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## time speed distance question

tagged by: shaparage

This topic has 1 expert reply and 2 member replies
shaparage Newbie | Next Rank: 10 Posts
Joined
31 Oct 2017
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#### time speed distance question

Tue Nov 21, 2017 9:46 am
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

### Top Member

regor60 Master | Next Rank: 500 Posts
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Wed Nov 22, 2017 9:39 am
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.

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Matt@VeritasPrep GMAT Instructor
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Thu Dec 07, 2017 3:56 pm
Fun follow up conceptual question: the harmonic mean of 20 and 30 just happens to be 24. Is this a shortcut to solve this problem, or just a coincidence? Explain your answer.

*Jeopardy music*

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shaparage Newbie | Next Rank: 10 Posts
Joined
31 Oct 2017
Posted:
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Sat Dec 16, 2017 10:10 pm
Matt@VeritasPrep wrote:
Fun follow up conceptual question: the harmonic mean of 20 and 30 just happens to be 24. Is this a shortcut to solve this problem, or just a coincidence? Explain your answer.

*Jeopardy music*
coz when distance is constant, speed is inversely proportional to time and if speed is in AP , time will be in HP. if buspeed -man speed , bus speed, bus speed +man speed are in AP the respective time will be in HP. please let me know if this logic is right .

Thanks

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