It is currently 21 Nov 2017, 10:48

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 26 Jun 2007
Posts: 104

Kudos [?]: 83 [0], given: 0

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

### Show Tags

22 Jan 2008, 04:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Kudos [?]: 83 [0], given: 0

Director
Joined: 14 Oct 2007
Posts: 751

Kudos [?]: 235 [0], given: 8

Location: Oxford
Schools: Oxford'10
Re: Shared Velocity Q from MGMAT [#permalink]

### Show Tags

22 Jan 2008, 04:49
$$S_L = 2$$ therefore $$d_L = 2t$$
$$S_T = 6$$ therefore $$d_T = 6(t-1)$$ since he sets off an hour later

$$d_T = \frac{d_L}{2}$$ since we need to find out when tom covers half the distance

therefore in terms if t:

$$6(t-1) = \frac{2t}{2}$$
i.e $$t = 6(t-1)$$

solving for t we get $$t= \frac{6}{5}$$

in minutes thats 72 mins.. ==> B

Kudos [?]: 235 [0], given: 8

CEO
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4669 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: Shared Velocity Q from MGMAT [#permalink]

### Show Tags

22 Jan 2008, 05:09
E

1. $$6*t_1=\frac12*(2+2*t_1)$$

$$t_1=\frac{2}{10} hour = 12 min$$

2. $$6*t_1=2*(2+2*t_1)$$

$$t_1=2 hour = 120 min$$

$$t=t_2-t_1=108 min$$
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Kudos [?]: 4669 [0], given: 360

Director
Joined: 14 Oct 2007
Posts: 751

Kudos [?]: 235 [0], given: 8

Location: Oxford
Schools: Oxford'10
Re: Shared Velocity Q from MGMAT [#permalink]

### Show Tags

22 Jan 2008, 05:34
walker wrote:
E

1. $$6*t_1=\frac12*(2+2*t_1)$$

$$t_1=\frac{2}{10} hour = 12 min$$

2. $$6*t_1=2*(2+2*t_1)$$

$$t_1=2 hour = 120 min$$

$$t=t_2-t_1=108 min$$

walker is correct (no surprises here),

i miss read the question, I actually calculated the time it takes for Toms to cover half the distance as Linda

Kudos [?]: 235 [0], given: 8

Director
Joined: 14 Oct 2007
Posts: 751

Kudos [?]: 235 [0], given: 8

Location: Oxford
Schools: Oxford'10
Re: Shared Velocity Q from MGMAT [#permalink]

### Show Tags

22 Jan 2008, 05:39
$$S_L = 2$$ therefore $$d_L = 2t$$
$$S_T = 6$$ therefore $$d_T = 6(t-1)$$ since he sets off an hour later

$$d_T = \frac{d_L}{2}$$ since we need to find out when tom covers half the distance

therefore in terms if t:

$$6(t-1) = \frac{2t}{2}$$
i.e $$t = 6(t-1)$$

solving for t we get $$t= \frac{6}{5}$$

in minutes thats 72 mins.. ==> B

let me finish the above

now lets calculate time that Tom takes to cover twice the distance as Linda

$$2d_L = d_T$$

substitute with expressions with t:
$$4t = 6(t-1)$$

simply fy get t=3 hours = 180 mins

therefore 180-72 (the difference) = 108.

E

if i keep reading the question incorrectly i am doomed!!!!

Kudos [?]: 235 [0], given: 8

Manager
Joined: 26 Jun 2007
Posts: 104

Kudos [?]: 83 [0], given: 0

Re: Shared Velocity Q from MGMAT [#permalink]

### Show Tags

22 Jan 2008, 13:23
OA= E

Guys, I remember I saw a solution to speed problems that involved the consept of "shared speed", which deals with those situations by adding speeds (when items are helping each other like in the above case) or subtructing speeds (like a boat against a stream direction).

Any takers would like solving the above with this consept?

Kudos [?]: 83 [0], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1076 [0], given: 4

Location: New York City
Re: Shared Velocity Q from MGMAT [#permalink]

### Show Tags

19 Feb 2008, 11:08

meeting TIME = Distance/ (Rate1 + Rate2)
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Kudos [?]: 1076 [0], given: 4

CEO
Joined: 29 Mar 2007
Posts: 2553

Kudos [?]: 528 [0], given: 0

Re: Shared Velocity Q from MGMAT [#permalink]

### Show Tags

20 Feb 2008, 21:46
GGUY 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 was a relatively confusing rate prob. to set up for a bit.

L: 2(t+1)=d
T: 6t=?

half ---> 6t=(2t+2)/2 ---> t=1/5 or 12minutes
2x---> 6t=2(2t+2) --> 6t=4t+4 --> 2t=4 --> t=2 or 120minutes

so 120-12= 108 minutes

E

Kudos [?]: 528 [0], given: 0

Director
Joined: 30 Jun 2007
Posts: 780

Kudos [?]: 190 [0], given: 0

Re: Shared Velocity Q from MGMAT [#permalink]

### Show Tags

21 Feb 2008, 01:49
Linda Tom
R 2 6
T T+60 T
D D ?
Tom: Half of Linda’s distance: D/2 = 6T
D = 12 T
Linda: D = 2T + 120

12 T = 2T + 120
T = 12 minutes

Twice distance:

2D = 6T
D = 3T

Equate with Linda’s distance: 3t = 2T + 120
T = 120
Time difference: 120 – 12 = 108

Kudos [?]: 190 [0], given: 0

Re: Shared Velocity Q from MGMAT   [#permalink] 21 Feb 2008, 01:49
Display posts from previous: Sort by

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

Moderator: chetan2u

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.