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John and Karen begin running at opposite ends of a trail until they me
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31 Aug 2022, 06:32
31 Aug 2022, 06:32
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50% (03:16) correct
50% (04:00) wrong based on 4 sessions
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John and Karen begin running at opposite ends of a trail until they meet somewhere in between their starting points. They each run at their respective constant rates until John gets a cramp and stops. If Karen runs 50% faster than John, who is only able to cover 25% of the distance before he stops, what percent longer would Karen have run than she would have had John been able to maintain his constant rate until they met.
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25 %
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John and Karen begin running at opposite ends of a trail until they me
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06 Sep 2022, 13:39
06 Sep 2022, 13:39
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Carcass wrote:
John and Karen begin running at opposite ends of a trail until they meet somewhere in between their starting points. They each run at their respective constant rates until John gets a cramp and stops. If Karen runs 50% faster than John, who is only able to cover 25% of the distance before he stops, what percent longer would Karen have run than she would have had John been able to maintain his constant rate until they met.
Show: ::
25 %
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
Let the distance to be travelled is 100m
Speed of John be 10m/s
So, speed to Karen is 15m/s
Relative Velocity (towards each other) = 15 + 10 = 25m/s
Time taken to meet in between = \(\frac{distance}{Velocity_{rel}} = 100/25 = 4\)sec
Distance travelled by John in 4 sec (Meeting point) = 10(4) = 40m
Distance travelled by Karen in 4 sec (Meeting point) = 15(4) = 60m
John was able to cover only 25% of the distance = 0.25(40) = 10m
Time required to do so = \(\frac{10}{10} = 1\)sec
Karen now has to run till 90m mark as John stopped at 10m mark (100m - 10m).
In that 1 sec, Karen already ran 15m. So she will run 90m - 15m = 75m more at 15m/s
Time required by Karen to do so = \(\frac{75}{15} = 5\) sec
Initially Karen would take 4 sec but not it will take her 5 sec to new Meeting point. Percent change = \(\frac{(5-4)}{4}(100) = 25\)%
Re: John and Karen begin running at opposite ends of a trail until they me
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07 Sep 2024, 03:03
07 Sep 2024, 03:03
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