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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 22 Jan 2008, 03:09
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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

Please Explain
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Re: Shared Velocity Q from MGMAT [#permalink] New post 22 Jan 2008, 03:49
\(S_L = 2\) 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
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Re: Shared Velocity Q from MGMAT [#permalink] New post 22 Jan 2008, 04:09
Expert's post
E

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

\(t_1=\frac{2}{10} hour = 12 min\)

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

\(t_1=2 hour = 120 min\)

\(t=t_2-t_1=108 min\)
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Re: Shared Velocity Q from MGMAT [#permalink] New post 22 Jan 2008, 04:34
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
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Re: Shared Velocity Q from MGMAT [#permalink] New post 22 Jan 2008, 04:39
buffdaddy wrote:
\(S_L = 2\) 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 :wall

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!!!!
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Re: Shared Velocity Q from MGMAT [#permalink] New post 22 Jan 2008, 12:23
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?
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Re: Shared Velocity Q from MGMAT [#permalink] New post 19 Feb 2008, 10:08
its additive rates.

meeting TIME = Distance/ (Rate1 + Rate2)
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Re: Shared Velocity Q from MGMAT [#permalink] New post 20 Feb 2008, 20:46
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

Please Explain


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
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Re: Shared Velocity Q from MGMAT [#permalink] New post 21 Feb 2008, 00:49
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

Answer: E
Re: Shared Velocity Q from MGMAT   [#permalink] 21 Feb 2008, 00:49
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Tom and Linda stand at point A. Linda begins to walk in a

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