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Tom and Linda stand at point A. Linda begins to walk in a

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Tom and Linda stand at point A. Linda begins to walk in a [#permalink] New post 02 May 2010, 00:11
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Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover half of the distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

A. 60
B. 72
C. 84
D. 90
E. 108
[Reveal] Spoiler: OA

Last edited by VeritasPrepKarishma on 11 Sep 2012, 20:07, edited 1 time in total.
(Edited the OA)
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Re: Tom and Linda stand at point A. [#permalink] New post 02 May 2010, 03:57
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neoreaves wrote:
Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover half of the distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

a)60
b)72
c)84
d)90
e)108


IMO E - 108

Case 1: 6*t1 = \frac{1}{2}* (t1+1) * 2 => t1 = \frac{1}{5} hour = 12 minutes

Case 2: 6*t2 = 2* (t2+1) * 2 => t1 = 2 = 120minutes

Difference = 120-12 = 108 minutes
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Re: Tom and Linda stand at point A. [#permalink] New post 05 May 2010, 03:20
I quite did not understand the solution. Could you please explain, Majorshareholder. Thanks,
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Re: Tom and Linda stand at point A. [#permalink] New post 03 Sep 2010, 13:12
When I solve this problem by picking numbers for total distance (not algebraically as gurpreet showed) I get different answer.....:(
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Re: Tom and Linda stand at point A. [#permalink] New post 04 Sep 2010, 23:27
E is the answer....
D = TS where D=distance, T=Time and S=Speed
To travel half distance, (2+2T) = 6T ==> T = 1/5 ==> 12 minutes
To travel double distance, 2(2+2T) = 6T ==> 2 ==> 120 minutes
Difference, 108 minutes
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Tom and Linda stand at point A. Linda begins to walk in a [#permalink] New post 05 Apr 2011, 18:29
Another version of the same question

Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover the exact distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

A. 60
B. 72
C. 84
D. 90
E. 120

Last edited by VeritasPrepKarishma on 11 Sep 2012, 20:09, edited 1 time in total.
Edited to avoid confusion
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Re: Rates & Work: Walk Away [#permalink] New post 05 Apr 2011, 18:44
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HelloKitty wrote:
Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover the exact distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

A) 60
B) 72
C) 84
D) 90
E) 120


My Solution:
Lrate: 2mph
Trate: 6mph

Ltime: t + 1 hour
Ttime: t hour

Ldistance: 2t + 2
Tdistance: 6t

T to cover L's distance: 2t + 2 = 6t, t = 1/2 hour
T to cover 2L distance: 2 (2t +2) = 6t, 2t = 4, t = 2 hours

2 - 1/2 = 1.5 hours = 90 minutes
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Re: Rates & Work: Walk Away [#permalink] New post 05 Apr 2011, 18:58
This approach is same as mine. But there seems to be a gap between our thinking and Ron's although the numerical answer is same. See this article - http://www.manhattangmat.com/forums/wal ... t6180.html. Couldn't put this in right perspective :(

Quote:
if you use that instead:

first situation:
2t = 6(t - 1)
2t = 6t - 6
6 = 4t
3/2 = t
(notice this is the same as above: the two times are t = 3/2 and (t - 1) = 1/2. in the above, they were t = 1/2 and (t + 1) = 3/2.)

second situation:
2(2t) = 6(t - 1)
4t = 6t - 6
6 = 2t
3 = t
(notice this is the same as above: the two times are t = 3 and (t - 1) = 2. in the above, they were t = 2 and (t + 1) = 3.)
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Re: Rates & Work: Walk Away [#permalink] New post 05 Apr 2011, 19:02
2t = 6(t-1)

=> t = 6/4 = 3/2 hrs

2* 2T = 6(T - 1)

=> 4T = 6T - 6

=> T = 3 hrs

So T - t = 3 - 3/2 = 3/2 hrs

Time in min = 90 min

Answer - D
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Re: Rates & Work: Walk Away [#permalink] New post 17 Apr 2011, 16:41
6t = 2(t+1) => t = (1/2) hr
6t = 2* 2(t+1) => t =2 hrs
Positive difference = 2-(1/2)
=(3/2) hrs
= 90 minutes
Answer is D.

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Re: Rates & Work: Walk Away [#permalink] New post 17 Apr 2011, 17:21
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Note that there are two different questions being discussed here:
One posted by neoreaves, the original poster - the answer to that is 108 mins;
the other posted by HelloKitty - the answer to that is 90 mins.

Both are based on the same logic but ask a different question.

Here I am discussing the logic and providing the answer to the question asked by HelloKitty.

HelloKitty wrote:
Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover the exact distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

A) 60
B) 72
C) 84
D) 90
E) 120


There is also a logical way to answer this without equations (you still may want to stick to equations in such questions during the exam but consider the logical solution an intellectual exercise)

Say Linda starts at 12:00. In an hour i.e. at 1:00, Linda has traveled 2 miles. Now Tom needs to cover the distance that Linda is covering now plus he has to cover the extra 2 miles to cover the same distance as Linda. Out of his speed of 6 mph, 2 mph is utilized in covering what Linda is covering right now (since Linda's speed is also 2 mph) and the rest 4 mph can be used to catch up the 2 miles. So it will take him half an hour (2miles/4mph) to cover as much distance as Linda has covered.

