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Old 06-29-2008, 10:57 PM   #1 (permalink)
ustado
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another comparison problem

http://img32.picoodle.com/img/img32/...pm_105c68e.png

I am not sure how the answer is A..
This is coming from the big book.
Can anyone explain?
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Old 06-30-2008, 02:22 AM   #2 (permalink)
ustado
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any takers?
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Old 06-30-2008, 03:15 AM   #3 (permalink)
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On solving, you will realize that 2n+r = s + 4/3*t

Comparing s + 4/3*t with 2s + t. we get 1/3t compared with s. Seeing that they can take just about any values, I feel that the answer is D.
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Old 06-30-2008, 05:26 AM   #4 (permalink)
geek_goddess
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You sure that the 8 under the 's' is an 8? And not a 3?
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Old 06-30-2008, 07:49 AM   #5 (permalink)
chestnut.cc
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After much tomfoolery... the ans is straightforward... to elaborate... two possible solutions are:
(2,4,4,3),(8,1,1,12)... you can see that for the first A<B... for the second A>B... D

Last edited by chestnut.cc : 06-30-2008 at 01:18 PM.
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Old 06-30-2008, 12:54 PM   #6 (permalink)
Jeanette
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Let's make all the fractions equal to 1.

Then n=4
r=8
s=8
t=6

2n+r=16
2s+t=16+6=22

Does the q say that the integers have to be different? Here is at least one case where B is greater, so the answer can't be A.
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Old 06-30-2008, 01:20 PM   #7 (permalink)
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yes... nowhere does it say that they cant be the same...
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Old 07-01-2008, 02:52 AM   #8 (permalink)
ivsham
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This is an older thread: http://www.urch.com/forums/gre-math/...ple-hours.html (This qc bugged me for a couple of hours)
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Old 07-02-2008, 09:45 AM   #9 (permalink)
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it seems that the answer is D.
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