A man covers a certain distance on bike. Had he moved 4 kmph faster, he would have taken 72 minutes less. If he moves slow by 4 kmph, he will take 2 hours more to cover the distance. What is the distance travelled by him?

Let the original speed be v kmph and distance be d km. Original time = d/v hours. When speed increases by 4 kmph: time = d/(v+4) = d/v - 72/60 = d/v - 6/5. When speed decreases by 4 kmph: time = d/(v-4) = d/v + 2. From the equations: d/v - d/(v+4) = 6/5 and d/(v-4) - d/v = 2. Solving these gives v = 20 kmph and d = 96 km. Verification: at 24 kmph, time = 96/24 = 4 hours (72 minutes less than 5 hours); at 16 kmph, time = 96/16 = 6 hours (2 hours more than 4 hours).