# A train covered a certain distance at a uniform speed.

Question:

A train covered a certain distance at a uniform speed. If the train could have been 10 km/hr. faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/hr; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

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$=(x-10)(y+3)x y=(x-10)(y+3)0=-10 y+3 x-303 x-10 y-30=0 \cdots(iii)$Thus, we obtain the following system of equations:$-x+5 y-10=03 x-10 y-30=0$By using cross multiplication, we have$\frac{x}{5 x-30-(-10) \times-10}=\frac{-y}{(-1 \times-30)-(3 x-10)}=\frac{1}{(-1 \times-10)-(3 \times 5)}\frac{x}{-150-100}=\frac{-y}{30+30}=\frac{1}{10-15}\frac{x}{-250}=\frac{-y}{60}=\frac{1}{-5}x=\frac{-250}{-5}x=50y=\frac{-60}{-5}y=12$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}\$