GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 28 Jan 2020, 01:53 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Tom and Linda stand at point A. Linda begins to walk in a

Author Message
TAGS:

### Hide Tags

Intern  Joined: 09 Apr 2010
Posts: 43
Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

### Show Tags

5
44 00:00

Difficulty:   95% (hard)

Question Stats: 52% (03:15) correct 48% (02:57) wrong based on 426 sessions

### HideShow timer Statistics

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

Originally posted by neoreaves on 02 May 2010, 01:11.
Last edited by VeritasKarishma on 11 Sep 2012, 21:07, edited 1 time in total.
(Edited the OA)
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10023
Location: Pune, India
Re: Rates & Work: Walk Away  [#permalink]

### Show Tags

12
1
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.
_________________
Karishma
Veritas Prep GMAT Instructor

SVP  Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2465
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Re: Tom and Linda stand at point A.  [#permalink]

### Show Tags

19
5
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
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html
##### General Discussion
Manager  Joined: 17 Mar 2010
Posts: 125
Re: Tom and Linda stand at point A.  [#permalink]

### Show Tags

1
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
Intern  Joined: 05 Feb 2011
Posts: 8
Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

### Show Tags

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

Originally posted by HelloKitty on 05 Apr 2011, 19:29.
Last edited by VeritasKarishma on 11 Sep 2012, 21:09, edited 1 time in total.
Edited to avoid confusion
Manager  Joined: 09 Aug 2010
Posts: 74
Re: Rates & Work: Walk Away  [#permalink]

### Show Tags

5
1
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
Director  Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 631
Re: Rates & Work: Walk Away  [#permalink]

### Show Tags

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.)
Retired Moderator B
Joined: 16 Nov 2010
Posts: 1219
Location: United States (IN)
Concentration: Strategy, Technology
Re: Rates & Work: Walk Away  [#permalink]

### Show Tags

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

_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings
Director  Joined: 01 Feb 2011
Posts: 528
Re: Rates & Work: Walk Away  [#permalink]

### Show Tags

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

Posted from my mobile device
Senior Manager  Joined: 08 Nov 2010
Posts: 264
Re: Rates & Work: Walk Away  [#permalink]

### Show Tags

11
3
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.
Manager  Joined: 21 May 2012
Posts: 86
Location: United States (CA)
Re: Tom and Linda stand at point A.  [#permalink]

### Show Tags

1
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.
Manager  Joined: 12 May 2012
Posts: 67
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

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.
Intern  Joined: 31 Oct 2011
Posts: 18
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

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!
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10023
Location: Pune, India
Re: Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

### Show Tags

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

Manager  Status: Perspiring
Joined: 15 Feb 2012
Posts: 79
Concentration: Marketing, Strategy
GPA: 3.6
WE: Engineering (Computer Software)
Re: Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

### Show Tags

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 ?
Intern  Joined: 30 Oct 2011
Posts: 31
Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

### Show Tags

1
1
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)
Manager  Joined: 03 Jul 2013
Posts: 81
Schools: ISB '17 (A), IIMC (A)
GMAT 1: 660 Q48 V32
Re: Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

### Show Tags

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.
_________________
Sometimes standing still can be, the best move you ever make......
Manager  Joined: 06 Mar 2014
Posts: 220
Location: India
GMAT Date: 04-30-2015
Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

### Show Tags

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.
Director  G
Joined: 26 Oct 2016
Posts: 599
Location: United States
Schools: HBS '19
GMAT 1: 770 Q51 V44
GPA: 4
WE: Education (Education)
Re: Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

### Show Tags

Attached the correct and precise solution to this problem.

Linda's rate = 2 mph
Linda's time = t hours

Tom's rate = 6 mph
Tom's time = (t - 1) hours (since Linda has a one hour headstart).

When they travel the SAME distance, 2t = 6*(t-1). This reduces to t = 3/2.

Now we need Tom to double up Linda. This is the same equation as before, except we DOUBLE Linda's distance. This gives us 2*(2t) = 6*(t-1), or t = 3.

The difference between the times is 3 - (3/2), or (3/2) of an hour. (3/2) of an hour = 90 minutes, and we're done.
_________________
Thanks & Regards,
Anaira Mitch
Intern  B
Joined: 10 Jan 2017
Posts: 4
Tom and Linda stand at Point A. Linda begins to walk in a straight  [#permalink]

### Show Tags

1
2
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 diference, in minutes, between the amount of time it takes tom to cover half the distance that Linda has covered and the amount of time it takes tom to cover twice the distance that linda covered?

A) 60 B) 72 C) 84 D) 90 E) 108

Can someone help me solve this problem using relative speeds?

Thank you. Tom and Linda stand at Point A. Linda begins to walk in a straight   [#permalink] 15 Feb 2017, 17:22

Go to page    1   2    Next  [ 26 posts ]

Display posts from previous: Sort by

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