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# Inequality question

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Manager
Joined: 28 Aug 2006
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Location: Albuquerque, NM
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03 Jan 2007, 22:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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SVP
Joined: 01 May 2006
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04 Jan 2007, 00:28
(E) for me

In p/k < 0, we must be sure that k is not equal to 0.

From 1
k could be 0, thus we could not conclude.

INSUFF.

From 2
k could be 0, thus we could not conclude.

INSUFF.

Both (1) and (2)
p - k > 0 (C)
<=> -p + k < 0 (A)

Since p + k < 0 (B), we have :
(A) + (B) <=> 2*k < 0 <=> k < 0.

Also,
(C) <=> p > k <=> p/k < 1 as k < 0
(B) <=> p < -k <=> p/k > -1 as k < 0

Thus,
-1 < p/k < 1 : it does reflect that p could be either a negative number or a positive one but superior to k.

INSUFF
Manager
Joined: 28 Aug 2006
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Location: Albuquerque, NM
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04 Jan 2007, 07:21
I had also got E, by picking numbers but OA is not E
SVP
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04 Jan 2007, 07:41
jainan24 wrote:
I had also got E, by picking numbers but OA is not E

So... What is the OA, perhaps OE?
Current Student
Joined: 06 Sep 2005
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05 Jan 2007, 09:26
it has to be E.

Statement 1 says that P + K < 0.

Both numbers can be negative, thus P/K > 0.
Or, one number can be negative and the other can be positive, thus P/K < )

Statement 2 essentially says that P > K. Both numbers can be positive, or one number can be positive.

If you combine them, you can say that P = 0 and K = any negative number to satisfy both constraints. That would mean that P + K = 0. But, if you use P = 2 and K = -3, this also satisfies both statements and makes P/K < 0.
Senior Manager
Joined: 23 Jun 2006
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05 Jan 2007, 09:55
don't care whatthe OA is, answer is E... here is the ultimative proof (that is, a counter example):

p=2 k =-5: st1 is true (2-5<0) and st2 is true (2-(-5)=7>0)
answer to stem is YES (p/k = -0.4 <0)

p=-2 k=-5: st1 is true (-2-5<0) and st2 is true (-2-(-5)=3>0)
answer to stem is NO (p/k = 0.4 > 0)

SVP
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05 Jan 2007, 15:26
jainan24,

Is this question comming from McGrawhill GMAT one more time?

Would u make give us the OE & OA officially given?... Thanks
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