Data sufficiancy

[magauovazamat]

Is x>1?

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

2) |x|<5

Please vote for the answer choices!!!

[tim]

Before we help with this question, we need you to show some effort of your own. What did you try on this question? Where did you get stuck?

[magauovazamat]

I don't get why answer choice is A.
I ended up with E because:

1) Let's get two cases:

Case a) x+1>0 and |x|-1>0
So, x>-1 and x>1; x<-1
This did not prove our answer.

Case b) x+1<0 and |x|-1<0
x<-1 and x<1; x>-1

Thus, 1 is insufficient

2) Is insufficient

[subbudrmz]

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

2) |x|<5

ANswer is (A).

(x+1)(|x|-1)>0 - For this relation to hold true, X cannot be 0. X cannot be <0. X cannot be between 0 and 1. Therefore X is greater than 1. hence sufficient.

|x|<5 - From this, -5 <x <5. So X may be or may not be greater than 1. Not sufficient.

[RonPurewal]

I don't get why answer choice is A.
I ended up with E because:

1) Let's get two cases:

Case a) x+1>0 and |x|-1>0
So, x>-1 and x>1; x<-1
This did not prove our answer.

your notation here is not easy to understand; i can't tell what you're indicating as the allowable values of x.
still -- in inequalities like this, you should actually TEST VALUES in the intervals that you find.
i.e., try actually plugging in a value that's less than -1; a value that's between -1 and 1; and a value that's greater than 1.
if you plug in anything less than -1, statement (1) will be false. (try it -- you'll get negative x positive.)
if you plug in anything between -1 and 1, statement (1) will also be false. (this time it's positive x negative.)
so you're left with only values that are greater than 1.

[magauovazamat]

I don't get why answer choice is A.
I ended up with E because:

1) Let's get two cases:

Case a) x+1>0 and |x|-1>0
So, x>-1 and x>1; x<-1
This did not prove our answer.

your notation here is not easy to understand; i can't tell what you're indicating as the allowable values of x.
still -- in inequalities like this, you should actually TEST VALUES in the intervals that you find.
i.e., try actually plugging in a value that's less than -1; a value that's between -1 and 1; and a value that's greater than 1.
if you plug in anything less than -1, statement (1) will be false. (try it -- you'll get negative x positive.)
if you plug in anything between -1 and 1, statement (1) will also be false. (this time it's positive x negative.)
so you're left with only values that are greater than 1.

Thanks!!! I found your solution very easy and fast!!!

[RonPurewal]

you're welcome.