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# Is x negative?

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Manager
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29 Jul 2011, 23:36
00:00

Difficulty:

55% (hard)

Question Stats:

55% (02:15) correct 45% (01:38) wrong based on 22 sessions

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Is x negative?

(1) x^3*(1 - x^2) < 0
(2) x^2 - 1 < 0
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Jul 2013, 02:09, edited 1 time in total.
RENAMED THE TOPIC.
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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, 23:38
If I go thru the OG explanation, and solve it how they do, this problem would take me 3 or 4 mins...

Can anyone give me any tips on what to do when you identify such a question?
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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, 23:54
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!

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

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30 Jul 2011, 00:01
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...
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Re: How do you slove this DS problem in less than 2 mins...? [#permalink]

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30 Jul 2011, 00:21
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MackyCee wrote:
From OG11 - 2nd last Question

Q154) Is x negative?

1) x^3.(1 - x^2) < 0
2) x^2 - 1 < 0

1)
$$x^3(1 - x^2) < 0$$
$$x(x^2-1) > 0$$
$$x(x+1)(x-1) > 0$$

Roots:$$-1, 0, +1$$

So, Valid range is:
$$x>1$$ OR $$-1<x<0$$

x can be -0.5 or x can be 2.
Not Sufficient.

2) $$x^2-1<0$$
$$(x-1)(x+1)<0$$

Roots: $$-1, +1$$
$$-1<x<1$$
Not Sufficient.

Together:
$$-1<x<0$$
Sufficient.

Ans: "C"
*********************************

Used the concept from this:
inequalities-trick-91482.html

Similar topics:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
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Re: How do you slove this DS problem in less than 2 mins...? [#permalink]

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30 Jul 2011, 01:05
1) x^3.(1 - x^2) < 0

This is possible when X^3 < zero or when (1-X^2) < 0

For X^3 to be negative X should be negative

For 1- X^2 to be be negative modulus X should be positive ( if negative, the result would cancel with X3 and give a positive result)

Insufficient

2) x^2 - 1 < 0

X^2 < 1

this is possible when modulus X is < 1 so X could be positive or negative.

Combining 1 and 2

When modulus X is < 1 and X is negative. That satisfies statement 1

When modulus X is < 1 and X is positive. That does not satisfy statement 1

Final result X is negative

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

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30 Jul 2011, 02:12

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

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30 Jul 2011, 02:21
MackyCee wrote:

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

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01 Aug 2011, 12:24
MackyCee wrote:
From OG11 - 2nd last Question

Q154) Is x negative?

1) x^3.(1 - x^2) < 0
2) x^2 - 1 < 0

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.
Re: How do you slove this DS problem in less than 2 mins...?   [#permalink] 01 Aug 2011, 12:24
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