DS #500 Test 20 - Q 20

f2001290 · Posted: Mon May 14, 2007 3:55 am Post subject: DS #500 Test 20 - Q 20

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

OA after few explanations

bww · Posted: Mon May 14, 2007 7:01 am Post subject:

I'm getting E, which, according to the answer key, is not the OA.

I) whatever n is, (1-x^2) must be the opposite, i.e. if n is positive, then n^3 is positive and thus (1-x^2) must be a negative value. if n is negative, then n^3 is negative and thus (1-x^2) must be a positive value for its product to be less than 0. regardless, we don't know what x is because x^2 always produces a positive value. I) alone is not sufficient.

II) with x^2-1<0, it is implied that x must be a fraction. again, x could be either negative or positive given the fact that x^2 is always positive. II) alone is not sufficient either.

combining the two statements, we learn that x is a fraction less than 1, but it can be either negative or positive. thus, both statements together are not sufficient, no? am i missing something?

you can also work the inequalities...for I) by multiplying you have n^3-n^3x^2<0 which ultimately produces x^2>0. for II) you have x^2-1<0 which produces x^2<1. so we have 0<x^2<1 when combining the two statements. x must be a fraction, but could be either negative or positive.

f2001290 · Posted: Mon May 14, 2007 7:26 am Post subject:

I think E is the right answer. I was stumped when I saw the OA.

bww · Posted: Mon May 14, 2007 7:33 am Post subject:

there are several erroneous answers in the answer key. let's hope this is one of them!