A train covered a certain distance at a uniform speed. If the train would 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 x km/hr  be the original speed of the train and y hrs be the time taken by the train to complete the journey.

Then, Distance covered = xy km

Case 1 : When  speed = (x + 10) km/hr

time taken = (y – 2) hr

Distance = (x + 10) (y – 2)

xy = (x + 10) (y – 2)

\(\implies\) 2x – 10y + 20 = 0           ……….(1)

Case 2 : When speed = (x – 10) km/hr

time taken = (y + 3) hr

Distance = (x – 10)(y + 3)

xy = (x – 10)(y + 3)

\(\implies\) 3x – 10y – 30 = 0       …….(2)

Now, subtract equation (1) from (2), we get

x – 50 = 0     \(\implies\)      x = 50

Put the value of x = 50  in equation (1), we get

100 – 10y + 20 = 0       \(\implies\)   y = 12

Hence, the distance covered by train is \(12\times 50\) = 600 km