multiple answers to multiple questions.
can you explain why you're only putting X>1 here? If X^2>1, shouldn't the result be X>1 and X<1?
The square root of x^2 is x. The square root of 1 is +/- 1. So x is either >1 or < -1 (not positive one, negative one). But the assumption I made when I started this one was that x is positive, so the x < -1 option is not a valid solution.
Shouldn't your last statement say X<0. If you've already assumed X is negative, then saying X<1 doesn't make sense (because X can't be positive if you're assuming it's negative. So from this statement, you should take away that X<0 or X>1, correct?
Sure! You can definitely take it that next step - that's great! (I just stopped thinking about it because I knew I didn't have sufficient info anyway.)
How does this make any sense? Shouldn't X be the same (either positive or negative) within each trial?
If the info is sufficient, sure. But it's not sufficient. What you've got are two scenarios:
EITHER (x+1) is pos, in which case x < 1, x > -1, which simplifies to -1<x<1 (note: you had an error here - you said x < -1, but it should be x > -1)
OR (x-1) is pos, in which case x > 1 or x < -1
Which is it? Don't know. (And even if we did know that it was one specific scenario, both scenarios leave open pos and neg possibilities.. so insufficient.)
Can you go one more step forward and show how the two statements together are sufficient?
Wow, this has gotten seriously messy. Let's see if we can tidy it up a bit into something that could actually be done in 2 min! From the top (and forgive the dots in the formatting - trying to get things to line up!):
is x < 0? y/n
Statement (1)
............(x^3)(1-x^2) < 0
............(x^3)(1-x)(1+x) < 0
if x=+....(pos)(neg)(pos) < 0
x^3>0, therefore x>0
1-x < 0, therefore x>1
1+x>0, therefore x>-1
We started with the assumption that x is pos, so x>1 (because all of the above statements are still true if x>1). So there's one possibility for statement 1: x is pos.
............(x^3)(1-x^2) < 0
............(x^3)(1-x)(1+x) < 0
if x=-....(neg)(pos)(pos) < 0
(Note: the last term must be pos because we've already got a neg in the mix - the first term.)
x^3 < 0, therefore x < 0
1-x > 0, therefore x < 1
1+x>0, therefore x>-1
We started with the assumption that x is neg, so -1 < x < 0. So here's another possibility: x is neg.
Insufficient. Cross off AD.
Statement (2)
............x^2 - 1 < 0
............(x+1)(x-1) < 0
if x=+...(pos)(neg) < 0
x+1 > 0, therefore x > -1
x-1 < 0, therefore x < 1
We started with the assumption that x is pos, so 0 < x < 1. So if this is the case, then x is pos.
............x^2 - 1 < 0
............(x+1)(x-1) < 0
if x=-....(pos)(neg) < 0
x+1 > 0, therefore x > -1
x-1 < 0, therefore x < 1
We started with the assumption that x is neg, so -1 < x < 0. So if this is the case, then x is neg.
Insufficient. Eliminate B.
Statements (1) and (2)
Now we only include the things that make both statements true.
statement 1 says either x > 1 or -1 < x < 0.
statement 2 says either 0 < x < 1 or -1 < x < 0.
x>1 is not true for statement 2. Eliminate it.
-1 < x < 0 is true for both statements. Keep it.
0 < x < 1 is not true for statement 1. Eliminate it.
The only thing you were allowed to keep says that x is neg. Sufficient. C.
