Is the answer B ?
A: y<x
case 1
x= 4 y = 1
|4-1|>|4|-|1| ?
NO
case 2
x=4 y=-1
|4-(-1)|>|4|-|-1|?
B: xy<0
case 1: x=4 y=-1
|4-(-1)|>|4|-|-1|? YES
case 2: x=-4 y=1
|-4-(1)|>|-4|-|1|? YES
5 > 3
Modulus and Inequalities
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phanideepak
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Anurag@Gurome
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(1) If Y = 3, X = 4, then |X - Y| = |4 - 3| = 1 and |X| - |Y| = 4 - 3 = 1. Here, |X - Y| = |X| - |Y|baladon99 wrote:Is |X-Y| > |X| -|Y| ?
1) Y < X
2) XY< 0
This is a GMAT prep problem.
If Y = -4, X = -3, then |X - Y| = |-3 - (-4)| = 1 and |X| - |Y| = 3 - 4 = -1. Here, |X - Y| > |X| - |Y|
No definite answer.
So, (1) is NOT SUFFICIENT.
(2) For XY < 0, either X should be positive, Y should be negative, or X should be negative, Y should be positive.
Generally, it can observed that |X - Y| will be the sum of X and Y, as one of them is negative and other is positive, while |X| - |Y| will be the difference in X and Y, which will obviously be less than |X - Y|. See the examples below:
If X = -3, Y = 4, then |X - Y| = |-3 - 4| = 7 and |X| - |Y| = 3 - 4 = -1. Here, |X - Y| > |X| - |Y|
If X = 8, Y = -4, then |X - Y| = |8 + 4| = 12 and |X| - |Y| = 8 - 4 = 4. Here, |X - Y| > |X| - |Y|
So, (2) is SUFFICIENT.
The correct answer is B.
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clock60
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hi guys
despite the problem elaborated, i want to ask, is it possible to square here both parts,thus avoid plugging numbers. i am not sure that it is legitimate way in inequalities
is
|x-y|>|x|-|y|, here i squared
is x^2-2xy+y^2>x^2-2|xy|+y^2.then i canceled all but 2xy, and 2|xy|, and left with
is |xy|>xy
as |xy|>0 if x or y or both does not equal to 0. the only way for xy to be less than |xy| is to be -ve
and 2 st gives us this relation
do you think guys my reasoning is valid?
thanks
despite the problem elaborated, i want to ask, is it possible to square here both parts,thus avoid plugging numbers. i am not sure that it is legitimate way in inequalities
is
|x-y|>|x|-|y|, here i squared
is x^2-2xy+y^2>x^2-2|xy|+y^2.then i canceled all but 2xy, and 2|xy|, and left with
is |xy|>xy
as |xy|>0 if x or y or both does not equal to 0. the only way for xy to be less than |xy| is to be -ve
and 2 st gives us this relation
do you think guys my reasoning is valid?
thanks
