A train covered a certain distance at a uniform speed. if the train had been 6 km/hr faster, it would have taken 4 hour less than the scheduled time. and, if the train were slower by 6 km/hr, the train would have taken 6 hr more than the scheduled time. the length of the journey is:
Answers
Let the actual speed of train be x km/hr and the actual time taken be y hours.
Then , Distance covered =(xy) km ................ (i){ Distance = speed × time}
If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hrs i.e., when speed is (x+6) km/hr, time of journey is (y₋4)hours.\
Therefore, Distance covered = (x+6)(y-4)
⇒xy=(x+6)(y-4)
⇒ -4x+6y-24=0
⇒ -2x+3y-12=0 .................(ii){by dividing eqn. by 2}
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)
⇒xy=(x-6)(y+6)
⇒6x-6y-36=0
⇒x-y-6=0..................(iii){by dividing equation by 6}
Thus, we obtain the following system of equations:
-2x+3y-12=0 & x-y-6=0
By using cross-multiplication, we have -
x/3×(-6)-(-1)×12 = -y/-2×-6-1×12 = 1/-2×-1-1×3
⇒x/ ₋30 = ₋y/24 = 1/1
⇒x= 30 and y= 24.
Putting values of x and y in equation (i), we obtain,
Distance = (30×24)km=720 km.
Hence, the length of journey is 720 km.
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