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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.


Solution

Let the actual speed of the train be $$x$$ km/hr and the actual time taken be $$y$$ hours. Then,
Distance covered $$=(xy)km$$ ..(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,
$$\dfrac{x}{3\times -6-(-1)\times -12}=\dfrac{-y}{-2\times -6-1\times -12}=\dfrac{1}{-2\times -1-1\times 3}$$

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

$$\Rightarrow x=30$$ 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.

Mathematics

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