A train travels 360 km at a uniform speed.

Solution:

Let the speed of the train be x km/hr.

We are given that 360 km distance is to be travelled at uniform speed of x km/hr.

Time taken to cover the distance of

$360 \mathrm{~km}=\frac{\mathbf{3 6 0}}{\mathbf{x}}$ hours.

In case, the speed is increased by $5 \mathrm{~km} / \mathrm{hr}$, the time required to cover $360 \mathrm{~km}=\frac{\mathbf{3 6 0}}{(\mathbf{x}+\mathbf{5})}$ hours.

Now, $\frac{\mathbf{3 C D}}{\mathbf{x}}-\frac{\mathbf{3 C D}}{\mathbf{x}+\mathbf{5}}=\mathbf{1}$

$\Rightarrow \mathbf{3 6 0} \times\left\{\frac{1}{\mathbf{x}}-\frac{1}{\mathbf{x}+\mathbf{5}}\right\}=1$

$\Rightarrow \frac{\mathbf{3 6 0} \times \mathbf{5}}{\mathbf{x} \times(\mathbf{x}+\mathbf{5})}=\mathbf{1}$

$\Rightarrow x \times(x+5)=360 \times 5$

$\Rightarrow x^{2}+5 x=1800 \Rightarrow x^{2}+5 x-1800=0$

$a=1, b=5, c=-1800$

$D=b^{2}-4 a c=(5)^{2}-4 \times 1 \times(-1800)$

$=25+7200=7225$

$\Rightarrow \sqrt{\mathbf{D}}=\sqrt{7225}=85$

Then $x=\frac{-\mathbf{b} \pm \sqrt{\mathbf{D}}}{\mathbf{2 a}}=\frac{-\mathbf{5} \pm \mathbf{8 5}}{\mathbf{2}}$

$\Rightarrow x=\frac{-\mathbf{5} \pm \mathbf{8 5}}{\mathbf{2}}$ or $\frac{-\mathbf{5}-\mathbf{8 5}}{\mathbf{2}}$

$\Rightarrow x=40$ or $-45$ We reject $x=-45$

$\Rightarrow x=40$