Is x negative? : GMAT Data Sufficiency (DS)
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# Is x negative?

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21 Dec 2009, 12:01
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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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Re: Need Help with OG 11 ed DS 154 [#permalink]

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21 Dec 2009, 14:12
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gi wrote:
Hi All,

I'm new to this forum. I'm working through some OG DS questions and I find that some of the explanations to ans are not clear or take too long to solve in 2 mins.

Q:DS154
Is x negative?
1) x^3(1-x^2) < 0
2) x^2-1 < 0

Thanks

(1) $$x^3(1-x^2)<0$$, two cases (one of the multiples negative, another positive):

A. $$x^3>0$$ and $$1-x^2<0$$;

$$x^3>0$$ --> $$x>0$$
----------------0----------------

$$1-x^2<0$$ --> $$x<-1$$ or $$x>1$$
-----(-1)------0------(1)-------

Hence the range for A is $$x>1$$.

B. $$x^3<0$$ and $$1-x^2>0$$

$$x^3<0$$ --> $$x<0$$
---------------0------------

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

Hence the range for B is $$-1<x<0$$.

Two ranges: $$x>1$$ OR $$-1<x<0$$. Not sufficient.

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

(1)+(2) Inersection of the ranges from (1) and (2) is $$-1<x<0$$, hence $$x<0$$ is true. Sufficient.

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Re: Need Help with OG 11 ed DS 154 [#permalink]

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21 Dec 2009, 14:21
IMO C:

A) X^2(1-X^2) <0 so X^3<0 or X^2>1
x could be positive or negative

B) X^2-1<0
(X-1)(X+1)<0 so X<1 or X<-1
x could be positive or negative

Combined X is negative for both statements to hold true.

OA?
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Re: Need Help with OG 11 ed DS 154 [#permalink]

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21 Dec 2009, 14:33
C; like Bunuel's details.
Inequalities without coffee can not be so much fun!!!

Bunuel,

What subject that you dont like in GMAT Maths?
Be warned, I may too know the answer...!
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Re: Need Help with OG 11 ed DS 154 [#permalink]

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21 Dec 2009, 14:52
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DestinyChild wrote:
C; like Bunuel's details.
Inequalities without coffee can not be so much fun!!!

Bunuel,

What subject that you dont like in GMAT Maths?
Be warned, I may too know the answer...!

Some word problems make me nervous, as I'm not a native English speaker.
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Re: Need Help with OG 11 ed DS 154 [#permalink]

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22 Dec 2009, 06:26
IMO C:

A) X^2(1-X^2) <0 so X^3<0 or X^2>1
x could be positive or negative

B) X^2-1<0
(X-1)(X+1)<0 so X<1 or X<-1
x could be positive or negative

Combined X is negative for both statements to hold true.

OA?

Yes the ans is C. Thanks
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10 Apr 2010, 09:37
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Is x negative?

(1) x^3(1-x^2) < 0
(2) x^2-1 < 0
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10 Apr 2010, 10:07
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IMO C.

1, is equivalent to x(x-1)(x+1) >0 ( x^2 is always>0 and i have inverted the sign)

this gives x>1 and -1<x<0 , thus not sufficient

2, gives -1< x<1 not sufficient.

if you combine the result we get -1<x<0 , thus x is negative and thus C.

Another way, since we know both are not sufficient individually.

2nd equation gives x^2 -1 <0 => 1- x^2 >0
now in first equation, if its -ve then x^3 must be -ve....thus both taken together sufficient.
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10 Apr 2010, 10:12
First I go to Stmt 2:

Stmt 2 shows that -1<x<1 so x can be negative or positive. Hence no sufficient.

Stmt1 not sufficient too!!

With the help of stmt 2, if we put the negative value of x, the stmt 1 becomes true!!

Hence "C" is the answer in my opinion!!!

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10 Apr 2010, 10:37
Thanks guys for the quick solution..Heres a question..

For st1, if I take a +ve number that is $$>=2$$ the inequality holds true, then why do we need to consider st2?

Look forward to the replies..
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10 Apr 2010, 10:41
abhi758 wrote:
Thanks guys for the quick solution..Heres a question..

For st1, if I take a +ve number that is $$>=2$$ the inequality holds true, then why do we need to consider st2?

Look forward to the replies..

Thats wrong approach towards DS question.

x>=2 satisfy the equation 1 , what about x = -1/2, that is also satisfying the equation.Then how we can be sure whether x is + or -... sufficiency comes when we are sure this statement as a whole ans the question and there is no other possibility which will be against it,.

Before starting another DS question do get this concept, else no use of DS questions.
letme know if you got it else i will try to explain in other way.
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11 Apr 2010, 22:41
Thanks for bringing this to my notice.

