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Is x negative?

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Source: — Data Sufficiency |
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by selango » Sun Aug 08, 2010 10:15 am
x is -ve?

stmt1,

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

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

x^3<0,x is -ve

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

x can be -ve or +ve

Insuff

stmt2,

x^2-1<0

x^2<1

x^2>=0

So x can be only 0.But o is neither +ve or -ve.

Combining 1 and 2,

x^2-1<0 or 1-x^2>0

-->x^3<0

x is -ve.

Suff

Pick C
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by sanp_l » Sun Aug 08, 2010 11:13 pm
Hi,

My understanding is as below:

Firstly, The value for x^3(1 - x^2) depends on
a) x^3 is +ve/-ve
b) x^2 is greater than or less than 1.

If x < 0, x^3 < 0. But (1-x^2) depends on the exact value of x. It may be that x < 1 or x < 0
If x > 0, (1- X^2) has to be < 0. Thus x < 1.

Hence Insufficient.


Secondly, x^2 - 1 < 0.
Thus -1 < x < 1.
Hence insufficient.

On Combining both, we have x^2 < 1. So, (1 - x ^2) > 0.
We know x^3(1 - x^2) < 0.
So x^3 is < 0.
Hence x < 0.

Both the answer choices together are sufficient.


Please do let me know if its correct.
Sandy
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by gmatmachoman » Sun Aug 08, 2010 11:26 pm
x^3(1-x^2)<0

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

when x^3 <0 :

x<0


when (1-x^2) <0

(1+x) ( 1-x) <0

So (1+x) <0 or ( 1-x) <0

(1+x) <0 leads to x<-1

( 1-x) <0 leads to x>1

Inconsistent . Insufficient

St2 : x^2-1<0

(x+1) (x-1) <0

So, x<-1 or x<1

Again st 2 in Insufficient.

Combining st1 & st2, we see that X is less than -1

Pick C
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by sanju09 » Mon Aug 09, 2010 12:09 am
greenwich wrote:Is x negative?

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

Please provide explanation as well.

If (1) is (x^3) × (1 - x^2) < 0, then either x^3 < 0, which promptly makes x < 0; or (1 - x^2) < 0 for some positive x. Insufficient

Statement (2) translates to either x + 1 < 0 and x - 1 > 0, which is an impossible case, OR x + 1 > 0 and x - 1 < 0. This can be bargained to -1 < x < 1, which include both positive and negative values for x. Insufficient

In concert, we realize (1 - x^2) > 0, hence x^3 < 0 or [spoiler]x < 0.

C[/spoiler]
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by greenwich » Mon Aug 09, 2010 3:41 am
greenwich wrote:Is x negative?

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

Please provide explanation as well.
OA is C.

Thanks everyone.
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by lunarpower » Wed Aug 11, 2010 3:39 am
greenwich wrote:Is x negative?

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

Please provide explanation as well.
these are standard inequalities with factored expressions; you should know how to solve these routinely.
the process:
(1) make sure that you have zero on one side of the inequality (this is already done here)
(2) factor the other side
(3) find the zeroes (roots) of the other side
(4) test the regions of the number line between those roots, to see if they satisfy the inequality

--

statement (1)
factor --> (x^3)(1-x)(1+x) < 0
roots are -1, 0, 1
test the number line:
<.................FALSE...........(-1)...........TRUE........(0)..........FALSE............(1)............TRUE.............>
there are both negative and non-negative numbers in this solution, so, insufficient.

--

statement (2)

factor --> (x-1)(x+1) < 0
roots are -1, 1
test the number line:
<.................FALSE...........(-1).........TRUE............(1)............FALSE.............>
there are both negative and non-negative numbers in this solution, so, insufficient.

--

together

look at both of the number lines above, and notice that the only values that are common to both of them are -1 < x < 0.
these are all negative.
sufficient.

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(c)
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by missrochelle » Mon Aug 23, 2010 11:02 am
this number line approach seems so tedious... i.e More than 2 minutes! Is it common on the GMAT these days?
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by lunarpower » Mon Aug 23, 2010 9:56 pm
missrochelle wrote:this number line approach seems so tedious... i.e More than 2 minutes! Is it common on the GMAT these days?
it shouldn't be overly tedious -- it's actually the standard protocol for solving inequalities like this. if it seems tedious or long, that's probably because you haven't practiced it enough yet.
same as with any other algebraic process. the first time you solved a quadratic equation, complete with factoring and rearrangement of terms, you might have had the same thought -- but after solving hundreds of them, you would no longer find this to be the case. therefore, if the problem is just that you haven't solved very many of these, you may want to grab a high school algebra book and get cracking.

--

incidentally, it's also the standard way of solving inequalities that contain fractional expressions, too. when you make that number line for a fractional inequality, you need to include two kinds of values as boundary points of the regions:
1) values that are zeros of the corresponding equation (this is the same as what's shown here)
2) values that make the denominators of the fractions equal to zero.

so, for instance, given the equation (x + 2)(x - 3)/(x + 5) < 0, you have to include -2, 3, and -5 on your number line, not just -2 and 3.
Ron has been teaching various standardized tests for 20 years.

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