shaparage wrote:While walking down the pavements of New York city, I notice that every 20 minutes there is a city bus coming in the opposite direction and every 30 minutes there is a city bus overtaking me from behind. What is the time gap between one city bus passing a stationary point known as Local Bus Stop beside the bus route and the immediately next city bus in the same direction passing the stationary point?
please solve this question , also if there is any other way to do it without equation . thanks
Recognize that the question is basically asking what the time interval is for bus dispatch.
Pick a convenient amount of time, say 60 minutes, which happens to be the least multiple of the two given times, 20 and 30 minutes.
Over that time, 2 buses will have overtaken from behind and 3 met from front. In fact, at that point in time, the 2nd and 3rd buses and the pedestrian will be in alignment.
Now, there is a bus out ahead in front of the pedestrian and of course one behind. Assuming the buses were first dispatched at the same time, follow the same interval and are going the same speeds, you can see that the pedestrian is halfway between the two or equidistant.
Since the pedestrians speed adds to the buses speed in approaching the bus ahead and subtracts from the one coming from behind
Distance = (RateBus + RatePed)*20 = (RateBus - RatePed)*30
Working through this and solving for RatePedestrian = RateBus/5
Plugging this back into either one of the two above formulas, but let's choose the first one:
Distance = (6*RateBus/5)*20 = 24 * RateBus
Since Distance = Rate * Time, 24 must be the time between buses.
