Question

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

• A. The length of the journey is $400km$.
• B. The length of the journey is $650km$.
• C. The length of the journey is $500km$.
• D. The length of the journey is $720km$.

Answer 1

Answer: (D)

The length of the journey is $720km$.

Let the actual speed of the train be $xkm/hr$ and the actual time taken by train is $y$ hours. Then,
Distance covered $=\left(xy\right)km...\left(i\right)$
[$\because$ Distance $=$ Speed $\times$ Time]
If the speed is increased by $6km/hr$, 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 $\left(i\right)$]
$\Rightarrow$ $-4x+6y-24=0$
$\Rightarrow$ $-2x+3y-12=0...\left(ii\right)$
When the speed is reduced by $6km/hr$, 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 $\left(i\right)$]
$\Rightarrow$ $6x-6y-36=0$
$\Rightarrow$ $x-y-6=\mathrm{0......}\left(iii\right)$
Thus, we obtain the following system of equations:
$-2x+3y-12=0$
$x-y-6=0$
By using cross-multiplication method, 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 $\left(i\right)$, we obtain
Distance $=(30\times 24)km=720km$
Hence, the length of the journey is $720km$.