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
DS - absolute value
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Frankenstein
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vikram4689
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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.
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ccassel
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Frankenstein, Why would you think of squaring both sides of the equation in statement 2?Frankenstein wrote: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
Thanks,
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Frankenstein
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Hi,ccassel wrote:Frankenstein, Why would you think of squaring both sides of the equation in statement 2?Frankenstein wrote: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
Thanks,
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.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
