Test your GRE quant concepts - February 1, 2022 (Relative speed problem)

GRE Quant Problem Solving
1

I am confused as to why for this question below we didn’t use the following approach because after coming equal to Jane at 675 meters, bib as to cover another 325 meters to overtake her twice so the time for that would be 325/4 but in the solution video Greg said that to find out how will he overtake her twice we will divide 1000/4 but why 1000 because both of them have already covered 675 meters ?

I have outlined my solution below so I want to understand where am I going wrong.

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  1. Track Length and Speeds:
  • The track is 1000 meters long.
  • Jane has a head start of 300 meters and runs at 5 meters per second (m/s).
  • Bob runs faster at 9 m/s.
  1. How Far Ahead is Jane?:
  • When Bob starts, Jane is already 300 meters ahead of him.
  1. Relative Speed:
  • Bob is faster than Jane by: 9 m/s−5 m/s=4 m/s
  • This means Bob gains 4 meters on Jane every second.
  1. First Catch:
  • To catch Jane for the first time, Bob needs to cover the 300 meters between them.
  • Time taken to catch up: 300 meters4 m/s=75 seconds
  1. Positions After 75 Seconds:
  • In 75 seconds, Jane runs: 5 m/s×75 s=375 meters5
  • So, she is now at: 300 meters (head start)+375 meters=675 meters300
  • Bob runs: 9 m/s×75 s=675 meters
  • Bob also reaches 675 meters at the same time.
  1. Second Catch:
  • Now, for Bob to pass Jane again, he has to go all the way around the track again (1000 meters) from the 675-meter point.
  • The distance he needs to cover to pass her again is: 1000 meters−675 meters=325 meters
  • Time to cover this distance: 325 meters4 m/s=81.25 seconds
  1. Total Time:
  • Total time for Bob to pass Jane twice: 75 seconds+81.25 seconds≈156.25 seconds
    Final Answer: It will take Bob about 156.25 seconds to pass Jane twice.
2

Hello again @Archi :joy: ,
So I think you are missing something in the step below.

As you corrected pointed out that the relative speed is 4m/s. When you want to catch the person again, you and the person will be moving together. This can be better understood if we chart the distance covered. Let’s consider 100 m track for simplicity.

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As you can bob will cross her only on his 3rd round and her 2nd round if they begin together. Here we get 100m / 4 m/s = 25 s. Distance/Speed = Time

Hence the formula is 1000m/4 = 250s

Hope this helps!

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3

Hello again @mohitraj !

Could you explain a bit more with the same numbers in the question? I am still not able to understand it properly :frowning: Sorry

4

wait…I think I got it. So, basically it is a circular track and the first time when Bob takes over Jane he gets a lead over her and is running ahead of her which means to overtake her second time he still has to complete 1000 meters lap again since they both are running at different speeds and both are in motion so he can take over again only when he completes full lap right?

5

Yes perfect!

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6

@Leaderboard Is it possible to find resources on the website to solve relative problems? I think I need more practice for such type of problems so are there any video on the website or a dedicated series that just deals with distance and speed problems?

Thank you!

7

PrepSwift and our problem-solving section (https://www.gregmat.com/problems/quant), filtering for the topics you want.

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