Is x negative? (1) x^3(1 - x^2)

Question

Is x negative?

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

(2)  x^2 - 1 < 0

Bunuel · Accepted Answer

Is x negative? (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 --> -11 OR -1 -1

GMATinsight · Answer

Question : Is x negative?

Please check the attachment for detailed solution.

This is how GMATINSIGHT teaches Data sufficiency

Answer: Option C

adamsmith2010 · Answer

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?

gurpreetsingh · Answer

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 1- x^2 >0 now in first equation, if its -ve then x^3 must be -ve....thus both taken together sufficient.

mads · Answer

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.

aiyedh · Answer

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

Bunuel · Answer

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 -11 and -1 -11 or -1 intersection of the ranges from (1) and (2) is -1 x^3>x^5 . Also x can be negative in the range -1

NinetyFour · Answer

My reasoning: Statement 1: either x^3 is negative or (1 - x^2) is negative. They are opposite signs. However we can have a positive x and a negative x in this case. Testing values of 2 and \frac{-1}{3} will prove this. Statement 2: x^2 - 1 < 0 -> x^2 < 1 so x > -1 & x < 1 Again inconclusive because we can have any positive or negative fraction. Combining the two statements only "negative fractions" are in our overlapping set. The answer is C. x MUST be less than 0

Gmatprep550 · Answer

Hi  chetan2u ,  Bunuel ,  Gladiator59

I am not able to figure out how we are able to find solution combining both the statement for above question, Could you please assist.

chetan2u · Answer

The question is Let us take both combined.. Statement II tells us x^2 - 1 < 0 , so when we multiply this with -1, the inequality sign will change => -1(x^2 - 1) > 0*(-1).....1-x^2>0 We did this to get the term of statement I.. Now take statement I.. x^3(1 - x^2) < 0 ... Here we see from statement II that 1-x^2>0 , so x^3(1 - x^2) < 0 => x^3(positive..value) < 0 , thus x^3<0...x<0

h31337u · Answer

Is x negative?

(1) x^3(1−x^2)<0

(2) x^2−1<0

_________________

IMO, it is C.

The question is x < 0?
(1) divide both RHS and LHS by x^2.
then you have, x (1-x^2) <0.
now, take "-1" out from the parenthesis. -x(x^2 - 1) < 0.
Divide both RHS and LHS by -1.
then you have x (x^2 - 1) > 0.
In other words, x ( x+1) (x-1) >0.
so -1<x<0 and x>1
insufficient.

(2) we can say ( x+1) (x-1) <0
so -1<x<1
insufficient.

(1) + (2) together
-1<x<0.
so x is negative.
sufficient to answer whether x is negative.

unraveled · Answer

Graphical solution: Screenshot.png 1) x^3(1 - x^2) < 0 Case I: x^3 < 0 and (1 - x^2) > 0 x < 0 and x^2 < 1 x < 0 and - 1 < x < 1 So, - 1 < x < 0 YES Case II: x^3 > 0 and (1 - x^2) < 0 x > 0 and x^2 > 1 x > 0 and x > 1 and x < -1 So, x > 1 YES INSUFFICIENT. 2) x^2 - 1 < 0 x^2 < 1 - 1 < x < 1 YES and NO both cases. INSUFFICIENT. Together 1 and 2 Common range is -1 < x < 0 SUFFICIENT. Answer C.

Basshead · Answer

Is x negative? Is x < 0 ? (1) x^3(1 - x^2) < 0 Either x^3 or (1 - x^2) is negative. If x^3 < 0 then 1 - x^2 > 0 x^2 < 1 -1 < x < 1 If x^3 > 0 then 1 - x^2 < 0 x^2 - 1 > 0 (X+1)(X-1) > 0 X > 1 and x < -1 INSUFFICIENT. (2) x^2 - 1 < 0 (x+1)(X-1) < 0 -1 < x < 1 INSUFFICIENT. (1&2) Combined, the range is -1 < x < 0 . Thus, x < 0 . SUFFICIENT.

2020prep2020 · Answer

Could someone explain how can we transform x^2-1<0 to -1<x<1. I mean I understand it logically but algebraically we have (x+1)(x-1)<0 => x<-1 or x<1

CEdward · Answer

It's a matter of rearranging

x^2 - 1 < 0
x^2 < 1
\sqrt{x^2} < \sqrt{1}
| x | < 1 <-----MEMORIZE this...anytime you see the square root of a variable raised to an even exponent, you end up with the absolute value of that variable

Hence, -1 < x < 1.

This might be a silly question, but suppose after combining both statements we find that there are no areas of overlap. Then the answer would be E?

Fdambro294 · Answer

Is X < 0 ?

(S1) (x)^3 * (1 - (x)^2) < 0

Case 1: (x)^3 < 0 ------and------- 1 - (x)^2 > 0

X < 0 ------ and -------(x)^2 < 1

X < 0 ------and------- (-)1 < X < +1

Summary of Case 1: if X < 0 -------- (-)1 < X < 0

Case 2: (x)^3 > 0 ------ and -------- 1 - (x)^2 < 0

X > 0 ----- and ------ (X)^2 > 1

Summary of Case 2: if X > 0 ------- X > +1

X can be Negative or Positive

S1 NOT Sufficient alone

(2)
(x)^2 < 1

[X] < 1

-1 < X < +1

X can be Negative or Positive
S2 NOT Sufficient Alone

Together:

If X < 0:

Statement 1 says: -1 < X < 0
and
Statement 2 says: -1 < X < 0

If X > 0

Statement 1 says: X > +1
and
Statement 2 says: 0 < X < +1

if X is a Positive Value, it can NOT Satisfy S1 and S2 TOGETHER at the Same Time

therefore, X < 0

and Definite YES: X is Negative

(C) Together Sufficient

Laksh47 · Answer

Bunuel   chetan2u   VeritasKarishma

Could you please explain why the below solution is incorrect?

chetan2u · Answer

When x is between 0 and -1, say x is -1/2.
 \frac{-1}{2}^3=\frac{-1}{8} 
 \frac{-1}{2}^5=\frac{-1}{32} 
 \frac{-1}{32}>\frac{-1}{8}

So x can be negative too.

KarishmaB · Answer

x^3 < x^5 does not imply that x is positive. This is satisfied for x = 2 as well as x = -1/2. Note that the number properties vary over 4 basic intervals of the number line x < -1 -1 < x< 0 0 < x < 1 and x > 1 Also, -1, 0 and 1 behave differently. It is best to use x^3 - x^5 < 0 x^5 - x^3 > 0 x^3 (x^2 - 1) > 0 x^3 (x + 1)(x - 1) > 0 Transition points are 0, -1, 1. x will be positive when x > 1 or -1 < x < 0.

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

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