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# Is p a negative number?

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Is p a negative number?  [#permalink]

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Updated on: 05 Jul 2013, 00:01
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00:00

Difficulty:

65% (hard)

Question Stats:

58% (02:03) correct 42% (02:08) wrong based on 141 sessions

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Is p a negative number?

(1) p^3(1 – p^2) < 0
(2) p^2 – 1 < 0

Originally posted by zest4mba on 14 Jul 2010, 07:52.
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
1
2
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.

Hope it's clear.
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Re: DS - inequality problem  [#permalink]

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14 Jul 2010, 08:18
I assume that stmt 1 is $$p^2-1<0$$ & 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]

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14 Jul 2010, 08:21
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]

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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 $$1-p^2$$ 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]

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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
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.

Hope it's clear.
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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.

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
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.

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]

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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 (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.
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Re: Is p a negative number?  [#permalink]

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Re: Is p a negative number?   [#permalink] 29 Aug 2017, 11:30
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# Is p a negative number?

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