MGMAT CAT Question - Absolute Value Inequalities

[mdavidm_531]

If x is not equal to 0, is |x| less than 1?

(1) (x/|x|) < x

(2) |x| > x

This is Greek to me.

[Frankenstein]

Hi,
When it is Greek to you, just start from basics
Basic rule:
|x| = x if x>0
|x| = -x if x<0
From(1):
(x/|x|) < x
case-1: x>0, So, |x| = x =>x/|x| = 1
So, (x/|x|) < x implies x>1.
So, |x| > 1
case-2: x<0. So, {x{ = -x => x/|x| = -1
So, (x/|x|) < x implies x>-1.
So, -1<x<0 => |x| is less than 1
So, we cannot definitely say whether |x| is less than 1
All we can say is |x| > 1 for x >0 and |x|<1 for x<0
Not sufficient

From(2):
|x|>x
If x is positive, |x| = x
So, |x|>x means x is negative.
Not sufficient

Both(1) and (2):
From st(1) we get : |x| > 1 for x >0 and |x|<1 for x<0
From st(2) we get : x < 0
So, combining we get
|x| < 1
Sufficient

Hence, C

[Whitney Garner]

If x is not equal to 0, is |x| less than 1?

(1) (x/|x|) < x

(2) |x| > x

This is Greek to me.

I often find that Absolute Value problems (particularly DS) can get some help through number lines. So first we want to rephrase the stem:

if x/=0, then is |x|<1?

Using a number line, you can visualize the numbers that make this absolute value expression true? Stuff less than one, BUT also bigger than -1. The new question becomes, "Is x in the shaded region of the number line?"



Statement (1): (x/|x|) < x

If I divide X by the absolute value of X, the X parts of the numbers cancel each other out, and I'm only left to deal with the sign. That means the following:

If x>0, then the left side is positive 1, and if x<0, the left side is negative 1. I have to combine that with the X on the other side:

x>0 ---> 1 < x --> x is either a positive number greater than 1
x<0 ---> -1 < x --> or x is a negative number greater than -1 (between -1 and 0)



If x is a negative number, the answer to our stem is YES, but if x is a positive number our answer is NO. We have shaded parts of the number line BOTH inside and outside of the range we want so the statement is INSUFF

Statement (2): |x| > x

For |x| to be greater than x, it must be the case that x is a negative number so that taking the absolute value makes it positive. If x were positive, then |x| would have to equal x. So our number line for statement (2) becomes:



Because the shaded area is both inside and outside of the range we want - INSUFF.

Statement (1+2):

When we combine the 2 graphs, we shade ONLY the areas included in both statements.



This means that we can only shade an area that is completely included in the desired region - the answer is SUFF.

I like using visuals whenever possible because it can really help clean up scratch paper (and help prevent you making careless errors in long strings of equations/expressions).


Whit

[srikanthb.69]

If x is not equal to 0, is |x| less than 1?

(1) (x/|x|) < x

(2) |x| > x

This is Greek to me.

I often find that Absolute Value problems (particularly DS) can get some help through number lines. So first we want to rephrase the stem:

if x/=0, then is |x|<1?

Using a number line, you can visualize the numbers that make this absolute value expression true? Stuff less than one, BUT also bigger than -1. The new question becomes, "Is x in the shaded region of the number line?"



Statement (1): (x/|x|) < x

If I divide X by the absolute value of X, the X parts of the numbers cancel each other out, and I'm only left to deal with the sign. That means the following:

If x>0, then the left side is positive 1, and if x<0, the left side is negative 1. I have to combine that with the X on the other side:

x>0 ---> 1 < x --> x is either a positive number greater than 1
x<0 ---> -1 < x --> or x is a negative number greater than -1 (between -1 and 0)



If x is a negative number, the answer to our stem is YES, but if x is a positive number our answer is NO. We have shaded parts of the number line BOTH inside and outside of the range we want so the statement is INSUFF

Statement (2): |x| > x

For |x| to be greater than x, it must be the case that x is a negative number so that taking the absolute value makes it positive. If x were positive, then |x| would have to equal x. So our number line for statement (2) becomes:



Because the shaded area is both inside and outside of the range we want - INSUFF.

Statement (1+2):

When we combine the 2 graphs, we shade ONLY the areas included in both statements.



This means that we can only shade an area that is completely included in the desired region - the answer is SUFF.

I like using visuals whenever possible because it can really help clean up scratch paper (and help prevent you making careless errors in long strings of equations/expressions).


Whit

I Think the Numbers are interchanged.. -1 should be in place of 1 because u are drawing number range of X. Correct me if i'm wrong

[amit2k9]

q is asking -1 < x < 1

a for x=0.5 | -0.5. different values for LHS and RHS. not sufficient.

b x < 0. not sufficient.


a+b -1<x<0. hence C it is. for x=-2 LHS > RHS.

[goalevan]

It helps to visualize with a graph too.

Statement 1 gives a range of -1 < x < 0 AND x > 1 (what I've shaded green in first graph)
Statement 2 gives a range of x < 0 (shaded green in second graph)

Together we know x lies in -1 < x < 0, so sufficient. C