GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Jun 2019, 18:06 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Is x<0 ?

Author Message
TAGS:

Hide Tags

Manager  Joined: 05 Mar 2011
Posts: 102

Show Tags

3
14 00:00

Difficulty:   65% (hard)

Question Stats: 63% (02:05) correct 38% (02:04) wrong based on 319 sessions

HideShow timer Statistics

Is x<0 ?

(1) x^3(1-x^2)<0
(2) x^2-1<0
Intern  Joined: 17 Oct 2011
Posts: 9

Show Tags

IMO it is C.

(1) By itself is not sufficient.
(2) tells us x is a +ve or -ve fraction. Not sufficient.

(Together)
If x is +ve fraction it does not satisfy st. (1) (plug in x = 1/2 for testing)
If x is -ve fraction it satifies (1). so x is a -ve fraction and hence x < 0. SUFFICIENT.
_________________
We will either find a way, or make one.

Please consider giving kudos if you find my post hepful Manager  Joined: 08 Sep 2011
Posts: 56
Concentration: Finance, Strategy

Show Tags

agree C

stmt 1: x<0 or x>1
stmt 2: x>-1 x<1

Together:
x>0 and x>-1
Director  Joined: 07 Aug 2011
Posts: 518
GMAT 1: 630 Q49 V27 Show Tags

How does 1) X^3(1-X^2)<0 led to x<0 or x>1
eg: X=-2
-2^3 (1-(-2)^2) = -8*-3 = 24
eg: X=-1 , result 0<0 , not true.
eg:X=0 result 0<0, not true
X=1 , 0<0, not true.

i feel it is only when X>1 , X^3(1-X^2)<0
Manager  Status: Employed
Joined: 17 Nov 2011
Posts: 79
Location: Pakistan
GMAT 1: 720 Q49 V40 GPA: 3.2
WE: Business Development (Internet and New Media)

Show Tags

1
Lucky2783 wrote:
How does 1) X^3(1-X^2)<0 led to x<0 or x>1
eg: X=-2
-2^3 (1-(-2)^2) = -8*-3 = 24
eg: X=-1 , result 0<0 , not true.
eg:X=0 result 0<0, not true
X=1 , 0<0, not true.

i feel it is only when X>1 , X^3(1-X^2)<0

Actually just re-write with factors:

x^3(1-x^2) < 0
So: (x^3)(1-x)(1+x) < 0

so either all three or negative or any one can be negative for the expression to be true.

If (x^3) is negative, x < 0
If (1-x) is negative, x > 1
If (1+x) is negative, x < -1

When you combine statement 1 and 2, the only overlap is x < -1 so we know for sure that x is negative

_________________
"Nowadays, people know the price of everything, and the value of nothing." Oscar Wilde
Director  Joined: 07 Aug 2011
Posts: 518
GMAT 1: 630 Q49 V27 Show Tags

i think we need to consider the three cases in parallel eg: if X<0 what will happen to other two expressions

If (x^3) is negative, x < 0
say X is -0.5, X^3 is -ive , 1-X is +ive and 1+X too is positive. expression holds.
Now say X=-2, X^3 is -ive , 1-X is +ive but 1+X is -ive . expression doesnt hold .
so we cannot say that if X<0 , X^3 (1-X) (1+X) < 0

1+X will be -ive only when X <-1 , in that case 1-X will be +ive and X^3 will be -ive , over all expression will be +ive
so this set of values of X also ruled out.

now
1-X will be -ive for all values of X greater than 1 and the other two X^3 and 1+X will be positive in this case.
we can safely say that for all X>1 the expression will hold.

may be i am wrong though in my thought process , pls advise.
Intern  Joined: 14 Sep 2010
Posts: 16

Show Tags

Is X<0 ?

1) X^3(1-X^2)<0
2) X^2-1<0

C

From 1)

-1 < x < 0 and 1 < x

From 2)

-1 < x < 1

Combining,

-1 < x < 0
Director  Joined: 07 Aug 2011
Posts: 518
GMAT 1: 630 Q49 V27 Show Tags

Study1 you sound promising...
Intern  Joined: 11 Jun 2011
Posts: 10

Show Tags

ashiima wrote:
Is X<0 ?

1) X^3(1-X^2)<0
2) X^2-1<0

Question asks if X is negative

Statement-1
If X is -ve: x^3 will be -ve, then (1-x^2) has to be positive, can't say
If X is +ve: x^3 will be +ve then (1-x^2) has to be negative, can't say
Insufficient

Statement-2
X^2-1<0, this is possible only when x is fraction. But fraction can be negative or positive, Insufficient

Both statements together

we get
when x is -ve, x^3 is negative, and 1-x^2 is positive ( as x is a fraction now eg. 1/2)
when x is +ve, not valid
Hence x is -ve, when both statements combined together.

Hence C

Hope I am clear.
_________________
_______________________________________________________________________________________________________________________________
If you like my solution kindly reward me with Kudos.
Math Expert V
Joined: 02 Sep 2009
Posts: 55670

Show Tags

3
1
Is x<0 ?

(1) x^3(1-x^2)<0 --> $$x^3(1-x)(1+x)<0$$ --> the roots are -1, 0, and 1 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: $$x<-1$$, $$-1<x<0$$, $$0<x<1$$, and $$x>1$$.

Now, test some extreme value: for example if $$x$$ is very large number then $$x$$ and $$1+x$$ are positive, while $$1-x$$ is negative, which gives negative product for the whole expression, so when $$x>1$$ the expression is negative. Now the trick: as in the 4th range expression is negative then in 3rd it'll be positive, in 2nd it'll be negative again and finally in 1st it'll be positive: + - + -. So, the ranges when the expression is negative are: $$-1<x<0$$ and $$x>1$$.

So, $$x$$ could be negative as well as positive. Not sufficient.

Or: $$x^3(1-x^2)<0$$ --> $$x(1-x^2)<0$$ --> $$x<x^3$$ --> $$-1<x<0$$ or $$x>1$$. Not sufficient.

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

(1)+(2) Intersection of the ranges from (1) and (2) is $$-1<x<0$$, hence the answer to the question whether $$x<0$$ is YES. Sufficient.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
_________________
Manager  Joined: 21 Oct 2013
Posts: 181
Location: Germany
GMAT 1: 660 Q45 V36 GPA: 3.51

Show Tags

(1): x³(1-x²) < 0 --> x³ < 0 or 1-x² < 0. If x³ < 0 then x < 0, BUT x² < 1 x could be both. IS.

(2): Clearly IS. x² < 1, x could be < or > 0.

Together: From 1 we know EITHER x³ <0 OR 1-x² < 0. From 2 we know that 1-x² < 0 hence x³ is > 0, thus x is > 0.
Non-Human User Joined: 09 Sep 2013
Posts: 11392

Show Tags

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 x<0 ?   [#permalink] 26 Sep 2018, 05:44
Display posts from previous: Sort by

Is x<0 ?  