x( s+6)$

$\Rightarrow 54s+324+63s=3s^{2} +18s$

$\Rightarrow 3s^{2} -117s+18s-324=0$

$\Rightarrow 3s^{2} -99s-324=0$

$\Rightarrow 3\left( s^{2} -33s-108\right) =0$

$\Rightarrow s^{2} -33s-108=0$

$\Rightarrow s^{2} -36s+3s-108=0$

$\Rightarrow s( s-36) +3( s-36) =0$

$\Rightarrow ( s+3)( s-36) =0$

$\Rightarrow s=-3,\ 36$

$\because $ Speed can't be negative, therefore we reject $s=-3$.

Therefore, speed of the train, $s= 36 \ km/h$.