Two Buses on the Ring Road
Two bus routes share a depot. Route A runs a 42 km circuit and Route B a 63 km circuit. A bus on Route A averages 36 km/h including stops; a bus on Route B averages 42 km/h. Both leave the depot at 07:00. Route A buses depart every 20 minutes, Route B every 30 minutes. A dispatcher claims that because Route B is longer, its buses spend more time out on the road per trip. A driver disputes this, saying Route A's slower average speed more than cancels the shorter distance. Both agree the depot needs at least one bus back before 08:30.
1. Calculate the round-trip time for one bus on each route and determine which route's buses are out longer per trip.
Read the worked answer
Route A: time = distance ÷ speed = 42 ÷ 36 = 1.1667 h = 1 h 10 min. Route B: time = 63 ÷ 42 = 1.5 h = 1 h 30 min.
Route B buses are out 20 minutes longer per trip. The dispatcher is correct on this specific point.
2. Determine which buses return to the depot before 08:30, and how many departures of each route occur before 08:30.
Read the worked answer
A bus leaving 07:00 on Route A returns at 08:10 - before 08:30. ✓ A bus leaving 07:00 on Route B returns at 08:30 exactly - not strictly before 08:30. ✗
Departures before 08:30: Route A: 07:00, 07:20, 07:40, 08:00, 08:20 = 5 departures. Route B: 07:00, 07:30, 08:00 = 3 departures.
Only Route A reliably meets the 'one bus back before 08:30' condition from the 07:00 wave.
3. The driver claims Route A's slower speed cancels its shorter distance. Using time-per-km, test this claim.
Read the worked answer
Time per km, Route A = 70 min ÷ 42 km = 1.667 min/km. Time per km, Route B = 90 min ÷ 63 km = 1.429 min/km.
Route A is slower per km, but its total trip is still shorter (70 vs 90 min). The driver's claim is wrong: the shorter distance more than compensates for the slower speed. Route A buses are out 20 min less per trip.