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Marla starts running around a circular track at the same time Nick sta
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16 Mar 2012, 06:55
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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?
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16 Mar 2012, 08:26
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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 32-12=20 more laps in 1 hour. Marla to complete 4 (20/5=4) more laps will need 1/5 hours, which is 12 minutes.
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29 Jan 2019, 09:58
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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
Let t = the time (in HOURS) that it takes Marla to complete 4 more laps than Nick. So, after t hours, we can write: (Marla's lap count) = (Nick's lap count) + 4
Now that we have a "word equation" we need only fill in the missing information
Marla completes 32 laps per hour We can think of 1 lap as being a unit of distance. So, 32 laps per hour is Marla's speed.
Distance = (speed)(time) So, after t hours, Marla's lap count = 32t
Nick completes 12 laps around the track per hour So, after t hours, Nick's lap count = 12t
We can now plug the above values into the word equation. We get: 32t = 12t + 4 Subtract 12t from both sides to get: 20t = 4 Solve: t = 4/20 = 1/5 HOURS
Re: Marla starts running around a circular track at the same time Nick sta
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12 Apr 2012, 03:35
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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
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16 Mar 2012, 10:00
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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((32-12)/60) = 4, x*20/60 = 4, x = 12 mins.
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16 Mar 2012, 15:53
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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.
_________________
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23 Apr 2012, 20:50
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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.
Re: Marla starts running around a circular track at the same time Nick sta
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27 Dec 2015, 11:57
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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
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26 Sep 2017, 09:43
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Given, the distance traveled by Marla in 60 minutes = 32 Distance traveled by Marla in 't' mins = \(\frac{32t}{60}\)
Also given, the distance traveled by Nick in 60 minutes = 12 Distance traveled by Nick in 't' mins = \(\frac{12t}{60}\)
Prompt says there will be a difference of 4 laps at some point in time between the above two distances.
Since Maria travels faster, the distance covered by Maria will be more than the distance covered by Nick. Therefore, subtract the distance traveled by Nick from the distance traveled by Maria in order to avoid a negative resultant.
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29 Sep 2017, 09:33
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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
We can use the following formula:
time = change in distance/change in rate
time = 4/20 = 1/5 hour = 12 minutes
Alternate Solution:
Note that Marla’s rate is 32/60 laps per minute and Nick’s rate is 12/60 laps per minute. Let’s say after t minutes, Marla completes 4 more laps than Nick. Then, in t minutes Marla completes 32t/60 laps and Nick completes 12t/60 laps. Since the number of laps completed by Marla is 4 more than the number completed by Nick, we have:
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26 Sep 2019, 00:45
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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?