Now, at 1:30, they are both 3 miles away from point A. Now, Tom has to cover twice the distance that Linda covers from now on and he has to cover another 3 miles (to double Linda's current distance of 3 miles). From now on, 4mph of his 6 mph speed will go in covering twice of what Linda is covering at 2mph and the rest 2 mph of his 6 mph speed will go in covering the extra 3 miles that he has to cover. So it will take him 1.5 hours (3miles/2mph) to cover double of what Linda covers.

Since it took him 1.5 hrs (90 mins) extra after covering the same distance as Linda, this is the required time difference.
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Re: Rates & Work: Walk Away [#permalink] New post 19 Apr 2011, 04:56
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i did it very simple similar to Karishma

after 1 hour - L=2, T=0
after 1.5 - L=3, T=3 (first timing point)
after 2 hours - L=4, T=6
after 2.5 hours - L=5, T=9
After 3 hours - L=6, T=12. DONE!
20 seconds! very safe way.
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Re: Tom and Linda stand at point A. [#permalink] New post 31 May 2012, 12:29
E

When Tom has covered 1/2 Linda's distance, the following equation will hold: 6T = 0.5(2(T + 1)). We can solve for T:
6T = 0.5(2(T + 1))
6T = 0.5(2T + 2)
6T = T+1
5T = 1
T = 1/5

So it will take Tom 1/5 hour, or 12 minutes, to cover 1/2 Linda's distance. When Tom has covered twice Linda's distance, the following equation will hold: 6T = 2(2(T + 1)). We can solve for T:
6T = 2(2(T + 1))
6T = 2(2T + 2)
6T = 4T + 4
2T = 4
T = 2

So it will take Tom 2 hours, or 120 minutes, to cover twice Linda's distance.
We need to find the positive difference between these times: 120 – 12 = 108.
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Re: Rates & Work: Walk Away [#permalink] New post 01 Jun 2012, 02:14
VeritasPrepKarishma wrote:
HelloKitty wrote:
Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover the exact distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered?

A) 60
B) 72
C) 84
D) 90
E) 120


There is also a logical way to answer this without equations (you still may want to stick to equations in such questions during the exam but consider the logical solution an intellectual exercise)

Say Linda starts at 12:00. In an hour i.e. at 1:00, Linda has traveled 2 miles. Now Tom needs to cover the distance that Linda is covering now plus he has to cover the extra 2 miles to cover the same distance as Linda. Out of his speed of 6 mph, 2 mph is utilized in covering what Linda is covering right now (since Linda's speed is also 2 mph) and the rest 4 mph can be used to catch up the 2 miles. So it will take him half an hour (2miles/4mph) to cover as much distance as Linda has covered.

Now, at 1:30, they are both 3 miles away from point A. Now, Tom has to cover twice the distance that Linda covers from now on and he has to cover another 3 miles (to double Linda's current distance of 3 miles). From now on, 4mph of his 6 mph speed will go in covering twice of what Linda is covering at 2mph and the rest 2 mph of his 6 mph speed will go in covering the extra 3 miles that he has to cover. So it will take him 1.5 hours (3miles/2mph) to cover double of what Linda covers.

Since it took him 1.5 hrs (90 mins) extra after covering the same distance as Linda, this is the required time difference.


Logic always beats everything.
It was beautifully explained.
U made it very simple to understand.
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Re: Tom and Linda stand at point A. Linda begins to walk in a [#permalink] New post 11 Sep 2012, 03:14
Really confusing!!


Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover the exact distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered? The answer is 90min

Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover half of the distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered? The answer is 108 min

The answer depends on the question stem! Therefore the OA is not correct!
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Re: Tom and Linda stand at point A. Linda begins to walk in a [#permalink] New post 11 Sep 2012, 20:10
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Maxswe wrote:
Really confusing!!


Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover the exact distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered? The answer is 90min

Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 6 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover half of the distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered? The answer is 108 min

The answer depends on the question stem! Therefore the OA is not correct!


Yes, there are two different versions and hence the different answers. I have edited the OA. Hope it sorts out the confusion.
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Re: Tom and Linda stand at point A. Linda begins to walk in a [#permalink] New post 28 Nov 2013, 06:56
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Re: Tom and Linda stand at point A. Linda begins to walk in a [#permalink] New post 10 Jan 2014, 06:08
hi bunuel,

this question has been troubling for a while could you please provide an alternative method for this question?

Also, I have been been trying to use the concept of relative speed in the context of this question, ( both bodies moving in the opposite direction) but I dont seem to reach the answer with that... do you have a way to solve with the relative speeds ? or the only way to solve is the way its been mentioned?

Please guide me. Thanks.
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Re: Rates & Work: Walk Away [#permalink] New post 03 May 2014, 05:54
144144 wrote:
i did it very simple similar to Karishma

after 1 hour - L=2, T=0
after 1.5 - L=3, T=3 (first timing point)
after 2 hours - L=4, T=6
after 2.5 hours - L=5, T=9
After 3 hours - L=6, T=12. DONE!
20 seconds! very safe way.


Indeed quick but relying on the premise that answer will be coming out early in the table
Re: Rates & Work: Walk Away   [#permalink] 03 May 2014, 05:54
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