A student travels to his school at a speed of 4 km/hour and reaches the school 15 minutes late. On travelling at a speed of 6 km/hour; he r…

Given, a student travels to his school at a speed of 4 km/hour and reaches the school 15 minutes late. On travelling at a speed of 6 km/hour; he reaches the school 5 minutes early.

Let the original speed and time be ‘s’ and ‘t’ respectively.

Distance = speed × time

Distance = s × t = 4 × (t + 15/60) = 6 × (t – 5/60)

Distance = st = 4t + 1 = 6t – 0.5

⇒ 4t + 1 = 6t – 0.5

⇒ 2t = 1.5

⇒ t = 0.75 hours

Now, st = 4t + 1

⇒ s = 4 + 1/0.75 = 16/3 = 5 1/3 km/hr