chipbmk wrote:Is X negative?

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

OA: C

Please let me know what the best way is to tackle this one.

Thanks!

Hi chipbmk,

By reasoning first, and alegebra second!

Statement 1:

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

So, the left hand side is negative. Instead of x^3(1-x^2)<0, try to look at it as: (number)(another number) < 0.

How can the product of two numbers be negative: only if one is positive and the other negative. Accordingly, as soon as we see one number can be either pos or neg we know the statement is not sufficient.

We know x^3 will have the same sign as x. So, like any unknown x, x^3 can be pos or neg.

Statement 1 is not sufficient.

Statement 2:

x^2-1<0

So, again, the left hand side is negative (and these two statements look suspiciusly similar).

We know squares are always positive, so, if x were 2 or 3 it will remain positive even if we subtracted 1 from it. Therefore, x must be a fraction (when we square a fraction we get an even smaller fraction). But it can be either a positive or negative fraction.

Insufficient.

combo:

The fact that these expressions are similar is telling us something.

We have a "1-x^2" in statement 1 and "x^2 -1". Notice that the 1 and x^2 have exactly opposite signs. This means:

(1-x^2) = (-1)*(x^2 -1) Or: (x^2-1) = (-1)*(1-x^2)

So, if from statement 2, x^2-1 is (-), then the (1-x^2) in statement 1 is positive. Then, we have:

x^3*(pos) < 0

This means that x^3 is negative. (Product of two numbers is negative only when one is positive and the other negative).

And that means that x is negative. (x^3 and x will have same sign).

So, x is negative, the answer to the question is "yes", and the statements, although insufficient in isolation, are sufficient in combination.

Choose C

chipbmk wrote:Is X negative?

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

OA: C

Please let me know what the best way is to tackle this one.

Thanks!

The key is getting rid of squares, if any, since a square is always positive.

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

(x+1)x(x-1) > 0

This can be +ve/-ve in only two of the all sets: (-inf, -1), (-1,0), (0,1) and (1, +inf).

Insufficient.

2. x^-1 < 0

(-1, 1), Insufficient.


Combining gives out the answer.

Given x^3(1-x^2)<0, we cannot conclude anything. Take x = 20, and x = -0.1, and the statement is inconclusive. Taking the second statement by itself, we have -1 < x < 1, which is inconclusive. However taking both together, we have

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

Let C = 1-x^2. We have

x^3 * C < 0

Since C != 0, divide by both sides to get

x^3 < 0

So x < 0, and is negative. The answer is thus C.

palvarez wrote:

chipbmk wrote:Is X negative?

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

OA: C

Please let me know what the best way is to tackle this one.

Thanks!

The key is getting rid of squares, if any, since a square is always positive.

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

(x+1)x(x-1) > 0

This can be +ve/-ve in only two of the all sets: (-inf, -1), (-1,0), (0,1) and (1, +inf).

Insufficient.

2. x^-1 < 0

(-1, 1), Insufficient.


Combining gives out the answer.

Palvarez,

Rather than taking difference of squares, it is much easier to see that because (x^2-1) is negative, (1-x^2) is positive. GMAT loves testing this one, where you just factor out -1.

And of course because 1-x^2 is positive, in order for the first statement to be true, x^3 will have to be nagtive. Therefore, x is negative, and we are done.

Note that, under this reasoning approach, there was no taking of difference of squares, solving writing down sets, etc.

Just a bit of critical thinking--the number one thing GMAT likes to test!