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

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Intern
Joined: 10 Jan 2017
Posts: 4
Re: Tom and Linda stand at Point A. Linda begins to walk in a straight  [#permalink]

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15 Feb 2017, 17:24
I am trying to use relative speed to solve this. So my thoughts are that Linda after 1 hour is 2 miles away, at which point Tom starts jogging. If I imagine Linda stationary, Tom's relative speed would be 8 miles per hour, and I would have to find the time it would take him to run 1 mile and 4 miles, but that doesn't work. Can somone explain to me why relative speed does not work in this case. Thank you!
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Tom and Linda stand at Point A. Linda begins to walk in a straight  [#permalink]

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Updated on: 16 Feb 2017, 18:47
2
Corrected....

D=RT
Linda's Distance: D = 2 + 2t, because she has a one hour advantage (of 2 miles) and moves at 2mph (or 2t).
Tom's Distance: D = 6t

To find the amount of time needed for Tom to reach half the distance Linda reached, we divide Linda's RT in half and set this equal to Tom's RT ----> (2+2t)/2 = 6t ----> t=1/5 ----> then multiply by 60 minutes and we get 12 minutes.

To find amount of time needed for Tom to reach double the distance Linda reached, we multiply Linda's RT by 2 and set this equal to Tom's RT ----> 2(2+2t) = 6t ----> t=2 ----> then multiply by 60 minutes and we get 120 minutes.

The difference between these two is 108 minutes. The answer is E.

Originally posted by mbah191 on 15 Feb 2017, 18:53.
Last edited by mbah191 on 16 Feb 2017, 18:47, edited 1 time in total.
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Re: Tom and Linda stand at Point A. Linda begins to walk in a straight  [#permalink]

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15 Feb 2017, 19:12
1
E. 108
rate Tom: Rt
rate Linda: Rl
distance covered by Tom: Dt
distance covered by Linda: Dl
We have in certain amount of time: Rt/Rl=Dt/Dl
When Tom covers twice the distance Linda has covered: (2+Dt/3)/Dt=1/2 => Dt=12 => it takes Tom 12/6= 2 hours= 120 mins
When Tom covers half the distance Linda covered: (2+Dt/3)/Dt=2 => Dt=6/5 => it takes Tom 12/(6/5)=1/5 hours=12 mins
=> the difference in time is 120-12=108 mins
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Re: Tom and Linda stand at Point A. Linda begins to walk in a straight  [#permalink]

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15 Feb 2017, 19:51
robypeter wrote:
I am trying to use relative speed to solve this. So my thoughts are that Linda after 1 hour is 2 miles away, at which point Tom starts jogging. If I imagine Linda stationary, Tom's relative speed would be 8 miles per hour, and I would have to find the time it would take him to run 1 mile and 4 miles, but that doesn't work. Can somone explain to me why relative speed does not work in this case. Thank you!

Hi,

The problem is that you are just working on initial 2 mile or 1 hour.... 1 mile or 4 miles
But the half distance and twice the distance involves travelled distance after initial 1 hour
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Tom and Linda stand at Point A. Linda begins to walk in a straight  [#permalink]

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16 Feb 2017, 01:00
lets say x is time (in hours) of jogger.
distance = time * rate

6*x is distance jogger jogs
x + 1 will be time first person walks

so now we will build equation

for half distance: 6x=1/2 * 2 *(x+1) => 6x = x+1 => x = 1/5 so 12 minutes
for twice the distance : 6x = 2 * 2 * (x+1) => 6x =4x +4 . x = 2 hours

difference = 120 min -12 min = 108
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Re: Tom and Linda stand at point A. Linda begins to walk in a  [#permalink]

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09 Jun 2019, 13:27
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Re: Tom and Linda stand at point A. Linda begins to walk in a   [#permalink] 09 Jun 2019, 13:27

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