Question

A train covered a certain distance at a uniform speed. If the train would have been $6$ km/h faster, it would have taken $4$ hours less than the scheduled time. And, if the train were slower by $6$ km/h, it would have taken $6$ hours more than the scheduled time. Find the length of the journey.

Answer 1

Let the actual speed of the train be $x$ km/hr and the actual time taken be $y$ hours. Then,
Distance covered $=\left(xy\right)km$ ..(i) $[\therefore$ Distance $=$ Speed $\times$ Time$]$

If the speed is increased by $6$ km/hr, then time of journey is reduced by $4$ hours i.e., when speed is $(x+6)$km/hr, time of journey is $(y-4)$ hours.

$\therefore$ Distance covered $=(x+6)(y-4)$
$\Rightarrow xy=(x+6)(y-4)$ [Using (i)]
$\Rightarrow -4x+6y-24=0$
$\Rightarrow -2x+3y-12=0$ ..(ii)

When the speed is reduced by $6$ km/hr, then the time of journey is increased by $6$ hours i.e., when speed is $(x-6)$ km/hr, time of journey is $(y-6)$ hours.

$\therefore$ Distance covered $=(x-6)(y+6)$
$\Rightarrow xy=(x-6)(y+6)$ [Using (i)]
$\Rightarrow 6x-6y-36=0$
$\Rightarrow x-y-6=0$ (iii)

Thus, we obtain the following system of equations:
$-2x+3y-12=0$
$x-y-6=0$

By using cross-multiplication, we have,
$\frac{x}{3\times -6-(-1)\times -12}=\frac{-y}{-2\times -6-1\times -12}=\frac{1}{-2\times -1-1\times 3}$

$\Rightarrow \frac{x}{-30}=\frac{-y}{24}=\frac{1}{-1}$

$\Rightarrow x=30$ and $y=24$

Putting the values of x and y in equation (i), we obtain
Distance $=(30\times 24)$km $=720$km.

Hence, the length of the journey is $720$km.