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Re: How do you slove this DS problem in less than 2 mins...? [#permalink]

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29 Jul 2011, 22:54

Not great at explanations but here goes...

Looking at 1) If x^2 > 1 then x is positive and if x^2 < 1 then x is negative. 2) gives says that x^2 < 1, hence, x is negative when both 1) and 2) are used!

Re: How do you slove this DS problem in less than 2 mins...? [#permalink]

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29 Jul 2011, 23:01

Wolverine9 wrote:

Not great at explanations but here goes...

Looking at 1) If x^2 > 1 then x is positive and if x^2 < 1 then x is negative. 2) gives says that x^2 < 1, hence, x is negative when both 1) and 2) are used!

Hence C!

with statement 1, if x is a fraction, but positive, then x^2 < 1 doesn't necessarily make it negative, just really really small (say x = \frac{1}{10} => 0.001 x 0.9 = 0.0009 < 0 => not true)

I think I've got my head around the method, and I'll just allocated 3 mins to such type questions on the GMAT, considering this is a 700-800 level question...
_________________

Re: How do you slove this DS problem in less than 2 mins...? [#permalink]

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30 Jul 2011, 01:12

fluke, the links are epic!

I'm starting to figure out all of the "roots" stuff, and how "[insert equation here] < 0" means "solution lies in between roots of equation", and "[insert equation here] > 0" means "solution lies outside of roots of equation"
_________________

Re: How do you slove this DS problem in less than 2 mins...? [#permalink]

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30 Jul 2011, 01:21

MackyCee wrote:

fluke, the links are epic!

I'm starting to figure out all of the "roots" stuff, and how "[insert equation here] < 0" means "solution lies in between roots of equation", and "[insert equation here] > 0" means "solution lies outside of roots of equation"

Glad you liked them; I copied those from one of Bunuel's posts!!! We all should be thankful to Bunuel, gurpreetsingh, Karishma, and many others who made inequalities appear a piece of cake. N'joy!!!
_________________

First - you don't always have to start with piece of information (1). I would start with (2) because it's less complex. It's insufficient - try x = 0.5 and x = -0.5. (1) is also insufficient for similar reasons - try x = -0.5 and x = 5. Now, together:

\(x^2 - 1 < 0\) implies that \(1-x^2 > 0\), so in order for \((x^3)(1 - x^2)\) to be negative, \(x^3\) would have to be negative, and, therefore, x is negative.

C

That should take less than 2 minutes, I think.

The take away is - you don't have to solve either inequality in order to answer the question.

gmatclubot

Re: How do you slove this DS problem in less than 2 mins...?
[#permalink]
01 Aug 2011, 11:24

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