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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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Question Stats:
70% (02:49) correct
30% (01:55) wrong based on 486 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? (A) 5 (B) 8 (C) 12 (D) 15 (E) 20 How to do this guys?
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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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Re: Marla starts running around a circular track at the same [#permalink]
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16 Mar 2012, 11:00
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Marla completes 32laps/60min, Nick completes 12laps/60mins. After x mins Marla would have completed 4 laps more than Nick had completed. x((3212)/60) = 4, x*20/60 = 4, x = 12 mins.



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Re: Marla starts running around a circular track at the same [#permalink]
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16 Mar 2012, 16:46
Bunuel  how did you get this? Marla to complete 4 (20/5=4) more laps will need 1/5 hours, which is 12 minutes.
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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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Re: Marla starts running around a circular track at the same [#permalink]
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17 Mar 2012, 07:06
Best way is to use plug in!
Always plug in C first, since this is the median value!
In 12 minutes she will have completed 32/5 rounds ( /5 because 12min is 1/5 hours) and he will have completed 12/5 rounds.
32/5 = 6 2/5 12/5 = 2 2/5
4 is the difference and hence C is th answer.
Sometimes you get lucky and plug in the right answer first! Will help you save heaps of time!



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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)  (32/6012/60)*t=40 (since Nick was chasing Maria) t=12 min needed Maria to run 4 laps



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Re: Marla starts running around a circular track at the same [#permalink]
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23 Apr 2012, 15:27
I say C. 12 mins.
Marla's speed : 32/60 laps per min Nick's speed : 12/60 laps per min
So by plugging in ans choices,
at 12 mins, Marla : 6.4 laps, Nick 2.4 laps.



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Re: Marla starts running around a circular track at the same [#permalink]
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23 Apr 2012, 21:50
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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.



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Re: Marla starts running around a circular track at the same [#permalink]
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23 Apr 2012, 22:28
In 15 min  Marla has completed 32/4 = 8 laps and Nick > 12/4 = 3 the difference at 15 mins is 5 laps. So plugging in C.
In 12 mins  Marla has completed 32/5 = 6.4laps while Nick has completed 12/5 = 2.4 laps , a difference of 4 laps.
So ans  C



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Re: Marla starts running around a circular track at the same [#permalink]
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23 Apr 2012, 23:32
In 1 min: Marla runs: 32/60 = 8/15 laps Nick walks: 12/60 = 1/5 laps
Let x be the number of minutes required for Marla to be 4 laps ahead of Nick. x * 1/5 + 4 = x * 8/15 x =12 mins



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Re: Marla starts running around a circular track at the same [#permalink]
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25 Apr 2012, 02:21
The answer is C use relative velocity to solve it
the relative velocity would be (3212) in 1 hour. hence, 20 laps in 1 hour therefore time to complete four laps is 1/5hr 12 mins



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Re: Marla starts running around a circular track at the same [#permalink]
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21 Apr 2014, 02:10
Let time be t 32/60t12/60t=4 so t=12



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Re: Marla starts running around a circular track at the same [#permalink]
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21 Apr 2014, 23:01
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Let the one lap distance = d Speed of Marla \(= \frac{32}{60}\) Speed of Nick \(= \frac{12}{60}\) Time t would be same in both the cases Setting up the equation \(d = \frac{32}{60}* t\) ............ (1) \(d4 = \frac{12}{60} * t\)........(2) Equating (1) & (2) \(\frac{12}{60} * t + 4 = \frac{32}{60} * t\) \(\frac{1}{3} * t = 4\) t = 12 Answer = C
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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



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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 8/153/15=1/3 laps per minute 4 laps/1/3 lap per minute=12 minutes



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Marla starts running around a circular track at the same [#permalink]
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23 Mar 2017, 14:39
Why to make it more difficult than it is?
Marla walks 32 laps in 1 hour and spends x hours on the road. Laps covered = 32x Nick walks 12 laps in 1 hour and spends x hours on the road. Laps covered = 12x (which must be 4 laps fewer than Marla)
32x12x=4 x=0.2h=12 min (C)




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