DS - absolute value

[ccassel]

How would you efficiently work through this question?

Is |x - y| > |x| - |y|?

I. y < x
II. xy < 0

Answer is B

[Frankenstein]

Hi,
From(1):
if x=4, y=2, |x - y| = |x| - |y|
if x=4, y=-2, |x - y| > |x| - |y|
Not sufficient

From(2):As xy<0, |xy| = -xy
case-1 : |x| - |y| < 0
As |x - y| > 0, |x - y| > |x| - |y| always holds

case-2 : |x| - |y| > 0
|x - y|^2 = x^2 + y^2 - 2xy = x^2 + y^2 + 2|xy|
(|x| - |y|)^2 = x^2 + y^2 - 2|xy|
So, |x - y|^2 > (|x| - |y|)^2
So, |x - y| > |x| - |y|
Sufficient

By picking x and y on the number line or by plugin values also, we can solve this.
Hence, B

[vikram4689]

A is not required for such a symm. eqn. (can always verify by plugging no.s). Since x & y are of opposite signs (x - y) will always ADD whereas |x| -|y| will always subtract, so B is sufficient.

[ccassel]

Hi,
From(2):As xy<0, |xy| = -xy
case-1 : |x| - |y| < 0
As |x - y| > 0, |x - y| > |x| - |y| always holds

case-2 : |x| - |y| > 0
|x - y|^2 = x^2 + y^2 - 2xy = x^2 + y^2 + 2|xy|
(|x| - |y|)^2 = x^2 + y^2 - 2|xy|
So, |x - y|^2 > (|x| - |y|)^2
So, |x - y| > |x| - |y|
Sufficient

B

Frankenstein, Why would you think of squaring both sides of the equation in statement 2?

Thanks,

[Frankenstein]

Hi,
From(2):As xy<0, |xy| = -xy
case-1 : |x| - |y| < 0
As |x - y| > 0, |x - y| > |x| - |y| always holds

case-2 : |x| - |y| > 0
|x - y|^2 = x^2 + y^2 - 2xy = x^2 + y^2 + 2|xy|
(|x| - |y|)^2 = x^2 + y^2 - 2|xy|
So, |x - y|^2 > (|x| - |y|)^2
So, |x - y| > |x| - |y|
Sufficient

B

Frankenstein, Why would you think of squaring both sides of the equation in statement 2?

Thanks,

Hi,
If you go by logic or by plugin method, you don't need to. But, by explaining mathematically, I need to somehow bring xy in the inequality. So, I squared both sides.