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Marla starts running around a circular track at the same [#permalink]

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16 Mar 2012, 07:55

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72% (03:05) correct
28% (01:57) wrong based on 325 sessions

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Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 32 laps around the track per hour and Nick completes 12 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick?

Re: Marla starts running around a circular track at the same [#permalink]

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16 Mar 2012, 09:26

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enigma123 wrote:

Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 32 laps around the track per hour and Nick completes 12 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick?

(A) 5 (B) 8 (C) 12 (D) 15 (E) 20

How to do this guys?

Marla completes 32-20=20 more laps in 1 hour. Marla to complete 4 (20/5=4) more laps will need 1/5 hours, which is 12 minutes.

Re: Marla starts running around a circular track at the same [#permalink]

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16 Mar 2012, 11:00

Marla completes 32laps/60min, Nick completes 12laps/60mins. After x mins Marla would have completed 4 laps more than Nick had completed. x((32-12)/60) = 4, x*20/60 = 4, x = 12 mins.

Re: Marla starts running around a circular track at the same [#permalink]

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16 Mar 2012, 16:53

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enigma123 wrote:

Bunuel - how did you get this?

Marla to complete 4 (20/5=4) more laps will need 1/5 hours, which is 12 minutes.

Since Marla completes 20 more laps in 1 hour, then to complete 1/5 th of 20 laps (4 laps) she will need 1/5 th of an hour, which is 12 minutes. _________________

Re: Marla starts running around a circular track at the same [#permalink]

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12 Apr 2012, 04:35

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Maria's rate - 32 laps per hour --> 32/60 laps/min Nick's rate - 12 laps per hour --> 12/60 laps/min

lets set equations:

32/60*t=4 (since Maria had to run 4 laps before Nick would start) 12/60*t=0 (Hick has just started and hasn't run any lap yet) ----------------------------------- (32/60-12/60)*t=4-0 (since Nick was chasing Maria) t=12 min needed Maria to run 4 laps

Re: Marla starts running around a circular track at the same [#permalink]

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23 Apr 2012, 21:50

I also got C (12).

t= time 32 laps per hour * t = 12 laps per hour * t + 4 laps

32t = 12t + 4 20t = 4 t= 1/5 hr

1/5(60min)= 12 mins

I believe the recommended strategy is to change to minutes right away but in this situation I found it easier to do it at the end because of how the rates simplified per minute.

Re: Marla starts running around a circular track at the same [#permalink]

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02 May 2014, 03:43

Bunuel wrote:

enigma123 wrote:

Bunuel - how did you get this?

Marla to complete 4 (20/5=4) more laps will need 1/5 hours, which is 12 minutes.

Since Marla completes 20 more laps in 1 hour, then to complete 1/5 th of 20 laps (4 laps) she will need 1/5 th of an hour, which is 12 minutes.

I believe you used the concept of relative speed over here as well Bunuel . Isn't it so ? Relative speed of Bunue is 20 laps/hr so 4 laps in 1/5 of an hr = 12 minutes

Re: Marla starts running around a circular track at the same [#permalink]

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26 Dec 2015, 03:10

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Re: Marla starts running around a circular track at the same [#permalink]

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27 Dec 2015, 12:57

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marla runs 32/60=8/15 laps per minute nick walks 12/60=3/15 laps per minute marla gains 8/15-3/15=1/3 laps per minute 4 laps/1/3 lap per minute=12 minutes

gmatclubot

Re: Marla starts running around a circular track at the same
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27 Dec 2015, 12:57

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