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Question -

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



Answer -

Let the original speed of the train be x km/h
and the time taken to complete the journey be y hours.            ‘
Then the distance covered = xy km
Case I: When speed = (x + 10) km/h and time taken = (y – 2) h
Distance = (x + 10) (y – 2) km
⇒   xy = (x + 10) (y – 2)
⇒ 10y – 2x = 20
⇒  5y – x = 10
⇒ -x + 5y = 10   …(i)

Case II: When speed = (x – 10) km/h and time taken = (y + 3) h
Distance = (x – 10) (y + 3) km
⇒  xy = (x – 10) (y + 3)
⇒ 3x- 10y = 30    …(ii)
Multiplying equation (i) by 3 and adding the result to equation (ii), we get
15y – 10y = 30 f 30
⇒ 5y = 60
⇒   y   = 12
Putting y = 12 in equation (ii), we get
3x- 10 x 12= 30
⇒  3x   = 150
⇒ x   = 50
∴  x = 50 and y =   12
Thus, original speed of train is 50 km/h and time taken by it is 12 h.
Distance covered by train = Speed x Time
=  50 x 12 = 600 km.


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