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

1) y<x
2) xy<0

i thought it С,but the answer is B

who can explain?

If xy < 0 , either x < 0 < y (case i) or y < 0 < x (case ii)


Remember that |x - y | = x - y if x > y but |x - y| = y - x if x < y
Also keep in mind that |x| = x if x > 0 but |x| = - x if x < 0


In case i , |x - y| = y - x and |x| - |y| = - x - y

Since y > 0 , y > - y and thus y - x > -x - y (answer yes)

In case ii, |x - y| = x - y and |x| - |y| = x + y

Since y < 0 , - y > y and thus x - y > x + y (answer yes)

SUFF

Alternatively, if x and y have opposite signs, x and - y have the same sign. x + (- y) will thus be farther from 0 than is x

Thus |x - y| > |x| > |x| - |y| since |y| > 0
SUFF


One more alternative!

If |x| - |y| < 0 , the answer is obviously yes, as | x - y| > 0 ( x and y are not the same number, as they have opposite signs.

If |x| - |y| >= 0 the answer will be yes if and only if the square of the left side is greater than the square of the right

i.e |x - y|^2 = (x - y)^2 = x^2 - 2xy + y^2 > (|x| - |y|)^2 = x^2 - 2|xy| + y^2

But since we know that xy < 0 xy < 0 < |xy| , thus -2xy > -2|xy| and the inequality above holds.


SUFF

kevinmarmstrong
thanks

Statement 1:
x > y
x = 5, y = 1, |x-y| = 4, |x| - |y| = 4 violated
x = 5, y = -1, |x-y| = 6, |x| - |y| = 4 Not violated
x = -1, y = -5, |x-y| = 4, |x| - |y| = -4 Not voilated

Not sufficient

Statement 2:
xy < 0
x = 5, y = -1, |x-y| = 6, |x| - |y| = 4 Not Violated
x = 5, y = -10, |x-y| = 15, |x| - |y| = -5 Not Violated
x = -1, y = 5, |x-y| = 6, |x| - |y| = -4 Not Violated
x = -10, y = 5, |x-y| = 15, |x| - |y| = 5 Not Violated

Only one opinion

sufficient

Ans B

good work!

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Stacey Koprince
Instructor
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ManhattanGMAT