grandh01 wrote:15. If x≠0, is |x| <1?
(1) x2<1
(2) |x| <1/x

thanks in advance

I'd going with D

My logic is that the only way for |x|<1... x must be a fraction because if x is negative than |x| will be greater than 1.

Statement 1: x2<1.
The only way this can happen is if x is a fraction, because x2 cannot be negative [if x=-2 (-2)^2 is 4, not -4). Therefore x must be a fraction, say 1/2.
(1/2)^2 = 1/4 < 1
Thus, x2 <1
So, for this logic than |x| is also < 1.

Statement 2: |x|<1/x.
If you plug in values will you find that the only way to make this work is if x is a fraction. If x=5. than |5|<1/5 does not work! If x=1/2 than |1/2| < 2, this works and it satisfies our original statement (|x|<1)

Does this help?
What is the OA?

grandh01 wrote:15. If x≠0, is |x| <1?
(1) x2<1
(2) |x| <1/x

thanks in advance

We need to find whether |x| < 1.
Square both sides, since both sides are positive.
Question becomes Is |x|^2 < 1^2 => Is x^2 < 1.

Now it should be easier to solve.

1) x^2 < 1. This is the same as our rephrased condition. Sufficient.

2) |x| < 1/x.

First off |x| is always positive here. Therefore, 1/x > 0 => x is positive.
But if x >0, |x| = x
So the condition becomes x < 1/x
=> x^2 < 1. Same as our rephrase. Sufficient.

D is correct.

Thanks guys, both of you are right. The OA is D. Great explanations much appreciated.

grandh01 wrote:Thanks guys, both of you are right. The OA is D. Great explanations much appreciated.

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