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ahistegt
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Post subject: Walk Away (2) Posted: Sun Feb 01, 2009 9:47 am |
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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 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
I don't understand why OE, below, uses "T" for both Tom and Linda. I'd thought it should be "T" and "T1+1".
I'd appreciate quick response; G-day is Tuesday :)
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OE:
The formula to calculate distance is Distance = (Rate)(Time). So at any given moment Tom's distance (let's call it DT) can be expressed as DT = 6T. So, at any given moment, Linda's distance (let's call it DL) can be expressed as DL = 2(T + 1) (remember, Linda's time is one hour more than Tom's). The question asks us to find the positive difference between the amount of time it takes Tom to cover Linda's distance and the time it takes him to cover twice her distance. Let's find each time separately first.
When Tom has covered Linda's distance, the following equation will hold: 6T = (2(T + 1)). We can solve for T:
6T = (2(T + 1))
6T = 2T + 2
4T = 2
T = 1/2
So it will take Tom 1/2 hour, or 30 minutes, to cover 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 – 30 = 90.
The correct answer is D.
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JonathanSchneider
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Post subject: Re: Walk Away (2) Posted: Thu Feb 19, 2009 2:45 pm |
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Posts: 480 Location: Durham, NC
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The solution is laid out this way because we want to first calculate the distance that Linda has traveled on her own. This = 2 miles. From there, we can say that they are traveling for the same amount of time, so that we can use "t" for both. We just need to add +2 to Linda's distance.
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craig
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Post subject: Re: Walk Away (2) Posted: Wed May 06, 2009 9:33 pm |
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Why do you take 6T = (2(T + 1)) for Linda. Linda depart one hour before Tom
instead 2T = (6(T - 1)) for Tom. Tom depart one hour after Linda
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RonPurewal
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Post subject: Re: Walk Away (2) Posted: Fri May 08, 2009 8:07 pm |
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craig wrote: Why do you take 6T = (2(T + 1)) for Linda. Linda depart one hour before Tom
instead 2T = (6(T - 1)) for Tom. Tom depart one hour after Linda that will also work. if you use that instead: 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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elaine1920
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Post subject: Re: Walk Away (2) Posted: Tue Aug 18, 2009 11:09 pm |
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I was trying to use another way to approach but got stuck in the middle. Could any expert please shed some lights on my approach? thank you.
If Linda has walked one hour ahead of Tom, then Linda has travelled 2 miles already ( 2 mile/hour * 1 hour) . So in this case, in order for Tom to cover the same distance Linda traveled, Tom needs to make up the difference for the 2 miles. Tom travelled 4 miiles faster than Linda, so he use 1/2 hours to walked the distance that Linda walked, which is 2miles / 4 miles per hour. So in this step, I got the first number for 30 mins.
But when come to calculate the second part, which is the amount of time for Tom to cover twice the distance that Linda travelled, I don't know how to continue this way. Please give me some pointers. thanks!
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Ben Ku
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Post subject: Re: Walk Away (2) Posted: Tue Aug 25, 2009 1:45 pm |
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Quote: If Linda has walked one hour ahead of Tom, then Linda has travelled 2 miles already ( 2 mile/hour * 1 hour) . So in this case, in order for Tom to cover the same distance Linda traveled, Tom needs to make up the difference for the 2 miles. Tom travelled 4 miiles faster than Linda, so he use 1/2 hours to walked the distance that Linda walked, which is 2miles / 4 miles per hour. So in this step, I got the first number for 30 mins.
Hi Elaine: I'm not sure that this approach is correct. There are two portions that need to be considered: the portion where Linda walked by herself, and the portion where they both traveled away from each other. Here, the 2 miles you used is ONLY the distance Linda walked by herself, while the 4miles per hour is the difference in rate when they traveled together. Even though the numerical answer is correct, the process is not correct. That is also why I think you are having trouble setting up the second part.
_________________ Ben Ku Instructor ManhattanGMAT
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karchopra
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Post subject: Re: Walk Away (2) Posted: Mon Dec 27, 2010 12:26 pm |
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Unfortunately, I still dont completely get the answer to this question.
For the first part we get T = 1/2
As Tom travels at 6 miles/hr. Distance = 3 miles. This is the same distance covered by Linda.
Now the second part asks us "amount of time it takes Tom to cover twice the distance that Linda has covered"
This informs us to double the distance that Linda has covered = 3 * 2 = 6 miles.
Now the new time is 6 miles/6 miles/hr = 1 hr!
Thus the difference should be 30 minutes!
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tim
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Post subject: Re: Walk Away (2) Posted: Mon Dec 27, 2010 8:20 pm |
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Posts: 2242 Location: Southwest Airlines, seat 21C
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Try again. By the time Tom has covered 6 miles, Linda has covered more than just 3. You need to adjust both of their distances upward as you run the clock forward..
_________________ Tim Sanders Manhattan GMAT Instructor
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