A person moved from A to B at 3 PM at a speed of 4.5 km/h, stayed at B for 1.5 hrs. and returned to place A at a speed of 6 km/h.  If he comes back to place at 8 PM, what is the distance b/w A and B?

9 km

Let distance b/w A and B = D km

Total time taken = 8pm - 3pm = 5 hrs

⇒ Time = \(\frac{Distance}{Speed}\)

⇒ Total time (T) = \(\frac{D}{4.5}\) + 1.5 + \(\frac{D}{6}\)

⇒ T = \(\frac{D × 2}{9}\) + \(\frac{3}{2}\) + \(\frac{D}{6}\)

⇒ 5 = \(\frac{4D + 27 + 3D}{18}\)

⇒ 5 = \(\frac{7D + 27}{18}\)

⇒ 90 = 7D + 27

⇒ 7D = 90 - 27 = 63

⇒ D = 9 km