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Absolute Inequalities

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by Rahul@gurome » Fri Dec 17, 2010 10:10 pm
Given: x² > y²
Implies |x| > |y|

Statement 1: x > |y|
Certainly x is positive. Now y can be positive or negative. If y is negative, x is obviously greater than y (as x is positive). If y is positive, then its absolute value that is y itself is less than x. Thus x > y always.

Sufficient.

Statement 2: |x| > y
If x is positive, this statement directly implies x > y.
If x is negative, then two possible cases
  • 1. y positive => y > x (Example: x = -5, y = 2)
    2. y negative => x < y as |x| > |y|
Not sufficient.

The correct answer is A.
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by gmatusa2010 » Fri Dec 17, 2010 10:22 pm
Rahul,

Here's my approach, is it correct?


X^2-Y^2>0 meaning X+Y and X-Y have to have the same sign. I'm just going to look at Statement 2 here:

2) |X|>Y means X>Y or X-Y>0 OR -X>Y meaning X+Y<0 (Is this correct re-statement?)

Since the signs have to be the same when X-Y then X+Y is also positive, that step is meaningless because it already answer our question of is X-Y>0.

The second condition of X+Y<0 means X-Y also has to be less than Zero. Now we have two scenarios that will satisfy X^2-Y^2 but gives different result of X-Y so insufficient.
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by Night reader » Fri Dec 17, 2010 10:23 pm
gmatusa2010 wrote:If X^2>Y^2 is X>y?

1) X>|Y|
2) |X|>Y


Let's see some different approaches....
simplifying problem => |x|>|y| => x>y and x<y

st(1) x>y and x>-y => x>y only if -x<-y, however st(1) defines only -x<y, Not sufficient
st(2) x>y and -x>y => x>y only if -x<-y, however st(2) defines only x<-y, Not sufficient

Combining st(1&2) => x>-y and -x>y we get |x|<|y| from where only possible x<y Satisfies - Answer No x<y

IOM C
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by Rahul@gurome » Fri Dec 17, 2010 10:38 pm
@gmatusa2010: Yes you are correct! :)
Night reader wrote:simplifying problem => |x|>|y| => x>y and x<y

st(1) x>y and x>-y => x>y only if -x<-y, however st(1) defines only -x<y, Not sufficient
st(2) x>y and -x>y => x>y only if -x<-y, however st(2) defines only x<-y, Not sufficient
These are not AND cases, they are OR cases.

x > |y| implies either x > y (if y is positive) or x > -y (if y is negative).
As of your consideration both of them are simultaneously true which is not possible. Same for statement 2.
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by Night reader » Fri Dec 17, 2010 10:42 pm
yes Rahul, I see my mistakes. I approached mods mechanically, better was plug-in to see that A is correct.
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by gmatusa2010 » Fri Dec 17, 2010 10:42 pm
Rahul,

I noticed that when u did your cases there were 3. X is positive, X is negative and Y negative, and X is negative and Y is positive. I got the statement correctly but am I missing things with just two cases?
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by SUHAILK » Fri Dec 17, 2010 10:53 pm
statement 1:

x > |y| ==> x > 0 ---(1)

now lets see whether x > y :

case 1) y > 0

==> x >|y| = x > y (if y > 0 |y| = y) -- yes x > y

case 2) y< 0

x > 0 from inequality (1) & y < 0 ...==> yes x > y

Hence Statement 1 is Sufficient


Statement 2:

|x| > y lets assume |x| = K (positive real no) , so now we can rewrite statement 2 as

y < K (a positive real no)

above inequality can be represented on number line as given below:

Image

means y can be positive or negative

with this knowledge lets check whether x > y

case 1 : x > 0

then
|x| > y = x > y ==> yes x > y

case 2 : x < 0

to understand x > y under this condition lets take some e.g

example 1: x = -4 => |x| = 4 => y < 4 => y = (....-5,-4, -3 , -2, -1, 0, 1, 2, 3)

=> for some cases x > y and for some cases x < y

hence Statement 2 is Insufficient

Answer A


Please let me know if my answer is correct or not.
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by anshumishra » Fri Dec 17, 2010 11:02 pm
X^2 > y^2 => |x| > |y|

Is x > y ?

Let us represent the relationship (|x| > |y|) on a number line :


--------------X'-----Y'-------0--------Y----X------------------


Statement 1:

x > |y| , Since |y| is always non-negative, that means x = X' in the diagram above
Hence, clearly X > Y ---- Sufficient


Statement 2 :

|x| > y
which is true for both X and X'
When x = X, then X > Y,
but when x = X' then x < Y

hence - Insufficient

Answer is A.
Last edited by anshumishra on Sat Dec 18, 2010 7:53 am, edited 1 time in total.
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by gmatusa2010 » Fri Dec 17, 2010 11:12 pm
Interesting approach... how did you get x^2=y^2?
anshumishra wrote:X^2 = y^2 => |x| > |y|

Is x > y ?

Let us represent the relationship (|x| > |y|) on a number line :


--------------X'-----Y'-------0--------Y----X------------------


Statement 1:

x > |y| , Since |y| is always non-negative, that means x = X' in the diagram above
Hence, clearly X > Y ---- Sufficient


Statement 2 :

|x| > y
which is true for both X and X'
When x = X, then X > Y,
but when x = X' then x < Y

hence - Insufficient

Answer is A.
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by anshumishra » Fri Dec 17, 2010 11:15 pm
gmatusa2010 wrote:Interesting approach... how did you get x^2=y^2?
anshumishra wrote:X^2 = y^2 => |x| > |y|

Is x > y ?

Let us represent the relationship (|x| > |y|) on a number line :


--------------X'-----Y'-------0--------Y----X------------------


Statement 1:

x > |y| , Since |y| is always non-negative, that means x = X' in the diagram above
Hence, clearly X > Y ---- Sufficient


Statement 2 :

|x| > y
which is true for both X and X'
When x = X, then X > Y,
but when x = X' then x < Y

hence - Insufficient

Answer is A.
Hey gmatusa2010, that was a typo, I have fixed it.

It is in the given question :

If X^2>Y^2 is X>y?

1) X>|Y|
2) |X|>Y
Last edited by anshumishra on Sat Dec 18, 2010 7:54 am, edited 1 time in total.
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by Rahul@gurome » Fri Dec 17, 2010 11:21 pm
gmatusa2010 wrote:Rahul,

I noticed that when u did your cases there were 3. X is positive, X is negative and Y negative, and X is negative and Y is positive. I got the statement correctly but am I missing things with just two cases?
No, you're not missing anything.
Note that my first case "If x is positive, this statement directly implies x > y" is same as your first case "X-Y>0". 2nd and 3rd cases of my analysis is also same as the 2nd one of yours (This equivalence is not so direct as the 1st one).
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