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Is p a negative number? [#permalink]
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14 Jul 2010, 07:52
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Is p a negative number? (1) p^3(1 – p^2) < 0 (2) p^2 – 1 < 0
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Last edited by Bunuel on 05 Jul 2013, 00:01, edited 1 time in total.
Edited the question and added the OA



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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 08:12
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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 08:18
I assume that stmt 1 is \(p^21<0\) & stmt 2 is \(p^3(1p^2)\) Stmt 1. \(p^3(1p^2)<0\) Either p\(^3<0\) or \(1p^2<0\) \(p<0\) or \(p^2>1\) \(p<0\) or \(p>1\) or \(p<1\) so not sufficient Stmt 2 \(p^21<0\) \((p1)(p+1)<0\) either \(p<1\) or \(p<1\) Not sufficient again Combing the two, again multiple answers. So "E" for me. I think Bunuel can clarify better, I can be wrong
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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 08:21



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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 08:28
The fastest way to solve this problem is to plot a graph. I would love to help but I lack right now the resources to post a picture. I believe that Bunuel will soon show up and help us. I will try to explain anyway: (1) Consider three parallel number lines (p) , one for \(p^3\), one for \(1p^2\) and a third one to represent the product of the these two functions. The "+" and "" represents the sign (Y) of the functions. A: \(p^3\):(1)0+++(+1)++++ B: \(1p^2\):(1)+++0+++(+1) A*B:++++++++(1)0+++(+1) > So p can be either positive or negative = Insuff.(2) \(p^2  1\) ++++(1)(0)(+1)++++++ Same as in (1), p can be either positive or negative = Insuff.(1) and (2) together shows a clear intersection when p < 0, so Suff.



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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 08:32
Wow, I was writting my reply and in the mean time you guys had already posted !



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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 08:46
Hussain15 wrote: Bunuel, If I put 2 in Stmt 1 then it comes true. So stmt 1 also shows p>1.
Where am I wrong? Statement (1) is true for: \(p>1\) (1.5, 2, 7, 12.5, ...) AND \(1<p<0\) (1/2, 3/4, ...). Statement (2) is true for: \(1<p<1\) (5/6, 2/9, 0, 1/2, 3/4, 7/8, ...). (1)(0)(1) (1)(0)(1)Combined: (1)(0)(1)
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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 11:50
shouldn't the question be
is p negative??
in the above solution we considered "p" in the range 1<p<0



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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 12:23



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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 13:08
Bunuel wrote: FedX wrote: shouldn't the question be
is p negative??
in the above solution we considered "p" in the range 1<p<0 Question: is \(p\) negative? Is \(p<0\)? When we considered statements together we've got that \(1<p<0\): every \(p\) from this range is negative (every \(p\) from this range is \(<0\)). Hence taken together statements are sufficient. Answer: C. Hope it's clear. This is fine..but the original poster has put it as "Is p a negative number"? Do they both mean the same??



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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 13:20



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Re: DS  inequality problem [#permalink]
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14 Jul 2010, 13:34
Thanks for the clarification Bunuel !!.



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Re: DS  inequality problem [#permalink]
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22 Jul 2010, 00:41
C should be the correct answer, just see the following counterexamples: statement 1: p=2 statement 2: p=0.1 Considering both statements, p3 must be negative then p negative
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Re: Is p a negative number? [#permalink]
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22 Sep 2014, 10:02
zest4mba wrote: Is p a negative number?
(1) p^3(1 – p^2) < 0 (2) p^2 – 1 < 0 1) the inequality holds true for +ve numbers as well as 1<p<0 so insufficient. 2) the inequality holds true for 0<p<1 & 1<p<0 or 1<p<1 so insufficient. (1)+(2) (2) says p^2  1 < 0 or (1p^2) > 0 plugging that in 1, we get p^3(positive quantity) < 0 or p^3 < 0 which means p < 0 hence sufficient.
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