  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 obtainDistance $$=(30\times 24)$$km $$=720$$km.Hence, the length of the journey is $$720$$km.Mathematics

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