Algebra Question

[davidforsberg]

x<0???

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

answer is C. Anyone care to elaborate?

[its_me07]

St 1 is INSUFF
bcoz we donot know whether x^3<0 or 1-x^2<0
as there are only two possibilities (-x)y<0 or x(-y)<0

St 2 IS INSUFF
as x^2<1
x^2-1<0 (as x can be positive or negative)

Combining both
x^2-1<0>0 as it is pos

so x^3<0>x<0.

Hope htis helps.

[Stuart@KaplanGMAT]

x<0???

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

answer is C. Anyone care to elaborate?

To elaborate on It'sme's solution (btw, if you disable HTML, the formatting issues will go away):

(1) if the product of two terms is negative, then exactly one of those terms must be negative. So, either:

(x^3) < 0
(therefore x < 0)

or

(1 - x^2) < 0

(therefore 1 < x^2)

We can pick numbers to make x positive or negative, so (1) is insufficient.

(2) x^2 < 1

To make this true, -1 < x < 1 (i.e. x is a fraction or 0). So, x could be positive, negative or 0: insufficient.

If we combine the statements, we know that:

x^2 < 1

AND

x^3 < 0

In other words, we've eliminated the "1 < x^2" possibility from statement (1), since it conflicts with statement (2).

If we know for sure that x^3 < 0, then x MUST be < 0. Therefore, together the statements are sufficient: choose (c).

[its_me07]

Disabled the HTML code
Thx Staurt.