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

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Manager
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Tom and Linda stand at point A. Linda begins to walk in a [#permalink]

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02 May 2010, 01:11
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Difficulty:

95% (hard)

Question Stats:

47% (02:17) correct 53% (01:49) wrong based on 764 sessions

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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, 21:07, edited 1 time in total.
(Edited the OA)

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Re: Tom and Linda stand at point A. [#permalink]

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02 May 2010, 04:57
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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: $$6*t2 = 2* (t2+1) * 2 => t1 = 2 = 120$$minutes

Difference = 120-12 = 108 minutes
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Re: Tom and Linda stand at point A. [#permalink]

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03 Sep 2010, 14:12
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]

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05 Sep 2010, 00:27
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]

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05 Apr 2011, 19:29
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, 21:09, edited 1 time in total.
Edited to avoid confusion

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Re: Rates & Work: Walk Away [#permalink]

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05 Apr 2011, 19:44
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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]

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05 Apr 2011, 19:58
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:

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]

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05 Apr 2011, 20:02
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

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Re: Rates & Work: Walk Away [#permalink]

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17 Apr 2011, 17:41
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

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Re: Rates & Work: Walk Away [#permalink]

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17 Apr 2011, 18:21
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18098 [9], given: 236 Senior Manager Joined: 08 Nov 2010 Posts: 388 Kudos [?]: 133 [10], given: 161 WE 1: Business Development Re: Rates & Work: Walk Away [#permalink] ### Show Tags 19 Apr 2011, 05:56 10 This post received KUDOS 3 This post was BOOKMARKED 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. _________________ Kudos [?]: 133 [10], given: 161 Current Student Joined: 21 May 2012 Posts: 97 Kudos [?]: 42 [0], given: 0 Location: United States (CA) Re: Tom and Linda stand at point A. [#permalink] ### Show Tags 31 May 2012, 13:29 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. Kudos [?]: 42 [0], given: 0 Manager Joined: 12 May 2012 Posts: 80 Kudos [?]: 121 [0], given: 14 Location: India Concentration: General Management, Operations GMAT 1: 650 Q51 V25 GMAT 2: 730 Q50 V38 GPA: 4 WE: General Management (Transportation) Re: Rates & Work: Walk Away [#permalink] ### Show Tags 01 Jun 2012, 03:14 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. Kudos [?]: 121 [0], given: 14 Intern Joined: 31 Oct 2011 Posts: 19 Kudos [?]: 35 [0], given: 2 Schools: ESSEC '15 (A) GMAT 1: 650 Q45 V35 Re: Tom and Linda stand at point A. Linda begins to walk in a [#permalink] ### Show Tags 11 Sep 2012, 04:14 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! Kudos [?]: 35 [0], given: 2 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7788 Kudos [?]: 18098 [0], given: 236 Location: Pune, India Re: Tom and Linda stand at point A. Linda begins to walk in a [#permalink] ### Show Tags 11 Sep 2012, 21:10 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Rates & Work: Walk Away [#permalink]

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03 May 2014, 06:54
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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

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Re: Tom and Linda stand at point A. Linda begins to walk in a [#permalink]

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02 Sep 2014, 09:21
I have a doubt. I took another approach & I am getting a different answer.
I want to know, whats wrong with the approach below...

After 1 hour :
L ------------------------------- T
<- 2miles ->
(2m/hr) (6m/hr)

Now suppose it takes time T for Tom to travel half the distance of Linda.
In this time T, distance traveled by Linda = d @ a speed of 2

Therefore : d/2 = [(2+d)/2]/6 ----------------------- (1) Dist/speed = time, Time is same when Linda moved d, & Tom moved half of (initial 2 + d)
Solving (1), d = 2/5
Now Time taken by Tom to travel above : (2/5)/6 -------------------------------- (2)

Similarly, d'/2 = [2(d+2)]/6 ----------------------- (3) Dist/speed = time, Time is same when Linda moved d', & Tom moved twice of (initial 2 + d')
Solving (2), d' = 4
Now Time taken by Tom to travel above : 4/6 -------------------------------- (4)

Taking the diff, (1) - (2)
Time = 3/5 Hrs = (3/5)*60 = 36 mins.

But no such option exists !!
Please let me know where am I going wrong ?

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Tom and Linda stand at point A. Linda begins to walk in a [#permalink]

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03 Oct 2014, 16:46
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scofield1521 wrote:
Cannot remember the above tricks for exam day. Need more solutions for this!!

It was difficult for me to understand what exactly the question needs and then how to use formula to solve the question. After too much thinking, I've found the below sol.

scenario 1: Tom and Linda cover same distance in different time i.e. here distance is same and the no. of hours taken are different by 1 hour:

Linda

Rate: 2
Distance: d
Time: (d/2)

Tom
Rate: 6
Distance: d
Time: (d/2) -1 or d/6

Solving, (d/2)-1 = d/6 => d=3

time taken by tom = (3/2) -1 = 0.5 hour or 3/6 = 0.5 hour

scenario 2: Here we need to find out in how much time distance covered by Tom would be double of the distance covered by Linda i.e here distance is different and the no. of hours taken are different by one hour

Linda
Rate: 2
Distance: d
Time: (d/2)

Tom
Rate: 6
Distance: 2d
Time: (d/2) -1 or 2d/6

Solving, (d/2)-1 = 2d/6 => (d-2)/2=d/3 => 3d-6=2d => d=6

time taken by tom = (6/2) -1 =2 or (2*6)/6 = 2 hours

Hence scenario 1 - scenario 2 = 1.5 hours = 90 min is the answer (D)

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Re: Tom and Linda stand at point A. Linda begins to walk in a [#permalink]

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08 Oct 2014, 02:10
My Approach:

Say linda covers 2 miles in 1 hour and stops. Tom's speed will become 6-2=4mph.

Now tom takes 30 minutes to cover 2 miles and 60 minutes to cover 4 miles. Answer should be 30 mins. Can someone please explain where i am wrong because this is the strategy i use in most of the speed questions.
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Tom and Linda stand at point A. Linda begins to walk in a [#permalink]

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10 Nov 2014, 11:43
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

This is perhaps the easiest OR one of the tough ones.

Everything depends on how one interprets the highlighted part.

The Question could have instead easily stated that if the distance travelled by Tom is equal to the distance travelled by Linda and if the distance travelled by Tom is equal to twice the distance travelled by Linda.

We could easily take the difference of the time taken by tom, only after first equating Distance travelled by tom to given both conditions and subsequently finding the value of time taken by tom in each case.

The language was the trick here.

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Tom and Linda stand at point A. Linda begins to walk in a   [#permalink] 10 Nov 2014, 11:43

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