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Re: DS - inequality problem [#permalink]
14 Jul 2010, 07:12

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

Re: DS - inequality problem [#permalink]
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. 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.

Re: DS - inequality problem [#permalink]
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.

Re: DS - inequality problem [#permalink]
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??

Re: DS - inequality problem [#permalink]
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"? _________________

Re: Is p a negative number? [#permalink]
20 Sep 2014, 02:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Illegitimi non carborundum.

gmatclubot

Re: Is p a negative number?
[#permalink]
22 Sep 2014, 09:02

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