A mother left her child at Bus stop 'P' in a bus station at 6:00 a.m. Four hours later, a bus departed from Bus stop 'Q' to Bus stop 'P' and passed the child after an additional 4 hours. Upon reaching Bus stop 'P', the bus promptly turned around and headed back to Bus stop 'Q'. If the bus and the child arrived at Bus stop 'Q' simultaneously, what was the time when the child reached Bus stop 'Q'?

Solution

ATQ, Let the distance between Bus stop 'A' and 'B' be 'k' km Since, the Child flew from Bust stop 'A' to Stop 'B', the bus is moving downhill from point 'A' to 'B'. Let the speed of the Bus be 'b' km/h and speed of the downhill road be 'a' km/h Distance covered by the child in first 4 hours = '4a' km {(k - 4a) /(a + b - a) } = 4 Or, k - 4a = 4k So, k = 4(b + a) ...... (I) Now, since the bus and the child both reach Bus stop 'Q' together, we have; {k/(a + b) } + {k/(b - a) } = (k/a) - 4 Or, {k/(k/4) } + {k/(b - a) } = (k/a) - 4 {Using equation (I) } Or, 4 + {k/(b - a) } = (d/a) - 4 Or, 8 = (k/a) - {k/(b - a) } Or, 8 = k × [(1/a) - {1/(b - a) }] Or, 8 = 4 × (a + b) × {(b - a - a) /(ab - a²) } Or, 2 × (ab - a²) = (a + b) × (b - 2a) Or, 2ab - 2a² = ab - 2a² + b²- 2ab Or, 2ab = b²- ab Or, 3ab = b² So, 3a = b So, k = 4 × (a + 3a) So, k = 16a So, time taken by the bus to reach Bust stop 'B' = (k/a) = 16a ÷ a = 16 hrs