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m03#28

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m03#28 [#permalink] New post 11 Jan 2011, 10:58
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Lionel left his house and walked towards Walt's house, 48 miles away. Two hours later, Walt left his house and ran towards Lionel's house. If Lionel's speed was 4 miles per hour and Walt's 6 miles per hour, how many miles had Lionel walked when he met Walt?


Let denote the time it took Lionel and Walt to meet after Walt left his house. The product of speed and time gives us the distance traveled. Construct the following equation in which each side respectively indicates the Lionel's and Walt's location:

2*4 +4t = 48-6t

how this equation is made ? :roll:
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Re: m03#28 [#permalink] New post 11 Jan 2011, 11:23
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This equation is based on thinking of the problem as a story about distance. How far has Lionel traveled? If we know that d=rt, then we know that first he traveled 4 miles an hour for 2 hours (that gives us the 2*4 term) and then 4 miles an hour for another unknown amount of time (that gives us the 2*t term). But how far does Lionel get? He doesn't quite go the whole 48 miles, because Walt meets him partway. So now we ask ourselves, "How far has Walt traveled?" to which the answer is, again, since d=rt, rate(6) times time (the same unknown t). This gives us 6t for Walt's distance. Since Lionel covers all of the 48 miles except for what Walt covers, his distance -- 2*4 + 4t -- equals 48 - 6t. That's where this equation comes from.

Another way of thinking about these problems that leads to virtually the same equation is to realize that, regardless of what else is going on, Walt's total distance and Lionel's total distance must add up to 48. Again, let "t" stand for the time when BOTH are traveling. So Lionel's total distance is 2*4 + 4t -- or, if you prefer, 4(t+2) since he is traveling for 2 hours longer due to his head start -- and Walt's is 6t. Since, when they meet, their total distance must be the whole 48, the equation is 4(t+2) + 6t = 48. Thus, you can quickly say, is basically the same equation.

While writing out equations is the key to success on most of the GMAT, when two things are moving, we can shortcut the process as follows. First, figure out where they are when BOTH are moving. Since Lionel gets a head start, he goes 8 miles in that time. This means that, at the first moment hen BOTH are moving, they are 40 miles apart. Now, *whenever two things are moving towards one another, you simply add their speeds*. This is because it doesn't matter who is moving at which speed; the total speed with which the two are getting *closer* is simply 4+6 = 10 mph. Now it's just simple mental math: How long does it take to cover 40 miles at 10 mph? 4 hours, of course! So t=4, whichever way you do the problem.

Finally, it's worth noting that this problem asks for the distance, not the time. On any such problem, you'll find the time first, and then plug in to find the distance, as we have here. t=4, so Lionel walks for 6 total hours, making for a total of 24 total miles. After all that, they meet half-way; who'd've thunk it?
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Re: m03#28 [#permalink] New post 12 Jan 2011, 07:44
Adam: That is brilliant +1
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Re: m03#28 [#permalink] New post 13 Jan 2011, 10:23
Wow , you are the best Adam :-D
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Re: m03#28 [#permalink] New post 05 Mar 2011, 05:11
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Another way to think about that question is:
first two hours he walked 4*2=8miles.
so right now they have 48-8=40 to walk towards each other

Now - we know their speeds is in a rate of 4:6 so we can say that each ratio size is 40/10=4

so Lionel walked 4*4=16
so 16+8 = 24. done.
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Re: m03#28 [#permalink] New post 02 Oct 2011, 00:15
144144 wrote:
Another way to think about that question is:
first two hours he walked 4*2=8miles.
so right now they have 48-8=40 to walk towards each other

Now - we know their speeds is in a rate of 4:6 so we can say that each ratio size is 40/10=4

so Lionel walked 4*4=16
so 16+8 = 24. done.



Great explanation. I was try to figure out how to solve it this way.
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Re: m03#28 [#permalink] New post 04 Oct 2011, 03:21
Plz, refer this thread with more explainations http://gmatclub.com/forum/distance-m03q28-59171.html
Re: m03#28   [#permalink] 04 Oct 2011, 03:21
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