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Is p a negative number?

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Is p a negative number? [#permalink] New post 14 Jul 2010, 06:52
00:00
A
B
C
D
E

Difficulty:

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Question Stats:

49% (02:37) correct 51% (01:24) wrong based on 72 sessions
Is p a negative number?

(1) p^3(1 – p^2) < 0
(2) p^2 – 1 < 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Jul 2013, 23:01, edited 1 time in total.
Edited the question and added the OA
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Re: DS - inequality problem [#permalink] New post 14 Jul 2010, 07:12
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zest4mba wrote:
Can someone explain how to solve this quickly

Is p a negative number?

(1) p3(1 – p2) < 0

(2) p2 – 1 < 0


Is p<0?

(1) p^3(1-p^2)<0, or which is the same p(1-p^2)<0 --> p<p^3 --> either p is more than 1, p>1 OR p is negative fraction -1<p<0.

So we have two ranges for p: p>1 or -1<p<0. Not sufficient.

(2) p^2-1<0 --> -1<p<1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is the range -1<p<0, so the answer to the question "is p<0" is YES. Sufficient.

Answer: C.

Hope it's clear.
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Re: DS - inequality problem [#permalink] New post 14 Jul 2010, 07:18
I assume that stmt 1 is p^2-1<0 & stmt 2 is p^3(1-p^2)

Stmt 1.
p^3(1-p^2)<0
Either p^3<0 or 1-p^2<0
p<0 or p^2>1
p<0 or p>1 or p<-1 so not sufficient

Stmt 2
p^2-1<0
(p-1)(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] New post 14 Jul 2010, 07:21
Bunuel, If I put 2 in Stmt 1 then it comes true. So stmt 1 also shows p>1.

Where am I wrong?
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Re: DS - inequality problem [#permalink] New post 14 Jul 2010, 07:28
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. :oops: 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 1-p^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: 1-p^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] New post 14 Jul 2010, 07:32
Wow, I was writting my reply and in the mean time you guys had already posted ! :o
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Re: DS - inequality problem [#permalink] New post 14 Jul 2010, 07:46
Expert's post
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] New post 14 Jul 2010, 10:50
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] New post 14 Jul 2010, 11:23
Expert's post
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.
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Re: DS - inequality problem [#permalink] New post 14 Jul 2010, 12:08
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] New post 14 Jul 2010, 12:20
Expert's post
FedX wrote:
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??


The questions "is p negative" and "is p a negative number" are the same. Maybe you confused "number" with "integer"?
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Re: DS - inequality problem [#permalink] New post 14 Jul 2010, 12:34
Thanks for the clarification Bunuel !!.
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Re: DS - inequality problem [#permalink] New post 21 Jul 2010, 23:41
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] New post 20 Sep 2014, 02:25
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Re: Is p a negative number? [#permalink] New post 22 Sep 2014, 09:02
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 (1-p^2) > 0
plugging that in 1, we get p^3(positive quantity) < 0
or p^3 < 0
which means p < 0
hence sufficient.
Re: Is p a negative number?   [#permalink] 22 Sep 2014, 09:02
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