A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hrs less than the scheduled time. And, if the train were slower by 6 km/h, it would have taken 6 hrs more than the scheduled time. Find the distance covered by the train.

Let original speed $= v$ km/h, original time $= t$ hours, and distance $= d$ km.

Distance $=$ Speed $\times$ Time

$d = vt$

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If the train is $6$ km/h faster, it takes $4$ hours less:

$d = (v + 6)(t - 4)$

If the train is $6$ km/h slower, it takes $6$ hours more:

$d = (v - 6)(t + 6)$

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Since distance is the same in all cases:

$vt = (v + 6)(t - 4) = (v - 6)(t + 6)$

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Using the first equality:

$vt = (v + 6)(t - 4)$

$vt = vt - 4v + 6t - 24$

$0 = -4v + 6t - 24$

$4v = 6t - 24$

$2v = 3t - 12$ ... equation $(1)$

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Using the second equality:

$vt = (v - 6)(t + 6)$

$vt = vt + 6v - 6t - 36$

$0 = 6v - 6t - 36$

$6v = 6t + 36$

$v = t + 6$ ... equation $(2)$

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From equation $(2)$: $v = t + 6$

Substitute into equation $(1)$:

$2(t + 6) = 3t - 12$

$2t + 12 = 3t - 12$

$24 = t$

Therefore $t = 24$ hours and $v = 24 + 6 = 30$ km/h

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Distance $=$ Speed $\times$ Time

$d = 30 \times 24$

$d = 720$ km

The distance covered by the train is $720$ km.