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1. Could someone solve the following DS prob If x≠0, is |x| <1? (1) x2 <1 == > (X Square)<1 (2) |x| < 1/x

Statement 1 is suff. I am able to solve this using statement 2.

2. Could someone share the strategy of solving questions involving |x|?

This one is tricky:

If \(x\neq{0}\), is \(|x| <1\)?

Basically we are asked: is x in the range (-1,1): is -\(1<x<1\) true?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(|x|=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient.

Please explain more why second condition is sufficient. I am not able to understand it

Bunuel wrote:

SMAbbas wrote:

1. Could someone solve the following DS prob If x≠0, is |x| <1? (1) x2 <1 == > (X Square)<1 (2) |x| < 1/x

Statement 1 is suff. I am able to solve this using statement 2.

2. Could someone share the strategy of solving questions involving |x|?

This one is tricky:

If \(x\neq{0}\), is \(|x| <1\)?

Basically we are asked: is x in the range (-1,1): is -\(1<x<1\) true?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(|x|=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient.

Please explain more why second condition is sufficient. I am not able to understand it

Bunuel wrote:

SMAbbas wrote:

1. Could someone solve the following DS prob If x≠0, is |x| <1? (1) x2 <1 == > (X Square)<1 (2) |x| < 1/x

Statement 1 is suff. I am able to solve this using statement 2.

2. Could someone share the strategy of solving questions involving |x|?

This one is tricky:

If \(x\neq{0}\), is \(|x| <1\)?

Basically we are asked: is x in the range (-1,1): is -\(1<x<1\) true?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(|x|=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient.

Answer D.

Hope it's clear.

You should specify what didn't you understand there. Thank you. _________________

Please explain more why second condition is sufficient. I am not able to understand it

Bunuel wrote:

SMAbbas wrote:

1. Could someone solve the following DS prob If x≠0, is |x| <1? (1) x2 <1 == > (X Square)<1 (2) |x| < 1/x

Statement 1 is suff. I am able to solve this using statement 2.

2. Could someone share the strategy of solving questions involving |x|?

This one is tricky:

If \(x\neq{0}\), is \(|x| <1\)?

Basically we are asked: is x in the range (-1,1): is -\(1<x<1\) true?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(2) \(|x| < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(|x|=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient.

Answer D.

Hope it's clear.

Dear piyushmnit,

the question asks whether -1<x<1 is true, so any range of x, that would fall into that one would be sufficient.

The second statement tells that 0<x<1, which fall into the the aforementioned range.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

It would be helpful if someone explains why we don't open the mod under the second option. As well, what's wrong with my algebra here?

\(|x| < \frac{1}{x}\) if \(x<0\), then \(-x*x<1\) and therefore, \(x\) could take all -ve number.

However, picking any negative number surely invalidates that claim. Obviously, having trouble reconciling my algebra with picking numbers.

Thanks for your help,

Question 1:

We did not open the mod because the LHS of the inequality (=|x|) is a positive quantity for ALL x ---> If 1/x is > a positive quantity ---> x > 0 . This is the reason the mod was 'not opened'.

\(1/x > a '+' quantity\) ---> x > 0

Question 2:

If \(x<0\) --> \(|x| = -x\) ---> \(-x<\frac{1}{x}\) ---> \(-x^2>1\) (reversed the sign of inequality as x<0 and multiplying an inequality by a negative number reverses the sign of inequality!)

---> This is not possible as \(x^2 = +\), \(-x^2 < 0\) and thus \(-x^2\) can never be > 1.

Thus , \(x<0\) is not a valid substitution. You did not reverse the sign of inequality when you multiplied the inequality by x as x<0 _________________

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