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Marla starts running around a circular track at the same [#permalink] New post 07 Oct 2011, 08:09
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: marla-starts-running-around-a-circular-track-at-the-same-129172.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 30 Jul 2012, 10:43, edited 2 times in total.
Edited the question.
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Re: Good Time and rate question [#permalink] New post 07 Oct 2011, 09:32
Marla 32 laps per hour==> 0.53 laps per min
Nick 12 laps per hour ==> 0.2 laps per min

for every min Marla is taking 0.33 laps advantage

to know the time when Marla takes 4 lap advantage====>4/0.33 = 12.12 ~ 12


Answer C
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Re: Good Time and rate question [#permalink] New post 07 Oct 2011, 09:40
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Let t be time when the condition required condition met,
then
32/60*t = 12/60*t+4 => t=12.
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Re: Good Time and rate question [#permalink] New post 07 Oct 2011, 10:12
I am not sure whether the approach mentioned by me earlier is right way to go.

Same time i just thought the below way

The difference between the distance covered by Marla and Nick is 20 in 60 mins.

to get the time in which Marla would be 4 km ahead of Nick (which is 20/5)

Need to divide 60 by 5 which will give 12 mins.

It stuck me only after knowing the answer but thought it is simple.
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Re: Good Time and rate question [#permalink] New post 08 Oct 2011, 02:14
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gmatcracker24 wrote:
Q.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

...


Thanks guys for sharing your approaches.

The way i solved it is :

In one hour Maria covers 20 extra laps than Nick.

So, in 3 minutes -> Maria would cover 1 extra lap.

Hence , for 4 extra lap -> 3*4 = 12 minutes.
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Re: Good Time and rate question [#permalink] New post 08 Oct 2011, 03:20
ans is C.
(32/60 -12/60 )* x = 4
x= 12 min
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Re: Good Time and rate question [#permalink] New post 08 Oct 2011, 13:56
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this can be solved in two ways.

method1
----rate---distance
N---12----x
M---32----x+4

x/12 = (x+4)/32 = > x = 12/5

time taken by M to cover 4 laps more than N = time taken by N to cover some laps = 12/(5*12) = 1/5 hours = 12 minutes


method 2
Difference in their rates = 20 laps per hour

=> M can cover 20 laps more than N's in 1 hour

=> M can cover 20 laps more than N's in 60 minutes

=> M can cover 1 lap more than N's in 3 minutes

=> M can cover 4 laps more than N's in 12 minutes

Answer is C
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Re: Good Time and rate question [#permalink] New post 09 Oct 2011, 09:04
C is the answer

60/(32-12) * 4 = 12.
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CR notes
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133
http://gmatclub.com/forum/gmat-prep-critical-reasoning-collection-106783.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html?hilit=chineseburned

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Re: Good Time and rate question [#permalink] New post 12 Oct 2011, 01:16
M speed=32 time=t distance =s+4
N speed=12 time=t distance=s

32t=s+4
12t=s

t=1/5 h or 12 min


p.s.have doubts that it is a 700 level question
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Re: Good Time and rate question [#permalink] New post 14 Oct 2011, 10:07
Howdy, I usually post just in the SC forum--find me there if you have SC questions--but I wanted to add something to the OP's excellent solution.

Whenever a difference in distances is described, you can subtract distances. The paradigm GMAT case is two people/cars/trains/whatever going in the same direction. Whenever a difference in distances is described, and both parties move simultaneously, you can also subtract rates: (difference in rates)*time=(difference in distance).

This question asks for a difference in the distances covered by two people walking simultaneously. The difference in rates is 20 laps/hour, or 1 lap/3 minutes. (1/3)t=4. t-12.

Many good test-takers find this formula, (difference in rates)*time=(difference in distance), so useful that they tweak problems with different times so that they can apply this formula. Let me give you an example, based on this question.

Q.Nick starts walking around a circular track at noon. Marla starts running around the same circular track at 12:20. Nick completes 12 laps per hour and Marla completes 32 laps per hour. At what time will Marla have completed as many laps as Nick?
(A) 12:25
(B) 12:28
(C) 12:32
(D) 12:35
(E) 12:40

Here's how to answer this as one hard question:
(Marla's time)(Marla's rate)=Marla's distance.
(Nick's time)(Nick's rate)=Nick's distance.

Nick's time is 20 minutes greater than Marla's.
You could say,
Let t= Marla's time in hours and t+1/3=Nick's time in hours.
But we'll go with,
Let t=Marla's time in minutes and t+20=Nick's time in minutes.
Marla's distance is Nick's.
Let d= Nick's distance and d=Marla's distance.
This yields two equations:
32t=d
12(t+20)=d
Solve for 12:20+t.

