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Marla starts running around a circular track at the same time Nick sta

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

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New post 16 Mar 2012, 07:55
6
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A
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E

Difficulty:

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Question Stats:

70% (02:17) correct 30% (02:41) wrong based on 877 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

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

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

Answer: C.
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Re: Marla starts running around a circular track at the same time Nick sta  [#permalink]

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New post 12 Apr 2012, 04:35
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2
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 time Nick sta  [#permalink]

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New post 16 Mar 2012, 11:00
2
4
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 time Nick sta  [#permalink]

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New post 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 time Nick sta  [#permalink]

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New post 16 Mar 2012, 16:53
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Re: Marla starts running around a circular track at the same time Nick sta  [#permalink]

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New post 17 Mar 2012, 07:06
1
1
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 time Nick sta  [#permalink]

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New post 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 time Nick sta  [#permalink]

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New post 25 Apr 2012, 02:21
1
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 time Nick sta  [#permalink]

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New post 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)

\(d-4 = \frac{12}{60} * t\)........(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 time Nick sta  [#permalink]

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New post 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
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Re: Marla starts running around a circular track at the same time Nick sta  [#permalink]

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New post 26 Sep 2017, 10:43
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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.

\(\frac{32t}{60}\) - \(\frac{12t}{60}\) = 4

Solving,
\(\frac{20t}{60}\) = 4
\(\frac{t}{3}\) = 4

t= 12 mins.
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Re: Marla starts running around a circular track at the same time Nick sta  [#permalink]

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New post 29 Sep 2017, 10:33
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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:

32t/60 = 12t/60 + 4

32t/60 - 12t/60 = 4

20t/60 = 4

t/3 = 4

t = 12

Answer: C
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Re: Marla starts running around a circular track at the same time Nick sta  [#permalink]

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New post 29 Jan 2019, 10:58
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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

1/5 hours = 12 minutes.

Answer: C

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

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New post 26 Sep 2019, 01:45
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


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Re: Marla starts running around a circular track at the same time Nick sta   [#permalink] 26 Sep 2019, 01:45
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