Answer :

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.

Rate this question :

Draw the graph ofNCERT - Mathematics