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

Let the original speed be v km/h and the scheduled time be t hours, so the distance is D = vt. Using the condition for the faster speed, (v+10)(t-2) = vt, which simplifies to -2v + 10t = 20 or 10t - 2v = 20. Using the condition for the slower speed, (v-10)(t+3) = vt, which simplifies to 3v - 10t = 30. Adding the two derived equations (10t - 2v = 20 and 3v - 10t = 30) eliminates t and gives v = 50 km/h. Substituting v = 50 into 10t - 2v = 20 gives 10t - 100 = 20, so t = 12 hours. The distance covered by the train is D = vt = 50 * 12 = 600 km.

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