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Marla starts running around a circular track at the same

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

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

How to do this guys?
[Reveal] Spoiler: OA

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Re: Marla starts running around a circular track at the same [#permalink] New post 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 32-20=20 more laps in 1 hour. Marla to complete 4 (20/5=4) more laps will need 1/5 hours, which is 12 minutes.

Answer: C.
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Re: Marla starts running around a circular track at the same [#permalink] New post 16 Mar 2012, 10:00
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.
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Re: Marla starts running around a circular track at the same [#permalink] New post 16 Mar 2012, 15:46
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] New post 16 Mar 2012, 15:53
Expert's post
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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Re: Marla starts running around a circular track at the same [#permalink] New post 17 Mar 2012, 06:06
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] New post 12 Apr 2012, 03:35
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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Re: Marla starts running around a circular track at the same [#permalink] New post 23 Apr 2012, 14:27
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] New post 23 Apr 2012, 20:50
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.
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Re: Marla starts running around a circular track at the same [#permalink] New post 23 Apr 2012, 21:28
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] New post 23 Apr 2012, 22:32
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] New post 25 Apr 2012, 01:21
The answer is C
use relative velocity to solve it

the relative velocity would be (32-12) 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] New post 21 Apr 2014, 01:10
Let time be t
32/60t-12/60t=4 so t=12
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Re: Marla starts running around a circular track at the same [#permalink] New post 21 Apr 2014, 22:01
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)

d-4 = \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] New post 02 May 2014, 02:43
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] 02 May 2014, 02:43
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