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


Let the actual speed of the train be x km/hr and the actual time taken be y km/hr.
Then, Distance = Speed × Time taken = (xy) km
If the speed is increased by 6 km /hr, then time of journey is reduced by 4 hours.

When speed is (x + 6) km / hr, time of journey is (y – 4) hours.

Distance = (x + 6) (y – 4)

xy = xy - 4x + 6y – 24

-4x + 6y + 24 = 0

-2x + 3y - 12 = 0 … (i)
When the speed is reduced by 6 km/hr, then the time taken for the journey is increased by 6 hours.
When the speed is (x – 6) km/hr, time of journey is (y + 6) hours
Distance = (x – 6) (y + 6)
xy = (x – 6) (y + 6)
xy = xy + 6x – 6y – 36

6x – 6y – 36 = 0

x – y – 6 = 0… (ii)
Solving (i) and (ii),
Using cross multiplication, we have,

x = 30 and y = 24
Putting the values of x and y in equation (i) we get,
Distance = 30 × 24 = 720 km

The length of the journey = 720 km.

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