Doubt on a modulus question

Question

13. x is a non-zero real number. Is |x| <1?
I. x^2 <1
II. |x|< 1/x

I is sufficient to answer the question.

How can we test the second condition?

J2S2019 · Answer

13. x is a non-zero real number. Is |x| <1?
I. x^2 <1
II. |x|< 1/x

From St 2, X cannot be zero, so only the condition holds when X is a positive fraction, i.e, 0<x<1,

but we are missing the other half, if Is |x| <1, then we basically needs to find -1<x<1 or not
St 1 sufficiently tells us that

ShankSouljaBoi · Answer

Reprhasing is -1 < x < 1

2---> try -1/2 ... No

try 1/2 ... Yes

Hence, insufficient

Real numbers = everything possible except infinity.

MahmoudFawzy · Answer

breaking down for the second statement is as follows:

1- As the LHS is always non negative, so x must be a non negative
2- x can't be zero as x is a denominator, so x must be positive
3- as x is positive, we can multiply both sides by x without flipping, so it is x^2 <1 --> x^2-1<0
so : (x-1)(x+1) < 1 ---> -1<x<1 , excluding zero
but because of our first conclusion, 0<x<1 so |x| <1 --> sufficient

Breaking down statement one would lead to (x-1)(x+1) < 1 ---> -1<x<1
so |x| <1 --> sufficient

D

CrackverbalGMAT · Answer

This is a ‘Yes-No’ type of DS question. To know whether |x| < 1, we need to find if -10, x^2 -1<0. The range that satisfies x^2 – 1< 0 is -10, the effective range would be 0

Bunuel · Answer

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Doubt on a modulus question

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