If you could explain the first st1 working in the solution provided by you:
Quote:
1, is equivalent to x(x-1)(x+1)>0 ( x^2 is always>0 and i have inverted the sign)

this gives x>1 and -1<x<0 , thus not sufficient

Unable to understand the working for $$x^3(1-x^2)<0$$.
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11 Apr 2010, 23:51
$$x^3(1-x^2) = x* x^2 * (x^2 -1)* (-1 ) < 0$$

since x^2 >0 , it will not affect the sign

$$x*(x^2-1) * (-1) < 0$$
=> $$x*(x^2-1) * > 0$$ when we multiple both sides by -1, > becomes <

eg. -1 > -2 , but 1 < 2

=> $$x*(x-1)*(x+1) * > 0$$
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12 Apr 2010, 06:38
Thanks gurpreet for your time and patience!

Just a ques on the last part..

After: $$x(x-1)(x+1)>0$$
you get => $$x>1$$
AND
you get => $$x>-1$$ OR $$-1<x$$
AND
next would you get => $$x>0$$ OR $$x<0$$??

This might sound a little basic but really confused with the whole thing..
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12 Apr 2010, 09:03
No, its not right.

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12 Apr 2010, 09:55
IMO E

ST 1 both a pos and a neg # satisfiy this equation. Not suf

ST 2....samething.

Cobined we can't conclude the sign of X

also, keep in mind that the ques doesn't states if X is 0 or 1

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12 Apr 2010, 10:36
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Alternative approach by taking numbers.
Statement1: $$x^3(1-x^2) < 0$$
From this we can say $$x <> 0$$
and $$x <>+1 or -1$$.

Consider x=2 Then the equation results in $$8*-3 <0$$. Satisfy the equation.
Consider x= 1/2 then the equation results in $$1/8*(3/4) >0$$. Does not satisfy the equation.

Consider x=-2 Then the equation results in $$-8*-3 >0$$. Does not satisfy the equation.
Consider x= -1/2 then the equation results in $$-1/8*(3/4) <0$$.Satisfy the equation.

This statement is not sufficient to conclude.

Statement2: $$x^2-1 <0$$
From the statement we can say that $$x <>=1 or -1$$
Consider x=2 then the equation results in $$3 >0$$. Does not satisfy the equation.
Consider x=1/2 then the equation results in $$-3/4 <0$$. Satisfy the equation.

Consider x=-2 then the equation results in $$3 >0$$. Does not satisfy the equation.
Consider x=-1/2 then the equation results in $$-3/4 <0$$. Satisfy the equation.

This statement alone is not sufficient.

Combining both the statements when x is -ve that is -1/2 , it satisfies both the equations.

Ans is C.
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14 Apr 2010, 01:18
1. x^3(1-x^2) <0
x^3- x^5<0
x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).

2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).

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14 Apr 2010, 02:09
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abhi758 wrote:
Is X negative?

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

Is $$x<0$$?

(1) $$x^3(1-x^2) < 0$$
Two cases:
A. $$x>0$$ and $$1-x^2<0$$, which means $$x<-1$$ or $$x>1$$ --> $$x>1$$;
B. $$x<0$$ and $$1-x^2>0$$, which means $$-1<x<1$$ --> $$-1<x<0$$.

So $$x^3(1-x^2) < 0$$ holds true for two ranges of $$x$$: $$x>1$$ and $$-1<x<0$$. If $$x$$ is in the first range answer to the question is NO, but if $$x$$ is in the second range answer to the question is YES. Two different answers. Not sufficient.

(2) $$x^2-1<0$$ --> $$-1<x<1$$. x can be positive as well as negative. Not sufficient.

(1)+(2) $$x>1$$ or $$-1<x<0$$ AND $$-1<x<1$$ --> intersection of the ranges from (1) and (2) is $$-1<x<0$$. $$x$$ is negative. Sufficient.

aiyedh wrote:
1. x^3(1-x^2) <0
x^3- x^5<0
x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).

2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).

Red part is not correct. $$x^3 < x^5$$ does not hold true for all positive $$x-es$$. For example $$x=0.5$$ --> $$x^3>x^5$$. Also $$x$$ can be negative in the range $$-1<x<0$$ and $$x^3 < x^5$$ will hold true.
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17 Apr 2010, 11:25
abhi758 wrote:
Is X negative?

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

here, is small note:
as soon as we see one number can be either positive or negative we know the statement is not sufficient.
so statement 1)
$$x^3(1-x^2) < 0$$ , here $$x^3$$ can +ve or -ve => not suff.
similarly; $$x^2-1 < 0$$ => (x-1)(x+1)<0 ; here also x-1 or x+1 can +ve or -ve not suff.

combining both : since $$x^2-1 < 0$$ or 1- $$x^2 > 0$$ hence from statement 1
$$x^3 < 0$$ => x< 0 => suff.
Re: Inequalities   [#permalink] 17 Apr 2010, 11:25

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