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 of 6 km/hr more than the first speed.

Solution:

Let the first speed of the train be x km/h.

$\therefore$ Time taken to cover $54 \mathrm{~km}=\frac{54}{x} \mathrm{~h}$          $\left(\right.$ Time $\left.=\frac{\text { Distance }}{\text { Speed }}\right)$

New speed of the train = (x + 6) km/h

$\therefore$ Time taken to cover $63 \mathrm{~km}=\frac{63}{x+6} \mathrm{~h}$

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h

$\therefore \frac{54}{x}+\frac{63}{x+6}=3$

$\Rightarrow \frac{54 x+324+63 x}{x(x+6)}=3$

$\Rightarrow 117 x+324=3\left(x^{2}+6 x\right)$

$\Rightarrow 117 x+324=3 x^{2}+18 x$

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

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

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

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

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

$\Rightarrow x-36=0$ or $x+3=0$

$\Rightarrow x=36$ or $x=-3$

∴ x = 36                  (Speed cannot be negative)

Hence, the first speed of the train is 36 km/h.