I'll trust you to solve algebraically.

Here's the same question rephrased as three easy questions:
What did Nick do before Marla started running? Walk 4 laps.
Once Marla started running--once they were moving simultaneously--how much faster did she go than Nick? 20 laps/hour, or 1 lap/3 minutes.
How long will it take her to gain 4 laps, if it takes her 3 minutes to gain 1 lap? 12 minutes.
Add 12 minutes to 12:20, the time from which they moved simultaneously. 12:32.

Notice that this second method would be harder to apply if we complicated the problem a bit. If we asked at what time Marla would have gone twice as far as Nick, the more formal algebraic method works just as neatly (just make Marla's distance 2d instead of d), but the alternative method gets much messier.

Okay, here's one more for you to solve. It's not a real GMAT question, and I've rigged it so that the second method described above will be much easier for most people. Answer below and I'll check in tomorrow with my explanation.

At 9:00, Nick started walking the 9 miles from A to B. At 9:40, Marla started walking the 9 miles from A to B. Until Marla caught up to Nick, Marla walked at a constant rate of 5 miles per hour, and Nick walked at a constant rate of 3 miles per hour. Once Marla caught Nick, they walked together at a constant rate of 4 miles per hour. At what time did they arrive together at B?
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Last edited by MichaelS on 14 Oct 2011, 11:50, edited 1 time in total.
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Re: Good Time and rate question [#permalink] New post 14 Oct 2011, 11:34
Well i think there is a problem with the solution and the answer choices mentioned in the question are also wrong.

Say at 12:32

Nick has covered 12/60 * 32 = 6.4 laps in 32 minutes.
Marla has covered 32/60 * 12 = 6.4 laps in 12 minutes.

So if the question is in what time Nick and Marla has covered same number of laps. The answer is 12:32.

But in the question it is asked that Marla has covered twice as nick
Then the equation becomes

12(t+20) = 32(t) /2
=> 12t + 12 * 20 = 16 t => 4t = 12*20 => t = 60 minutes

i.e At 1:30 PM Marla has covered twice the number of laps than Nick.
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MGMAT 6 650 (51,31) on 31/8/11
MGMAT 1 670 (48,33) on 04/9/11
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MGMAT 3 680 (47,35) on 18/9/11
GMAT Prep1 680 ( 50, 31) on 10/11/11

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
CR notes
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133
http://gmatclub.com/forum/gmat-prep-critical-reasoning-collection-106783.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html?hilit=chineseburned


Last edited by catfreak on 14 Oct 2011, 12:15, edited 1 time in total.
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Re: Good Time and rate question [#permalink] New post 14 Oct 2011, 11:44
Thanks catfreak, you're exactly right. I wrote the problem, wrote the answers, wrote the explanation, then changed the problem to make it harder, without making other changes. I'll edit it right now. Good catch.
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Re: Good Time and rate question [#permalink] New post 14 Oct 2011, 12:18
Thanks catfreak !! good explanation .. I got 12 minutes as well.
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Re: Good Time and rate question [#permalink] New post 18 Oct 2011, 19:59
MichaelS wrote:
Howdy, I usually post just in the SC forum--find me there if you have SC questions--but I wanted to add something to the OP's excellent solution.

Whenever a difference in distances is described, you can subtract distances. The paradigm GMAT case is two people/cars/trains/whatever going in the same direction. Whenever a difference in distances is described, and both parties move simultaneously, you can also subtract rates: (difference in rates)*time=(difference in distance).

This question asks for a difference in the distances covered by two people walking simultaneously. The difference in rates is 20 laps/hour, or 1 lap/3 minutes. (1/3)t=4. t-12.

Many good test-takers find this formula, (difference in rates)*time=(difference in distance), so useful that they tweak problems with different times so that they can apply this formula. Let me give you an example, based on this question.

Q.Nick starts walking around a circular track at noon. Marla starts running around the same circular track at 12:20. Nick completes 12 laps per hour and Marla completes 32 laps per hour. At what time will Marla have completed as many laps as Nick?
(A) 12:25
(B) 12:28
(C) 12:32
(D) 12:35
(E) 12:40

Here's how to answer this as one hard question:
(Marla's time)(Marla's rate)=Marla's distance.
(Nick's time)(Nick's rate)=Nick's distance.

