If a man increase his speed by 6 km/hr, he cover a certain distance 4 hours earlier than the usual time and if he reduce his speed by 6 km/hr, he takes 6 hours more to cover the same distance. Find the distance.

720 km

Let the original speed is S km/h & time taken to cover the distance TR

ATQ,         D  = S × T

           (S+6) (T-4) = ST

   ST + 6T - 4S - 24 = ST

                        6T = 4S + 24  ------- (i)

 (S-6) (T+6) = ST

ST + 6S - 6T - 36 = ST

6T = 6S - 36   ----- (ii)

From eqn (i) & eqn (ii)

4S  + 24 = 6S - 36

         2S = 60

⇒ S = 30 km/h

From eqn (i)

           6T = 4S + 24

           6T = 4 × 30 + 24

           6T = 144

             T = 24 hours

Distance = speed × time

             30 × 24 = 720 km