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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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. _________________

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. _________________

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.

(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.

Answer: C.

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).

the answer is A

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. _________________

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.

while solving stmt 1 X^3(1-X^2)<0 X^3(X+1)(1-X)<0 Then i took the roots as -1,0,+1 on the number line to find the range as per vertias prep graph approach but am getting ranges as 0>X<1 and x<-1 which is wrong .please do correct me _________________

"Giving kudos" is a decent way to say "Thanks" and motivate contributors. Please use them, it won't cost you anything

while solving stmt 1 X^3(1-X^2)<0 X^3(X+1)(1-X)<0 Then i took the roots as -1,0,+1 on the number line to find the range as per vertias prep graph approach but am getting ranges as 0>X<1 and x<-1 which is wrong .please do correct me

You will have to re-arrrange the equation to read x(x+1)(x-1)>0. You can drop the x^2 as it is always positive and x is not equal to zero. Now if you plot the roots, it will give you -1,0 and 1. Thus, the valid ranges will be

x>1 OR -1<x<0. Insufficient.

From F.S 2, we know that |x|<1 --> -1<x<1. Insufficient.

On taking both fact statements together, for -1<x<0, both the conditions are fulfilled. Sufficient.

Re: Is x negative? [#permalink]
03 May 2013, 00:11

abhi758 wrote:

Is x negative?

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

Hi all,

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

let x=10....==>therefore 10^3(1-100)=-99*10^3<0 satisfying for positive X now let x= -0.1===> therefore (-0.1)^3(1-0.01)= -0.99*0.1^3<0 satisfying when X is negative... so X can be positive and negative both so not definite answer.

(2) x^2-1 < 0 this holds for -1<x<1

again not sufficient

now combining from statement 2 we got 2 things: ==> x^2-1<0....therefore 1-x^2>0-----(1) ==>range of x...-1<x<1-----(2)

now coming to statement 1... we know 1-x^2>0.....from (1) therefore for x^3(1-x^2) < 0....to hold true...x^3 <0....since we know 1-x^2 is positive from (1) now we have x^3<0....therfeore x<0----(3)

now using (2) and (3) x belongs to-1<x<0 menas x is negative.hence sufficient

therefore (C)

hope it helps.

SKM _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...