A train covered a certain distance at a uniform speed.

Solution:

Let the actual speed of the train be Xkm\h mathrm{~km} / \mathrm{hr}$ and the actual time taken by y hours. Then,

Dis $\tan c e=$ Speed $\times$ Time

Distance covered $=(x y) \mathrm{km} \cdots(i)$

If the speed is increased by $10 \mathrm{Km} / \mathrm{hr}$, then time of journey is reduced by 2 hours

when speed is $(x+10) \mathrm{km} / h r$, time of journey is $(y-2)$ hours

- Distance covered $=(x+10)(y-2)$

$x y=(x+10)(y-2)$

$x y=x y+10 y-2 x-20$

$-2 x+10 y-20=0$

$-2 x+3 y-12=0$....(iii)

When the speed is reduced by $10 \mathrm{Km} / \mathrm{hr}$, then the time of journey is increased by 3 hours when speed is $(x-10) \mathrm{Km} / \mathrm{hr}$, time of journey is $(y+3)$ hours

$=$ Distance covered $=(x-10)(y+3)$

$x y=(x-10)(y+3)$

$0=-10 y+3 x-30$

$3 x-10 y-30=0 \cdots(iii)$

Thus, we obtain the following system of equations:

$-x+5 y-10=0$

$3 x-10 y-30=0$

By using cross multiplication, we have

$\frac{x}{5 x-30-(-10) \times-10}=\frac{-y}{(-1 \times-30)-(3 x-10)}=\frac{1}{(-1 \times-10)-(3 \times 5)}$

$\frac{x}{-150-100}=\frac{-y}{30+30}=\frac{1}{10-15}$

$\frac{x}{-250}=\frac{-y}{60}=\frac{1}{-5}$

$x=\frac{-250}{-5}$

$x=50$

$y=\frac{-60}{-5}$

$y=12$

Putting the values of x and y in equation (i), we obtain

Distance $=x y \mathrm{~km}$

$=50 \times 12$

$=600 \mathrm{~km}$

Hence, the length of the journey is $600 \mathrm{~km}$