Nick's time is 20 minutes greater than Marla's.
You could say,
Let t= Marla's time in hours and t+1/3=Nick's time in hours.
But we'll go with,
Let t=Marla's time in minutes and t+20=Nick's time in minutes.
Marla's distance is Nick's.
Let d= Nick's distance and d=Marla's distance.
This yields two equations:
32t=d
12(t+20)=d
Solve for 12:20+t.

I'll trust you to solve algebraically.

Here's the same question rephrased as three easy questions:
What did Nick do before Marla started running? Walk 4 laps.
Once Marla started running--once they were moving simultaneously--how much faster did she go than Nick? 20 laps/hour, or 1 lap/3 minutes.
How long will it take her to gain 4 laps, if it takes her 3 minutes to gain 1 lap? 12 minutes.
Add 12 minutes to 12:20, the time from which they moved simultaneously. 12:32.

Notice that this second method would be harder to apply if we complicated the problem a bit. If we asked at what time Marla would have gone twice as far as Nick, the more formal algebraic method works just as neatly (just make Marla's distance 2d instead of d), but the alternative method gets much messier.

Okay, here's one more for you to solve. It's not a real GMAT question, and I've rigged it so that the second method described above will be much easier for most people. Answer below and I'll check in tomorrow with my explanation.

At 9:00, Nick started walking the 9 miles from A to B. At 9:40, Marla started walking the 9 miles from A to B. Until Marla caught up to Nick, Marla walked at a constant rate of 5 miles per hour, and Nick walked at a constant rate of 3 miles per hour. Once Marla caught Nick, they walked together at a constant rate of 4 miles per hour. At what time did they arrive together at B?


Hi:

Is the answer to the highlighted practice problem above 11:40?
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Re: Good Time and rate question [#permalink] New post 21 Oct 2011, 23:53
4=(32/60)t-(12/60)t
60*4=20t
t=12
Answer c
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Re: Good Time and rate question [#permalink] New post 22 Oct 2011, 01:11
At 9:00, Nick started walking the 9 miles from A to B. At 9:40, Marla started walking the 9 miles from A to B. Until Marla caught up to Nick, Marla walked at a constant rate of 5 miles per hour, and Nick walked at a constant rate of 3 miles per hour. Once Marla caught Nick, they walked together at a constant rate of 4 miles per hour. At what time did they arrive together at B?


Is the answer 11:40 for this ???

Explanation:
Suppose they both meet at point C somewhere in the path from A to B.
Nick reached point C at time t and Marla reached point C at time t+40/60 (as she started 40 mins late so equating time taken on both sides). Suppose distance between A and C is d. Then both of them covered the distance d in time t (for Nick) and t+40/60 (for Marla).

Using t=d/s, we have the equation:

d/3=d/5+40/60
-> d=5.

so, d=5 distance was covered by Marla in 1 hour.

Remaining distance is 9-5=4, which they cover with constant speed of 4, Another 1 hour to reach C.
So total of 2 hours from when Marla started.

Hence, they reach B at 11:40.
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Re: Good Time and rate question [#permalink] New post 31 Oct 2011, 11:26
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Thanks for letting me hijack your thread gmatcracker.

Here's the question I asked:

At 9:00, Nick started walking the 9 miles from A to B. At 9:40, Marla started walking the 9 miles from A to B. Until Marla caught up to Nick, Marla walked at a constant rate of 5 miles per hour, and Nick walked at a constant rate of 3 miles per hour. Once Marla caught Nick, they walked together at a constant rate of 4 miles per hour. At what time did they arrive together at B?

Here's a relatively easy way to solve it:

What happened from 9:00 to 9:40?
Nick walked 2 miles (3mph*2/3 of a mile)

What was the difference in the rates from that point till Marla caught Nick?

2mph (5mph-3mph)

How long did it take for Marla to catch Nick?

1 hour (2mph*1 hour=2 miles)

What time was it then?
10:40

How long had each walked?
5 miles each (Marla had walked for 1 hour at 5mph)

How many miles remained?
4 miles (9 miles-5 miles)

How long did it take them to cover that distance?
1 hour

What time was it then?
11:40

I think that that is substantially the same as ashimood's response. Certainly it's the same result.
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Re: Good Time and rate question [#permalink] New post 02 Nov 2011, 07:41
Marlas speed is 32 lph and Nicks speed is 12 lph. So in x hours they'll each over 32x and 12x laps. (Don't forget to convert x to minutes).

From the problem: 32x = 12x + 4 -> x = 1/5 hours or 12 minutes.
Re: Good Time and rate question   [#permalink] 02 Nov 2011, 07:41
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