A train, travelling at a uniform speed for 360 km,

Let the original speed of the train = x km/h

Then, the increased speed of the train = (x + 5) km/h                   [by given condition]

and distance = 360 km

According to the question,

$\frac{360}{x}-\frac{360}{x+5}=\frac{4}{5}$ $\left[\begin{array}{l}\because \text { time }=\frac{\text { Distance }}{\text { Speed }} \\ \text { and } 48 \mathrm{~min}=\frac{48}{60} \mathrm{~h}=\frac{4}{5} \mathrm{~h}\end{array}\right]$

$\Rightarrow \quad \frac{360(x+5)-360 x}{x(x+5)}=\frac{4}{5}$ $\left[\because 48 \mathrm{~min}=\frac{48}{60} \mathrm{~h}=\frac{4}{5} \mathrm{~h}\right]$

$\Rightarrow \quad \frac{360 x+1800-360 x}{x^{2}+5 x}=\frac{4}{5}$

$\Rightarrow$ $\frac{1800}{x^{2}+5 x}=\frac{4}{5}$

$\Rightarrow$ $x^{2}+5 x=\frac{1800 \times 5}{4}=2250$

$\Rightarrow$ $x^{2}+5 x-2250=0$

$x^{2}+(50 x-45 x)-2250=0$

$\Rightarrow \quad x^{2}+50 x-45 x-2250=0 \quad$ [by factorisation method]

$\Rightarrow \quad x(x+50)-45(x+50)=0$

$\Rightarrow \quad(x+50)(x-45)=0$

Now, x+50=0⇒x=-50

which is not possible because speed cannot be negative and x – 45 = 0⇒x = 45. Hence, the original speed of the train = 45 km/h