A local bus leaves the stations at 6:00 a.m. and returns to pick up new passengers every 40 minutes. Providing there are no delays and the bus maintains its schedule, what time could the bus NOT return to the station?
3a=2b+1
3b=5a-3
To determine the time the bus could not return to the station, we can find the possible arrival times of the bus and identify the time when there would be a gap of 40 minutes.
Since the bus leaves the station at 6:00 a.m. and returns every 40 minutes, we can create a series of times when the bus arrives by adding 40 minutes to the previous time.
Starting with 6:00 a.m., we can calculate the following arrival times:
6:00 a.m. + 40 minutes = 6:40 a.m.
6:40 a.m. + 40 minutes = 7:20 a.m.
7:20 a.m. + 40 minutes = 8:00 a.m.
8:00 a.m. + 40 minutes = 8:40 a.m.
8:40 a.m. + 40 minutes = 9:20 a.m.
9:20 a.m. + 40 minutes = 10:00 a.m.
The bus continues to arrive every 40 minutes, and so on.
To find the time the bus could not return to the station, we look for a gap of 40 minutes where the bus could not arrive. In this case, there is no such gap, as the bus maintains its schedule. Therefore, the bus can potentially return to the station at any time.